Step 1: Understanding the Question:

We are given a 3x3 matrix A and asked to determine its properties from the given options.

The key properties to check are whether the matrix is singular, invertible, or symmetric.


Step 2: Key Formula or Approach:

A matrix is singular if its determinant is zero (\(\det(A) = 0\)).

A matrix is invertible (or non-singular) if its determinant is non-zero (\(\det(A) \neq 0\)).

A matrix is symmetric if it is equal to its transpose (\(A = A^T\)).

We will calculate the determinant of A to check if it is singular or invertible.


Step 3: Detailed Explanation:

The given matrix is:
\[ A = \begin{bmatrix} 1 & 2 & 3
4 & 5 & 6
7 & 8 & 9 \end{bmatrix} \]

Let's calculate the determinant of A:
\[ \det(A) = 1 \begin{vmatrix} 5 & 6
8 & 9 \end{vmatrix} - 2 \begin{vmatrix} 4 & 6
7 & 9 \end{vmatrix} + 3 \begin{vmatrix} 4 & 5
7 & 8 \end{vmatrix} \]
\[ \det(A) = 1(5 \times 9 - 6 \times 8) - 2(4 \times 9 - 6 \times 7) + 3(4 \times 8 - 5 \times 7) \]
\[ \det(A) = 1(45 - 48) - 2(36 - 42) + 3(32 - 35) \]
\[ \det(A) = 1(-3) - 2(-6) + 3(-3) \]
\[ \det(A) = -3 + 12 - 9 \]
\[ \det(A) = 0 \]

Since the determinant of A is 0, the matrix is singular.

This also means the matrix is not invertible.


Now, let's check if the matrix is symmetric.

The transpose of A is:
\[ A^T = \begin{bmatrix} 1 & 4 & 7
2 & 5 & 8
3 & 6 & 9 \end{bmatrix} \]

Since \(A \neq A^T\), the matrix is not symmetric.


Step 4: Final Answer:

The determinant of the matrix A is 0. Therefore, the matrix is singular.
Quick Tip: For a 3x3 matrix with elements in an arithmetic progression like this one, the determinant is always zero. Notice that the elements in each row (1,2,3), (4,5,6), (7,8,9) and each column (1,4,7), (2,5,8), (3,6,9) are in AP. A property of determinants states that if we perform the operation \(C_2 \rightarrow C_2 - C_1\) and \(C_3 \rightarrow C_3 - C_2\), the new columns will be identical, making the determinant zero.

Step 1: Understanding the Question:

We are given a 2x2 matrix A with its determinant equal to 5.

We need to find the determinant of the matrix 2A, which is obtained by multiplying every element of A by the scalar 2.


Step 2: Key Formula or Approach:

For a square matrix A of order \(n\) and a scalar \(k\), the determinant of the matrix \(kA\) is given by the property:
\[ \det(kA) = k^n \det(A) \]


Step 3: Detailed Explanation:

We are given the following information:

The matrix A is of order \(n=2\).

The determinant of A is \(\det(A) = 5\).

We need to find the determinant of the matrix 2A.

Here, the scalar \(k\) is 2.


Using the formula from Step 2:
\[ \det(2A) = 2^n \det(A) \]

Substitute the values \(n=2\) and \(\det(A)=5\):
\[ \det(2A) = 2^2 \times 5 \]
\[ \det(2A) = 4 \times 5 \]
\[ \det(2A) = 20 \]


Step 4: Final Answer:

The determinant of the matrix 2A is 20.
Quick Tip: Remember that when a matrix is multiplied by a scalar \(k\), every element gets multiplied by \(k\). When calculating the determinant, each of the \(n\) rows (or columns) has a common factor of \(k\), which can be taken out. This results in the factor \(k\) being taken out \(n\) times, leading to the formula \(\det(kA) = k^n \det(A)\). This is a common source of error where students might mistakenly think \(\det(kA) = k \det(A)\).

Step 1: Understanding the Question:

We are dealing with a system of linear equations represented in matrix form as \(AX = B\), where A is a 3x3 coefficient matrix.

The question states that this system has a unique solution and asks about the property of the determinant of matrix A.


Step 2: Key Formula or Approach:

According to the Cramer's rule and matrix inversion method for solving systems of linear equations, a system \(AX = B\) has a unique solution if and only if the coefficient matrix A is invertible (non-singular).

A matrix is invertible if and only if its determinant is non-zero.


Step 3: Detailed Explanation:

The system of linear equations is given by \(AX = B\).

If the matrix A is invertible, we can find its inverse, \(A^{-1}\).

Multiplying the equation by \(A^{-1}\) on the left, we get:
\[ A^{-1}(AX) = A^{-1}B \]
\[ (A^{-1}A)X = A^{-1}B \]
\[ IX = A^{-1}B \]
\[ X = A^{-1}B \]

This equation gives a unique solution for the variable matrix X.


The condition for the existence of \(A^{-1}\) is that the determinant of A must be non-zero (\(\det(A) \neq 0\)).

If \(\det(A) = 0\), the matrix A is singular, and its inverse does not exist. In this case, the system of equations will have either no solution or infinitely many solutions, but not a unique solution.


Since the problem states that the system has a unique solution, it is necessary that the determinant of A is non-zero.

The value can be any non-zero real number; it is not restricted to be 1 or only positive.


Step 4: Final Answer:

For the system of equations \(AX = B\) to have a unique solution, the determinant of the coefficient matrix A must be non-zero.
Quick Tip: For a system of linear equations \(AX=B\): If \(\det(A) \neq 0\), there is a unique solution. If \(\det(A) = 0\) and \((adj A)B \neq 0\), there is no solution (inconsistent system). If \(\det(A) = 0\) and \((adj A)B = 0\), there are infinitely many solutions (consistent system). This summary is crucial for solving problems related to the nature of solutions of linear equations.

Step 1: Understanding the Question:

We are given a 2x2 matrix A containing an unknown variable \(x\), and its inverse matrix A\(^{-1}\).

We need to find the value of \(x\).


Step 2: Key Formula or Approach:

The product of a matrix and its inverse is the identity matrix, i.e., \(A A^{-1} = I\).

For a 2x2 matrix \(A = \begin{bmatrix} a & b
c & d \end{bmatrix}\), its inverse is given by \(A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} d & -b
-c & a \end{bmatrix}\).

We can either use the property \(A A^{-1} = I\) or calculate the inverse of A and compare it with the given A\(^{-1}\).


Step 3: Detailed Explanation:

Method 1: Using the formula for the inverse

First, calculate the determinant of A:
\[ \det(A) = (x)(4) - (3)(2) = 4x - 6 \]

Now, find the inverse of A using the formula:
\[ A^{-1} = \frac{1}{4x-6} \begin{bmatrix} 4 & -3
-2 & x \end{bmatrix} = \begin{bmatrix} \frac{4}{4x-6} & \frac{-3}{4x-6}
\frac{-2}{4x-6} & \frac{x}{4x-6} \end{bmatrix} \]

We are given:
\[ A^{-1} = \begin{bmatrix} -2 & 1.5
1 & -0.5 \end{bmatrix} \]

By comparing the corresponding elements of the calculated inverse and the given inverse, we can set up equations. Let's compare the element in the first row, first column:
\[ \frac{4}{4x-6} = -2 \]
\[ 4 = -2(4x - 6) \]
\[ 4 = -8x + 12 \]
\[ 8x = 12 - 4 \]
\[ 8x = 8 \]
\[ x = 1 \]


Method 2: Using \(A A^{-1} = I\)
\[ A A^{-1} = \begin{bmatrix} x & 3
2 & 4 \end{bmatrix} \begin{bmatrix} -2 & 1.5
1 & -0.5 \end{bmatrix} = \begin{bmatrix} 1 & 0
0 & 1 \end{bmatrix} \]

Let's compute the element in the first row, first column of the product:
\[ (x)(-2) + (3)(1) = 1 \]
\[ -2x + 3 = 1 \]
\[ -2x = 1 - 3 \]
\[ -2x = -2 \]
\[ x = 1 \]

Both methods yield the same result.


Step 4: Final Answer:

The value of \(x\) is 1.
Quick Tip: Using the property \(A A^{-1} = I\) is often faster than calculating the inverse from scratch. You only need to calculate one or two elements of the product matrix to form an equation and solve for the unknown, saving valuable time in an exam.

Step 1: Understanding the Question:

We are given two 2x2 matrices, A and B.

We need to find the transpose of their product, \((AB)^T\).


Step 2: Key Formula or Approach:

There are two ways to solve this:

1. First, calculate the product matrix \(C = AB\). Then, find the transpose of C, which is \(C^T\).

2. Use the property of transpose: \((AB)^T = B^T A^T\). First, find the transposes of A and B, then multiply them in reverse order.

We will use the first method as it is more direct.


Step 3: Detailed Explanation:

Step 3a: Calculate the product AB
\[ A = \begin{bmatrix} 1 & 2
3 & 4 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 0
1 & 0 \end{bmatrix} \]
\[ AB = \begin{bmatrix} (1)(1) + (2)(1) & (1)(0) + (2)(0)
(3)(1) + (4)(1) & (3)(0) + (4)(0) \end{bmatrix} \]
\[ AB = \begin{bmatrix} 1 + 2 & 0 + 0
3 + 4 & 0 + 0 \end{bmatrix} \]
\[ AB = \begin{bmatrix} 3 & 0
7 & 0 \end{bmatrix} \]


Step 3b: Find the transpose of AB

The transpose of a matrix is found by interchanging its rows and columns.

Let \(C = AB = \begin{bmatrix} 3 & 0
7 & 0 \end{bmatrix}\).

Then the transpose of C is:
\[ C^T = (AB)^T = \begin{bmatrix} 3 & 7
0 & 0 \end{bmatrix} \]


Step 4: Final Answer:

The resulting matrix \((AB)^T\) is \(\begin{bmatrix} 3 & 7
0 & 0 \end{bmatrix}\).
Quick Tip: Remember the "reversal law" for the transpose of a product: \((AB)^T = B^T A^T\). This property is very useful and extends to more matrices, e.g., \((ABC)^T = C^T B^T A^T\). While direct multiplication worked well here, knowing this rule is essential for more complex problems.

Step 1: Understanding the Question:

The problem asks us to find the sum of the constants A and B in the partial fraction decomposition of the given rational expression.


Step 2: Key Formula or Approach:

To find the values of A and B, we first combine the terms on the right-hand side and then equate the numerators.
\[ \frac{A}{x-1} + \frac{B}{x+3} = \frac{A(x+3) + B(x-1)}{(x-1)(x+3)} \]

So, we have the identity:
\[ 2x+5 = A(x+3) + B(x-1) \]

We can find A and B by substituting strategic values for \(x\) (the "cover-up" method) or by comparing coefficients.


Step 3: Detailed Explanation:

We start with the identity:
\[ 2x+5 = A(x+3) + B(x-1) \]


Method 1: Cover-up Method

To find A, we substitute \(x=1\) to make the B term zero:
\[ 2(1) + 5 = A(1+3) + B(1-1) \]
\[ 7 = A(4) + B(0) \]
\[ 7 = 4A \implies A = \frac{7}{4} \]


To find B, we substitute \(x=-3\) to make the A term zero:
\[ 2(-3) + 5 = A(-3+3) + B(-3-1) \]
\[ -6 + 5 = A(0) + B(-4) \]
\[ -1 = -4B \implies B = \frac{1}{4} \]


Now, we calculate the sum A+B:
\[ A+B = \frac{7}{4} + \frac{1}{4} = \frac{8}{4} = 2 \]


Method 2: Comparing Coefficients

Expand the right side of the identity:
\[ 2x+5 = Ax + 3A + Bx - B \]
\[ 2x+5 = (A+B)x + (3A-B) \]

Now, compare the coefficients of \(x\) and the constant terms on both sides.

Comparing coefficients of \(x\):
\[ A+B = 2 \]

Comparing constant terms:
\[ 3A-B = 5 \]

From the first equation, we directly get the required value \(A+B=2\). We don't even need to solve for A and B individually.


Step 4: Final Answer:

The value of A+B is 2.
Quick Tip: When asked for a sum or combination of the constants (like A+B), always try the method of comparing coefficients first. As seen in Method 2, comparing the coefficients of the highest power of the variable (in this case, \(x\)) can directly give the answer without needing to calculate the individual values of A and B. This is a very efficient exam strategy.

Step 1: Understanding the Question:

We need to find the values of the constants A, B, and C in the partial fraction decomposition of the given rational function.


Step 2: Key Formula or Approach:

The problem is based on partial fraction decomposition. We start by setting up the identity:
\[ 3x-1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2) \]

The most efficient way to find A, B, and C is the "cover-up" method, where we substitute the roots of the denominator (\(x=1, x=2, x=3\)) into this identity.


Step 3: Detailed Explanation:

We use the identity: \(3x-1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)\).


To find A, set \(x=1\):

Substitute \(x=1\) into the identity. The terms with B and C will become zero.
\[ 3(1) - 1 = A(1-2)(1-3) + B(0) + C(0) \]
\[ 2 = A(-1)(-2) \]
\[ 2 = 2A \]
\[ A = 1 \]


To find B, set \(x=2\):

Substitute \(x=2\) into the identity. The terms with A and C will become zero.
\[ 3(2) - 1 = A(0) + B(2-1)(2-3) + C(0) \]
\[ 6 - 1 = B(1)(-1) \]
\[ 5 = -B \]
\[ B = -5 \]


To find C, set \(x=3\):

Substitute \(x=3\) into the identity. The terms with A and B will become zero.
\[ 3(3) - 1 = A(0) + B(0) + C(3-1)(3-2) \]
\[ 9 - 1 = C(2)(1) \]
\[ 8 = 2C \]
\[ C = 4 \]


Step 4: Final Answer:

The values are \(A=1\), \(B=-5\), and \(C=4\). So, (A, B, C) = (1, -5, 4).
Quick Tip: The "cover-up" method is extremely fast for finding coefficients in partial fractions when the denominator has distinct linear factors. To find the coefficient A for a term \(\frac{A}{x-a}\), "cover-up" the \((x-a)\) factor in the original fraction's denominator and substitute \(x=a\) into the rest of the expression. For example, to find B: cover up \((x-2)\) in \(\frac{3x-1}{(x-1)(x-2)(x-3)}\) to get \(\frac{3x-1}{(x-1)(x-3)}\), and substitute \(x=2\): \(\frac{3(2)-1}{(2-1)(2-3)} = \frac{5}{(1)(-1)} = -5\).

Step 1: Understanding the Question:

We are given the value of \(\sin \theta\) and asked to find the possible values of \(\cos \theta\). The question does not specify the quadrant in which \(\theta\) lies.


Step 2: Key Formula or Approach:

The fundamental trigonometric identity relating sine and cosine is:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]

We will use this identity to solve for \(\cos \theta\).


Step 3: Detailed Explanation:

We are given \(\sin \theta = \frac{3}{5}\).

Substitute this value into the Pythagorean identity:
\[ \left(\frac{3}{5}\right)^2 + \cos^2 \theta = 1 \]
\[ \frac{9}{25} + \cos^2 \theta = 1 \]

Now, solve for \(\cos^2 \theta\):
\[ \cos^2 \theta = 1 - \frac{9}{25} \]
\[ \cos^2 \theta = \frac{25 - 9}{25} \]
\[ \cos^2 \theta = \frac{16}{25} \]

Take the square root of both sides to find \(\cos \theta\):
\[ \cos \theta = \pm \sqrt{\frac{16}{25}} \]
\[ \cos \theta = \pm \frac{4}{5} \]

Since the quadrant of \(\theta\) is not specified, both positive and negative values are possible.

If \(\theta\) is in the first quadrant, \(\cos \theta = \frac{4}{5}\).

If \(\theta\) is in the second quadrant, \(\cos \theta = -\frac{4}{5}\).

Therefore, \(\cos \theta\) can be either \(\frac{4}{5}\) or \(-\frac{4}{5}\).


Step 4: Final Answer:

The possible values for \(\cos \theta\) are \(\frac{4}{5}\) or \(-\frac{4}{5}\).
Quick Tip: Always be careful when taking the square root in trigonometric problems. Unless the quadrant of the angle is specified, you must consider both the positive and negative roots. A positive \(\sin \theta\) value means the angle can be in Quadrant I (where cosine is positive) or Quadrant II (where cosine is negative).

Step 1: Understanding the Question:

We are given a trigonometric equation and the quadrant in which the angle \(\theta\) lies. We need to find the specific value of \(\cos \theta\).


Step 2: Key Formula or Approach:

First, simplify the given trigonometric equation using the reciprocal identity for cosecant: \(\csc \theta = \frac{1}{\sin \theta}\).

Then, use the simplified equation to determine the value of \(\theta\) or a trigonometric ratio related to \(\theta\).

Finally, use the information about the quadrant to find the correct value of \(\cos \theta\).


Step 3: Detailed Explanation:

Start by simplifying the given equation:
\[ \cos \theta \csc \theta = -1 \]

Substitute \(\csc \theta = \frac{1}{\sin \theta}\):
\[ \cos \theta \left(\frac{1}{\sin \theta}\right) = -1 \]
\[ \frac{\cos \theta}{\sin \theta} = -1 \]

We know that \(\frac{\cos \theta}{\sin \theta} = \cot \theta\).

So, the equation becomes:
\[ \cot \theta = -1 \]

We need to find the angle \(\theta\) in the second quadrant for which \(\cot \theta = -1\).

The reference angle for \(\cot \alpha = 1\) is \(\alpha = \frac{\pi}{4}\) or 45\(^{\circ}\).

In the second quadrant, the angle \(\theta\) is given by \(\theta = \pi - \alpha\) (in radians) or \(\theta = 180^{\circ} - \alpha\) (in degrees).
\[ \theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \quad or \quad \theta = 180^{\circ} - 45^{\circ} = 135^{\circ} \]

Now, we need to find the value of \(\cos \theta\) for \(\theta = \frac{3\pi}{4}\).
\[ \cos\left(\frac{3\pi}{4}\right) = \cos\left(\pi - \frac{\pi}{4}\right) \]

Using the identity \(\cos(\pi - x) = -\cos(x)\):
\[ \cos\left(\frac{3\pi}{4}\right) = -\cos\left(\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}} \]

Rationalizing the denominator gives:
\[ \cos \theta = -\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = -\frac{\sqrt{2}}{2} \]

This matches the condition that cosine is negative in the second quadrant.


Step 4: Final Answer:

The value of \(\cos \theta\) is \(-\frac{\sqrt{2}}{2}\).
Quick Tip: When you find a simple trigonometric ratio like \(\cot \theta = -1\), you can quickly visualize the angle. Cotangent is the ratio \(x/y\). For it to be -1, \(x = -y\). In the second quadrant, x is negative and y is positive, which fits. This corresponds to the line \(y = -x\) in the second quadrant, which makes a 45\(^{\circ}\) angle with the negative x-axis. The coordinates on the unit circle would be \((-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})\), so \(\cos \theta\) is the x-coordinate, which is \(-\frac{1}{\sqrt{2}}\) or \(-\frac{\sqrt{2}}{2}\).

Step 1: Understanding the Question:

We are given the value of \(\sin \theta\) and asked to find the value of a more complex trigonometric expression.

The condition \(5 \sin \theta = 4\) implies \(\sin \theta = 4/5\), which is positive. So, \(\theta\) can be in the first or second quadrant.


Step 2: Key Formula or Approach:

First, find the values of \(\csc \theta\) and \(\cot \theta\) using \(\sin \theta = 4/5\).
\(\csc \theta = \frac{1}{\sin \theta}\).

To find \(\cot \theta\), we first need \(\cos \theta\) from the identity \(\cos^2 \theta = 1 - \sin^2 \theta\). Then \(\cot \theta = \frac{\cos \theta}{\sin \theta}\).

A simpler approach is to simplify the given expression first.


Step 3: Detailed Explanation:

Given \(\sin \theta = \frac{4}{5}\).

From this, we can immediately find \(\csc \theta\):
\[ \csc \theta = \frac{1}{\sin \theta} = \frac{1}{4/5} = \frac{5}{4} \]

Next, we find \(\cos \theta\):
\[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25} \]
\[ \cos \theta = \pm \sqrt{\frac{9}{25}} = \pm \frac{3}{5} \]

This gives two possible values for \(\cot \theta\):
\[ \cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\pm 3/5}{4/5} = \pm \frac{3}{4} \]


Let's evaluate the expression for both cases.

Case 1: \(\theta\) is in the first quadrant (\(\cot \theta = 3/4\))
\[ \frac{\csc \theta - \cot \theta}{\csc \theta + \cot \theta} = \frac{5/4 - 3/4}{5/4 + 3/4} = \frac{(5-3)/4}{(5+3)/4} = \frac{2/4}{8/4} = \frac{2}{8} = \frac{1}{4} \]

Case 2: \(\theta\) is in the second quadrant (\(\cot \theta = -3/4\))
\[ \frac{\csc \theta - \cot \theta}{\csc \theta + \cot \theta} = \frac{5/4 - (-3/4)}{5/4 + (-3/4)} = \frac{5/4 + 3/4}{5/4 - 3/4} = \frac{8/4}{2/4} = \frac{8}{2} = 4 \]

Since \(1/4\) is an option and \(4\) is not, we choose \(1/4\). Typically, in such problems without quadrant specification, the angle is assumed to be acute.


Alternative Simplification Method:

The expression can be simplified as follows:
\[ \frac{\csc \theta - \cot \theta}{\csc \theta + \cot \theta} = (\csc \theta - \cot \theta) \times \frac{1}{\csc \theta + \cot \theta} \]

Multiply numerator and denominator by \((\csc \theta - \cot \theta)\):
\[ \frac{(\csc \theta - \cot \theta)^2}{(\csc \theta + \cot \theta)(\csc \theta - \cot \theta)} = \frac{(\csc \theta - \cot \theta)^2}{\csc^2 \theta - \cot^2 \theta} \]

Using the identity \(\csc^2 \theta - \cot^2 \theta = 1\), the expression becomes:
\[ (\csc \theta - \cot \theta)^2 = \left(\frac{5}{4} - \left(\pm \frac{3}{4}\right)\right)^2 \]

If \(\cot \theta = 3/4\), we get \((5/4 - 3/4)^2 = (2/4)^2 = (1/2)^2 = 1/4\).

If \(\cot \theta = -3/4\), we get \((5/4 - (-3/4))^2 = (8/4)^2 = 2^2 = 4\).

The answer is \(1/4\).


Step 4: Final Answer:

Assuming the standard case of an acute angle, the value of the expression is \(1/4\).
Quick Tip: Simplifying the trigonometric expression before substituting values can often make the calculation easier. Here, converting the expression to \((\csc \theta - \cot \theta)^2\) simplifies the problem. Also, recognizing Pythagorean triples (like 3-4-5) can help you quickly find the values of other trig ratios. If \(\sin \theta = 4/5\) (Opp/Hyp), then Adjacent must be 3, so \(\cos \theta = \pm 3/5\) and \(\cot \theta = \pm 3/4\).

Step 1: Understanding the Question:

We are given an equation that relates a real number \(x\) to a trigonometric function \(\cos \theta\). We need to find the possible values of \(\cos \theta\).


Step 2: Key Formula or Approach:

The problem requires understanding the range of the function \(f(x) = x + \frac{1}{x}\) for real \(x\), and the range of the function \(g(\theta) = 2 \cos \theta\). The equality can only hold where their ranges overlap.


The range of \(f(x) = x + \frac{1}{x}\) can be found using the AM-GM inequality or calculus.

For \(x > 0\), by AM-GM, \(\frac{x + 1/x}{2} \ge \sqrt{x \cdot \frac{1}{x}} \implies x + \frac{1}{x} \ge 2\).

For \(x < 0\), let \(x = -y\) where \(y > 0\). Then \(x + \frac{1}{x} = -y - \frac{1}{y} = -(y + \frac{1}{y})\). Since \(y + \frac{1}{y} \ge 2\), we have \( -(y + \frac{1}{y}) \le -2\).

So, the range of \(x + \frac{1}{x}\) is \((-\infty, -2] \cup [2, \infty)\).


The range of \(\cos \theta\) is \([-1, 1]\). Therefore, the range of \(2 \cos \theta\) is \([-2, 2]\).


Step 3: Detailed Explanation:

We have the equation:
\[ x + \frac{1}{x} = 2 \cos \theta \]

Let's analyze the possible values for both sides of the equation.

The Left-Hand Side (LHS): \(y_1 = x + \frac{1}{x}\). As established, for any real \(x \neq 0\), the value of \(y_1\) must satisfy \(y_1 \ge 2\) or \(y_1 \le -2\). So, \(|y_1| \ge 2\).


The Right-Hand Side (RHS): \(y_2 = 2 \cos \theta\). Since the range of \(\cos \theta\) is \([-1, 1]\), the range of \(y_2\) is \([-2, 2]\). So, \(|y_2| \le 2\).


For the equation to hold, we must have \(y_1 = y_2\). The only values that are in both the range of \(y_1\) and the range of \(y_2\) are the boundary points where \(|y_1| \ge 2\) and \(|y_2| \le 2\) meet.

This can only happen when the value is exactly 2 or -2.

So, we must have:
\[ 2 \cos \theta = 2 \quad or \quad 2 \cos \theta = -2 \]

Solving for \(\cos \theta\):
\[ \cos \theta = 1 \quad or \quad \cos \theta = -1 \]

This can be written concisely as \(\cos \theta = \pm 1\).


Step 4: Final Answer:

The only possible values for \(\cos \theta\) are 1 and -1.
Quick Tip: This is a classic problem that tests your knowledge of the ranges of functions. Whenever you see the expression \(x + \frac{1}{x}\) for a real number \(x\), immediately recall its range: \((-\infty, -2] \cup [2, \infty)\). Comparing this with the bounded range of a trigonometric function like cosine or sine will quickly lead you to the solution at the boundary points.

Step 1: Understanding the Question:

We need to simplify the given trigonometric expression. The expression is a known identity.


Step 2: Key Formula or Approach:

We will use the algebraic identity \(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\) or \(a^3 + b^3 = (a+b)^3 - 3ab(a+b)\).

Let \(a = \sin^2 \theta\) and \(b = \cos^2 \theta\).

Then the expression \(\sin^6 \theta + \cos^6 \theta\) becomes \(a^3 + b^3\).

We also know the fundamental trigonometric identity: \(\sin^2 \theta + \cos^2 \theta = 1\).


Step 3: Detailed Explanation:

Let's rewrite the expression using \(a = \sin^2 \theta\) and \(b = \cos^2 \theta\).

The expression is \(a^3 + b^3 + 3ab\). This does not immediately simplify. Let's use the second identity from Step 2.


We know that \(a+b = \sin^2 \theta + \cos^2 \theta = 1\).

Now, let's use the identity \(a^3 + b^3 = (a+b)^3 - 3ab(a+b)\).

Substitute \(a = \sin^2 \theta\) and \(b = \cos^2 \theta\):
\[ \sin^6 \theta + \cos^6 \theta = (\sin^2 \theta + \cos^2 \theta)^3 - 3(\sin^2 \theta)(\cos^2 \theta)(\sin^2 \theta + \cos^2 \theta) \]

Since \(\sin^2 \theta + \cos^2 \theta = 1\), this simplifies to:
\[ \sin^6 \theta + \cos^6 \theta = (1)^3 - 3\sin^2 \theta \cos^2 \theta (1) \]
\[ \sin^6 \theta + \cos^6 \theta = 1 - 3\sin^2 \theta \cos^2 \theta \]

Now, substitute this result back into the original expression given in the question:

Original Expression = \((\sin^6 \theta + \cos^6 \theta) + 3\sin^2 \theta \cos^2 \theta\)
\[ = (1 - 3\sin^2 \theta \cos^2 \theta) + 3\sin^2 \theta \cos^2 \theta \]
\[ = 1 \]


Step 4: Final Answer:

The value of the expression is 1.
Quick Tip: The identities for \(\sin^4 \theta + \cos^4 \theta\) and \(\sin^6 \theta + \cos^6 \theta\) are very common in competitive exams. It's useful to remember them: \(\sin^4 \theta + \cos^4 \theta = 1 - 2\sin^2 \theta \cos^2 \theta\) \(\sin^6 \theta + \cos^6 \theta = 1 - 3\sin^2 \theta \cos^2 \theta\) Memorizing these can save you derivation time during an exam.

Step 1: Understanding the Question:

We need to find the maximum possible value of the expression \(3 \cos \theta + 4 \sin \theta\).


Step 2: Key Formula or Approach:

An expression of the form \(a \cos \theta + b \sin \theta\) can be converted into the form \(R \cos(\theta - \alpha)\) or \(R \sin(\theta + \beta)\).

The range of such an expression is \([-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}]\).

The maximum value is \(\sqrt{a^2+b^2}\) and the minimum value is \(-\sqrt{a^2+b^2}\).


Step 3: Detailed Explanation:

The given expression is \(3 \cos \theta + 4 \sin \theta\).

This is in the form \(a \cos \theta + b \sin \theta\) with \(a=3\) and \(b=4\).

Using the formula, the maximum value is \(\sqrt{a^2+b^2}\).

Calculate the value:
\[ \sqrt{3^2 + 4^2} = \sqrt{9 + 16} \]
\[ = \sqrt{25} \]
\[ = 5 \]

The minimum value would be -5. The entire expression ranges from -5 to 5.

Therefore, the maximum value is 5.


Derivation (for understanding):

Let \(3 = R \cos \alpha\) and \(4 = R \sin \alpha\).

Squaring and adding these gives:
\(3^2 + 4^2 = R^2 \cos^2 \alpha + R^2 \sin^2 \alpha\)
\(25 = R^2(\cos^2 \alpha + \sin^2 \alpha)\)
\(25 = R^2 \implies R = 5\) (since R is a magnitude).

The original expression becomes:
\(R \cos \alpha \cos \theta + R \sin \alpha \sin \theta = R(\cos \theta \cos \alpha + \sin \theta \sin \alpha)\)

Using the identity \(\cos(A-B) = \cos A \cos B + \sin A \sin B\), this is:
\(R \cos(\theta - \alpha) = 5 \cos(\theta - \alpha)\).

Since the maximum value of any cosine function is 1, the maximum value of \(5 \cos(\theta - \alpha)\) is \(5 \times 1 = 5\).


Step 4: Final Answer:

The maximum value of the expression \(3 \cos \theta + 4 \sin \theta\) is 5.
Quick Tip: For any expression of the form \(a \sin x + b \cos x\), immediately calculate \(\sqrt{a^2+b^2}\). This value is the maximum value (amplitude) of the resulting single trigonometric function. This is a very frequent question type, so knowing the formula for the maximum and minimum values is a must.

Step 1: Understanding the Question:

We need to solve the trigonometric equation \(\sin 5x + \sin 3x + \sin x = 0\) for a non-zero value of \(x\) in the interval \([0, \frac{\pi}{2}]\).


Step 2: Key Formula or Approach:

We will use the sum-to-product trigonometric formula:
\[ \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]

It's strategic to apply this formula to \(\sin 5x\) and \(\sin x\) to get a term involving \(3x\), which can then be factored with the middle term \(\sin 3x\).


Step 3: Detailed Explanation:

The given equation is \(\sin 5x + \sin 3x + \sin x = 0\).

Rearrange the terms: \((\sin 5x + \sin x) + \sin 3x = 0\).

Apply the sum-to-product formula to the terms in the parenthesis with \(A=5x\) and \(B=x\):
\[ \sin 5x + \sin x = 2 \sin\left(\frac{5x+x}{2}\right) \cos\left(\frac{5x-x}{2}\right) \]
\[ = 2 \sin\left(\frac{6x}{2}\right) \cos\left(\frac{4x}{2}\right) \]
\[ = 2 \sin(3x) \cos(2x) \]

Substitute this back into the equation:
\[ 2 \sin(3x) \cos(2x) + \sin(3x) = 0 \]

Factor out the common term \(\sin(3x)\):
\[ \sin(3x) (2 \cos(2x) + 1) = 0 \]

This gives two possibilities:

1) \(\sin(3x) = 0\)

2) \(2 \cos(2x) + 1 = 0 \implies \cos(2x) = -\frac{1}{2}\)


Let's solve each case for \(x\) in the interval \([0, \frac{\pi}{2}]\).

Case 1: \(\sin(3x) = 0\)

The general solution is \(3x = n\pi\), where \(n\) is an integer. So, \(x = \frac{n\pi}{3}\).

For \(n=0\), \(x = 0\). (This is a solution, but the question asks for a value other than zero).

For \(n=1\), \(x = \frac{\pi}{3}\). This value is in the interval \([0, \frac{\pi}{2}]\).

For \(n=2\), \(x = \frac{2\pi}{3}\), which is outside the interval.

So, from this case, we get the solution \(x = \frac{\pi}{3}\).


Case 2: \(\cos(2x) = -\frac{1}{2}\)

The principal value for which cosine is \(-\frac{1}{2}\) is \(\frac{2\pi}{3}\).

The general solution is \(2x = 2n\pi \pm \frac{2\pi}{3}\). So, \(x = n\pi \pm \frac{\pi}{3}\).

For \(n=0\), \(x = \pm \frac{\pi}{3}\). \(x = \frac{\pi}{3}\) is a valid solution.

For \(n=1\), \(x = \pi - \frac{\pi}{3} = \frac{2\pi}{3}\) (outside the interval) and \(x = \pi + \frac{\pi}{3} = \frac{4\pi}{3}\) (outside the interval).

This case also gives the solution \(x = \frac{\pi}{3}\).


Step 4: Final Answer:

The non-zero value of \(x\) in the given interval that satisfies the equation is \(\frac{\pi}{3}\).
Quick Tip: When solving equations with three or more sine or cosine terms, look for pairs of terms that can be combined using sum-to-product formulas to create a common factor. The choice of pairing is important. Pairing \(\sin 5x\) and \(\sin x\) was strategic because \((5x+x)/2 = 3x\), which matched the remaining term.

Step 1: Understanding the Question:

We need to find the general solution for the trigonometric equation \(\tan^2 x = 1\). This means finding all possible values of \(x\) that satisfy the equation.


Step 2: Key Formula or Approach:

The general solution for \(\tan^2 x = \tan^2 \alpha\) is given by the formula:
\[ x = n\pi \pm \alpha \]

where \(n\) is any integer.

First, we will find the principal value \(\alpha\) for which \(\tan^2 \alpha = 1\), and then apply the formula.


Step 3: Detailed Explanation:

The given equation is \(\tan^2 x = 1\).

We need to find an angle \(\alpha\) such that \(\tan^2 \alpha = 1\).

Taking the tangent of \(\frac{\pi}{4}\), we have \(\tan\left(\frac{\pi}{4}\right) = 1\).

Therefore, \(\tan^2\left(\frac{\pi}{4}\right) = (1)^2 = 1\).

So, we can choose our principal value \(\alpha = \frac{\pi}{4}\).


Now, we use the general solution formula for \(\tan^2 x = \tan^2 \alpha\):
\[ x = n\pi \pm \alpha \]

Substituting \(\alpha = \frac{\pi}{4}\), we get:
\[ x = n\pi \pm \frac{\pi}{4} \]

where \(n\) is any integer.


Alternative Method:

We can solve \(\tan^2 x = 1\) by taking the square root:
\[ \tan x = \pm \sqrt{1} \implies \tan x = 1 \quad or \quad \tan x = -1 \]

The general solution for \(\tan x = \tan \alpha\) is \(x = n\pi + \alpha\).

For \(\tan x = 1\), \(\alpha = \frac{\pi}{4}\). The solution is \(x = n\pi + \frac{\pi}{4}\).

For \(\tan x = -1\), \(\alpha = -\frac{\pi}{4}\). The solution is \(x = n\pi - \frac{\pi}{4}\).

Combining these two sets of solutions gives \(x = n\pi \pm \frac{\pi}{4}\).


Step 4: Final Answer:

The general solution of the equation is \(x = n\pi \pm \frac{\pi}{4}\), where \(n \in \mathbb{Z}\).
Quick Tip: Remember the general solution formulas for squared trigonometric functions, as they are often simpler than solving the two linear cases separately: If \(\sin^2 x = \sin^2 \alpha\), then \(x = n\pi \pm \alpha\). If \(\cos^2 x = \cos^2 \alpha\), then \(x = n\pi \pm \alpha\). If \(\tan^2 x = \tan^2 \alpha\), then \(x = n\pi \pm \alpha\). Notice that the format is the same for all three.

Step 1: Understanding the Question:

We need to evaluate a trigonometric expression involving cosine functions with arguments that are multiples of \(\frac{\pi}{17}\).


Step 2: Key Formula or Approach:

We will use the product-to-sum formula:
\[ 2 \cos A \cos B = \cos(A+B) + \cos(A-B) \]

And the property \(\cos(\pi - \theta) = -\cos(\theta)\).


Step 3: Detailed Explanation:

Let the given expression be \(E\).
\[ E = \cos\frac{5\pi}{17} + \cos\frac{7\pi}{17} + 2\cos\frac{11\pi}{17}\cos\frac{\pi}{17} \]

First, apply the product-to-sum formula to the third term:
\[ 2\cos\frac{11\pi}{17}\cos\frac{\pi}{17} = \cos\left(\frac{11\pi}{17} + \frac{\pi}{17}\right) + \cos\left(\frac{11\pi}{17} - \frac{\pi}{17}\right) \]
\[ = \cos\left(\frac{12\pi}{17}\right) + \cos\left(\frac{10\pi}{17}\right) \]

Now substitute this back into the expression for \(E\):
\[ E = \cos\frac{5\pi}{17} + \cos\frac{7\pi}{17} + \cos\frac{12\pi}{17} + \cos\frac{10\pi}{17} \]

We can rewrite the arguments of the last two terms using \(\pi\).

Notice that \(\frac{12\pi}{17} = \pi - \frac{5\pi}{17}\).

And \(\frac{10\pi}{17} = \pi - \frac{7\pi}{17}\).

Let's apply the identity \(\cos(\pi - \theta) = -\cos(\theta)\).
\[ \cos\left(\frac{12\pi}{17}\right) = \cos\left(\pi - \frac{5\pi}{17}\right) = -\cos\left(\frac{5\pi}{17}\right) \]
\[ \cos\left(\frac{10\pi}{17}\right) = \cos\left(\pi - \frac{7\pi}{17}\right) = -\cos\left(\frac{7\pi}{17}\right) \]

Now substitute these simplified terms back into the expression for \(E\):
\[ E = \cos\frac{5\pi}{17} + \cos\frac{7\pi}{17} - \cos\frac{5\pi}{17} - \cos\frac{7\pi}{17} \]

All the terms cancel each other out.
\[ E = 0 \]


Step 4: Final Answer:

The value of the expression is 0.
Quick Tip: When dealing with trigonometric sums involving fractions of \(\pi\), always look for pairs of angles that add up to \(\pi\) or \(\pi/2\). This allows you to use identities like \(\cos(\pi - \theta) = -\cos(\theta)\) or \(\cos(\pi/2 - \theta) = \sin(\theta)\) to simplify the expression. Here, recognizing \(12\pi/17\) as \(\pi - 5\pi/17\) was the key to solving the problem.

Step 1: Understanding the Question:

We are given the value of \(\sin \theta - \cos \theta\) and we need to find the value of \(\sin \theta + \cos \theta\).


Step 2: Key Formula or Approach:

This is a standard problem type that uses the relationship between \((\sin \theta - \cos \theta)^2\) and \((\sin \theta + \cos \theta)^2\).

We know that:
\((\sin \theta - \cos \theta)^2 = \sin^2 \theta - 2\sin \theta \cos \theta + \cos^2 \theta = 1 - 2\sin \theta \cos \theta\)
\((\sin \theta + \cos \theta)^2 = \sin^2 \theta + 2\sin \theta \cos \theta + \cos^2 \theta = 1 + 2\sin \theta \cos \theta\)

By adding these two equations, we get a direct relationship:
\((\sin \theta - \cos \theta)^2 + (\sin \theta + \cos \theta)^2 = 2\)


Step 3: Detailed Explanation:

Let \(x = \sin \theta + \cos \theta\). We need to find the value of \(x\).

We are given \(\sin \theta - \cos \theta = \frac{4}{5}\).

Using the identity from Step 2:
\[ (\sin \theta - \cos \theta)^2 + (\sin \theta + \cos \theta)^2 = 2 \]

Substitute the given values into this identity:
\[ \left(\frac{4}{5}\right)^2 + x^2 = 2 \]
\[ \frac{16}{25} + x^2 = 2 \]

Now, solve for \(x^2\):
\[ x^2 = 2 - \frac{16}{25} \]
\[ x^2 = \frac{50 - 16}{25} \]
\[ x^2 = \frac{34}{25} \]

Take the square root of both sides:
\[ x = \pm \sqrt{\frac{34}{25}} = \pm \frac{\sqrt{34}}{5} \]

Since one of the options is \(\frac{\sqrt{34}}{5}\), we select this value. The question doesn't provide enough information to determine the sign, but only the positive value is listed as a primary option choice.


Step 4: Final Answer:

The value of \(\sin \theta + \cos \theta\) is \(\frac{\sqrt{34}}{5}\).
Quick Tip: The identity \((\sin\theta \pm \cos\theta)^2 = 1 \pm 2\sin\theta\cos\theta\) is extremely useful. Memorizing the combined identity \((\sin\theta - \cos\theta)^2 + (\sin\theta + \cos\theta)^2 = 2\) provides a direct shortcut for problems where one expression is given and the other is asked.

Step 1: Understanding the Question:

We are asked to find the real part of a given complex number. The complex number is given as a fraction.


Step 2: Key Formula or Approach:

To find the real part, we first need to simplify the expression and write it in the standard form \(a+bi\).

First, expand the denominator \((2-i)^2\).

Then, multiply the numerator and the denominator by the conjugate of the resulting denominator to eliminate the imaginary part from the denominator.


Step 3: Detailed Explanation:

Let the complex number be \(z = \frac{1+2i}{(2-i)^2}\).

Step 3a: Simplify the denominator
\[ (2-i)^2 = 2^2 - 2(2)(i) + i^2 \]
\[ = 4 - 4i - 1 \quad (since i^2 = -1) \]
\[ = 3 - 4i \]

So, the expression becomes:
\[ z = \frac{1+2i}{3-4i} \]

Step 3b: Rationalize the denominator

To convert this into the form \(a+bi\), multiply the numerator and denominator by the conjugate of the denominator, which is \(3+4i\).
\[ z = \frac{1+2i}{3-4i} \times \frac{3+4i}{3+4i} \]
\[ z = \frac{(1+2i)(3+4i)}{(3-4i)(3+4i)} \]

Numerator:
\[ (1)(3) + (1)(4i) + (2i)(3) + (2i)(4i) = 3 + 4i + 6i + 8i^2 = 3 + 10i - 8 = -5 + 10i \]

Denominator:
\[ (3)^2 - (4i)^2 = 9 - 16i^2 = 9 - 16(-1) = 9 + 16 = 25 \]

So, the complex number is:
\[ z = \frac{-5 + 10i}{25} \]

Step 3c: Write in standard form
\[ z = \frac{-5}{25} + \frac{10i}{25} = -\frac{1}{5} + \frac{2}{5}i \]

The standard form is \(a+bi\), where \(a\) is the real part and \(b\) is the imaginary part.

Here, the real part is \(a = -\frac{1}{5}\).


Step 4: Final Answer:

The real part of the complex number is \(-\frac{1}{5}\).
Quick Tip: When simplifying complex fractions, always handle the denominator first. Expand any powers and then multiply by the conjugate to make the denominator a real number. Remember the conjugate of \(a+bi\) is \(a-bi\) and their product is always a real number: \((a+bi)(a-bi) = a^2 + b^2\).

Step 1: Understanding the Question:

We need to find the modulus of a complex number which is given in a fractional form with powers.


Step 2: Key Formula or Approach:

We will use the properties of modulus:

1. \(|z_1/z_2| = |z_1|/|z_2|\)

2. \(|z^n| = |z|^n\)

3. The modulus of a complex number \(z = a+bi\) is \(|z| = \sqrt{a^2+b^2}\).

Using these properties, we can find the modulus of the numerator and denominator separately without actually computing the complex number itself.


Step 3: Detailed Explanation:

Let \(z = \frac{(1+i)^{10}}{(2i-4)^4}\). We need to find \(|z|\).

Using the properties of modulus:
\[ |z| = \left| \frac{(1+i)^{10}}{(-4+2i)^4} \right| = \frac{|(1+i)^{10}|}{|(-4+2i)^4|} = \frac{|1+i|^{10}}{|-4+2i|^4} \]


Step 3a: Calculate the modulus of the numerator term

First, find the modulus of \(1+i\):
\[ |1+i| = \sqrt{1^2 + 1^2} = \sqrt{2} \]

Now, raise it to the power of 10:
\[ |1+i|^{10} = (\sqrt{2})^{10} = (2^{1/2})^{10} = 2^5 = 32 \]


Step 3b: Calculate the modulus of the denominator term

First, find the modulus of \(-4+2i\):
\[ |-4+2i| = \sqrt{(-4)^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} \]

We can simplify \(\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}\).

Now, raise it to the power of 4:
\[ |-4+2i|^4 = (\sqrt{20})^4 = (20^{1/2})^4 = 20^2 = 400 \]

Alternatively, \((2\sqrt{5})^4 = 2^4 \times (\sqrt{5})^4 = 16 \times 5^2 = 16 \times 25 = 400\).


Step 3c: Combine the results
\[ |z| = \frac{32}{400} \]

Simplify the fraction:
\[ |z| = \frac{16}{200} = \frac{8}{100} = \frac{2}{25} \]


Step 4: Final Answer:

The modulus of the complex number is \(\frac{2}{25}\). Note that modulus must be a non-negative real number, so options (B) and (D) are incorrect by definition.
Quick Tip: When asked for the modulus of a complex expression involving products, divisions, or powers, never try to simplify the complex number first. It's much faster and easier to use the properties of modulus (\(|z_1 z_2| = |z_1||z_2|\), \(|z_1/z_2| = |z_1|/|z_2|\), \(|z^n| = |z|^n\)) to calculate the modulus of each part separately and then combine them.

Step 1: Understanding the Question:

We are given the length of a chord and its perpendicular distance from the center of a circle. We need to find the diameter of the circle.


Step 2: Key Formula or Approach:

The key geometric property is that the perpendicular from the center of a circle to a chord bisects the chord. This forms a right-angled triangle with the radius of the circle as the hypotenuse, the perpendicular distance from the center as one leg, and half the length of the chord as the other leg. We can use the Pythagorean theorem: \((radius)^2 = (distance)^2 + (half of chord)^2\).


Step 3: Detailed Explanation:

Let \(r\) be the radius of the circle.

Length of the chord, \(L = 6\) cm.

Distance from the center to the chord, \(d = 4\) cm.


The perpendicular from the center bisects the chord. So, the length of half the chord is \(\frac{L}{2} = \frac{6}{2} = 3\) cm.


Now, we have a right-angled triangle with sides:

Hypotenuse = \(r\)

One leg = \(d = 4\) cm

Other leg = \(\frac{L}{2} = 3\) cm


Using the Pythagorean theorem (\(a^2 + b^2 = c^2\)):
\[ d^2 + \left(\frac{L}{2}\right)^2 = r^2 \]
\[ 4^2 + 3^2 = r^2 \]
\[ 16 + 9 = r^2 \]
\[ 25 = r^2 \]
\[ r = \sqrt{25} = 5 cm \]

The radius of the circle is 5 cm.


The question asks for the length of the diameter.

Diameter \(D = 2 \times r\).
\[ D = 2 \times 5 = 10 cm \]


Step 4: Final Answer:

The length of the diameter is 10 cm.
Quick Tip: This problem uses the very common 3-4-5 Pythagorean triple. Whenever you see a right-angled triangle with two sides as 3 and 4 in a geometry problem, the hypotenuse is almost always 5. Recognizing this can save you calculation time. Also, be careful to read the question fully; it asks for the diameter, not the radius, which is a common mistake.

Step 1: Understanding the Question:

We need to find the length of the tangent drawn from an external point \((x_1, y_1) = (5, 1)\) to a given circle.


Step 2: Key Formula or Approach:

The length of the tangent, \(L\), from an external point \((x_1, y_1)\) to a circle with the equation \(S \equiv x^2 + y^2 + 2gx + 2fy + c = 0\) is given by the formula:
\[ L = \sqrt{S_1} \]

where \(S_1\) is the value of the circle's expression when the coordinates of the point are substituted into it, i.e.,
\[ S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c \]


Step 3: Detailed Explanation:

The given equation of the circle is \(S \equiv x^2 + y^2 + 6x - 4y - 3 = 0\).

The external point is \((x_1, y_1) = (5, 1)\).

First, we calculate \(S_1\) by substituting \(x=5\) and \(y=1\) into the equation of the circle:
\[ S_1 = (5)^2 + (1)^2 + 6(5) - 4(1) - 3 \]
\[ S_1 = 25 + 1 + 30 - 4 - 3 \]
\[ S_1 = 56 - 7 \]
\[ S_1 = 49 \]

Now, we find the length of the tangent using the formula \(L = \sqrt{S_1}\).
\[ L = \sqrt{49} \]
\[ L = 7 \]


Step 4: Final Answer:

The length of the tangent from the point (5, 1) to the circle is 7.
Quick Tip: To find the length of the tangent from a point to a circle, simply substitute the point's coordinates into the circle's equation (make sure the RHS is 0) and take the square root of the result. This is a direct and quick formula-based question common in exams.

Step 1: Understanding the Question:

We are given the length of a tangent from an external point and the distance from that point to the circle's center. We need to find the area of the circle.


Step 2: Key Formula or Approach:

The radius of a circle (\(r\)), the length of the tangent from an external point (\(L\)), and the distance from the external point to the center of the circle (\(d\)) form a right-angled triangle. The distance \(d\) is the hypotenuse.

By the Pythagorean theorem:
\[ d^2 = r^2 + L^2 \]

The area of a circle is given by the formula \(A = \pi r^2\).


Step 3: Detailed Explanation:

We are given:

Length of the tangent, \(L = 8\) cm.

Distance from the point to the center, \(d = 11\) cm.

Using the Pythagorean theorem, we can find the radius \(r\):
\[ 11^2 = r^2 + 8^2 \]
\[ 121 = r^2 + 64 \]

Solve for \(r^2\):
\[ r^2 = 121 - 64 \]
\[ r^2 = 57 \]

Now, we calculate the area of the circle, \(A = \pi r^2\).
\[ A = \pi \times 57 = 57\pi \]

To find the numerical value, we use the approximation \(\pi \approx 3.14159\).
\[ A \approx 57 \times 3.14159 \]
\[ A \approx 179.07063 cm^2 \]

This value is closest to the option 179.14 cm.


Step 4: Final Answer:

The area of the circle is \(57\pi\) cm\(^2\), which is approximately 179.14 cm\(^2\).
Quick Tip: Always visualize the geometry. The radius from the center to the point of tangency is perpendicular to the tangent line. This creates a right-angled triangle, which is key to solving many problems involving tangents to circles.

Step 1: Understanding the Question:

We are given the coordinates of the vertex and the focus of a parabola. We need to find its equation.


Step 2: Key Formula or Approach:

First, determine the orientation and the parameter 'a' of the parabola.

The vertex is \((h, k)\). The focus is \((h+a, k)\) for a parabola opening right.

The standard equation for a parabola opening horizontally with vertex at \((h, k)\) is:
\[ (y-k)^2 = 4a(x-h) \]


Step 3: Detailed Explanation:

The vertex is given as \(V(h, k) = (1, 0)\).

The focus is given as \(S = (2, 0)\).

Since the y-coordinates of the vertex and focus are the same, the axis of symmetry is horizontal (the x-axis, y=0).

The focus (x=2) is to the right of the vertex (x=1), so the parabola opens to the right.

The distance 'a' is the distance between the vertex and the focus.
\[ a = \sqrt{(2-1)^2 + (0-0)^2} = \sqrt{1^2} = 1 \]

Now we use the standard equation \((y-k)^2 = 4a(x-h)\) with \(h=1, k=0, a=1\).
\[ (y-0)^2 = 4(1)(x-1) \]
\[ y^2 = 4(x-1) \]

Expanding the equation, we get:
\[ y^2 = 4x - 4 \]


Step 4: Final Answer:

The equation of the parabola is \(y^2 = 4x - 4\).
Quick Tip: The position of the focus relative to the vertex determines the direction the parabola opens. If the focus is to the right of the vertex, it opens right. If left, it opens left. If above, it opens up. If below, it opens down. This directly tells you which standard form of the equation to use.

Step 1: Understanding the Question:

We are given the vertex and the directrix of a parabola and we need to find its focus.


Step 2: Key Formula or Approach:

The vertex of a parabola is the midpoint between its focus and its directrix.

The axis of the parabola is perpendicular to the directrix and passes through the vertex.

The distance from the vertex to the directrix is equal to the distance from the vertex to the focus. Let this distance be 'a'.


Step 3: Detailed Explanation:

The vertex is \(V(h, k) = (2, 0)\).

The directrix is the y-axis, which is the line \(x=0\).

Since the directrix is a vertical line (\(x=0\)), the axis of symmetry must be a horizontal line. Since the axis passes through the vertex (2,0), the axis is the x-axis (\(y=0\)).

The parabola opens away from the directrix. The vertex is at \(x=2\) and the directrix is at \(x=0\), so the parabola opens to the right.

The distance 'a' from the vertex to the directrix is the horizontal distance between the point (2,0) and the line \(x=0\).
\[ a = |2 - 0| = 2 \]

The focus lies on the axis of symmetry (\(y=0\)) and is at a distance 'a' from the vertex, inside the curve.

Since the parabola opens to the right, the focus will be at \((h+a, k)\).

Focus \(S = (2+2, 0) = (4, 0)\).


Step 4: Final Answer:

The focus of the parabola is at (4, 0).
Quick Tip: A quick way to find the focus is to think about the vertex as the "center". The directrix is on one side, and the focus is on the other, at an equal distance. Here, the vertex is at x=2, directrix at x=0 (2 units to the left). So, the focus must be 2 units to the right, at x=2+2=4.

Step 1: Understanding the Question:

We are given the equation of an ellipse and we need to calculate its eccentricity.


Step 2: Key Formula or Approach:

First, convert the equation of the ellipse to the standard form: \(\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\) (for a vertical ellipse) or \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) (for a horizontal ellipse), where \(a > b\).

The eccentricity \(e\) is given by the formula:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]


Step 3: Detailed Explanation:

The given equation is \(16x^2 + 7y^2 = 112\).

To convert it to standard form, divide the entire equation by 112:
\[ \frac{16x^2}{112} + \frac{7y^2}{112} = \frac{112}{112} \]
\[ \frac{x^2}{7} + \frac{y^2}{16} = 1 \]

This is the standard form of an ellipse. We can identify \(a^2\) and \(b^2\). By convention, \(a^2\) is the larger denominator.

Here, \(a^2 = 16\) and \(b^2 = 7\).

Since \(a^2\) is under the \(y^2\) term, the major axis of the ellipse is vertical.

Now we calculate the eccentricity \(e\):
\[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{7}{16}} \]
\[ e = \sqrt{\frac{16-7}{16}} = \sqrt{\frac{9}{16}} \]
\[ e = \frac{3}{4} \]


Step 4: Final Answer:

The eccentricity of the ellipse is \(\frac{3}{4}\).
Quick Tip: Eccentricity of an ellipse is always between 0 and 1. If you get a value greater than or equal to 1 (like option A), you have made a calculation error, likely by swapping \(a^2\) and \(b^2\) in the formula.

Step 1: Understanding the Question:

We need to evaluate the limit of a rational function as \(x\) approaches infinity.


Step 2: Key Formula or Approach:

For a rational function \( \lim_{x\to\infty} \frac{P(x)}{Q(x)} \), where \(P(x)\) and \(Q(x)\) are polynomials, the limit is determined by the ratio of the leading terms (the terms with the highest power of x).

A more formal method is to divide both the numerator and the denominator by the highest power of \(x\) in the denominator.


Step 3: Detailed Explanation:

First, let's simplify the denominator of the expression:

Denominator = \(x^2 - 4x(1-x^2) = x^2 - 4x + 4x^3\).

So the limit is:
\[ \lim_{x\to\infty} \frac{4x^3 - x + 1}{4x^3 + x^2 - 4x} \]

The degree of the numerator polynomial is 3, and the degree of the denominator polynomial is also 3. Since the degrees are equal, the limit is the ratio of the leading coefficients.

Leading coefficient of numerator = 4.

Leading coefficient of denominator = 4.
\[ Limit = \frac{4}{4} = 1 \]

Formal Method:

The highest power of \(x\) in the expression is \(x^3\). Divide the numerator and denominator by \(x^3\):
\[ \lim_{x\to\infty} \frac{\frac{4x^3}{x^3} - \frac{x}{x^3} + \frac{1}{x^3}}{\frac{4x^3}{x^3} + \frac{x^2}{x^3} - \frac{4x}{x^3}} = \lim_{x\to\infty} \frac{4 - \frac{1}{x^2} + \frac{1}{x^3}}{4 + \frac{1}{x} - \frac{4}{x^2}} \]

As \(x \to \infty\), all terms with \(x\) in the denominator approach 0.
\[ Limit = \frac{4 - 0 + 0}{4 + 0 - 0} = \frac{4}{4} = 1 \]


Step 4: Final Answer:

The value of the limit is 1.
Quick Tip: For limits at infinity of rational functions: If degree of numerator \(<\) degree of denominator, limit is 0. If degree of numerator \(>\) degree of denominator, limit is \(\pm\infty\). If degree of numerator = degree of denominator, limit is the ratio of leading coefficients. This shortcut can save a lot of time.

Step 1: Understanding the Question:

We need to evaluate the limit of a function as \(x\) approaches 1. Direct substitution of \(x=1\) gives the indeterminate form \(\frac{0}{0}\).


Step 2: Key Formula or Approach:

We can solve this using one of three common methods for indeterminate forms:

1. Factorization: Factor the numerator and cancel common terms. The relevant identity is \(a^3-b^3 = (a-b)(a^2+ab+b^2)\).

2. L'Hôpital's Rule: Differentiate the numerator and the denominator separately and then take the limit.

3. Standard Limit Formula: Use the formula \( \lim_{x\to a} \frac{x^n - a^n}{x - a} = na^{n-1} \).


Step 3: Detailed Explanation:

Method 1: Factorization
\[ \lim_{x\to 1} \frac{x^3 - 1^3}{x - 1} = \lim_{x\to 1} \frac{(x-1)(x^2 + x \cdot 1 + 1^2)}{x - 1} \]

Cancel the \((x-1)\) term:
\[ \lim_{x\to 1} (x^2 + x + 1) \]

Now substitute \(x=1\):
\[ (1)^2 + 1 + 1 = 1 + 1 + 1 = 3 \]


Method 2: L'Hôpital's Rule

Since we have the \(\frac{0}{0}\) form, we can differentiate the numerator and denominator:
\[ \lim_{x\to 1} \frac{\frac{d}{dx}(x^3 - 1)}{\frac{d}{dx}(x - 1)} = \lim_{x\to 1} \frac{3x^2}{1} \]

Now substitute \(x=1\):
\[ \frac{3(1)^2}{1} = 3 \]


Method 3: Standard Limit Formula

The limit is in the form \( \lim_{x\to a} \frac{x^n - a^n}{x - a} \) with \(n=3\) and \(a=1\).

The result is \(na^{n-1}\):
\[ 3 \cdot (1)^{3-1} = 3 \cdot 1^2 = 3 \]


Step 4: Final Answer:

The value of the limit is 3.
Quick Tip: Recognizing the standard limit form \( \lim_{x\to a} \frac{x^n - a^n}{x - a} = na^{n-1} \) is the fastest way to solve this type of problem. It's a fundamental limit that's worth memorizing for competitive exams.

Step 1: Understanding the Question:

We need to find the derivative of the function \(y = x^x\), which is a function raised to the power of a function.


Step 2: Key Formula or Approach:

For functions of the form \(y = [f(x)]^{g(x)}\), we use logarithmic differentiation.

1. Take the natural logarithm (\(\ln\)) of both sides.

2. Use logarithm properties to simplify the expression.

3. Differentiate both sides implicitly with respect to \(x\).

4. Solve for \(\frac{dy}{dx}\).


Step 3: Detailed Explanation:

Let \(y = x^x\).

Take the natural logarithm of both sides:
\[ \ln y = \ln(x^x) \]

Using the logarithm power rule, \(\ln(a^b) = b \ln a\):
\[ \ln y = x \ln x \]

Now, differentiate both sides with respect to \(x\). We use the product rule on the right side.
\[ \frac{d}{dx}(\ln y) = \frac{d}{dx}(x \ln x) \]
\[ \frac{1}{y} \cdot \frac{dy}{dx} = \left(\frac{d}{dx}(x)\right) \cdot \ln x + x \cdot \left(\frac{d}{dx}(\ln x)\right) \]
\[ \frac{1}{y} \frac{dy}{dx} = (1) \cdot \ln x + x \cdot \left(\frac{1}{x}\right) \]
\[ \frac{1}{y} \frac{dy}{dx} = \ln x + 1 \]

To find \(\frac{dy}{dx}\), multiply both sides by \(y\):
\[ \frac{dy}{dx} = y (1 + \ln x) \]

Finally, substitute back \(y = x^x\):
\[ \frac{dy}{dx} = x^x (1 + \ln x) \]

Assuming \(\log x\) in the options represents the natural logarithm, this matches option (D).


Step 4: Final Answer:

The derivative of \(x^x\) is \(x^x(1 + \log x)\).
Quick Tip: The derivative of \(x^x\) is a standard result that's good to remember. Whenever you see a function in the form of (variable)\textsuperscript{(variable)}, logarithmic differentiation is the method to use.

Step 1: Understanding the Question:

We need to find the derivative of the inverse tangent function \(\tan^{-1}\left(\frac{x}{a}\right)\).


Step 2: Key Formula or Approach:

We will use the chain rule along with the standard derivative of the arctangent function.

Standard Derivative: \(\frac{d}{du}(\tan^{-1} u) = \frac{1}{1+u^2}\).

Chain Rule: If \(y = f(u)\) and \(u = g(x)\), then \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\).


Step 3: Detailed Explanation:

Let \(y = \tan^{-1}\left(\frac{x}{a}\right)\).

Let \(u = \frac{x}{a}\). Then \(y = \tan^{-1}(u)\).

First, find the derivative of \(y\) with respect to \(u\):
\[ \frac{dy}{du} = \frac{1}{1+u^2} \]

Next, find the derivative of \(u\) with respect to \(x\):
\[ \frac{du}{dx} = \frac{d}{dx}\left(\frac{x}{a}\right) = \frac{1}{a} \]

Now, apply the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \left(\frac{1}{1+u^2}\right) \cdot \left(\frac{1}{a}\right) \]

Substitute \(u = \frac{x}{a}\) back into the expression:
\[ \frac{dy}{dx} = \left(\frac{1}{1 + \left(\frac{x}{a}\right)^2}\right) \cdot \left(\frac{1}{a}\right) \]

Simplify the expression:
\[ \frac{dy}{dx} = \left(\frac{1}{1 + \frac{x^2}{a^2}}\right) \cdot \left(\frac{1}{a}\right) = \left(\frac{1}{\frac{a^2+x^2}{a^2}}\right) \cdot \left(\frac{1}{a}\right) \]
\[ \frac{dy}{dx} = \left(\frac{a^2}{a^2+x^2}\right) \cdot \left(\frac{1}{a}\right) \]
\[ \frac{dy}{dx} = \frac{a}{a^2+x^2} \]


Step 4: Final Answer:

The derivative is \(\frac{a}{a^2+x^2}\).
Quick Tip: The derivative of \(\tan^{-1}\left(\frac{x}{a}\right)\) is a standard formula in calculus. It's highly recommended to memorize this result, \(\frac{a}{a^2+x^2}\), and the similar one for \(\sin^{-1}\left(\frac{x}{a}\right)\) to save time in exams.

Step 1: Understanding the Question:

We are given an infinitely nested radical function and asked to find its derivative.


Step 2: Key Formula or Approach:

The key to solving this is to recognize the repeating pattern. The entire expression under the first square root is simply the original function \(y\) itself. This allows us to write a simple implicit equation for \(y\), which can then be differentiated.


Step 3: Detailed Explanation:

The given function is:
\[ y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \dots \infty}}} \]

We can rewrite this as:
\[ y = \sqrt{\sin x + y} \]

To remove the square root, we square both sides of the equation:
\[ y^2 = \sin x + y \]

Now, we differentiate this equation implicitly with respect to \(x\):
\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(\sin x) + \frac{d}{dx}(y) \]
\[ 2y \frac{dy}{dx} = \cos x + \frac{dy}{dx} \]

Now, we need to solve for \(\frac{dy}{dx}\). Group all the \(\frac{dy}{dx}\) terms on one side:
\[ 2y \frac{dy}{dx} - \frac{dy}{dx} = \cos x \]

Factor out \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} (2y - 1) = \cos x \]

Isolate \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{\cos x}{2y - 1} \]

To match the given options, we can multiply the numerator and denominator by -1:
\[ \frac{dy}{dx} = \frac{-\cos x}{-(2y - 1)} = \frac{-\cos x}{1 - 2y} \]


Step 4: Final Answer:

The derivative \(\frac{dy}{dx}\) is \(-\frac{\cos x}{1 - 2y}\).
Quick Tip: For any function of the form \(y = \sqrt{f(x) + \sqrt{f(x) + \dots}}\), the derivative follows the pattern \(\frac{dy}{dx} = \frac{f'(x)}{2y-1}\). Memorizing this general form can lead to an instant answer for such problems.

Step 1: Understanding the Question:

We need to find the slope of the normal line to the given curve (an astroid) at the specified point (1, 1).


Step 2: Key Formula or Approach:

1. Find the derivative \(\frac{dy}{dx}\) of the curve's equation using implicit differentiation. This will give the slope of the tangent (\(m_T\)).

2. Evaluate \(\frac{dy}{dx}\) at the given point (1, 1) to find the specific slope of the tangent at that point.

3. The slope of the normal (\(m_N\)) is the negative reciprocal of the slope of the tangent: \(m_N = -\frac{1}{m_T}\).


Step 3: Detailed Explanation:

The equation of the curve is \(x^{2/3} + y^{2/3} = 2\).

Differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}(x^{2/3}) + \frac{d}{dx}(y^{2/3}) = \frac{d}{dx}(2) \]
\[ \frac{2}{3}x^{(2/3 - 1)} + \frac{2}{3}y^{(2/3 - 1)} \cdot \frac{dy}{dx} = 0 \]
\[ \frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3} \frac{dy}{dx} = 0 \]

Divide the entire equation by \(\frac{2}{3}\):
\[ x^{-1/3} + y^{-1/3} \frac{dy}{dx} = 0 \]

Solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = -\frac{x^{-1/3}}{y^{-1/3}} = -\left(\frac{y}{x}\right)^{1/3} \]

This is the slope of the tangent, \(m_T\). Now, evaluate \(m_T\) at the point (1, 1):
\[ m_T = -\left(\frac{1}{1}\right)^{1/3} = -1 \]

The slope of the normal, \(m_N\), is the negative reciprocal of \(m_T\).
\[ m_N = -\frac{1}{m_T} = -\frac{1}{-1} = 1 \]


Step 4: Final Answer:

The slope of the normal to the curve at (1, 1) is 1.
Quick Tip: Be careful not to confuse the slope of the tangent with the slope of the normal. After finding \(\frac{dy}{dx}\), always remember to take the negative reciprocal if the question asks for the normal's slope. A common mistake is to stop after finding the tangent's slope.

Step 1: Understanding the Question:

We need to find the equation of the line that is tangent to the curve \(y=x^3\) at the point (1, 1).


Step 2: Key Formula or Approach:

1. Find the derivative of the function, \(\frac{dy}{dx}\), to get the formula for the slope of the tangent.

2. Evaluate the derivative at the given point to find the numerical slope, \(m\).

3. Use the point-slope form of a linear equation, \(y - y_1 = m(x - x_1)\), to find the equation of the tangent line.


Step 3: Detailed Explanation:

The curve is given by \(y = x^3\).

First, find the derivative with respect to \(x\):
\[ \frac{dy}{dx} = 3x^2 \]

This gives the slope of the tangent at any point \(x\). We need the slope at the point (1, 1). Substitute \(x=1\) into the derivative:
\[ m = 3(1)^2 = 3 \]

The slope of the tangent at (1, 1) is 3.

Now, use the point-slope form with \(m=3\) and the point \((x_1, y_1) = (1, 1)\):
\[ y - y_1 = m(x - x_1) \]
\[ y - 1 = 3(x - 1) \]

Simplify the equation:
\[ y - 1 = 3x - 3 \]

Rearrange it into the general form \(Ax+By+C=0\):
\[ 3x - y - 3 + 1 = 0 \]
\[ 3x - y - 2 = 0 \]


Step 4: Final Answer:

The equation of the tangent to the curve at (1, 1) is \(3x - y - 2 = 0\).
Quick Tip: To quickly check your answer, ensure two things: 1. The point (1,1) must satisfy the final equation. For option (C), \(3(1) - 1 - 2 = 3-3 = 0\). It works. 2. The slope of the line from the equation must match your calculated slope. For \(3x - y - 2 = 0\), the slope is \(-A/B = -3/(-1) = 3\), which matches.

Step 1: Understanding the Question:

We need to find the x-coordinate of the point where the given polynomial function has a local minimum value.


Step 2: Key Formula or Approach:

We use the second derivative test to find local minima and maxima.

1. Find the first derivative, \(f'(x)\).

2. Find the critical points by solving the equation \(f'(x) = 0\).

3. Find the second derivative, \(f''(x)\).

4. For each critical point \(c\), evaluate \(f''(c)\).

- If \(f''(c) > 0\), the function has a local minimum at \(x=c\).

- If \(f''(c) < 0\), the function has a local maximum at \(x=c\).


Step 3: Detailed Explanation:

The function is \(f(x) = 2x^3 + 3x^2 - 36x + 10\).

1. Find the first derivative:
\[ f'(x) = \frac{d}{dx}(2x^3 + 3x^2 - 36x + 10) = 6x^2 + 6x - 36 \]

2. Find the critical points by setting \(f'(x) = 0\):
\[ 6x^2 + 6x - 36 = 0 \]

Divide by 6 to simplify:
\[ x^2 + x - 6 = 0 \]

Factor the quadratic equation:
\[ (x+3)(x-2) = 0 \]

The critical points are \(x = -3\) and \(x = 2\).

3. Find the second derivative:
\[ f''(x) = \frac{d}{dx}(6x^2 + 6x - 36) = 12x + 6 \]

4. Apply the second derivative test for each critical point:

For \(x = -3\):
\[ f''(-3) = 12(-3) + 6 = -36 + 6 = -30 \]

Since \(f''(-3) < 0\), there is a local maximum at \(x = -3\).

For \(x = 2\):
\[ f''(2) = 12(2) + 6 = 24 + 6 = 30 \]

Since \(f''(2) > 0\), there is a local minimum at \(x = 2\).


Step 4: Final Answer:

The function has a minimum at the value \(x = 2\).
Quick Tip: For a cubic polynomial with two critical points, the smaller x-value (more negative) typically corresponds to the local maximum, and the larger x-value corresponds to the local minimum, assuming a positive leading coefficient. This can be a quick check.

Step 1: Understanding the Question:

We are given a function \(z\) of two variables, \(x\) and \(y\), and we need to evaluate an expression involving its partial derivatives.


Step 2: Key Formula or Approach:

1. Find the partial derivative of \(z\) with respect to \(x\), \(\frac{\partial z}{\partial x}\), by treating \(y\) as a constant.

2. Find the partial derivative of \(z\) with respect to \(y\), \(\frac{\partial z}{\partial y}\), by treating \(x\) as a constant.

3. Substitute these derivatives into the given expression and simplify.


Step 3: Detailed Explanation:

The given function is \(z = x^2 - y^2\).

1. Calculate the partial derivative with respect to \(x\):
\[ \frac{\partial z}{\partial x} = \frac{\partial}{\partial x}(x^2 - y^2) = 2x - 0 = 2x \]

2. Calculate the partial derivative with respect to \(y\):
\[ \frac{\partial z}{\partial y} = \frac{\partial}{\partial y}(x^2 - y^2) = 0 - 2y = -2y \]

3. Now, substitute these results into the given expression:
\[ \frac{1}{x} \frac{\partial z}{\partial x} + \frac{1}{y} \frac{\partial z}{\partial y} = \frac{1}{x} (2x) + \frac{1}{y} (-2y) \]

Simplify the expression:
\[ = 2 - 2 = 0 \]


Step 4: Final Answer:

The value of the expression is 0.
Quick Tip: This problem type is a straightforward application of partial differentiation rules. Remember that when differentiating with respect to one variable, all other variables are treated as constants.

Step 1: Understanding the Question:

We need to find the sum of the second partial derivatives of the function \(u=e^{xy}\) with respect to \(x\) and \(y\), and then evaluate this sum at the point (1, 1).


Step 2: Key Formula or Approach:

1. Find the first partial derivative \(\frac{\partial u}{\partial x}\).

2. Find the second partial derivative \(\frac{\partial^2 u}{\partial x^2}\) by differentiating \(\frac{\partial u}{\partial x}\) with respect to \(x\).

3. Find the first partial derivative \(\frac{\partial u}{\partial y}\).

4. Find the second partial derivative \(\frac{\partial^2 u}{\partial y^2}\) by differentiating \(\frac{\partial u}{\partial y}\) with respect to \(y\).

5. Add the two second derivatives and substitute \(x=1, y=1\).


Step 3: Detailed Explanation:

The function is \(u = e^{xy}\).

Derivatives with respect to x:

First partial derivative w.r.t. x (using chain rule, treat y as constant):
\[ \frac{\partial u}{\partial x} = e^{xy} \cdot \frac{\partial}{\partial x}(xy) = y e^{xy} \]

Second partial derivative w.r.t. x (treat y as constant):
\[ \frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x}(y e^{xy}) = y \cdot \frac{\partial}{\partial x}(e^{xy}) = y \cdot (y e^{xy}) = y^2 e^{xy} \]


Derivatives with respect to y:

First partial derivative w.r.t. y (using chain rule, treat x as constant):
\[ \frac{\partial u}{\partial y} = e^{xy} \cdot \frac{\partial}{\partial y}(xy) = x e^{xy} \]

Second partial derivative w.r.t. y (treat x as constant):
\[ \frac{\partial^2 u}{\partial y^2} = \frac{\partial}{\partial y}(x e^{xy}) = x \cdot \frac{\partial}{\partial y}(e^{xy}) = x \cdot (x e^{xy}) = x^2 e^{xy} \]


Sum of second derivatives:
\[ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = y^2 e^{xy} + x^2 e^{xy} = (x^2 + y^2)e^{xy} \]

Now, evaluate this expression at the point (1, 1):
\[ \left. \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} \right|_{(1,1)} = (1^2 + 1^2)e^{(1)(1)} = (1+1)e^1 = 2e \]


Step 4: Final Answer:

The value of the expression at (1, 1) is 2e.
Quick Tip: When performing partial differentiation, be methodical. Write down each step clearly to avoid confusion, especially when calculating second-order derivatives. Pay close attention to which variable you are treating as a constant in each step.

Step 1: Understanding the Question:

We need to find the indefinite integral of the function \((\log \sec x) \tan x\).


Step 2: Key Formula or Approach:

This integral can be solved using the method of substitution. We look for a function and its derivative within the integrand.

Let's test the substitution \(t = \log \sec x\). We need to find its derivative, \(\frac{dt}{dx}\).


Step 3: Detailed Explanation:

Let \(I = \int (\log \sec x) \tan x \,dx\).

Let's use the substitution \(t = \log \sec x\).

Now, we differentiate \(t\) with respect to \(x\) to find \(dt\):
\[ \frac{dt}{dx} = \frac{d}{dx}(\log \sec x) \]

Using the chain rule, let \(u = \sec x\), so \(t = \log u\).
\[ \frac{dt}{dx} = \frac{dt}{du} \cdot \frac{du}{dx} = \frac{1}{u} \cdot (\sec x \tan x) = \frac{1}{\sec x} \cdot (\sec x \tan x) = \tan x \]

So, we have \(\frac{dt}{dx} = \tan x\), which means \(dt = \tan x \,dx\).

Now we can substitute \(t\) and \(dt\) back into the integral:
\[ I = \int (\log \sec x) (\tan x \,dx) = \int t \,dt \]

This is a standard integral:
\[ \int t \,dt = \frac{t^2}{2} + c \]

Finally, substitute back \(t = \log \sec x\):
\[ I = \frac{(\log \sec x)^2}{2} + c = \frac{1}{2}(\log \sec x)^2 + c \]


Step 4: Final Answer:

The value of the integral is \(\frac{1}{2}(\log \sec x)^2 + c\).
Quick Tip: In substitution integrals, look for a composite function. The derivative of the "inner" function is often present as a factor in the integrand. Here, the inner function is \(\log(\sec x)\), and its derivative, \(\tan x\), is conveniently the other factor. Recognizing this pattern is key.

Step 1: Understanding the Question:

We need to find the indefinite integral of \(\sin^2 x\).


Step 2: Key Formula or Approach:

We cannot integrate \(\sin^2 x\) directly. We must first reduce the power using a trigonometric identity. The relevant power-reduction formula comes from the double-angle identity for cosine:
\[ \cos(2x) = 1 - 2\sin^2 x \]

Rearranging this formula to solve for \(\sin^2 x\), we get:
\[ \sin^2 x = \frac{1 - \cos(2x)}{2} \]


Step 3: Detailed Explanation:

Let \(I = \int \sin^2 x \,dx\).

Substitute the power-reduction formula into the integral:
\[ I = \int \frac{1 - \cos(2x)}{2} \,dx \]

We can split this into two separate integrals:
\[ I = \frac{1}{2} \int (1 - \cos(2x)) \,dx = \frac{1}{2} \left( \int 1 \,dx - \int \cos(2x) \,dx \right) \]

Integrate each term:
\[ \int 1 \,dx = x \]
\[ \int \cos(2x) \,dx = \frac{\sin(2x)}{2} \]

Now, substitute these back into the expression for \(I\):
\[ I = \frac{1}{2} \left( x - \frac{\sin(2x)}{2} \right) + c \]

Distribute the \(\frac{1}{2}\):
\[ I = \frac{x}{2} - \frac{\sin(2x)}{4} + c \]


Step 4: Final Answer:

The integral of \(\sin^2 x\) is \(\frac{x}{2} - \frac{\sin 2x}{4} + c\).
Quick Tip: Memorize the power-reduction formulas for \(\sin^2 x\) and \(\cos^2 x\), as they are fundamental for integrating even powers of sine and cosine. \(\sin^2 x = \frac{1 - \cos(2x)}{2}\) \(\cos^2 x = \frac{1 + \cos(2x)}{2}\) These are direct applications of the \(\cos(2x)\) double-angle identities.

Step 1: Understanding the Question:

We need to evaluate the indefinite integral of \(\frac{1}{25 - x^2}\).


Step 2: Key Formula or Approach:

This integral matches the standard integration formula:
\[ \int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \ln \left| \frac{a+x}{a-x} \right| + C \]

Alternatively, we can use partial fraction decomposition.


Step 3: Detailed Explanation:

Method 1: Using the Standard Formula

The integral is \(\int \frac{dx}{25 - x^2}\).

We can write this as \(\int \frac{dx}{5^2 - x^2}\).

This matches the standard form with \(a=5\).

Applying the formula:
\[ \int \frac{dx}{5^2 - x^2} = \frac{1}{2(5)} \ln \left| \frac{5+x}{5-x} \right| + c \]
\[ = \frac{1}{10} \ln \left| \frac{5+x}{5-x} \right| + c \]

Assuming \(\log\) in the options means natural logarithm (\(\ln\)), this matches option (C).


Method 2: Partial Fraction Decomposition

Factor the denominator: \(25 - x^2 = (5-x)(5+x)\).

Decompose the fraction:
\[ \frac{1}{(5-x)(5+x)} = \frac{A}{5-x} + \frac{B}{5+x} \]

Using the cover-up method:

To find A, cover \((5-x)\) and set \(x=5\): \(A = \frac{1}{5+5} = \frac{1}{10}\).

To find B, cover \((5+x)\) and set \(x=-5\): \(B = \frac{1}{5-(-5)} = \frac{1}{10}\).

So the integral becomes:
\[ \int \left( \frac{1/10}{5-x} + \frac{1/10}{5+x} \right) dx = \frac{1}{10} \int \frac{dx}{5-x} + \frac{1}{10} \int \frac{dx}{5+x} \]
\[ = \frac{1}{10} (-\ln|5-x|) + \frac{1}{10} (\ln|5+x|) + c \]
\[ = \frac{1}{10} (\ln|5+x| - \ln|5-x|) + c \]

Using the logarithm property \(\ln a - \ln b = \ln(a/b)\):
\[ = \frac{1}{10} \ln \left| \frac{5+x}{5-x} \right| + c \]


Step 4: Final Answer:

The result of the integration is \(\frac{1}{10} \log \left| \frac{5+x}{5-x} \right| + c\).
Quick Tip: Memorizing the standard integral forms for \(\frac{1}{a^2-x^2}\), \(\frac{1}{x^2-a^2}\), and \(\frac{1}{a^2+x^2}\) is essential for speed and accuracy in competitive exams. They appear very frequently.

Step 1: Understanding the Question:

We need to evaluate a definite integral. The integrand has a term \((1-x)^9\) which is difficult to expand.


Step 2: Key Formula or Approach:

We can use a property of definite integrals or substitution.

Property: \(\int_0^a f(x) \,dx = \int_0^a f(a-x) \,dx\).

Substitution: Let \(u = 1-x\).

We will use the property as it is often faster.


Step 3: Detailed Explanation:

Let \(I = \int_0^1 x(1-x)^9 \,dx\).

Using the property \(\int_0^a f(x) \,dx = \int_0^a f(a-x) \,dx\) with \(a=1\):

Here, \(f(x) = x(1-x)^9\).

So, \(f(a-x) = f(1-x) = (1-x)(1-(1-x))^9 = (1-x)(x)^9 = x^9(1-x)\).

Therefore, the integral becomes:
\[ I = \int_0^1 x^9(1-x) \,dx \]

This is much easier to integrate as we just need to expand the integrand:
\[ I = \int_0^1 (x^9 - x^{10}) \,dx \]

Now, perform the integration:
\[ I = \left[ \frac{x^{10}}{10} - \frac{x^{11}}{11} \right]_0^1 \]

Evaluate at the limits:
\[ I = \left( \frac{1^{10}}{10} - \frac{1^{11}}{11} \right) - \left( \frac{0^{10}}{10} - \frac{0^{11}}{11} \right) \]
\[ I = \left( \frac{1}{10} - \frac{1}{11} \right) - (0) \]

Find a common denominator:
\[ I = \frac{11 - 10}{110} = \frac{1}{110} \]


Step 4: Final Answer:

The value of the definite integral is \(\frac{1}{110}\).
Quick Tip: The property \(\int_0^a f(x) \,dx = \int_0^a f(a-x) \,dx\) is extremely useful for definite integrals where the integrand involves a term like \((a-x)^n\). It often simplifies the integrand significantly, making integration straightforward.

Step 1: Understanding the Question:

We need to evaluate the definite integral of the absolute value function, \(|x|\), over a symmetric interval \([-a, a]\).


Step 2: Key Formula or Approach:

The function \(f(x) = |x|\) is an even function, because \(f(-x) = |-x| = |x| = f(x)\).

For any even function, we have the property:
\[ \int_{-a}^{a} f(x) \,dx = 2 \int_{0}^{a} f(x) \,dx \]

We also need the definition of \(|x|\): \(|x| = x\) for \(x \ge 0\).


Step 3: Detailed Explanation:

Let \(I = \int_{-a}^{a} |x| \,dx\).

Since \(|x|\) is an even function, we can simplify the integral:
\[ I = 2 \int_{0}^{a} |x| \,dx \]

In the interval \([0, a]\), \(x\) is non-negative, so \(|x| = x\).

The integral becomes:
\[ I = 2 \int_{0}^{a} x \,dx \]

Now, we evaluate this simple integral:
\[ I = 2 \left[ \frac{x^2}{2} \right]_0^a \]
\[ I = 2 \left( \frac{a^2}{2} - \frac{0^2}{2} \right) \]
\[ I = 2 \left( \frac{a^2}{2} \right) = a^2 \]


Alternative Method (Splitting the integral):

We can split the integral based on the definition of \(|x|\):
\(|x| = -x\) for \(x < 0\) and \(|x| = x\) for \(x \ge 0\).
\[ I = \int_{-a}^{0} |x| \,dx + \int_{0}^{a} |x| \,dx \]
\[ I = \int_{-a}^{0} (-x) \,dx + \int_{0}^{a} x \,dx \]
\[ I = \left[ -\frac{x^2}{2} \right]_{-a}^{0} + \left[ \frac{x^2}{2} \right]_{0}^{a} \]
\[ I = \left( -\frac{0^2}{2} - \left(-\frac{(-a)^2}{2}\right) \right) + \left( \frac{a^2}{2} - \frac{0^2}{2} \right) \]
\[ I = \left( 0 + \frac{a^2}{2} \right) + \left( \frac{a^2}{2} - 0 \right) = \frac{a^2}{2} + \frac{a^2}{2} = a^2 \]


Step 4: Final Answer:

The value of the integral is \(a^2\).
Quick Tip: Recognizing whether a function is even or odd is a powerful shortcut for definite integrals over symmetric intervals like \([-a, a]\). If \(f(x)\) is even, \(\int_{-a}^{a} f(x) \,dx = 2 \int_{0}^{a} f(x) \,dx\). If \(f(x)\) is odd, \(\int_{-a}^{a} f(x) \,dx = 0\). This can simplify the calculation significantly.

Step 1: Understanding the Question:

We need to evaluate a definite integral involving trigonometric functions.


Step 2: Key Formula or Approach:

The integrand looks complicated. We should try to simplify it using trigonometric identities. The key is to relate the numerator \(\cos 2x\) to the denominator \(\sin x + \cos x\).

The relevant identity for the numerator is the difference of squares form:
\[ \cos 2x = \cos^2 x - \sin^2 x \]

This can be factored as \((\cos x - \sin x)(\cos x + \sin x)\).


Step 3: Detailed Explanation:

Let \(I = \int_{0}^{\pi/2} \frac{\cos 2x}{\sin x + \cos x} \,dx\).

Substitute the identity for \(\cos 2x\) into the integral:
\[ I = \int_{0}^{\pi/2} \frac{\cos^2 x - \sin^2 x}{\sin x + \cos x} \,dx \]

Factor the numerator as a difference of squares:
\[ I = \int_{0}^{\pi/2} \frac{(\cos x - \sin x)(\cos x + \sin x)}{\sin x + \cos x} \,dx \]

Assuming \(\sin x + \cos x \neq 0\) in the interval \((0, \pi/2)\), we can cancel the common term:
\[ I = \int_{0}^{\pi/2} (\cos x - \sin x) \,dx \]

Now, we can integrate this simplified expression:
\[ I = \left[ \sin x - (-\cos x) \right]_{0}^{\pi/2} \]
\[ I = \left[ \sin x + \cos x \right]_{0}^{\pi/2} \]

Evaluate at the upper and lower limits:
\[ I = \left( \sin\left(\frac{\pi}{2}\right) + \cos\left(\frac{\pi}{2}\right) \right) - (\sin(0) + \cos(0)) \]
\[ I = (1 + 0) - (0 + 1) \]
\[ I = 1 - 1 = 0 \]


Step 4: Final Answer:

The value of the definite integral is 0.
Quick Tip: When faced with a trigonometric fraction in an integral, always look for identities that can simplify the expression. The double angle formulas, especially \(\cos 2x\), are very versatile and have multiple forms (\(\cos^2x - \sin^2x\), \(2\cos^2x - 1\), \(1 - 2\sin^2x\)). Choosing the right form is key. Here, the difference of squares form was perfect for cancellation.

Step 1: Understanding the Question:

We need to find the area of the region enclosed by the parabola \(y=4x^2\), the x-axis (\(y=0\)), and the vertical lines \(x=0\) and \(x=1\).


Step 2: Key Formula or Approach:

The area under a curve \(y=f(x)\) from \(x=a\) to \(x=b\) is given by the definite integral:
\[ A = \int_{a}^{b} f(x) \,dx \]

We must ensure that \(f(x) \ge 0\) in the interval \([a,b]\).


Step 3: Detailed Explanation:

The function is \(f(x) = 4x^2\).

The boundaries are \(a=0\) and \(b=1\).

In the interval \([0, 1]\), \(x^2\) is always non-negative, so the curve \(y=4x^2\) is above the x-axis.

We set up the definite integral for the area:
\[ A = \int_{0}^{1} 4x^2 \,dx \]

Now, we evaluate the integral:
\[ A = 4 \int_{0}^{1} x^2 \,dx \]
\[ A = 4 \left[ \frac{x^3}{3} \right]_0^1 \]

Evaluate at the limits:
\[ A = 4 \left( \frac{1^3}{3} - \frac{0^3}{3} \right) \]
\[ A = 4 \left( \frac{1}{3} - 0 \right) = \frac{4}{3} \]

The area is \(\frac{4}{3}\) square units.


Step 4: Final Answer:

The area of the bounded region is \(\frac{4}{3}\).
Quick Tip: Finding the area under a curve is a direct application of definite integration. Always check if the function is above or below the x-axis in the given interval. If the function dips below the x-axis, you'll need to split the integral and take the absolute value of the negative parts to get the total area.

Step 1: Understanding the Question:

We need to calculate the Root Mean Square (RMS) value of the function \(f(x) = x^2\) over the interval \([0, 1]\).


Step 2: Key Formula or Approach:

The RMS value of a function \(f(x)\) over the interval \([a, b]\) is given by the formula:
\[ RMS = \sqrt{\frac{1}{b-a} \int_a^b [f(x)]^2 \,dx} \]


Step 3: Detailed Explanation:

Here, the function is \(f(x) = x^2\) and the interval is \([a, b] = [0, 1]\).

First, we find the square of the function:
\[ [f(x)]^2 = (x^2)^2 = x^4 \]

Next, we calculate the mean square value, which is the average of the squared function over the interval.
\[ Mean Square = \frac{1}{1-0} \int_0^1 x^4 \,dx = \int_0^1 x^4 \,dx \]

Evaluate the integral:
\[ \int_0^1 x^4 \,dx = \left[ \frac{x^5}{5} \right]_0^1 = \frac{1^5}{5} - \frac{0^5}{5} = \frac{1}{5} \]

Finally, the RMS value is the square root of the mean square value:
\[ RMS = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} \]


Step 4: Final Answer:

The RMS value of \(x^2\) in the interval [0, 1] is \(\frac{1}{\sqrt{5}}\).
Quick Tip: Remember the three steps for RMS: \textbf{S}quare the function, find the \textbf{M}ean (average) of the result over the interval, and then take the square \textbf{R}oot. Following the steps in reverse order (S-M-R) helps in remembering the process.

Step 1: Understanding the Question:

We need to find the degree of the given differential equation. The degree is the highest power of the highest order derivative after the equation has been made free of radicals and fractions with respect to its derivatives.


Step 2: Key Formula or Approach:

To find the degree, we must first clear any fractions or radicals involving the derivatives. The given equation has a derivative \(y'\) in the denominator.


Step 3: Detailed Explanation:

The given differential equation is:
\[ y' + y = \frac{5}{y'} \]

To eliminate the fraction, we multiply the entire equation by \(y'\):
\[ y'(y' + y) = y'\left(\frac{5}{y'}\right) \]
\[ (y')^2 + y \cdot y' = 5 \]

The equation is now a polynomial in terms of its derivatives.

First, identify the order of the equation. The highest order derivative present is \(y'\) (or \(\frac{dy}{dx}\)), so the order is 1.

Next, identify the degree. The degree is the highest power of the highest order derivative. In this equation, the highest power of \(y'\) is 2.

Therefore, the degree of the differential equation is 2.


Step 4: Final Answer:

The degree of the given differential equation is 2.
Quick Tip: Don't be tempted to state the degree by just looking at the initial form of the equation. Always clear denominators and radicals involving any derivative terms before determining the degree. The order can be found from the original equation, but the degree requires this simplification step.

Step 1: Understanding the Question:

We need to find the order of the differential equation that corresponds to the given general solution.


Step 2: Key Formula or Approach:

The order of a differential equation is equal to the number of independent arbitrary constants in its general solution. To form the differential equation, we need to differentiate the solution as many times as there are constants and then eliminate the constants.


Step 3: Detailed Explanation:

The given general solution is:
\[ y = a \sin x + b \cos x \]

This solution contains two independent arbitrary constants, \(a\) and \(b\).

According to the rule, the order of the differential equation must be equal to the number of these constants.

Therefore, the order of the differential equation is 2.


Derivation (for verification):

1. Differentiate the solution with respect to \(x\):
\[ \frac{dy}{dx} = a \cos x - b \sin x \]

2. Differentiate a second time to eliminate the constants:
\[ \frac{d^2y}{dx^2} = -a \sin x - b \cos x \]

Notice that the right side is the negative of the original expression for \(y\).
\[ \frac{d^2y}{dx^2} = -(a \sin x + b \cos x) = -y \]

So, the differential equation is:
\[ \frac{d^2y}{dx^2} + y = 0 \]

The highest order derivative in this equation is the second derivative, so the order is 2. This confirms our initial conclusion.


Step 4: Final Answer:

The order of the differential equation is 2.
Quick Tip: A very quick way to solve this type of problem is to simply count the number of independent arbitrary constants in the general solution. This count directly gives you the order of the differential equation.

Step 1: Understanding the Question:

We need to classify the given first-order differential equation into one of the standard types.


Step 2: Key Formula or Approach:

We need to check if the equation can be rearranged into the standard forms for each type:

- Variable Separable: Can it be written as \(f(y)dy = g(x)dx\)?

- Homogeneous: Can \(\frac{dy}{dx}\) be expressed as a function of \(\frac{y}{x}\)?

- Linear: Can it be written as \(\frac{dy}{dx} + P(x)y = Q(x)\)?

- Exact: Can it be written as \(M(x,y)dx + N(x,y)dy = 0\) where \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\)?


Step 3: Detailed Explanation:

The given equation is \(\frac{dy}{dx} = -\frac{x+y}{1+x^2}\).

Let's split the fraction on the right-hand side:
\[ \frac{dy}{dx} = -\frac{x}{1+x^2} - \frac{y}{1+x^2} \]

Now, let's rearrange the terms to see if it matches the linear form by moving the term with y to the left side:
\[ \frac{dy}{dx} + \frac{1}{1+x^2}y = -\frac{x}{1+x^2} \]

This equation is exactly in the standard form of a first-order linear differential equation, \(\frac{dy}{dx} + P(x)y = Q(x)\), where:

- \(P(x) = \frac{1}{1+x^2}\)

- \(Q(x) = -\frac{x}{1+x^2}\)

It is not variable separable as we cannot group all \(x\) terms with \(dx\) and all \(y\) terms with \(dy\). It is also not homogeneous. Therefore, the correct classification is a first-order linear equation.


Step 4: Final Answer:

The differential equation is a First order Linear equation.
Quick Tip: When classifying a differential equation, always try to rearrange it into the standard linear form \(\frac{dy}{dx} + P(x)y = Q(x)\) first, as it's a very common type. If terms can be separated into functions of only x and only y, it's linear.

Step 1: Understanding the Question:

We need to find the general solution of the given first-order differential equation.


Step 2: Key Formula or Approach:

The equation is of the variable separable type. We will separate the terms involving \(y\) and \(dy\) from the terms involving \(x\) and \(dx\), and then integrate both sides. The key integral required is \(\int \frac{1}{1+y^2} dy = \tan^{-1}(y)\).


Step 3: Detailed Explanation:

The differential equation is:
\[ \frac{dy}{dx} = 1 + y^2 \]

Separate the variables by multiplying by \(dx\) and dividing by \((1+y^2)\):
\[ \frac{dy}{1+y^2} = dx \]

Now, integrate both sides of the equation:
\[ \int \frac{1}{1+y^2} \,dy = \int 1 \,dx \]

Performing the integration gives:
\[ \tan^{-1}(y) = x + c \]

where \(c\) is the constant of integration.

To find the explicit solution for \(y\), we take the tangent of both sides:
\[ y = \tan(x+c) \]


Step 4: Final Answer:

The solution of the differential equation is \(y = \tan(x+c)\).
Quick Tip: When solving differential equations, remember that the constant of integration \(c\) is added immediately after integrating. In this case, `tan⁻¹(y) = x + c` becomes `y = tan(x+c)`, which is different from `y = tan(x) + c`. The position of the constant is crucial.

Step 1: Understanding the Question:

We need to solve a first-order linear differential equation with a given initial condition (an Initial Value Problem).


Step 2: Key Formula or Approach:

The equation is in the linear form \(\frac{dy}{dx} + P(x)y = Q(x)\). We solve it using an integrating factor (I.F.).

1. Identify \(P(x)\) and \(Q(x)\).

2. Calculate the Integrating Factor: I.F. = \(e^{\int P(x)dx}\).

3. The general solution is given by: \(y \cdot (I.F.) = \int Q(x) \cdot (I.F.) \,dx + C\).

4. Use the initial condition \(y(1)=1\) to find the value of the constant \(C\).


Step 3: Detailed Explanation:

The given equation is \(\frac{dy}{dx} + \frac{1}{x}y = x^2\).

1. Here, \(P(x) = \frac{1}{x}\) and \(Q(x) = x^2\).

2. Calculate the integrating factor:
\[ I.F. = e^{\int \frac{1}{x}dx} = e^{\ln x} = x \quad (for x>0) \]

3. Find the general solution:
\[ y \cdot x = \int x^2 \cdot x \,dx + C \]
\[ xy = \int x^3 \,dx + C \]
\[ xy = \frac{x^4}{4} + C \]

4. Apply the initial condition \(y(1) = 1\) (when \(x=1\), \(y=1\)):
\[ (1)(1) = \frac{(1)^4}{4} + C \]
\[ 1 = \frac{1}{4} + C \]
\[ C = 1 - \frac{1}{4} = \frac{3}{4} \]

Substitute the value of \(C\) back into the general solution:
\[ xy = \frac{x^4}{4} + \frac{3}{4} \]

To match the format of the options, multiply the entire equation by 4:
\[ 4xy = x^4 + 3 \]


Step 4: Final Answer:

The solution of the initial value problem is \(4xy = x^4 + 3\).
Quick Tip: The integrating factor method is a standard procedure for first-order linear DEs. Remember the three key steps: find P(x) and Q(x), calculate the I.F., and then apply the solution formula \(y \cdot (I.F.) = \int Q(x) \cdot (I.F.) \,dx + C\).

Step 1: Understanding the Question:

We need to find the general solution for a third-order homogeneous linear differential equation with constant coefficients.


Step 2: Key Formula or Approach:

We solve this by finding the roots of the auxiliary (or characteristic) equation.

1. Form the auxiliary equation by replacing \(\frac{d^ny}{dx^n}\) with \(m^n\).

2. Find the roots of the resulting polynomial equation.

3. The form of the general solution depends on the nature of these roots (real and distinct, real and repeated, or complex).


Step 3: Detailed Explanation:

The given differential equation is \(y''' + 3y'' + 2y' = 0\).

1. The auxiliary equation is:
\[ m^3 + 3m^2 + 2m = 0 \]

2. Factor the polynomial to find the roots:
\[ m(m^2 + 3m + 2) = 0 \]

Factor the quadratic part:
\[ m(m+1)(m+2) = 0 \]

The roots are \(m_1 = 0\), \(m_2 = -1\), and \(m_3 = -2\).

3. Since we have three distinct real roots, the general solution is of the form:
\[ y = c_1 e^{m_1 x} + c_2 e^{m_2 x} + c_3 e^{m_3 x} \]

Substituting the roots we found:
\[ y = c_1 e^{0x} + c_2 e^{-1x} + c_3 e^{-2x} \]

Since \(e^{0x} = 1\), the solution is:
\[ y = c_1(1) + c_2 e^{-x} + c_3 e^{-2x} \]

Using the arbitrary constants \(a, b, c\) from the options, this becomes:
\[ y = a + be^{-x} + ce^{-2x} \]


Step 4: Final Answer:

The general solution is \(y = a + be^{-x} + ce^{-2x}\).
Quick Tip: For homogeneous linear DEs with constant coefficients, the process is always the same: form the auxiliary equation, find its roots, and write the solution based on the type of roots. A root of \(m=0\) always contributes a simple constant term to the solution.

Step 1: Understanding the Question:

We need to find the particular integral (PI) for a second-order non-homogeneous linear differential equation with constant coefficients.


Step 2: Key Formula or Approach:

We use the operator method. The particular integral \(y_p\) is given by \(y_p = \frac{1}{f(D)} R(x)\), where \(D = \frac{d}{dx}\), \(f(D)\) is the differential operator, and \(R(x)\) is the function on the right-hand side.

For \(R(x) = e^{ax}\), we evaluate \(\frac{1}{f(a)}e^{ax}\). If \(f(a)=0\) (case of failure), the rule is \(y_p = x \frac{1}{f'(a)}e^{ax}\).


Step 3: Detailed Explanation:

The differential equation is \((D^2 + 3D + 2)y = e^{-2x}\).

So, \(f(D) = D^2 + 3D + 2\) and \(R(x) = e^{-2x}\).

The particular integral is:
\[ y_p = \frac{1}{D^2 + 3D + 2} e^{-2x} \]

We have the form \(e^{ax}\) with \(a=-2\). Let's evaluate \(f(a) = f(-2)\):
\[ f(-2) = (-2)^2 + 3(-2) + 2 = 4 - 6 + 2 = 0 \]

This is a "case of failure" because the denominator becomes zero. This happens because \(a=-2\) is a root of the auxiliary equation \(m^2+3m+2=0\).

When failure occurs, we apply the rule: multiply by \(x\) and differentiate the denominator with respect to \(D\).
\[ f'(D) = \frac{d}{dD}(D^2 + 3D + 2) = 2D + 3 \]

Now, the particular integral is given by:
\[ y_p = x \frac{1}{f'(D)} e^{-2x} = x \frac{1}{2D + 3} e^{-2x} \]

Substitute \(D=-2\) into the new operator:
\[ y_p = x \frac{1}{2(-2) + 3} e^{-2x} = x \frac{1}{-4 + 3} e^{-2x} = x \frac{1}{-1} e^{-2x} \]
\[ y_p = -xe^{-2x} \]


Step 4: Final Answer:

The particular integral of the differential equation is \(-xe^{-2x}\).
Quick Tip: When finding the particular integral for \(e^{ax}\) using the operator method, always first check if 'a' is a root of the auxiliary equation. If it is (i.e., if \(f(a)=0\)), you know it's a case of failure and you must apply the rule of multiplying by x and differentiating the denominator.

Step 1: Understanding the Question:

We are asked to find the dimensional formula for Power (P) using Force (F), Velocity (V), and Length (L) as the fundamental units instead of Mass (M), Length (L), and Time (T).


Step 2: Key Formula or Approach:

We will use the method of dimensional analysis.

1. Write the dimensions of Power and the new fundamental quantities in terms of the standard M, L, T system.

- Power \([P] = [ML^2T^{-3}]\)

- Force \([F] = [MLT^{-2}]\)

- Velocity \([V] = [LT^{-1}]\)

- Length \([L] = [L]\)

2. Assume that Power is related to F, V, and L by the equation \(P = k F^a V^b L^c\), where k is a dimensionless constant and a, b, c are the powers we need to find.

3. Equate the dimensions on both sides and solve for a, b, and c.


Step 3: Detailed Explanation:

Set up the dimensional equation:
\[ [P] = [F]^a [V]^b [L]^c \]

Substitute the standard dimensions:
\[ [ML^2T^{-3}] = [MLT^{-2}]^a [LT^{-1}]^b [L]^c \]
\[ [M^1L^2T^{-3}] = [M^a L^a T^{-2a}] [L^b T^{-b}] [L^c] \]

Combine the powers on the right side:
\[ [M^1L^2T^{-3}] = [M^a L^{a+b+c} T^{-2a-b}] \]

Now, equate the powers of M, L, and T from both sides:

- For M: \(a = 1\)

- For T: \(-2a - b = -3\)

- For L: \(a + b + c = 2\)

Solve the system of equations:

From the M equation, we have \(a = 1\).

Substitute \(a=1\) into the T equation:
\[ -2(1) - b = -3 \implies -2 - b = -3 \implies b = 1 \]

Substitute \(a=1\) and \(b=1\) into the L equation:
\[ 1 + 1 + c = 2 \implies 2 + c = 2 \implies c = 0 \]

So, the powers are \(a=1, b=1, c=0\). The expression for Power is \(F^1 V^1 L^0\).


Step 4: Final Answer:

Power can be expressed as \(F^1 L^0 V^1\).
Quick Tip: A much faster method is to use known physical relationships. We know that Power is the dot product of Force and Velocity: \(P = \vec{F} \cdot \vec{v}\). Dimensionally, this is simply \([P] = [F][V]\). This directly gives the answer as \(F^1 V^1 L^0\).

Step 1: Understanding the Question:

We need to check the dimensional formulas for each pair of physical quantities listed in the options and find the pair with identical dimensions.


Step 2: Key Formula or Approach:

We will derive the dimensional formula for each quantity based on its physical definition or formula. The fundamental dimensions are Mass (M), Length (L), and Time (T).


Step 3: Detailed Explanation:

Let's analyze each option:

(A) Torque and momentum:

- Torque (\(\tau\)) = Force \(\times\) perpendicular distance = \([MLT^{-2}] \times [L] = [ML^2T^{-2}]\).

- Momentum (\(p\)) = mass \(\times\) velocity = \([M] \times [LT^{-1}] = [MLT^{-1}]\).

The dimensions are not the same.


(B) Surface tension and tension:

- Surface Tension = Force per unit length = \([MLT^{-2}] / [L] = [MT^{-2}]\).

- Tension is a type of force, so its dimension is \([MLT^{-2}]\).

The dimensions are not the same.


(C) Pressure and modulus of elasticity:

- Pressure (\(P\)) = Force / Area = \([MLT^{-2}] / [L^2] = [ML^{-1}T^{-2}]\).

- Modulus of Elasticity (\(E\)) = Stress / Strain.

- Stress = Force / Area = \([MLT^{-2}] / [L^2] = [ML^{-1}T^{-2}]\).

- Strain = Change in dimension / Original dimension = \([L]/[L] = [M^0L^0T^0]\) (dimensionless).

- Therefore, the dimension of Modulus of Elasticity is the same as Stress: \([ML^{-1}T^{-2}]\).

The dimensions of Pressure and Modulus of Elasticity are the same.


(D) Force constant and Planck's constant:

- Force constant (\(k\)) from Hooke's Law (F=kx) = Force / distance = \([MLT^{-2}] / [L] = [MT^{-2}]\).

- Planck's constant (\(h\)) from (E=h\(\nu\)) = Energy / frequency = \([ML^2T^{-2}] / [T^{-1}] = [ML^2T^{-1}]\).

The dimensions are not the same.


Step 4: Final Answer:

The pair with the same dimensional formula is Pressure and modulus of elasticity.
Quick Tip: Remember that quantities like Pressure, Stress, and any Modulus of Elasticity (Young's, Bulk, Shear) all share the same dimension of Force/Area, which is \([ML^{-1}T^{-2}]\). This is because strain is always dimensionless.

Step 1: Understanding the Question:

We are given a vector relationship \(\vec{A} + \vec{B} = \vec{C}\) and a scalar relationship between their magnitudes, \(A^2 + B^2 = C^2\). We need to find the angle \(\theta\) between vectors \(\vec{A}\) and \(\vec{B}\).


Step 2: Key Formula or Approach:

The magnitude of the resultant vector \(\vec{C} = \vec{A} + \vec{B}\) is given by the law of cosines for vectors:
\[ C = |\vec{C}| = \sqrt{A^2 + B^2 + 2AB \cos\theta} \]

Squaring both sides gives:
\[ C^2 = A^2 + B^2 + 2AB \cos\theta \]


Step 3: Detailed Explanation:

We have two expressions for \(C^2\):

1. From the magnitude of the vector sum: \(C^2 = A^2 + B^2 + 2AB \cos\theta\)

2. From the given information: \(C^2 = A^2 + B^2\)

Equating these two expressions for \(C^2\):
\[ A^2 + B^2 = A^2 + B^2 + 2AB \cos\theta \]

Subtract \(A^2 + B^2\) from both sides:
\[ 0 = 2AB \cos\theta \]

Assuming the vectors \(\vec{A}\) and \(\vec{B}\) are non-zero vectors (so their magnitudes \(A\) and \(B\) are non-zero), the only way for the product to be zero is if \(\cos\theta = 0\).
\[ \cos\theta = 0 \]

The angle \(\theta\) for which \(\cos\theta = 0\) is \(90^{\circ}\) or \(\frac{\pi}{2}\) radians.


Step 4: Final Answer:

The angle between vectors \(\vec{A}\) and \(\vec{B}\) is \(90^{\circ}\).
Quick Tip: The condition \(A^2 + B^2 = C^2\) for the sum \(\vec{C} = \vec{A} + \vec{B}\) is the vector equivalent of the Pythagorean theorem. It holds true only when the vectors \(\vec{A}\) and \(\vec{B}\) are perpendicular to each other.

Step 1: Understanding the Question:

We are asked to find the area of a rectangle. The vectors \(\vec{A}\) and \(\vec{B}\) are given to represent the sides. The area of a rectangle is the product of the lengths of its adjacent sides. The wording implies that the lengths of the sides of the rectangle are given by the magnitudes of the vectors \(\vec{A}\) and \(\vec{B}\).


Step 2: Key Formula or Approach:

Area of a rectangle = length \(\times\) width.

The length of a vector \(\vec{V} = x\hat{i} + y\hat{j}\) is its magnitude, given by \(|\vec{V}| = \sqrt{x^2 + y^2}\).


Step 3: Detailed Explanation:

First, we find the length of the side represented by vector \(\vec{A}\).
\[ |\vec{A}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

Next, we find the length of the side represented by vector \(\vec{B}\).
\[ |\vec{B}| = \sqrt{1^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10} \]

Now, we calculate the area of the rectangle:
\[ Area = |\vec{A}| \times |\vec{B}| = 5 \times \sqrt{10} = 5\sqrt{10} units \]

Note: For the vectors to represent the sides of a rectangle, they should be perpendicular (\(\vec{A} \cdot \vec{B} = 0\)). Let's check: \(\vec{A} \cdot \vec{B} = (3)(1) + (4)(3) = 15 \neq 0\). The vectors are not perpendicular. The question is poorly phrased, but the intended meaning is to use the magnitudes of the given vectors as the lengths of the rectangle's sides.


Step 4: Final Answer:

The area of the rectangle is \(5\sqrt{10}\) units.
Quick Tip: When a question about a geometric shape provides vectors for its sides, it usually implies that the magnitudes of those vectors should be used as the lengths. If the shape was a parallelogram, the area would be given by the magnitude of the cross product, \(|\vec{A} \times \vec{B}|\).

Step 1: Understanding the Question:

We are given the initial upward velocity of a pebble thrown from a tower, the total time of flight, and the acceleration due to gravity. We need to find the final velocity just before it hits the ground.


Step 2: Key Formula or Approach:

We can use the first equation of motion for an object under constant acceleration:
\[ v = u + at \]

where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time.


Step 3: Detailed Explanation:

Let's establish a sign convention. We will consider the upward direction as positive and the downward direction as negative.

- Initial velocity, \(u = +5\) m/s (since it's thrown upwards).

- Acceleration, \(a = -g = -10\) m/s\(^2\) (gravity acts downwards).

- Time of flight, \(t = 3\) s.

Now, substitute these values into the equation of motion:
\[ v = 5 + (-10)(3) \]
\[ v = 5 - 30 \]
\[ v = -25 m/s \]

The negative sign indicates that the final velocity is in the downward direction. The question asks for the velocity with which it strikes, and the options are all positive, implying we need to find the speed.

The speed of striking the ground is \(|v| = 25\) m/s.


Step 4: Final Answer:

The pebble strikes the ground with a velocity of 25 m/s.
Quick Tip: For projectile motion problems, consistently applying a sign convention is crucial. Choosing 'up' as positive and 'down' as negative is a standard convention that helps avoid confusion with the signs of velocity and acceleration.

Step 1: Understanding the Question:

A body falls from rest from a height H, taking time T. We need to find its height from the ground at time T/2.


Step 2: Key Formula or Approach:

We use the equation of motion for distance traveled under constant acceleration, starting from rest:
\[ s = ut + \frac{1}{2}at^2 \]

Since the body is released from rest, \(u=0\). Let's take the downward direction as positive, so \(a=g\). The distance fallen from the top is \(s = \frac{1}{2}gt^2\).


Step 3: Detailed Explanation:

First, relate the total height H to the total time T. In time T, the body falls a distance H.
\[ H = \frac{1}{2}gT^2 \quad (Equation 1) \]

Next, find the distance the body has fallen from the top at time \(t = T/2\). Let's call this distance \(s_{T/2}\).
\[ s_{T/2} = \frac{1}{2}g\left(\frac{T}{2}\right)^2 = \frac{1}{2}g\frac{T^2}{4} = \frac{1}{4} \left(\frac{1}{2}gT^2\right) \]

From Equation 1, we know that \(\frac{1}{2}gT^2 = H\). So, we can substitute H into the expression for \(s_{T/2}\):
\[ s_{T/2} = \frac{H}{4} \]

This is the distance fallen from the top of the tower. The question asks for the position (height) of the body from the ground.
\[ Height from ground = Total Height - Distance fallen \]
\[ Height from ground = H - s_{T/2 = H - \frac{H}{4} = \frac{3H}{4} \]


Step 4: Final Answer:

At time T/2, the body is at a height of \(\frac{3H}{4}\) from the ground.
Quick Tip: For an object in free fall from rest, the distance covered is proportional to the square of the time (\(s \propto t^2\)). This means in half the total time, it covers \((1/2)^2 = 1/4\) of the total distance. Therefore, the remaining distance to the ground is \(1 - 1/4 = 3/4\) of the total height.

Step 1: Understanding the Question:

We are given that a body starts from rest with constant acceleration. We need to find the relationship between the distance covered in the first 2 seconds and the distance covered in the subsequent 2 seconds.


Step 2: Key Formula or Approach:

We use the equation of motion for displacement: \(s = ut + \frac{1}{2}at^2\).

Since the body starts from rest, the initial velocity \(u=0\). The formula simplifies to \(s = \frac{1}{2}at^2\).


Step 3: Detailed Explanation:

Let the uniform acceleration be \(a\).

The distance covered in the first 2 seconds (\(t_1 = 2\) s) is \(x\).
\[ x = \frac{1}{2}a(t_1)^2 = \frac{1}{2}a(2)^2 = \frac{1}{2}a(4) = 2a \]

The distance covered in the "next 2 seconds" means the distance traveled between \(t=2\) s and \(t=4\) s. This can be found by calculating the total distance in 4 seconds and subtracting the distance covered in the first 2 seconds.

Total time for both intervals is \(t_2 = 4\) s.

Total distance covered in 4 seconds is \(s_{total}\).
\[ s_{total} = \frac{1}{2}a(t_2)^2 = \frac{1}{2}a(4)^2 = \frac{1}{2}a(16) = 8a \]

The distance covered in the next 2 seconds, \(y\), is:
\[ y = s_{total} - x = 8a - 2a = 6a \]

Now, we find the relationship between \(y\) and \(x\):

We have \(x = 2a\) and \(y = 6a\).
\[ y = 6a = 3 \times (2a) = 3x \]


Step 4: Final Answer:

The relationship between y and x is \(y = 3x\).
Quick Tip: According to Galileo's law of odd numbers, the distances traversed during equal intervals of time by a body falling from rest stand to one another in the same ratio as the odd numbers beginning with unity (1:3:5:7...). Since the time intervals are equal (2s each), the ratio of distances \(x:y\) will be 1:3, which means \(y=3x\).

Step 1: Understanding the Question:

A juggler throws \(n\) balls per second. The time interval between throws is the time it takes for a ball to reach its maximum height. We need to find this maximum height.


Step 2: Key Formula or Approach:

1. Determine the time of flight to the highest point.

2. Use kinematic equations to relate this time to the initial velocity (\(v=u+at\)).

3. Use another kinematic equation to relate the initial velocity to the maximum height (\(v^2=u^2+2as\)).


Step 3: Detailed Explanation:

If the juggler throws \(n\) balls each second, the time interval between two consecutive throws is \(\Delta t = \frac{1}{n}\) seconds.

The problem states this is the time for a ball to reach its highest point. Let's call this time \(t_{up}\).
\[ t_{up} = \frac{1}{n} \]

At the maximum height, the final vertical velocity \(v\) is 0. Using \(v = u + at\) with \(a = -g\) (upwards as positive):
\[ 0 = u - g \cdot t_{up} \]
\[ u = g \cdot t_{up} = g \cdot \frac{1}{n} = \frac{g}{n} \]

This is the initial velocity with which each ball is thrown.

Now, to find the maximum height \(H\), we use the equation \(v^2 = u^2 + 2as\):
\[ 0^2 = u^2 + 2(-g)H \]
\[ u^2 = 2gH \]
\[ H = \frac{u^2}{2g} \]

Substitute the expression for \(u\) we found:
\[ H = \frac{(g/n)^2}{2g} = \frac{g^2/n^2}{2g} = \frac{g}{2n^2} \]


Step 4: Final Answer:

The balls rise to a height of \(\frac{g}{2n^2}\).
Quick Tip: This problem connects rate (\(n\) balls per second) to time (\(t = 1/n\)). Once the time to reach the peak is known, the kinematics of the projectile can be fully determined. Breaking down the problem statement into physical quantities is the first crucial step.

Step 1: Understanding the Question:

We need to find the minimum force required to push a block up an inclined plane, overcoming both gravity and friction.


Step 2: Key Formula or Approach:

We will use a free-body diagram and apply Newton's First Law (for the condition of impending motion, acceleration is zero). The main forces are the applied force, gravity, normal force, and friction.


Step 3: Detailed Explanation:

Let's analyze the forces acting on the block along axes parallel and perpendicular to the inclined plane.

1. Gravitational Force (Weight): \(mg\), acting vertically downwards.
- Component parallel to the incline: \(mg \sin \theta\) (acting down the incline).
- Component perpendicular to the incline: \(mg \cos \theta\) (acting into the incline).

2. Normal Force (N): Acts perpendicular to the surface, outwards. From equilibrium in the perpendicular direction, \(N = mg \cos \theta\).

3. Frictional Force (f): Opposes the motion (or impending motion) up the plane, so it acts down the plane. The maximum static friction (or kinetic friction) is \(f = \mu N = \mu mg \cos \theta\).

4. Applied Force (F): The force required to move the block up the plane, acting parallel to the incline, upwards.

For the block to move up, the applied force F must overcome the sum of the forces pulling it down the incline.
\[ F = (Gravitational component down the incline) + (Frictional force down the incline) \]
\[ F = mg \sin \theta + f \]

Substitute \(f = \mu mg \cos \theta\):
\[ F = mg \sin \theta + \mu mg \cos \theta \]


Step 4: Final Answer:

The force required to move the block up the inclined plane is \(mg \sin \theta + \mu mg \cos \theta\).
Quick Tip: When an object is pushed \textbf{up} an incline, both the parallel component of gravity (\(mg \sin \theta\)) and friction (\(\mu mg \cos \theta\)) act in the same direction (down the incline) and must be overcome. Thus, they add up. If the object were sliding \textbf{down}, friction would act up the incline, opposing the gravitational component.

Step 1: Understanding the Question:

A body starts from rest on a smooth (frictionless) incline. Given the time for the full journey, we need to find the time it takes to cover the first quarter of the distance.


Step 2: Key Formula or Approach:

For an object starting from rest (\(u=0\)) and moving with constant acceleration (\(a\)), the distance covered (\(s\)) in time (\(t\)) is given by the kinematic equation:
\[ s = \frac{1}{2}at^2 \]

From this, we can see that the distance is proportional to the square of the time (\(s \propto t^2\)).


Step 3: Detailed Explanation:

Let \(L\) be the total length of the inclined plane and \(T = 4\) s be the total time to slide down.

Let \(t\) be the time taken to slide a distance of \(s = L/4\).

Using the proportionality \(s \propto t^2\), we can set up a ratio:
\[ \frac{s_1}{s_2} = \frac{t_1^2}{t_2^2} \]

Let \(s_1 = L\), \(t_1 = T = 4\) s.

Let \(s_2 = L/4\), \(t_2 = t\).
\[ \frac{L}{L/4} = \frac{4^2}{t^2} \]
\[ 4 = \frac{16}{t^2} \]

Rearrange to solve for \(t^2\):
\[ t^2 = \frac{16}{4} = 4 \]
\[ t = \sqrt{4} = 2 s \]


Step 4: Final Answer:

The time taken to slide 1/4th of the length is 2 seconds.
Quick Tip: The relationship \(s \propto t^2\) for motion from rest is very powerful. It implies that \(t \propto \sqrt{s}\). To cover 1/4 of the distance, it will take \(\sqrt{1/4} = 1/2\) of the total time. Half of the total time of 4 seconds is 2 seconds.

Step 1: Understanding the Question:

We are given the mass and the initial and final velocity vectors of a body. We need to calculate the change in its kinetic energy.


Step 2: Key Formula or Approach:

The change in kinetic energy (\(\Delta KE\)) is the final kinetic energy minus the initial kinetic energy.
\[ \Delta KE = KE_{final} - KE_{initial} \]

The kinetic energy is given by \(KE = \frac{1}{2}mv^2\), where \(v\) is the speed (magnitude of the velocity vector).


Step 3: Detailed Explanation:

The mass of the body is \(m = 2\) Kg.

Initial velocity, \(\vec{v}_i = 3\hat{i} - 4\hat{j}\).

Final velocity, \(\vec{v}_f = 6\hat{j} + 2\hat{k}\).


First, calculate the initial speed squared (\(v_i^2\)):
\[ v_i^2 = |\vec{v}_i|^2 = (3)^2 + (-4)^2 = 9 + 16 = 25 \, (m/s)^2 \]

Now, calculate the initial kinetic energy (\(KE_i\)):
\[ KE_i = \frac{1}{2}mv_i^2 = \frac{1}{2}(2)(25) = 25 J \]


Next, calculate the final speed squared (\(v_f^2\)):
\[ v_f^2 = |\vec{v}_f|^2 = (6)^2 + (2)^2 = 36 + 4 = 40 \, (m/s)^2 \]

Now, calculate the final kinetic energy (\(KE_f\)):
\[ KE_f = \frac{1}{2}mv_f^2 = \frac{1}{2}(2)(40) = 40 J \]


Finally, calculate the change in kinetic energy:
\[ \Delta KE = KE_f - KE_i = 40 J - 25 J = 15 J \]


Step 4: Final Answer:

The change in kinetic energy of the body is 15 J.
Quick Tip: According to the Work-Energy Theorem, the net work done on an object equals its change in kinetic energy. This calculation gives the net work done on the body to change its velocity. Remember to find the magnitude (speed) from the velocity vector before calculating KE.

Step 1: Understanding the Question:

We need to find the maximum velocity of a girl on a swing, given her heights at the highest and lowest points of her swing. The maximum velocity occurs at the lowest point.


Step 2: Key Formula or Approach:

We will use the principle of conservation of mechanical energy. The total mechanical energy (Kinetic Energy + Potential Energy) at the highest point is equal to the total mechanical energy at the lowest point, assuming no air resistance.
\[ KE_{top} + PE_{top} = KE_{bottom} + PE_{bottom} \]
\[ \frac{1}{2}mv_{top}^2 + mgh_{top} = \frac{1}{2}mv_{bottom}^2 + mgh_{bottom} \]


Step 3: Detailed Explanation:

Let's define the given values:

- Height at the top, \(h_{top} = 3\) m.

- Height at the bottom, \(h_{bottom} = 2\) m.

- At the maximum height (top), the swing momentarily stops, so \(v_{top} = 0\).

- At the lowest point (bottom), the velocity is maximum, so \(v_{bottom} = v_{max}\).

- We use the standard value for acceleration due to gravity, \(g = 9.8\) m/s\(^2\).

Substitute these into the conservation of energy equation:
\[ \frac{1}{2}m(0)^2 + mgh_{top} = \frac{1}{2}mv_{max}^2 + mgh_{bottom} \]
\[ 0 + mgh_{top} = \frac{1}{2}mv_{max}^2 + mgh_{bottom} \]

The mass \(m\) cancels from all terms:
\[ gh_{top} = \frac{1}{2}v_{max}^2 + gh_{bottom} \]

Rearrange to solve for \(v_{max}^2\):
\[ \frac{1}{2}v_{max}^2 = gh_{top} - gh_{bottom} = g(h_{top} - h_{bottom}) \]
\[ v_{max}^2 = 2g(h_{top} - h_{bottom}) \]

Substitute the numerical values:
\[ v_{max}^2 = 2(9.8)(3 - 2) = 2(9.8)(1) = 19.6 \]
\[ v_{max} = \sqrt{19.6} m/s \]


Step 4: Final Answer:

Her maximum velocity is \(\sqrt{19.6}\) m/s.
Quick Tip: In energy conservation problems, the change in kinetic energy is equal to the negative of the change in potential energy: \(\Delta KE = -\Delta PE\). Here, the loss in potential energy, \(mg(h_{top} - h_{bottom})\), is converted into a gain in kinetic energy, \(\frac{1}{2}mv_{max}^2\).

Step 1: Understanding the Question:

We are given the output power of an engine and its efficiency. We need to calculate the power that is supplied to the engine, which is the input power.


Step 2: Key Formula or Approach:

The efficiency (\(\eta\)) of an engine is defined as the ratio of the useful output power to the total input power, usually expressed as a percentage.
\[ \eta = \frac{Output Power}{Input Power} \]

To find the input power, we can rearrange this formula:
\[ Input Power = \frac{Output Power}{\eta} \]


Step 3: Detailed Explanation:

Given values are:

- Output Power = 1000 W

- Efficiency, \(\eta = 80% = \frac{80}{100} = 0.8\)

Using the rearranged formula to find the Input Power:
\[ Input Power = \frac{1000 W}{0.8} \]
\[ Input Power = \frac{10000}{8} W \]
\[ Input Power = 1250 W \]


Step 4: Final Answer:

The input power required for the engine is 1250 W.
Quick Tip: Remember that efficiency is always less than 1 (or 100%). Therefore, the input power must always be greater than the output power, as some energy is always lost (usually as heat). This can help you eliminate options like 800 W and 1000 W immediately.

Step 1: Understanding the Question:

We are considering a "seconds pendulum," which has a specific time period on Earth. We need to find its new time period on a planet with different gravity.


Step 2: Key Formula or Approach:

A seconds pendulum is defined as a pendulum having a time period of exactly 2 seconds on Earth.

The formula for the time period (\(T\)) of a simple pendulum is:
\[ T = 2\pi\sqrt{\frac{L}{g}} \]

where \(L\) is the length of the pendulum and \(g\) is the acceleration due to gravity. From this formula, we can see that the time period is inversely proportional to the square root of gravity (\(T \propto \frac{1}{\sqrt{g}}\)).


Step 3: Detailed Explanation:

Let \(T_E\) and \(g_E\) be the time period and gravity on Earth.

Let \(T_P\) and \(g_P\) be the time period and gravity on the planet.

We are given:

- \(T_E = 2\) s (definition of a seconds pendulum).

- \(g_P = \frac{g_E}{2}\).

Using the proportionality \(T \propto \frac{1}{\sqrt{g}}\), we can set up a ratio:
\[ \frac{T_P}{T_E} = \frac{1/\sqrt{g_P}}{1/\sqrt{g_E}} = \sqrt{\frac{g_E}{g_P}} \]

Substitute the given relationship for gravity:
\[ \frac{T_P}{T_E} = \sqrt{\frac{g_E}{g_E/2}} = \sqrt{2} \]

Now, solve for the time period on the planet, \(T_P\):
\[ T_P = T_E \times \sqrt{2} \]
\[ T_P = 2 \times \sqrt{2} = 2\sqrt{2} sec \]


Step 4: Final Answer:

The time period of the pendulum on the planet is \(2\sqrt{2}\) seconds.
Quick Tip: Remember the definition of a "seconds pendulum" (its period is 2s, not 1s, because one "tick" or half-period is 1s). Also, recall the inverse square root relationship between period and gravity. Lower gravity means a longer (slower) period.

Step 1: Understanding the Question:

We need to find the position (\(x\)) of a particle in Simple Harmonic Motion (SHM) when its velocity (\(v\)) is half of its maximum possible velocity (\(v_{max}\)).


Step 2: Key Formula or Approach:

The velocity of a particle in SHM as a function of its position \(x\) is given by:
\[ v = \omega \sqrt{A^2 - x^2} \]

where \(\omega\) is the angular frequency and \(A\) is the amplitude.

The maximum velocity occurs at the equilibrium position (\(x=0\)) and is given by:
\[ v_{max} = A\omega \]


Step 3: Detailed Explanation:

We are given the condition that \(v = \frac{1}{2}v_{max}\).

Substitute the formulas for \(v\) and \(v_{max}\) into this condition:
\[ \omega \sqrt{A^2 - x^2} = \frac{1}{2}(A\omega) \]

The angular frequency \(\omega\) cancels from both sides:
\[ \sqrt{A^2 - x^2} = \frac{A}{2} \]

To solve for \(x\), square both sides of the equation:
\[ A^2 - x^2 = \left(\frac{A}{2}\right)^2 = \frac{A^2}{4} \]

Now, isolate \(x^2\):
\[ x^2 = A^2 - \frac{A^2}{4} = \frac{4A^2 - A^2}{4} = \frac{3A^2}{4} \]

Take the square root of both sides to find the position \(x\):
\[ x = \pm \sqrt{\frac{3A^2}{4}} = \pm \frac{\sqrt{3}A}{2} \]

The question asks for the position, and the positive value is given in the options.


Step 4: Final Answer:

The position of the particle is at \(\frac{\sqrt{3}A}{2}\).
Quick Tip: The key relationship \(v = \omega \sqrt{A^2 - x^2}\) is fundamental to SHM. It's derived from the conservation of energy in the oscillator system. Memorizing this formula is essential for solving problems that relate position and velocity in SHM.

Step 1: Understanding the Question:

We are given an equation for the displacement of a particle which is a sum of a sine and a cosine function with the same frequency. We need to find the amplitude of the resulting Simple Harmonic Motion (SHM).


Step 2: Key Formula or Approach:

An expression of the form \(x = a \sin(\omega t) + b \cos(\omega t)\) represents an SHM. The amplitude \(A\) of this resultant motion is given by:
\[ A = \sqrt{a^2 + b^2} \]

The expression can be rewritten as \(x = A \sin(\omega t + \phi)\) or \(x = A \cos(\omega t + \delta)\).


Step 3: Detailed Explanation:

The given equation for displacement is:
\[ x = 3 \sin(2t) + 4 \cos(2t) \]

This matches the standard form \(x = a \sin(\omega t) + b \cos(\omega t)\) with:

- \(a = 3\)

- \(b = 4\)

- \(\omega = 2\) rad/s

Now, we can calculate the resultant amplitude \(A\) using the formula:
\[ A = \sqrt{a^2 + b^2} = \sqrt{3^2 + 4^2} \]
\[ A = \sqrt{9 + 16} = \sqrt{25} \]
\[ A = 5 \]

The unit of the amplitude would be the same as the unit of displacement \(x\).


Step 4: Final Answer:

The amplitude of the particle is 5.
Quick Tip: This is a direct application of the superposition of two perpendicular vectors (or phasors in this context). If you have a vector with components 3 and 4, its magnitude is 5. This is a classic 3-4-5 Pythagorean triple, which often appears in physics problems. Recognizing it can provide an instant answer.

Step 1: Understanding the Question:

We are comparing the maximum intensity during the phenomenon of beats with the intensity of a single sound source. The sources have equal amplitudes.


Step 2: Key Formula or Approach:

The intensity (\(I\)) of a wave is proportional to the square of its amplitude (\(A\)).
\[ I \propto A^2 \]

When two waves interfere, the resultant amplitude depends on the phase difference. For beats, the waves cyclically go in and out of phase.

- At constructive interference (maximum loudness), the amplitudes add up.

- At destructive interference (minimum loudness), the amplitudes subtract.


Step 3: Detailed Explanation:

Let the amplitude of each individual sound source be \(A_0\).

The intensity of a single source, \(I_0\), is proportional to \(A_0^2\).
\[ I_0 = k A_0^2 \] (where k is a proportionality constant)

During the formation of beats, the maximum intensity occurs at points of constructive interference. At these points, the amplitudes of the two waves add.

The maximum resultant amplitude, \(A_{max}\), is:
\[ A_{max} = A_0 + A_0 = 2A_0 \]

The maximum intensity, \(I_{max}\), is proportional to the square of this maximum amplitude:
\[ I_{max} = k (A_{max})^2 = k (2A_0)^2 = k (4A_0^2) \]

Now, let's compare the maximum intensity \(I_{max}\) with the intensity of one source \(I_0\):
\[ I_{max} = 4 (k A_0^2) = 4 I_0 \]

This means the maximum intensity is four times the intensity of a single source.


Step 4: Final Answer:

The maximum intensity of beats will be four times the intensity of one source.
Quick Tip: A common mistake is to think that if amplitude doubles, intensity also doubles. Remember that intensity is proportional to the \textbf{square} of the amplitude. So, if amplitude becomes \(n\) times, intensity becomes \(n^2\) times. Here, amplitude doubles (from \(A_0\) to \(2A_0\)), so intensity becomes \(2^2 = 4\) times.

Step 1: Understanding the Question:

This is a problem on the Doppler effect for sound waves. The sound source is moving away from a stationary listener, and we need to find the apparent frequency heard by the listener.


Step 2: Key Formula or Approach:

The general formula for the Doppler effect is:
\[ f' = f \left( \frac{v \pm v_L}{v \mp v_S} \right) \]

where \(f'\) is the apparent frequency, \(f\) is the source frequency, \(v\) is the speed of sound, \(v_L\) is the speed of the listener, and \(v_S\) is the speed of the source.

In our case:

- The listener is static, so \(v_L = 0\).

- The source is moving away from the listener. This should cause the apparent frequency to decrease. To make the fraction smaller, we use a '+' sign in the denominator.

So the formula becomes:
\[ f' = f \left( \frac{v}{v + v_S} \right) \]


Step 3: Detailed Explanation:

We are given the following values:

- Source frequency, \(f = 800\) Hz.

- Speed of sound, \(v = 340\) m/s.

- Speed of the source, \(v_S = 30\) m/s.

Substitute these values into the formula:
\[ f' = 800 \left( \frac{340}{340 + 30} \right) \]
\[ f' = 800 \left( \frac{340}{370} \right) \]
\[ f' = 800 \times \frac{34}{37} \]
\[ f' = \frac{27200}{37} \approx 735.13 Hz \]

This value is closest to 733.3 Hz. The small discrepancy might be due to rounding in the problem's intended answer or a slightly different value for the speed of sound used. 733.3 Hz is the most plausible answer.

Let's re-calculate with 733.3 Hz as the target. \(800 * (340/370) = 735.13...\). \(800 * (330/360) = 733.33\). It seems the problem might have intended to use v=330m/s. However, based on the given values, 735.13 Hz is the calculated answer, and 733.3 Hz is the closest option.


Step 4: Final Answer:

The frequency of sound heard by the listener is approximately 733.3 Hz.
Quick Tip: To remember the signs in the Doppler formula, think logically. If the source and listener are moving closer, the frequency should increase (so make the numerator larger and/or the denominator smaller). If they are moving apart, the frequency should decrease (so make the numerator smaller and/or the denominator larger).

Step 1: Understanding the Question:

We need to analyze the thermodynamics of ice melting at its standard melting point and pressure. Specifically, we need to determine the nature of the work done and the change in internal energy.


Step 2: Key Formula or Approach:

1. Anomalous Expansion of Water: Water is one of the few substances that is denser in its liquid state than in its solid state. This means when ice melts, its volume decreases.

2. Work Done: The work done by a system on its surroundings at constant pressure is given by \(W_{by} = P \Delta V = P(V_{final} - V_{initial})\). The work done on the system is \(W_{on} = -W_{by}\).

3. First Law of Thermodynamics: The change in internal energy is \(\Delta U = Q - W_{by}\), where \(Q\) is the heat added to the system.


Step 3: Detailed Explanation:

When the slab of ice melts, it turns into water. Due to the anomalous property of water, the volume of the resulting water is less than the volume of the initial ice.

- Initial volume = \(V_{ice}\)

- Final volume = \(V_{water}\)

- \(V_{water} < V_{ice}\)

Therefore, the change in volume of the system is negative:
\[ \Delta V = V_{final} - V_{initial} = V_{water} - V_{ice} < 0 \]

Now, let's analyze the work done. The work done by the system on the atmosphere is:
\[ W_{by} = P \Delta V \]

Since \(P\) is positive and \(\Delta V\) is negative, \(W_{by}\) is negative. This means the system does negative work on the atmosphere. Option (A) is incorrect.

The work done on the system by the atmosphere is:
\[ W_{on} = -W_{by} = - (P \Delta V) \]

Since \(P \Delta V\) is negative, \(W_{on}\) is positive. This means positive work is done on the ice-water system by the atmosphere. Option (B) is correct and Option (C) is incorrect.

Let's check the internal energy. Melting is a phase change that requires the absorption of heat (latent heat of fusion), so the heat added to the system, \(Q\), is positive.

From the first law of thermodynamics:
\[ \Delta U = Q - W_{by} \]

Since \(Q > 0\) and \(W_{by} < 0\), we have:
\[ \Delta U = (positive) - (negative) > 0 \]

The internal energy of the ice-water system increases. Therefore, Option (D) is incorrect.


Step 4: Final Answer:

Positive work is done on the ice-water system by the atmosphere.
Quick Tip: The key to this problem is remembering that ice is less dense than water. This means ice melting is a process where the volume contracts. When a system's volume contracts, the surroundings do positive work on the system.

Step 1: Understanding the Question:

We are analyzing a thermodynamic process where a gas is compressed and then heated. We need to find the total change in its internal energy.


Step 2: Key Formula or Approach:

We will use the First Law of Thermodynamics, which states that the change in internal energy (\(\Delta U\)) of a system is equal to the heat added to the system (\(Q\)) minus the work done by the system (\(W\)).
\[ \Delta U = Q - W \]

The work done by the gas during a constant pressure (isobaric) process is given by:
\[ W = P \Delta V = P(V_{final} - V_{initial}) \]


Step 3: Detailed Explanation:

First, let's identify the given quantities:

- Heat added to the gas, \(Q = +100\) J.

- Constant pressure, \(P = 50\) N/m\(^2\).

- Initial volume, \(V_{initial} = 10\) m\(^3\).

- Final volume, \(V_{final} = 4\) m\(^3\).

Next, calculate the work done by the gas:
\[ W = P(V_{final} - V_{initial}) = 50 N/m^2 \times (4 m^3 - 10 m^3) \]
\[ W = 50 \times (-6) = -300 J \]

The negative sign indicates that work is not done by the gas, but rather work is done on the gas during compression.

Now, apply the First Law of Thermodynamics to find the change in internal energy:
\[ \Delta U = Q - W \]
\[ \Delta U = 100 J - (-300 J) \]
\[ \Delta U = 100 + 300 = 400 J \]

Since \(\Delta U\) is positive, the internal energy increases.


Step 4: Final Answer:

The internal energy of the gas increases by 400 J.
Quick Tip: Be very careful with the sign conventions in thermodynamics. \(Q\) is positive when heat is added to the system. \(W\) is positive when work is done by the system. In this case, compression means the system's volume decreases (\(\Delta V < 0\)), so the work done by the system is negative.

Step 1: Understanding the Question:

An ideal gas initially in one container is allowed to expand into an empty (evacuated) container. We need to find the final pressure of the gas.


Step 2: Key Formula or Approach:

This process is a free expansion of an ideal gas. Since no heat is exchanged and no work is done, the temperature of the ideal gas remains constant. Therefore, we can apply Boyle's Law.

Boyle's Law states that for a fixed amount of gas at constant temperature, the pressure and volume are inversely proportional:
\[ P_1 V_1 = P_2 V_2 \]


Step 3: Detailed Explanation:

Let's define the initial and final states of the gas.

Initial State:

- Initial pressure, \(P_1 = 760\) mm of Hg.

- Initial volume, \(V_1 = 10\) liters.

Final State:

- The gas expands to occupy both vessels. So, the final volume is the sum of the volumes of the two vessels.

- Final volume, \(V_2 = 10 liters + 9 liters = 19\) liters.

- Final pressure, \(P_2\), is what we need to find.

Apply Boyle's Law:
\[ P_1 V_1 = P_2 V_2 \]
\[ (760 mm of Hg) \times (10 L) = P_2 \times (19 L) \]

Solve for \(P_2\):
\[ P_2 = \frac{760 \times 10}{19} \]
\[ P_2 = \frac{7600}{19} \]
\[ P_2 = 400 mm of Hg \]


Step 4: Final Answer:

The resultant pressure is 400 mm of Hg.
Quick Tip: In problems where a gas expands into an evacuated container, the final volume is the total volume of all connected containers. Assuming the temperature is constant (which is usually the case for ideal gas free expansion), Boyle's law is the direct way to find the final pressure.

Step 1: Understanding the Question:

We are asked to predict the outcome when a sealed jar completely filled with water is cooled to its freezing point.


Step 2: Key Formula or Approach:

The solution relies on the physical property of water known as anomalous expansion. Unlike most substances, water expands when it freezes into ice.


Step 3: Detailed Explanation:

Water exhibits a unique behavior regarding its density and temperature. Most substances contract upon cooling and solidifying. However, water contracts as it cools from higher temperatures down to 4°C, where it reaches its maximum density. As it cools further from 4°C to 0°C, it begins to expand slightly.

The most significant change occurs during the phase transition from liquid water to solid ice at 0°C. During freezing, the water molecules arrange themselves into a crystalline lattice structure (hexagonal) which is less dense than liquid water. This results in a significant increase in volume, approximately by 9%.

Since the glass jar is sealed and completely full of water, there is no empty space to accommodate this expansion. The expanding ice exerts an immense pressure on the inner walls of the glass jar. This pressure, known as frost wedging or cryostatic pressure, is strong enough to overcome the tensile strength of the glass, causing the jar to crack and break.


Step 4: Final Answer:

The glass jar breaks due to the formation of ice.
Quick Tip: The anomalous expansion of water upon freezing is a fundamental concept with many real-world consequences, such as pipes bursting in winter, the weathering of rocks, and the fact that ice floats on water, which is crucial for aquatic life in cold climates.

Step 1: Understanding the Question:

An air bubble rises from the bottom of a lake to the surface. We need to find the factor by which its volume increases.


Step 2: Key Formula or Approach:

As the bubble rises, the external pressure on it decreases, causing it to expand. Assuming the temperature of the lake water is constant, we can apply Boyle's Law: \(P_1 V_1 = P_2 V_2\). We need to find the pressures at the bottom and at the surface. The pressure is conveniently given in terms of 'meters of water'.


Step 3: Detailed Explanation:

Let the state at the bottom be 1 and at the surface be 2.

Pressure at the surface (\(P_2\)):

This is just the atmospheric pressure.
\[ P_2 = P_{atm} = 10 m of water \]

Pressure at the bottom (\(P_1\)):

This is the sum of the atmospheric pressure and the gauge pressure due to the water column.
\[ P_1 = P_{atm} + P_{gauge} = P_{atm} + h \]
\[ P_1 = 10 m of water + 90 m of water = 100 m of water \]

Let the volume at the bottom be \(V_1\) and at the surface be \(V_2\).

According to Boyle's Law:
\[ P_1 V_1 = P_2 V_2 \]
\[ (100) \times V_1 = (10) \times V_2 \]

We want to find the ratio \(\frac{V_2}{V_1}\), which tells us how many times the volume becomes.
\[ \frac{V_2}{V_1} = \frac{100}{10} = 10 \]

So, \(V_2 = 10 V_1\). The volume becomes 10 times its original volume.


Step 4: Final Answer:

The volume of the bubble becomes 10 times larger.
Quick Tip: When pressure is given in "meters of water," it simplifies calculations. The total pressure at a depth 'h' is simply (Atmospheric pressure in m of water + h). This avoids having to use the formula \(P=\rho g h\) explicitly.

Step 1: Understanding the Question:

The question asks for the underlying physics principle of an endoscope.


Step 2: Key Formula or Approach:

This is a knowledge-based question about the application of optical phenomena. An endoscope uses optical fibers to transmit images from inside the body to an external viewer. We need to identify the principle that allows light to be guided along a curved fiber.


Step 3: Detailed Explanation:

An endoscope consists of a bundle of flexible optical fibers. These fibers are designed to guide light over long distances, even along curved paths. The principle that makes this possible is Total Internal Reflection (TIR).

An optical fiber consists of a core material with a high refractive index (\(n_1\)) surrounded by a cladding material with a slightly lower refractive index (\(n_2\)). Light is introduced into one end of the fiber. As the light travels down the fiber, it strikes the core-cladding boundary at an angle of incidence that is greater than the critical angle.

When the angle of incidence is greater than the critical angle, the light does not refract out of the core into the cladding. Instead, it is completely reflected back into the core. This process repeats itself along the length of the fiber, trapping the light and guiding it to the other end with very minimal loss of intensity. This allows a clear image of the internal organs to be transmitted to the physician's eyepiece or a camera.

- Refraction (A) is the bending of light, but it's TIR (a specific case of refraction and reflection) that is the key.

- Reflection (B) is too general.

- Dispersion (C) is the splitting of light into colors, which is not the primary principle here.


Step 4: Final Answer:

The endoscope is based on the principle of total internal reflection.
Quick Tip: Total Internal Reflection (TIR) is the principle behind several important technologies, including optical fibers (used in endoscopy and telecommunications), sparkling diamonds, and reflecting prisms used in binoculars and periscopes.

Step 1: Understanding the Question:

We are given the wavelength of incident light and the work function of a metal. We need to find the kinetic energy of the photoelectrons emitted.


Step 2: Key Formula or Approach:

We use Einstein's photoelectric equation:
\[ KE_{max} = E_{photon} - \phi \]

where \(KE_{max}\) is the maximum kinetic energy of the emitted electron, \(E_{photon}\) is the energy of the incident photon, and \(\phi\) is the work function of the material.

The energy of a photon can be calculated from its wavelength \(\lambda\). A very useful shortcut formula for this is:
\[ E_{photon} (in eV) = \frac{12400}{\lambda (in Angstroms)} \]


Step 3: Detailed Explanation:

Given values are:

- Wavelength of light, \(\lambda = 5000\) Å.

- Work function, \(\phi = 1.9\) eV.

First, calculate the energy of the incident photons in eV using the shortcut formula:
\[ E_{photon} = \frac{12400}{5000} eV \]
\[ E_{photon} = \frac{12.4}{5} = 2.48 eV \]

Now, use the photoelectric equation to find the maximum kinetic energy:
\[ KE_{max} = E_{photon} - \phi \]
\[ KE_{max} = 2.48 eV - 1.9 eV \]
\[ KE_{max} = 0.58 eV \]


Step 4: Final Answer:

The kinetic energy of the emitted photoelectron will be 0.58 eV.
Quick Tip: The formula \(E(eV) = \frac{12400}{\lambda(\AA)}\) is a lifesaver in exams for photoelectric effect problems. It avoids the need to use \(E = hc/\lambda\) with fundamental constants and the conversion from Joules to electron-volts, saving significant time and reducing calculation errors.

Step 1: Understanding the Question:

We need to examine the ground state electron configurations of the first 20 elements (from Hydrogen to Calcium) and count how many of them have exactly one unpaired electron.


Step 2: Key Formula or Approach:

We will list the elements and their electron configurations, then inspect the outermost orbital to count the unpaired electrons. An unpaired electron is one that occupies an orbital by itself.


Step 3: Detailed Explanation:

Let's list the elements whose configurations result in one unpaired electron:

1. Z=1, Hydrogen (H): \(1s^1\). The single electron in the 1s orbital is unpaired. (1)

2. Z=3, Lithium (Li): \([He] 2s^1\). The single electron in the 2s orbital is unpaired. (2)

3. Z=5, Boron (B): \([He] 2s^2 2p^1\). The single electron in the 2p subshell is unpaired. (3)

4. Z=9, Fluorine (F): \([He] 2s^2 2p^5\). The 2p subshell has three orbitals. The configuration is \(\uparrow\downarrow, \uparrow\downarrow, \uparrow\). There is one unpaired electron. (4)

5. Z=11, Sodium (Na): \([Ne] 3s^1\). The single electron in the 3s orbital is unpaired. (5)

6. Z=13, Aluminum (Al): \([Ne] 3s^2 3p^1\). The single electron in the 3p subshell is unpaired. (6)

7. Z=17, Chlorine (Cl): \([Ne] 3s^2 3p^5\). The 3p subshell has the configuration \(\uparrow\downarrow, \uparrow\downarrow, \uparrow\). There is one unpaired electron. (7)

8. Z=19, Potassium (K): \([Ar] 4s^1\). The single electron in the 4s orbital is unpaired. (8)


Other elements:

- Noble gases (He, Ne, Ar) have 0 unpaired electrons.

- Alkaline earth metals (Be, Mg, Ca) have 0 unpaired electrons (\(ns^2\)).

- Carbon (Z=6, 2p²) has 2 unpaired electrons.

- Nitrogen (Z=7, 2p³) has 3 unpaired electrons.

- Oxygen (Z=8, 2p⁴) has 2 unpaired electrons.

- Silicon (Z=14, 3p²) has 2 unpaired electrons.

- Phosphorus (Z=15, 3p³) has 3 unpaired electrons.

- Sulfur (Z=16, 3p⁴) has 2 unpaired electrons.

- Scandium (Z=21) is outside the range.

Counting the elements with exactly one unpaired electron, we have a total of 8.


Step 4: Final Answer:

There are 8 elements with only one unpaired electron in their ground state from Z=1 to Z=20.
Quick Tip: The elements with one unpaired electron are typically in Group 1 (alkali metals), Group 13 (Boron group), and Group 17 (halogens). Quickly identifying these groups can help you count the elements faster.

Step 1: Understanding the Question:

We need to identify which of the given atomic orbitals has its electron density lobes located between the coordinate axes, rather than directly along them.


Step 2: Key Formula or Approach:

This requires knowledge of the standard shapes and orientations of p and d atomic orbitals.

- p orbitals (\(p_x, p_y, p_z\)): These are dumbbell-shaped, and their lobes lie directly along the corresponding axis (x, y, or z).

- d orbitals: There are five d orbitals with two main groups:

- Axial orbitals (\(d_{z^2}, d_{x^2-y^2}\)): Their lobes lie along the axes. \(d_{x^2-y^2}\) has lobes on the x and y axes. \(d_{z^2}\) has a main lobe along the z-axis and a torus in the xy-plane.

- Non-axial orbitals (\(d_{xy}, d_{yz}, d_{xz}\)): These are cloverleaf-shaped, and their lobes lie in the planes indicated by their subscripts, but positioned *between* the axes.


Step 3: Detailed Explanation:

Let's analyze the options:

- (A) \(p_x\): The two lobes of the \(p_x\) orbital lie directly on the x-axis.

- (B) \(p_y\): The two lobes of the \(p_y\) orbital lie directly on the y-axis.

- (C) \(d_{x^2-y^2}\): The four lobes of this orbital lie directly on the x and y axes.

- (D) \(d_{yz}\): The four lobes of this orbital lie in the yz-plane, but they are oriented at 45° to the y and z axes, i.e., between the axes.

Therefore, the \(d_{yz}\) orbital has lobes that are not oriented on the axes.


Step 4: Final Answer:

The orbital \(d_{yz}\) has lobes not orienting on the axis.
Quick Tip: A simple mnemonic for d-orbitals: if the subscript has two different letters (like xy, yz, xz), the lobes are *between* those axes. If the subscript involves squares (like \(x^2-y^2\), \(z^2\)), the lobes are *on* the axes.

Step 1: Understanding the Question:

We are given four sets of quantum numbers (n, l, m, s) and we need to identify which set violates the rules governing these numbers.


Step 2: Key Formula or Approach:

The rules for the quantum numbers are:

1. Principal quantum number (n): Can be any positive integer (1, 2, 3, ...).

2. Azimuthal quantum number (l): Can be any integer from 0 to n-1.

3. Magnetic quantum number (m): Can be any integer from -l to +l, including 0.

4. Spin quantum number (s): Can be +1/2 or -1/2.


Step 3: Detailed Explanation:

Let's check each set against the rules:

- (A) n=2, l=0, m=-1, s=+1/2:

- n=2 is valid.

- l=0 is valid (since \(0 \le 0 \le 2-1\)).

- m=-1 is invalid. For l=0, the only possible value for m is 0. Since the rule is violated, this set is impossible.

- (B) n=3, l=0, m=0, s=-1/2:

- n=3 is valid.

- l=0 is valid (since \(0 \le 0 \le 3-1\)).

- m=0 is valid (since for l=0, m must be 0).

- s=-1/2 is valid. This set is possible (it describes an electron in the 3s orbital).

- (C) n=4, l=1, m=+1, s=+1/2:

- n=4 is valid.

- l=1 is valid (since \(0 \le 1 \le 4-1\)).

- m=+1 is valid (since for l=1, m can be -1, 0, +1).

- s=+1/2 is valid. This set is possible (it describes an electron in a 4p orbital).

- (D) n=3, l=2, m=-1, s=-1/2:

- n=3 is valid.

- l=2 is valid (since \(0 \le 2 \le 3-1\)).

- m=-1 is valid (since for l=2, m can be -2, -1, 0, +1, +2).

- s=-1/2 is valid. This set is possible (it describes an electron in a 3d orbital).


Step 4: Final Answer:

The impossible quantum number set is (2, 0, -1, +1/2).
Quick Tip: The most common errors in quantum number sets involve the 'l' and 'm' values. Always check them sequentially: first, is 'l' valid for the given 'n'? Second, is 'm' valid for the given 'l'? This structured check helps to quickly spot the error.

Step 1: Understanding the Question:

We are given a set of atomic numbers and need to find how many unique binary ionic compounds can be formed between them. An ionic compound is typically formed between a metal (cation) and a non-metal (anion).


Step 2: Key Formula or Approach:

1. Identify the elements from their atomic numbers.

2. Classify each element as a metal or a non-metal. Metals tend to lose electrons (form cations), while non-metals tend to gain electrons (form anions).

3. Count the number of possible combinations between one metal and one non-metal.


Step 3: Detailed Explanation:

1. Identify the elements:

- Z = 8 is Oxygen (O)

- Z = 9 is Fluorine (F)

- Z = 11 is Sodium (Na)

- Z = 19 is Potassium (K)

- Z = 20 is Calcium (Ca)

2. Classify the elements:

- Non-metals (form anions): Oxygen (O\(^{2-}\)), Fluorine (F\(^{-}\)) -- [2 non-metals]

- Metals (form cations): Sodium (Na\(^{+}\)), Potassium (K\(^{+}\)), Calcium (Ca\(^{2+}\)) -- [3 metals]

3. Count the combinations:

We can form an ionic compound by pairing each metal with each non-metal.

- Combinations with Sodium (Na):

- Na and O \(\rightarrow\) Na\(_2\)O (Sodium oxide)

- Na and F \(\rightarrow\) NaF (Sodium fluoride)

- Combinations with Potassium (K):

- K and O \(\rightarrow\) K\(_2\)O (Potassium oxide)

- K and F \(\rightarrow\) KF (Potassium fluoride)

- Combinations with Calcium (Ca):

- Ca and O \(\rightarrow\) CaO (Calcium oxide)

- Ca and F \(\rightarrow\) CaF\(_2\) (Calcium fluoride)

The total number of possible ionic compounds is the number of metals multiplied by the number of non-metals: \(3 \times 2 = 6\).


Step 4: Final Answer:

The number of possible ionic compounds is 6.
Quick Tip: To solve this quickly, just classify the elements into metals and non-metals. The total number of binary ionic compounds will be (number of metals) \(\times\) (number of non-metals).

Step 1: Understanding the Question:

We need to find the molecule from the given options where the ratio of the total number of lone pairs of electrons to the total number of bonding pairs of electrons is 2:3.


Step 2: Key Formula or Approach:

For each molecule, we will draw the Lewis structure to determine the number of lone pairs and bonding pairs. A bonding pair is a pair of electrons shared between two atoms (a single bond is 1 BP, a double bond is 2 BP, a triple bond is 3 BP). A lone pair is a pair of valence electrons that is not shared.


Step 3: Detailed Explanation:

Let's analyze each molecule:

- (A) Cl\(_2\): The Lewis structure is :Cl-Cl:. Each chlorine atom has 3 lone pairs, and there is 1 single bond between them.

- Total Lone Pairs (LP) = 3 + 3 = 6

- Total Bonding Pairs (BP) = 1

- Ratio LP:BP = 6:1.

- (B) O\(_2\): The Lewis structure is :O=O:. Each oxygen atom has 2 lone pairs, and there is 1 double bond between them.

- Total LP = 2 + 2 = 4

- Total BP = 2 (a double bond counts as two pairs)

- Ratio LP:BP = 4:2 = 2:1.

- (C) HCl: The Lewis structure is H-Cl:. The chlorine atom has 3 lone pairs, hydrogen has none. There is 1 single bond.

- Total LP = 3

- Total BP = 1

- Ratio LP:BP = 3:1.

- (D) N\(_2\): The Lewis structure is :N\(\equiv\)N:. Each nitrogen atom has 1 lone pair, and there is 1 triple bond between them.

- Total LP = 1 + 1 = 2

- Total BP = 3 (a triple bond counts as three pairs)

- Ratio LP:BP = 2:3.

This matches the required ratio.


Step 4: Final Answer:

The molecule with a lone pair to bond pair ratio of 2:3 is N\(_2\).
Quick Tip: Drawing the Lewis structure is essential for this type of question. Remember to count bonding pairs based on the bond order: a single bond is 1 pair, a double bond is 2 pairs, and a triple bond is 3 pairs.

Step 1: Understanding the Question:

We are given the molarity and volume of a urea solution and asked to calculate the number of moles of urea.


Step 2: Key Formula or Approach:

Molarity (M) is defined as the number of moles of solute per liter of solution.
\[ Molarity (M) = \frac{moles of solute}{Volume of solution in Liters (L)} \]

We can rearrange this formula to solve for the moles of solute:
\[ moles = Molarity \times Volume (L) \]


Step 3: Detailed Explanation:

Given values are:

- Molarity = 0.2 M (which means 0.2 moles/liter)

- Volume = 250 ml

First, we must convert the volume from milliliters (ml) to liters (L):
\[ Volume (L) = 250 ml \times \frac{1 L}{1000 ml} = 0.250 L \]

Now, use the rearranged formula to calculate the number of moles:
\[ moles = 0.2 \frac{mol}{L} \times 0.250 L \]
\[ moles = 0.05 mol \]


Step 4: Final Answer:

There are 0.05 moles of urea present in the solution.
Quick Tip: A common mistake in molarity calculations is forgetting to convert the volume to liters. Always ensure your units are consistent before multiplying. Molarity is always in moles per liter.

Step 1: Understanding the Question:

This is a dilution problem. A concentrated solution of NaOH is diluted with water to a larger volume and lower concentration. We need to find the initial volume of the concentrated solution.


Step 2: Key Formula or Approach:

When a solution is diluted, the amount (moles) of solute remains constant. This is expressed by the dilution equation:
\[ M_1 V_1 = M_2 V_2 \]

where \(M_1\) and \(V_1\) are the initial molarity and volume, and \(M_2\) and \(V_2\) are the final molarity and volume.


Step 3: Detailed Explanation:

Let's identify the given values:

- Initial Molarity, \(M_1 = 0.1\) M

- Initial Volume, \(V_1 = x\) ml (this is what we need to find)

- Final Molarity, \(M_2 = 0.01\) M

- Final Volume, \(V_2 = 250\) ml

Now, substitute these values into the dilution equation:
\[ M_1 V_1 = M_2 V_2 \]
\[ (0.1 M) \times (x ml) = (0.01 M) \times (250 ml) \]

Now, solve for \(x\):
\[ 0.1 \cdot x = 2.5 \]
\[ x = \frac{2.5}{0.1} = 25 \]

The value of x is 25 ml.


Step 4: Final Answer:

The value of x is 25 ml.
Quick Tip: The dilution formula \(M_1V_1 = M_2V_2\) is fundamental for solving problems involving the dilution of solutions. As long as the units of volume (\(V_1\) and \(V_2\)) are the same (e.g., both in ml or both in L), you don't need to convert them.

Step 1: Understanding the Question:

We need to calculate the normality of a sodium carbonate solution, given the number of molecules, volume, and molecular weight.


Step 2: Key Formula or Approach:

1. Calculate the number of moles of Na\(_2\)CO\(_3\).

2. Calculate the Molarity (M) of the solution.

3. Calculate the Normality (N) using the formula \(N = M \times n-factor\).


Step 3: Detailed Explanation:

1. Calculate moles of Na\(_2\)CO\(_3\):

The number of moles is the number of molecules divided by Avogadro's number (\(N_A\)).
\[ moles = \frac{Number of molecules}{N_A} = \frac{3 \times 10^{22}}{6 \times 10^{23}} = \frac{3}{60} = \frac{1}{20} = 0.05 mol \]

2. Calculate Molarity (M):

Molarity is moles of solute per liter of solution. The volume is 500 ml = 0.5 L.
\[ M = \frac{moles}{Volume (L)} = \frac{0.05 mol}{0.5 L} = 0.1 M \]

3. Calculate Normality (N):

Normality is Molarity times the n-factor. For a salt like Na\(_2\)CO\(_3\), the n-factor is the total positive (or negative) charge of the ions it dissociates into.

Na\(_2\)CO\(_3 \rightarrow 2Na^+ + CO_3^{2-}\)

The total positive charge is \(2 \times (+1) = 2\). The total negative charge is \(-2\). So, the n-factor is 2.
\[ N = M \times n-factor = 0.1 M \times 2 = 0.2 N \]


Step 4: Final Answer:

The normality of the solution is 0.2 N.
Quick Tip: The n-factor is crucial for converting between molarity and normality. For acids, it's the number of H⁺ ions; for bases, the number of OH⁻ ions; and for salts, it's the total charge on the cations (or anions).

Step 1: Understanding the Question:

We need to identify the pair of chemical species where both members are Lewis acids.


Step 2: Key Formula or Approach:

- A Lewis acid is a chemical species that can accept a pair of electrons. Common examples include cations, molecules with an incomplete octet, and molecules where the central atom can expand its octet.

- A Lewis base is a chemical species that can donate a pair of electrons. Common examples include anions and molecules with lone pairs of electrons on the central atom.


Step 3: Detailed Explanation:

Let's analyze each species in the options:

- BF\(_3\): Boron trifluoride. The boron atom has only 6 valence electrons (an incomplete octet), so it can accept an electron pair. It is a Lewis acid.

- NH\(_3\): Ammonia. The nitrogen atom has a lone pair of electrons that it can donate. It is a Lewis base.

- H\(^+\): A proton. It has an empty 1s orbital and readily accepts an electron pair. It is a Lewis acid.

- F\(^-\): A fluoride ion. It is an anion with four lone pairs of electrons, making it an electron pair donor. It is a Lewis base.

- H\(_2\)O: Water. The oxygen atom has two lone pairs of electrons that it can donate. It is a Lewis base.

- NH\(_4^+\): Ammonium ion. Nitrogen has a full octet and no lone pairs to donate. It is generally not considered a Lewis base. While it is a Brønsted-Lowry acid (can donate a proton), it is not a typical Lewis acid as it cannot accept an electron pair directly without first losing a proton. However, compared to Lewis bases, it is acidic.

Now let's evaluate the pairs:

- (A) BF\(_3\) (acid), NH\(_3\) (base). Incorrect.

- (B) H\(^+\) (acid), BF\(_3\) (acid). Both are Lewis acids. Correct.

- (C) F\(^-\) (base), H\(_2\)O (base). Both are Lewis bases. Incorrect.

- (D) NH\(_4^+\) (Brønsted-Lowry acid), NH\(_3\) (base). Incorrect.


Step 4: Final Answer:

The pair containing only Lewis acids is H\(^+\), BF\(_3\).
Quick Tip: To quickly identify Lewis acids, look for three main categories: 1. Positive ions (like \(H^+\), \(Mg^{2+}\)). 2. Molecules with an incomplete octet on the central atom (like \(BF_3\), \(AlCl_3\)). 3. Molecules with a central atom that can accommodate more than an octet and is bonded to electronegative atoms (like \(SiF_4\), \(PCl_5\)).

Step 1: Understanding the Question:

We are given the mass of sodium hydroxide (NaOH), a strong base, dissolved in a specific volume of solution.

We need to calculate the pH of this solution.


Step 2: Key Formula or Approach:

1. Calculate the molar mass of NaOH.

2. Calculate the number of moles of NaOH from the given mass.

3. Calculate the molarity of the NaOH solution. Since NaOH is a strong base, this concentration is equal to the hydroxide ion concentration, \([OH^-]\).

4. Calculate the pOH using the formula: \( pOH = -\log_{10}[OH^-] \).

5. Calculate the pH using the relationship: \( pH + pOH = 14 \).


Step 3: Detailed Explanation:

1. Molar Mass of NaOH:

The atomic masses are Na = 23, O = 16, H = 1.

Molar Mass of NaOH = \(23 + 16 + 1 = 40\) g/mol.


2. Moles of NaOH:
\[ moles = \frac{mass}{molar mass} = \frac{4 g}{40 g/mol} = 0.1 mol \]


3. Molarity of NaOH solution:

The volume of the solution is 1.0 L.
\[ Molarity [NaOH] = \frac{moles}{Volume (L)} = \frac{0.1 mol}{1.0 L} = 0.1 M \]

Since NaOH is a strong base, it dissociates completely: NaOH \(\rightarrow\) Na\(^+\) + OH\(^-\).

Therefore, the concentration of hydroxide ions is \([OH^-] = 0.1\) M or \(10^{-1}\) M.


4. Calculate pOH:
\[ pOH = -\log_{10}[OH^-] = -\log_{10}(10^{-1}) = -(-1) = 1 \]


5. Calculate pH:
\[ pH = 14 - pOH = 14 - 1 = 13 \]


Step 4: Final Answer:

The pH of the solution is 13.
Quick Tip: For strong bases like NaOH, the concentration of OH\(^-\) is equal to the molarity of the solution.
A quick way to calculate pH for simple concentrations like 0.1 M, 0.01 M, etc., is to first find the pOH, which will be an integer, and then subtract from 14.

Step 1: Understanding the Question:

This question asks for the value of the total charge carried by one mole of electrons.

This quantity is a fundamental constant in chemistry and physics known as the Faraday constant (F).


Step 2: Key Formula or Approach:

The Faraday constant is calculated by multiplying the charge of a single electron (the elementary charge, \(e\)) by Avogadro's number (\(N_A\)).
\[ F = e \times N_A \]


Step 3: Detailed Explanation:

The standard values for the constants are:

- Charge of one electron, \(e \approx 1.602 \times 10^{-19}\) Coulombs (C).

- Avogadro's number, \(N_A \approx 6.022 \times 10^{23}\) mol\(^{-1}\).

Now, we calculate the Faraday constant:
\[ F = (1.602 \times 10^{-19} C) \times (6.022 \times 10^{23} mol^{-1}) \]
\[ F \approx 9.6485 \times 10^4 C/mol \]

This value is commonly approximated for calculations as 96500 C/mol.

Writing this approximation in scientific notation gives:
\[ 96500 = 9.65 \times 10^4 \]

This matches option (B).


Step 4: Final Answer:

The number of coulombs corresponding to 1 mole of electrons is approximately \(9.65 \times 10^4\).
Quick Tip: The value of the Faraday constant, approximately 96500 C/mol, is one of the most important constants in electrochemistry.
It is essential to memorize this value for quick calculations in electrolysis and electrochemical cell problems.

Step 1: Understanding the Question:

We need to identify which of the given substances, when dissolved in water, does not form an electrically conducting solution.

Such a substance is known as a non-electrolyte.


Step 2: Key Formula or Approach:

An electrolyte is a substance that produces ions when dissolved in a solvent (like water), allowing the solution to conduct electricity.

- Electrolytes are typically ionic compounds (salts), acids, and bases.

- Non-electrolytes are typically molecular covalent compounds that do not ionize in solution, such as sugars, alcohols, and urea.


Step 3: Detailed Explanation:

Let's analyze each option:

- (A) Urea \((CO(NH_2)_2)\): Urea is a covalent organic molecule. When it dissolves in water, it disperses as neutral molecules. It does not dissociate into ions. Therefore, its aqueous solution does not conduct electricity, making it a non-electrolyte.

- (B) Copper Sulphate \((CuSO_4)\): This is an ionic salt. It dissolves in water and dissociates into mobile ions: \(Cu^{2+}(aq)\) and \(SO_4^{2-}(aq)\). It is a strong electrolyte.

- (C) Silver Nitrate (AgNO\(_3\)): This is an ionic salt. It dissolves and dissociates into Ag\(^+\)(aq) and NO\(_3^-\)(aq) ions. It is a strong electrolyte.

- (D) Sodium Chloride (NaCl): This is a common ionic salt that dissolves and dissociates completely into Na\(^+\)(aq) and Cl\(^-\)(aq) ions. It is a strong electrolyte.


Step 4: Final Answer:

The aqueous solution of Urea does not act as an electrolyte.
Quick Tip: To identify non-electrolytes, look for molecular covalent compounds that are not acids or bases.
Common examples include sugars (like glucose, sucrose), alcohols (like ethanol), and urea.
Ionic salts are almost always strong electrolytes.

Step 1: Understanding the Question:

This is an electrolysis problem where we need to calculate the mass of silver metal deposited by a specific amount of electric charge.


Step 2: Key Formula or Approach:

We will use Faraday's Laws of Electrolysis.

1. Write the reduction half-reaction for silver ions to determine the number of electrons involved per ion.

2. Calculate the moles of electrons corresponding to the given charge using the Faraday constant.

3. Use the stoichiometry from the half-reaction to find the moles of silver deposited.

4. Convert the moles of silver to mass, and then convert the units to milligrams.


Step 3: Detailed Explanation:

The reduction half-reaction for silver ions at the cathode is:
\[ Ag^+(aq) + e^- \rightarrow Ag(s) \]

This shows that 1 mole of electrons (e\(^-\)) deposits 1 mole of silver (Ag).


First, calculate the moles of electrons for the given charge Q = 9.65 C.
\[ Moles of electrons = \frac{Total Charge (Q)}{Faraday Constant (F)} = \frac{9.65 C}{96500 C/mol} = 0.0001 mol = 10^{-4} mol \]


From the 1:1 stoichiometry, the moles of Ag deposited equals the moles of electrons.
\[ Moles of Ag = 10^{-4} mol \]


Next, convert the moles of silver to mass in grams using the molar mass (108 g/mol).
\[ Mass of Ag (g) = moles \times molar mass = 10^{-4} mol \times 108 g/mol = 0.0108 g \]


Finally, convert the mass from grams (g) to milligrams (mg) by multiplying by 1000.
\[ Mass of Ag (mg) = 0.0108 g \times 1000 mg/g = 10.8 mg \]


Step 4: Final Answer:

The amount of silver deposited is 10.8 mg.
Quick Tip: A useful proportion to remember for electrolysis is:
\( \frac{Mass deposited}{Molar mass} = \frac{Charge passed}{n \times F} \)
where 'n' is the number of moles of electrons per mole of substance from the half-reaction.

Step 1: Understanding the Question:

We are given standard reduction potentials (E°) for three metals. We must determine which statement about their spontaneous redox behavior is correct.


Step 2: Key Formula or Approach:

The standard reduction potential (E°) measures the tendency for a species to be reduced.

- A more negative E° indicates a stronger reducing agent (the metal is more easily oxidized).

- A more positive E° indicates a stronger oxidizing agent (the ion is more easily reduced).

A metal can spontaneously reduce the ions of another metal that has a more positive E°.


Step 3: Detailed Explanation:

Let's list the potentials in increasing order to create an electrochemical series:

E°(Zn\(^{2+}\)/Zn) = -0.76 V

E°(Cu\(^{2+}\)/Cu) = +0.34 V

E°(Ag\(^+\)/Ag) = +0.80 V


This order establishes the relative strengths:

- Strength as reducing agents (metals): Zn \(>\) Cu \(>\) Ag.

- Strength as oxidizing agents (ions): Ag\(^+\) \(>\) Cu\(^{2+}\) \(>\) Zn\(^{2+}\).


Now, let's evaluate each statement:

- (A) Ag can oxidize Zn and Cu: Incorrect. The metal Ag is a reducing agent. Its ion, Ag\(^+\), is an oxidizing agent.

- (B) Ag can reduce Zn\(^{2+}\) and Cu\(^{2+}\): Incorrect. Ag is the weakest reducing agent and cannot reduce ions of metals with lower (more negative) E° values.

- (C) Zn can reduce Ag\(^+\) and Cu\(^{2+}\): Correct. Zn is the strongest reducing agent. Its E° is the most negative, so it can spontaneously reduce the ions of both Cu and Ag, which have more positive E° values.

- (D) Cu can oxidize Zn and Ag: Incorrect. The metal Cu is a reducing agent. Its ion, Cu\(^{2+}\), can oxidize Zn but not Ag.


Step 4: Final Answer:

The correct statement is "Zn can reduce Ag\(^+\) and Cu\(^{2+}\)".
Quick Tip: A simple rule to remember is: a metal can "displace" or reduce the ions of any metal below it in the reactivity series (or above it in the standard potential series).
Here, Zn is the most reactive, so it can reduce both Cu\(^{2+}\) and Ag\(^+\).

Step 1: Understanding the Question:

The question asks about the ion-exchange mechanism for softening hard water using the permutit process, specifically identifying the ions removed from the water.


Step 2: Key Formula or Approach:

This question requires factual knowledge of water chemistry.

1. Define the ions responsible for permanent hardness in water.

2. Understand the principle of the permutit (zeolite) ion-exchange process.


Step 3: Detailed Explanation:

1. Hardness of Water: The hardness of water is defined by the presence of dissolved divalent cations, primarily calcium (Ca\(^{2+}\)) and magnesium (Mg\(^{2+}\)). Permanent hardness is caused by their chloride and sulfate salts.


2. Permutit Process: The permutit process uses a synthetic zeolite, which is a hydrated sodium aluminum silicate (often represented as Na\(_2\)Z), as an ion-exchange resin. When hard water passes through a column of this material, the Ca\(^{2+}\) and Mg\(^{2+}\) ions are exchanged for the Na\(^+\) ions in the permutit. The zeolite has a higher affinity for the divalent Ca\(^{2+}\) and Mg\(^{2+}\) ions than for the monovalent Na\(^+\) ions.

The exchange reactions are:
\[ Ca^{2+}(aq) + Na_2Z(s) \rightarrow CaZ(s) + 2Na^+(aq) \]
\[ Mg^{2+}(aq) + Na_2Z(s) \rightarrow MgZ(s) + 2Na^+(aq) \]

The hardness-causing ions are thus trapped by the resin, and sodium ions are released into the water, resulting in soft water.


Step 4: Final Answer:

In the permutit process, Na\(^+\) ions are exchanged with Ca\(^{2+}\) and Mg\(^{2+}\) ions.
Quick Tip: Remember the simple fact: Water hardness is caused by Ca\(^{2+}\) and Mg\(^{2+}\) ions.
Therefore, any water softening process must be designed to remove these two specific ions.

Step 1: Understanding the Question:

We need to calculate the hardness of a water sample in ppm, expressed in terms of CaCO\(_3\) equivalents, given the mass of MgCl\(_2\) in a known mass of water.


Step 2: Key Formula or Approach:

1. Convert the mass of the hardness-causing salt (MgCl\(_2\)) to its equivalent mass of CaCO\(_3\) using the ratio of their molar masses. The molar mass of CaCO\(_3\) is 100 g/mol.

\[ Mass of CaCO_3 equiv. = (Mass of salt) \times \frac{Molar mass of CaCO_3}{Molar mass of salt} \]

2. Calculate the concentration in ppm (parts per million), defined as mg of CaCO\(_3\) equivalent per kg of water.


Step 3: Detailed Explanation:

1. Calculate CaCO\(_3\) equivalent mass:

- Mass of MgCl\(_2\) = 19 mg.

- Molar mass of MgCl\(_2\) = 95 g/mol.

- Molar mass of CaCO\(_3\) = 100 g/mol.

Using the equivalence formula:
\[ Mass of CaCO_3 equiv. = 19 mg \times \frac{100}{95} = \frac{1900}{95} mg = 20 mg \]

This means 19 mg of MgCl\(_2\) creates the same hardness as 20 mg of CaCO\(_3\).


2. Calculate hardness in ppm:

The mass of the water sample is 2 kg.
\[ Hardness (ppm) = \frac{Mass of CaCO_3 equiv. (mg)}{Mass of water (kg)} \]
\[ Hardness (ppm) = \frac{20 mg}{2 kg} = 10 mg/kg \]

Since 1 mg/kg is equivalent to 1 ppm, the hardness is 10 ppm.


Step 4: Final Answer:

The degree of hardness is 10 ppm.
Quick Tip: Hardness calculations always involve converting the given salt into its CaCO\(_3\) equivalent.
The conversion factor is always \(\frac{100}{Molar mass of the salt}\).
Then, remember that ppm = mg/kg (or mg/L for dilute solutions).

Step 1: Understanding the Question:

We need to identify the pair of dissolved chlorides that are responsible for causing permanent hardness in water.


Step 2: Key Formula or Approach:

This question is based on the chemical definition of water hardness.

- Water Hardness: Caused by dissolved divalent cations, primarily Calcium (Ca\(^{2+}\)) and Magnesium (Mg\(^{2+}\)).

- Permanent Hardness: Caused by the chlorides and sulfates of these ions; this type is not removed by boiling.

- Temporary Hardness: Caused by the bicarbonates of these ions.


Step 3: Detailed Explanation:

Let's analyze the pairs in the options:

- (A) NaCl, KCl: Sodium and potassium ions are monovalent and do not cause hardness.

- (B) CaCl\(_2\), KCl: Calcium chloride (CaCl\(_2\)) causes permanent hardness, but potassium chloride (KCl) does not.

- (C) AlCl\(_3\), MgCl\(_2\): Magnesium chloride (MgCl\(_2\)) causes permanent hardness, but aluminum chloride is not considered a cause of common water hardness.

- (D) MgCl\(_2\), CaCl\(_2\): Both magnesium chloride and calcium chloride are the definitive salts responsible for permanent hardness.


Step 4: Final Answer:

The pair of chlorides responsible for permanent hardness of water is MgCl\(_2\), CaCl\(_2\).
Quick Tip: A simple rule to remember for water hardness:
Hardness Ions: Ca\(^{2+}\) and Mg\(^{2+}\).
Anions for Temporary Hardness: Bicarbonate (HCO\(_3^-\)).
Anions for Permanent Hardness: Chloride (Cl\(^-\)) and Sulfate (SO\(_4^{2-}\)).

Step 1: Understanding the Question:

The question asks to identify the type of electrochemical corrosion cell formed in a bent metal pipe.


Step 2: Key Formula or Approach:

This question is about electrochemical corrosion. Corrosion often happens when different areas on a metal surface develop a potential difference, creating a tiny galvanic cell. The cell is named based on the cause of this potential difference.


Step 3: Detailed Explanation:

When a metal pipe is bent, it is subjected to non-uniform mechanical stress.

- The outer curve is under tensile stress (stretched).

- The inner curve is under compressive stress.

A region of a metal under higher stress is more energetically active and thus has a more negative electrode potential. This makes the highly stressed region (the bend) anodic, while the less stressed regions become cathodic.

This potential difference, arising from differences in mechanical stress across the metal surface, creates a galvanic cell known as a Stress Cell. This leads to accelerated corrosion at the stressed areas.


Step 4: Final Answer:

The cell formed in bent pipes is an example of a Stress Cell.
Quick Tip: Remember that non-uniformity is a major cause of corrosion.
- Difference in metal type \(\rightarrow\) Galvanic/Composition Cell.
- Difference in oxygen concentration \(\rightarrow\) Differential Aeration Cell.
- Difference in mechanical stress \(\rightarrow\) Stress Cell.

Step 1: Understanding the Question:

We are asked to identify the chemical compound that forms on silver, causing it to tarnish (turn black).


Step 2: Key Formula or Approach:

This is a factual question about the corrosion of silver. Tarnishing is the chemical reaction between silver and sulfur compounds present in the atmosphere.


Step 3: Detailed Explanation:

Silver (Ag) is a relatively unreactive metal but is susceptible to reaction with sulfur compounds. The atmosphere contains trace amounts of hydrogen sulfide (H\(_2\)S), which originates from pollution and biological decay.

Silver reacts with H\(_2\)S in the presence of oxygen to form a black layer of silver sulfide (Ag\(_2\)S). This layer is the tarnish.

The reaction is:
\[ 4Ag(s) + 2H_2S(g) + O_2(g) \rightarrow 2Ag_2S(s) + 2H_2O(l) \]

This black silver sulfide layer obscures the metal's luster.


Step 4: Final Answer:

Tarnishing of silver is due to the formation of its sulphide layer.
Quick Tip: Remember the colors of common corrosion products:
- Iron \(\rightarrow\) Reddish-brown rust (hydrated iron(III) oxide).
- Copper \(\rightarrow\) Greenish-blue patina (copper carbonate/sulfate).
- Silver \(\rightarrow\) Black tarnish (silver sulfide).

Step 1: Understanding the Question:

We must distinguish between a co-polymer and a homopolymer among the given options.


Step 2: Key Formula or Approach:

- A Homopolymer is a polymer formed from the polymerization of a single type of monomer.

- A Co-polymer is a polymer formed from two or more different types of monomers.

We need to know the monomer(s) for each polymer listed.


Step 3: Detailed Explanation:

Let's analyze the monomers of each polymer:

- (A) Buna-S rubber: This is a co-polymer. Its name is an abbreviation for its monomers: Bu (for 1,3-Butadiene) and Na-S (for Styrene, with Na indicating the original sodium catalyst).

- (B) Neoprene rubber: This is a homopolymer. It is formed by the addition polymerization of a single monomer: chloroprene (2-chloro-1,3-butadiene). Its systematic name is polychloroprene.

- (C) Bakelite: This is a co-polymer. It is a condensation polymer made from two different monomers: Phenol and Formaldehyde.

- (D) Urea - Formaldehyde: As its name clearly indicates, this is a co-polymer made from Urea and Formaldehyde monomers via condensation polymerization.

Therefore, Neoprene rubber is the only homopolymer in the list.


Step 4: Final Answer:

Neoprene rubber is not a co-polymer.
Quick Tip: The names of some polymers give clues to their composition.
"Buna-S" stands for Butadiene-Styrene.
Names like "Urea-Formaldehyde" explicitly state the two monomers.
When a polymer is named after a single monomer, like Polystyrene or Neoprene (polychloroprene), it's a homopolymer.

Step 1: Understanding the Question:

We need to identify the chemical structure of the monomer unit that polymerizes to form polystyrene.


Step 2: Key Formula or Approach:

The name "polystyrene" indicates that it is a polymer ("poly-") made from the monomer "styrene".

We need to know the structure of styrene. Styrene is also known as vinylbenzene or phenylethene.

It consists of a vinyl group (\(-CH=CH_2\)) attached to a phenyl group (a benzene ring, \(C_6H_5\)).


Step 3: Detailed Explanation:

Let's analyze the chemical structures given in the options:

- (A) \(CH_2=CH-Cl\): This is vinyl chloride, the monomer for polyvinyl chloride (PVC).

- (B) \(CH_2=CH-CN\): This is acrylonitrile, the monomer for polyacrylonitrile (PAN), used in fibers like Orlon.

- (C) \(CH_2=CH-C_6H_5\): This is styrene (vinylbenzene). The addition polymerization of this monomer across the double bond results in polystyrene.

- (D) \(CH_2=CH-CH_3\): This is propene (or propylene), the monomer for polypropylene.

Therefore, the correct monomer for polystyrene is styrene, \(CH_2=CH-C_6H_5\).


Step 4: Final Answer:

The monomer involved in the formation of polystyrene is \(CH_2=CH-C_6H_5\).
Quick Tip: Many common polymer names are simply the prefix "poly-" added to the name of the monomer.
For example: poly(ethene), poly(propene), poly(styrene), poly(vinyl chloride).
Recognizing this naming convention can help you quickly identify the monomer.

Step 1: Understanding the Question:

The question asks about the process used to improve the physical properties of natural rubber, such as its strength and elasticity.


Step 2: Key Formula or Approach:

This is a factual question about a specific industrial chemical process. The process of heating natural rubber with a cross-linking agent to improve its properties is known as vulcanization.


Step 3: Detailed Explanation:

Natural rubber is a polymer of isoprene. In its raw form, it has several undesirable properties: it is soft and sticky, has low tensile strength, becomes brittle at low temperatures, and is not very elastic.

To overcome these issues, natural rubber is heated with Sulphur. This process, discovered by Charles Goodyear, is called vulcanization.

During vulcanization, the sulphur atoms form cross-links (disulfide or polysulfide bridges) between the long polyisoprene chains. These cross-links tie the polymer chains together, preventing them from slipping past one another.

This modification drastically improves the rubber's properties, making it stronger, tougher, more elastic, and less susceptible to temperature changes.

Carbon (in the form of carbon black) is often added as a reinforcing filler, but Sulphur is the key vulcanizing agent.


Step 4: Final Answer:

The undesirable properties of natural rubber are overcome by heating it with Sulphur.
Quick Tip: The process of heating rubber with sulfur is called Vulcanization.
Remembering this term and its association with sulfur is key to answering questions about improving rubber's properties.

Step 1: Understanding the Question:

We need to identify the primary chemical components of Liquefied Petroleum Gas (LPG).


Step 2: Key Formula or Approach:

This is a factual question about common fuels. LPG is a byproduct of petroleum refining and natural gas processing. Its composition varies but is primarily composed of low-molecular-weight alkanes.


Step 3: Detailed Explanation:

Liquefied Petroleum Gas (LPG) is a flammable mixture of hydrocarbon gases used as fuel.

The main components of LPG are propane (\(C_3H_8\)) and butane (\(C_4H_{10}\)). Butane itself exists as two isomers: n-butane and isobutane. Commercial LPG is often a mix of these gases.

Let's analyze the options:

- (A) Methane, Ethane: Methane (\(CH_4\)) is the primary component of natural gas, not LPG.

- (B) Ethane, Propane: Propane is a major component, but ethane is typically a minor component. This is less accurate than other options.

- (C) Butane, Isobutane: Both n-butane and its isomer, isobutane, are major components of LPG, often blended with propane. This is the best description among the choices.

- (D) Ethene, Ethyne: These are unsaturated hydrocarbons (alkenes and alkynes) and are not the main constituents of LPG.


Step 4: Final Answer:

LPG mainly contains Butane and Isobutane (along with Propane).
Quick Tip: Remember the main components of common fuel gases:
- \textbf{Natural Gas:} Primarily Methane (C1).
- \textbf{LPG:} Primarily Propane (C3) and Butane (C4).
- \textbf{Gasoline:} A mix of hydrocarbons from C4 to C12.

Step 1: Understanding the Question:

We need to identify which of the given gases is a primary cause of the greenhouse effect.


Step 2: Key Formula or Approach:

The greenhouse effect is the process by which certain gases in the atmosphere trap heat by absorbing infrared radiation emitted from the Earth's surface.

These gases are known as greenhouse gases. To be a greenhouse gas, a molecule must have a changing dipole moment when it vibrates.


Step 3: Detailed Explanation:

Let's analyze the gases:

- The most important greenhouse gases are Water Vapor (\(H_2O\)), Carbon Dioxide (\(CO_2\)), Methane (\(CH_4\)), Nitrous Oxide (\(N_2O\)), and Ozone (\(O_3\)).

- (D) \(CO_2\): Carbon dioxide is a major greenhouse gas. It is the most significant long-lived greenhouse gas contributing to modern climate change.

- (A) \(NO_2\): Nitrogen dioxide is a greenhouse gas but its contribution is less significant than \(CO_2\).

- (B) CO, (C) NO: Carbon monoxide and nitric oxide are not significant greenhouse gases themselves, although they can indirectly affect atmospheric chemistry.

Comparing the options, \(CO_2\) is the most prominent and direct cause of the enhanced greenhouse effect.


Step 4: Final Answer:

The greenhouse effect is caused by \(CO_2\).
Quick Tip: While many gases are present in the atmosphere, the key greenhouse gases to remember are \(H_2O\), \(CO_2\), \(CH_4\), and \(N_2O\).
Symmetrical diatomic molecules like \(N_2\) and \(O_2\) are not greenhouse gases.

Step 1: Understanding the Question:

We need to identify the compound from the list that is the primary cause of the destruction of the stratospheric ozone layer.


Step 2: Key Formula or Approach:

Ozone layer depletion is primarily caused by chemical compounds containing chlorine and bromine that are stable enough to reach the stratosphere, known as chlorofluorocarbons (CFCs).


Step 3: Detailed Explanation:

The stratospheric ozone layer is depleted by catalytic reactions involving free radicals, primarily chlorine (\(Cl\cdot\)) and bromine (\(Br\cdot\)) radicals.

The main source of these halogen radicals is a class of man-made compounds called Chlorofluorocarbons (CFCs).

Let's look at the options:

- (A) \(CO_2\): A greenhouse gas, but does not deplete ozone.

- (B) \(CH_4\): A greenhouse gas, but does not directly deplete ozone.

- (C) \(CH_3OH\) (Methanol): Decomposes in the lower atmosphere and does not significantly affect the ozone layer.

- (D) \(CF_2Cl_2\) (Dichlorodifluoromethane): This is a classic example of a CFC (Freon-12). In the stratosphere, UV radiation breaks its C-Cl bonds, releasing chlorine free radicals that catalytically destroy ozone.


Step 4: Final Answer:

The compound mainly responsible for the depletion of the ozone layer is \(CF_2Cl_2\).
Quick Tip: Remember the distinction:
- \textbf{Greenhouse Effect} (Global Warming) \(\rightarrow\) CO\(_2\), CH\(_4\).
- \textbf{Ozone Depletion} \(\rightarrow\) CFCs (compounds with Chlorine and Fluorine).

Step 1: Understanding the Question:

The question asks for the term that describes a material's behavior when subjected to repeated or cyclic loading.


Step 2: Detailed Explanation:

Let's define the given material properties:


(A) Fatigue: This is the weakening of a material caused by repeatedly applied loads (cyclic loading). It is the progressive and localized structural damage that occurs when a material is subjected to cyclic loading. Fatigue failure often occurs at a stress level significantly lower than the ultimate tensile strength of the material. This exactly matches the description in the question.

(B) Hardness: This is a measure of a material's resistance to localized plastic deformation such as scratching or indentation. It is typically measured under a static load.

(C) Resilience: This is the ability of a material to absorb energy when it is deformed elastically and release that energy upon unloading.

(D) Toughness: This is the ability of a material to absorb energy and plastically deform without fracturing. It is a measure of a material's resistance to fracture when a crack is present.



The key phrase in the question is "repeated loading," which is the definition of the conditions that lead to fatigue.


Step 3: Final Answer:

The ability of materials to develop a characteristic behaviour under repeated loading is known as fatigue.
Quick Tip: Associate key loading conditions with material properties:
- \textbf{Repeated/Cyclic Load} \(\rightarrow\) \textbf{Fatigue}.
- \textbf{Static Indentation/Scratch} \(\rightarrow\) \textbf{Hardness}.
- \textbf{Energy Absorption (Elastic)} \(\rightarrow\) \textbf{Resilience}.
- \textbf{Energy Absorption (Plastic, before fracture)} \(\rightarrow\) \textbf{Toughness}.

Step 1: Understanding the Question:

The question asks to calculate the total elongation (change in length) of a bar under a given tensile load.


Step 2: Key Formula or Approach:

The elongation (\(\delta L\)) of an axially loaded bar is given by the formula:
\[ \delta L = \frac{PL}{AE} \]
where:
\(P\) = Axial load
\(L\) = Original length of the bar
\(A\) = Cross-sectional area of the bar
\(E\) = Modulus of Elasticity (Young's Modulus) of the material


Step 3: Detailed Explanation:

First, identify and convert all the given values to a consistent set of units (e.g., Newtons and millimeters).


Load (\(P\)) = 60 kN = \(60 \times 10^3\) N

Length (\(L\)) = 400 mm

Side of square cross section = 10 mm

Cross-sectional area (\(A\)) = side \(\times\) side = \(10 mm \times 10 mm = 100 mm^2\)

Modulus of Elasticity (\(E\)) = 200 GPa = \(200 \times 10^9\) Pa = \(200 \times 10^9\) N/m\(^2\).

To convert GPa to N/mm\(^2\): 1 GPa = \(10^3\) N/mm\(^2\).

So, \(E = 200 \times 10^3\) N/mm\(^2\).



Now, substitute these values into the elongation formula:
\[ \delta L = \frac{(60 \times 10^3 N) \times (400 mm)}{(100 mm^2) \times (200 \times 10^3 N/mm^2)} \] \[ \delta L = \frac{24 \times 10^6}{20 \times 10^6} mm \] \[ \delta L = \frac{24}{20} mm = 1.2 mm \]

Step 4: Final Answer:

The elongation of the bar is 1.2 mm.
Quick Tip: The formula \(\delta L = PL/AE\) is fundamental in mechanics of materials.
A common source of error is inconsistent units.
It's often easiest to convert everything to N and mm:
- 1 kN = \(10^3\) N
- 1 GPa = \(10^3\) N/mm\(^2\)
- 1 MPa = 1 N/mm\(^2\)

Step 1: Understanding the Question:

The question asks for the ratio of the Modulus of Rigidity (\(G\)) to the Modulus of Elasticity (\(E\)), given the Poisson's ratio (\(\nu\)).


Step 2: Key Formula or Approach:

For an isotropic elastic material, the three main elastic constants are related by the following formula:
\[ E = 2G(1 + \nu) \]
where:
\(E\) = Modulus of Elasticity (Young's Modulus)
\(G\) = Modulus of Rigidity (Shear Modulus)
\(\nu\) = Poisson's ratio


Step 3: Detailed Explanation:

We are given the Poisson's ratio, \(\nu = 0.25\).

We need to find the ratio \(G/E\). We can rearrange the formula to solve for this ratio.
\[ \frac{E}{G} = 2(1 + \nu) \]
Taking the reciprocal of both sides:
\[ \frac{G}{E} = \frac{1}{2(1 + \nu)} \]
Now, substitute the given value of \(\nu\):
\[ \frac{G}{E} = \frac{1}{2(1 + 0.25)} \] \[ \frac{G}{E} = \frac{1}{2(1.25)} \] \[ \frac{G}{E} = \frac{1}{2.5} \] \[ \frac{G}{E} = 0.40 \]

Step 4: Final Answer:

The ratio of modulus of rigidity to modulus of elasticity is 0.40.
Quick Tip: Memorize the key relationships between elastic constants (\(E\), \(G\), \(K\), \(\nu\)):
1. \(E = 2G(1 + \nu)\)
2. \(E = 3K(1 - 2\nu)\)
These two formulas are essential for solving many problems in strength of materials.

Step 1: Understanding the Question:

The question describes a composite bar made of two materials, perfectly bonded together, with the same length and area. It asks which quantity will be the same for both materials when the bar is subjected to an axial load.


Step 2: Detailed Explanation:

This describes a "parallel composite" system. When an axial load is applied to the composite bar, the two materials must deform together.


Since the bars are rigidly fixed together and have the same initial length (\(L\)), their change in length (\(\delta L\)) must be identical to maintain compatibility.

Strain (\(\epsilon\)) is defined as the change in length divided by the original length (\(\epsilon = \delta L / L\)). Since both \(\delta L\) and \(L\) are the same for both materials, the strain must be the same in both.

Stress (\(\sigma\)) is related to strain by Hooke's Law: \(\sigma = E \epsilon\). Since the two materials are different, their moduli of elasticity (\(E_1\) and \(E_2\)) will be different. Because the strain (\(\epsilon\)) is the same for both, but the moduli are different, the stresses in the two materials will be different (\(\sigma_1 \neq \sigma_2\)).

Load sharing: The total load (\(P\)) is shared between the two materials (\(P = P_1 + P_2\)). Since the stresses are different and the areas are the same, the loads carried by each material (\(P_1 = \sigma_1 A\) and \(P_2 = \sigma_2 A\)) will be different.

Poisson's ratio is an intrinsic material property and will be different for the two different materials.




Step 3: Final Answer:

The strain will be the same for both materials in the composite bar.
Quick Tip: For composite structures, remember the key compatibility condition for each arrangement:
- \textbf{Parallel bars} (loaded axially): \textbf{Strain is the same}.
- \textbf{Series bars} (loaded axially): \textbf{Stress (and Load) is the same}.

Step 1: Understanding the Question:

The question asks for the specific term for the maximum amount of strain energy a material can absorb within its elastic limit.


Step 2: Detailed Explanation:

Let's define the energy-related material properties:


(C) Resilience: This is the general ability of a material to absorb energy when deformed elastically and to release this energy upon unloading. The area under the elastic portion of the stress-strain curve is known as the modulus of resilience.

(D) Proof Resilience: This is the maximum strain energy that can be stored in a material. It corresponds to the total strain energy stored at the elastic limit. It is the area under the stress-strain curve up to the elastic limit point. This precisely matches the question's description.

(B) Toughness: This is the ability of a material to absorb energy and plastically deform before fracturing. It is represented by the total area under the entire stress-strain curve, up to the point of fracture.

(A) Hardness: This is the resistance to localized plastic deformation like scratching. It is not an energy absorption property in this context.



The key phrases are "maximum energy" and "up to elastic limit," which specifically define Proof Resilience.


Step 3: Final Answer:

The maximum energy that a given component can absorb without undergoing any permanent deformation up to the elastic limit is known as proof resilience.
Quick Tip: Differentiate the energy terms using the stress-strain curve:
- \textbf{Modulus of Resilience:} Area under the curve up to the yield point (a material property).
- \textbf{Proof Resilience:} Total energy stored up to the elastic limit (for a specific component).
- \textbf{Toughness:} Total area under the entire curve up to fracture.

Step 1: Understanding the Question:

The question asks for the ratio of the support reactions (\(R_A / R_B\)) for a simply supported beam with a UDL on its left half.


Step 2: Key Formula or Approach:

To find the reactions, we use the equations of static equilibrium:

1. Sum of vertical forces is zero (\(\Sigma F_y = 0\)).

2. Sum of moments about any point is zero (\(\Sigma M = 0\)).


Step 3: Detailed Explanation:

Let the beam be AB, with support A on the left and support B on the right. The span is \(L = 8\) m.

The uniformly distributed load (UDL) is \(w = 24\) kN/m and it acts over the left half, from \(x=0\) to \(x=4\) m.

First, find the total load from the UDL. It can be represented as a single point load for calculation purposes.

Total Load (\(W_{UDL}\)) = \(w \times\) length of load = \(24 kN/m \times 4 m = 96\) kN.

This equivalent point load acts at the centroid of the UDL, which is at half the length of the loaded portion, i.e., at \(x = 4/2 = 2\) m from support A.


Now, apply the equilibrium equations. Let \(R_A\) and \(R_B\) be the reactions at A and B.

Take moments about support A to find \(R_B\):
\(\Sigma M_A = 0\) (Clockwise moments are positive)
\[ (W_{UDL} \times 2 m) - (R_B \times 8 m) = 0 \] \[ (96 kN \times 2 m) = R_B \times 8 m \] \[ 192 = 8 R_B \] \[ R_B = \frac{192}{8} = 24 kN \]

Now, apply the vertical force equilibrium to find \(R_A\):
\(\Sigma F_y = 0\) (Upward forces are positive)
\[ R_A + R_B - W_{UDL} = 0 \] \[ R_A + 24 kN - 96 kN = 0 \] \[ R_A = 96 - 24 = 72 kN \]

Finally, calculate the required ratio:
\[ Ratio = \frac{R_A}{R_B} = \frac{72 kN}{24 kN} = 3 \]

Step 4: Final Answer:

The ratio of the reactions of the left support to the right support is 3.
Quick Tip: For calculating reactions, always replace a UDL with its equivalent point load.
The magnitude of the point load is the area of the UDL rectangle (\(w \times l\)).
The location of the point load is at the centroid of the UDL.
Then apply the standard equilibrium equations \(\Sigma M = 0\) and \(\Sigma F_y = 0\).

Step 1: Understanding the Question:

The question asks for the maximum bending moment in a cantilever beam with a point load applied at a specific location.


Step 2: Key Formula or Approach:

For a cantilever beam, the bending moment at any section is the algebraic sum of the moments of the forces acting to one side of the section. The maximum bending moment for a cantilever beam always occurs at the fixed support.

The bending moment (\(M\)) caused by a point load (\(P\)) at a distance (\(a\)) from the point of interest is \(M = P \times a\).


Step 3: Detailed Explanation:

The beam is a cantilever, fixed at A and free at B. The span is 3 m.

A point load of \(P = 50\) kN is applied at a distance of 2 m from the fixed end A.

The bending moment diagram for this beam will be as follows:

- From the free end B to the point of the load (from \(x=3\) m to \(x=2\) m), there are no forces, so the bending moment is zero.

- From the point of the load to the fixed end A (from \(x=2\) m to \(x=0\) m), the bending moment increases linearly.

The maximum bending moment will occur at the fixed support A (\(x=0\)).

To calculate the moment at A, we consider the load of 50 kN acting at a distance of 2 m from A.
\[ M_{max} = M_A = P \times (distance from A) \] \[ M_{max} = 50 kN \times 2 m \] \[ M_{max} = 100 kNm \]
This is a hogging (negative) bending moment. The magnitude is 100 kNm, and it occurs at the fixed support A. The bending moment under the load itself is also 100 kNm because the moment is constant from the load to the support in this specific problem setup (this is a mistake in reasoning, moment increases linearly from load to support). The bending moment at the point of the load (looking from the free end) is zero, the moment at the support is the maximum. The bending moment at a distance 'x' from the free end is \(M(x) = -50(x-1)\) for \(x>1\). The moment at the support A (x=3) is not well-defined. Let's assume the distance is from the fixed end.
Let the fixed end be at x=0. The load is at x=2. The free end is at x=3.
The bending moment at any section x is the moment of forces to its right.
For \(x > 2\), there are no forces, so \(M(x)=0\).
For \(x < 2\), the moment is \(M(x) = -50 \times (2-x)\).
The maximum magnitude of this moment occurs at the fixed end, x=0. \(M_{max} = |M(0)| = |-50 \times (2-0)| = 100\) kNm. This maximum moment is at the support A.

Step 4: Final Answer:

The maximum bending moment in the beam is 100 kNm at the fixed end A.
Quick Tip: For any cantilever beam, the maximum bending moment and maximum shear force will \textbf{always} occur at the fixed support.
Simply calculate the total moment caused by all loads about the fixed support to find the maximum bending moment.

Step 1: Understanding the Question:

The question asks to describe the shape of the bending moment diagram (BMD) for the unloaded portion of a cantilever beam that has a UDL on its outer half.


Step 2: Key Formula or Approach:

We need to find the equation for the bending moment, \(M(x)\), in the left half of the beam (the part near the fixed support). We can use the fundamental relationship between shear force (\(V\)) and bending moment (\(M\)):
\[ \frac{dM}{dx} = V(x) \]
And the relationship between load (\(w\)) and shear force (\(V\)):
\[ \frac{dV}{dx} = -w(x) \]

Step 3: Detailed Explanation:

Let the cantilever be fixed at \(x=0\) and free at \(x=L\).

The UDL of \(w\) kN/m is applied from \(x=L/2\) to \(x=L\).

The left half of the beam is the section from \(x=0\) to \(x=L/2\).


Let's find the expression for shear force and bending moment by considering a section at a distance \(x\) from the fixed end, looking at the forces to the right of the section.


Case 1: Right half (\(L/2 \le x \le L\))

The shear force is \(V(x) = w(L-x)\). This is linear.

The bending moment is \(M(x) = -\frac{w(L-x)^2}{2}\). This is a parabola.


Case 2: Left half (\(0 \le x \le L/2\))

Now consider a section in the unloaded left half. The entire UDL is to the right of this section.

The total load from the UDL is \(W_{UDL} = w \times (L/2)\).

The shear force at any point in this section is constant and equal to the total load:
\[ V(x) = W_{UDL} = \frac{wL}{2} \quad (for 0 \le x \le L/2) \]
Since the shear force \(V(x)\) is constant in this region, and \(\frac{dM}{dx} = V(x)\), the bending moment \(M(x)\) must be a linear function of \(x\).

Wait, let's re-calculate using moments. The moment at a section \(x\) in the left half is the moment caused by the entire UDL on the right half. The UDL from \(L/2\) to \(L\) has a total force of \(w(L/2)\) and its centroid is at a distance of \(L/2 + L/4 = 3L/4\) from the free end, or \(L - 3L/4 = L/4\) from the fixed support. No, that's the centroid of the UDL itself. Let's integrate.

The moment at section \(x\) (for \(0 \le x \le L/2\)) is:
\[ M(x) = -\int_{L/2}^L w(z-x) dz \] \[ M(x) = -w \left[ \frac{(z-x)^2}{2} \right]_{L/2}^L \] \[ M(x) = -w \left( \frac{(L-x)^2}{2} - \frac{(L/2-x)^2}{2} \right) \] \[ M(x) = -\frac{w}{2} \left( (L^2 - 2Lx + x^2) - (L^2/4 - Lx + x^2) \right) \] \[ M(x) = -\frac{w}{2} \left( \frac{3L^2}{4} - Lx \right) \]
This expression shows that \(M(x)\) is a linear function of \(x\) in the left half.


There seems to be a contradiction between the calculated result and the provided answer key. Let's re-evaluate the standard relationships.

- Where load is zero, SFD is constant, BMD is linear.

- Where load is UDL (constant), SFD is linear, BMD is parabolic.

- Where load is linear, SFD is parabolic, BMD is cubic.

In our problem, for the left half of the beam (from the fixed end to mid-span), there is no applied load. Therefore, the shear force diagram in this region must be constant, and the bending moment diagram must be linear.

The provided answer is "Parabola". This would be correct if the question asked for the variation of bending moment in the right half of the beam (under the UDL). It is incorrect for the left (unloaded) half. Following the given answer key requires us to assume the question is flawed and meant to ask about the right half.


Step 4: Final Answer:

Based on the provided answer key, the variation is a Parabola. However, a rigorous analysis shows that for the unloaded left half of the beam, the variation of the bending moment is linear. The variation is parabolic for the right half, where the load is applied.
Quick Tip: Remember the graphical relationship between load, shear, and moment:
The shape of the BMD is always one order higher than the SFD, which is one order higher than the load diagram.
- No Load (0 order) \(\rightarrow\) Constant Shear (0 order) \(\rightarrow\) \textbf{Linear Moment (1st order).
- UDL (0 order) \(\rightarrow\) Linear Shear (1st order) \(\rightarrow\) \textbf{Parabolic Moment} (2nd order).

Step 1: Understanding the Question:

The question asks for the location of the maximum bending moment in a simply supported beam subjected to a single, off-center point load.


Step 2: Key Formula or Approach:

A fundamental principle of beam theory states that the maximum bending moment occurs at a point where the shear force is zero or changes sign.


Step 3: Detailed Explanation:

Let the beam have length \(L=6\) m. The load \(W\) is at \(a=2\) m from A and \(b=4\) m from B.

1. Calculate the Reactions:

Take moments about A: \(R_B \times 6 - W \times 2 = 0 \implies R_B = W/3\).

Take moments about B: \(R_A \times 6 - W \times 4 = 0 \implies R_A = 2W/3\).


2. Analyze the Shear Force Diagram (SFD):

- From A to the load (0 \(< x <\) 2 m): The shear force is constant and equal to \(V = R_A = +2W/3\).

- At the point of the load (\(x=2\) m): The shear force drops by the magnitude of the load, from \(+2W/3\) to \(+2W/3 - W = -W/3\).

- From the load to B (2 \(< x <\) 6 m): The shear force is constant and equal to \(V = -W/3 = -R_B\).


3. Locate Maximum Bending Moment:

The shear force diagram changes sign (from positive to negative) exactly at the point where the concentrated load \(W\) is applied. Therefore, the bending moment is maximum at this location.

The value of the maximum bending moment is \(M_{max} = R_A \times a = (2W/3) \times 2 = 4W/3\).


Step 4: Final Answer:

For a simply supported beam with a single concentrated load, the maximum bending moment always occurs at the point of application of the load.
Quick Tip: Remember this critical rule: \textbf{Maximum bending moment occurs where shear force is zero}.
For a simply supported beam with only point loads, the SFD is a series of horizontal lines.
The shear force will change sign at one of the load points, and that's where the BMD will have its peak.

Step 1: Understanding the Question:

The question asks what type of loading causes a sudden vertical jump (an instantaneous increase or decrease) in a shear force diagram (SFD).


Step 2: Detailed Explanation:

Let's review the relationship between the loading diagram and the shear force diagram for a beam:


(A) No loading: If there is no load between two points, the shear force remains constant. The SFD is a horizontal line.

(C) Uniformly distributed load (UDL): A UDL causes the shear force to change linearly. The SFD is a straight line with a constant, non-zero slope.

(D) Uniformly varying load (UVL): A UVL (triangular load) causes the shear force to change parabolically. The SFD is a parabolic curve.

(B) Concentrated load (Point load): A point load is a force applied at a single point. As we move across this point from left to right, the shear force diagram shows a sudden, vertical jump. The magnitude of the jump is equal to the magnitude of the concentrated load. An upward force causes an upward jump, and a downward force causes a downward jump.




Step 3: Final Answer:

A sudden increase or decrease in the shear force diagram indicates the presence of a concentrated load at that point.
Quick Tip: Visual rules for SFDs:
- \textbf{Vertical Jump} \(\rightarrow\) Point Load.
- \textbf{Sloped Straight Line} \(\rightarrow\) Uniformly Distributed Load (UDL).
- \textbf{Horizontal Line} \(\rightarrow\) No Load.
- \textbf{Curve} \(\rightarrow\) Varying Load.

Step 1: Understanding the Question:

The question asks for the maximum bending moment in a fixed-end beam with a concentrated load at the center.


Step 2: Key Formula or Approach:

For standard loading cases on fixed beams, there are standard formulas for the fixed-end moments and the mid-span moment. For a fixed beam of span \(L\) with a concentrated load \(P\) at the center:

1. The fixed-end moments (hogging) at the supports (A and B) are:
\[ M_A = M_B = -\frac{PL}{8} \]
2. The bending moment at the mid-span (C) under the load (sagging) is:
\[ M_C = +\frac{PL}{8} \]
The maximum bending moment in the beam is the maximum of the absolute values of the moments at the supports and at mid-span. In this symmetric case, they are equal in magnitude.


Step 3: Detailed Explanation:

We are given the following values:


Load (\(P\)) = 72 kN

Span (\(L\)) = 10 m



Now, calculate the magnitude of the moments:
\[ |M| = \frac{PL}{8} = \frac{(72 kN) \times (10 m)}{8} \] \[ |M| = \frac{720}{8} kNm \] \[ |M| = 90 kNm \]
The bending moment at the supports is -90 kNm (hogging) and at the center is +90 kNm (sagging). The maximum magnitude is therefore 90 kNm.


Step 4: Final Answer:

The maximum bending moment in the beam is 90 kNm.
Quick Tip: Memorize the key formulas for fixed beams:
- \textbf{Point Load at Center:} Max Moment = \(PL/8\) (at center and supports).
- \textbf{UDL over entire span:} Max Moment = \(wL^2/12\) (at supports). The center moment is \(wL^2/24\).

Step 1: Understanding the Question:

The question asks about the values of bending stress and shear stress at the centroid (neutral axis) of a beam's cross-section. The provided answer is (A), which suggests a potential error in the question or options as shear stress is maximum at the centroid, not zero, and shear stress is minimum at the extreme fibers, not minimum at the centroid. Let's analyze based on bending stress.


Step 2: Key Formula or Approach:

1. Bending Stress (\(\sigma\)): The theory of simple bending gives the flexure formula:
\[ \sigma = \frac{My}{I} \]
where \(M\) is the bending moment, \(I\) is the moment of inertia about the neutral axis, and \(y\) is the perpendicular distance from the neutral axis. The neutral axis passes through the centroid of the cross-section.

2. Shear Stress (\(\tau\)): The shear stress formula is:
\[ \tau = \frac{VA\bar{y}}{Ib} \]
where \(V\) is the shear force, \(A\) is the area above (or below) the point of interest, \(\bar{y}\) is the distance from the neutral axis to the centroid of area \(A\), \(I\) is the moment of inertia, and \(b\) is the width of the section at that point.


Step 3: Detailed Explanation:


Bending Stress at the Centroid: The centroid of the cross-section lies on the neutral axis. At the neutral axis, the distance \(y\) is, by definition, equal to zero. Substituting \(y=0\) into the flexure formula gives:

\[ \sigma = \frac{M(0)}{I} = 0 \]
Therefore, the bending stress is always zero at the centroid (neutral axis) of a beam. This makes option (A) correct. Option (B) is incorrect; bending stress is maximum at the extreme fibers (farthest from the neutral axis).

Shear Stress at the Centroid: The term \(A\bar{y}\) in the shear stress formula is the first moment of area. This term is maximum at the neutral axis for most common shapes (rectangle, I-section, circle). Therefore, the shear stress is typically maximum at the centroid, not zero or minimum. This makes options (C) and (D) incorrect.




Step 4: Final Answer:

At the centroid of a loaded beam cross section, the bending stress is equal to zero. The selected answer 'A' is correct, but it is important to note that the shear stress is maximum at this location.
Quick Tip: Remember the stress distribution across a beam's cross-section:
- \textbf{Bending Stress (\(\sigma\)):} \textbf{Zero} at the neutral axis (centroid), \textbf{Maximum} at the top and bottom (extreme fibers).
- \textbf{Shear Stress (\(\tau\)):} \textbf{Maximum} at the neutral axis (centroid), \textbf{Zero} at the top and bottom (extreme fibers).

Step 1: Understanding the Question:

The question asks for the maximum bending stress in a plate that is bent into a circular arc.


Step 2: Key Formula or Approach:

This problem uses the theory of simple bending. The fundamental bending equation is:
\[ \frac{M}{I} = \frac{\sigma}{y} = \frac{E}{R} \]
where:
\(\sigma\) = Bending stress at a distance \(y\) from the neutral axis
\(E\) = Modulus of Elasticity
\(R\) = Radius of curvature of the bent beam

We need to find the maximum bending stress, \(\sigma_{max}\). From the equation, we can write:
\[ \sigma = \frac{E y}{R} \]
The maximum stress occurs at the extreme fiber, where \(y\) is maximum. For a rectangular plate of thickness \(t\), the neutral axis is at the center, so \(y_{max} = t/2\).
\[ \sigma_{max} = \frac{E (t/2)}{R} \]

Step 3: Detailed Explanation:

First, identify and convert all values to consistent units (N and mm).


Plate thickness (\(t\)) = 10 mm.

Maximum distance from neutral axis (\(y_{max}\)) = \(t/2 = 10 mm / 2 = 5\) mm.

Radius of curvature (\(R\)) = 10 m = \(10 \times 1000 = 10,000\) mm.

Modulus of Elasticity (\(E\)) = 200 GPa = \(200 \times 10^3\) N/mm\(^2\).

The width of the plate (100 mm) is not needed to calculate the stress.



Now, substitute these values into the formula:
\[ \sigma_{max} = \frac{E \cdot y_{max}}{R} \] \[ \sigma_{max} = \frac{(200 \times 10^3 N/mm^2) \times (5 mm)}{10,000 mm} \] \[ \sigma_{max} = \frac{1,000,000}{10,000} N/mm^2 \] \[ \sigma_{max} = 100 N/mm^2 \]
Note that 1 N/mm\(^2\) is equal to 1 MPa. So the stress is 100 MPa.


Step 4: Final Answer:

The maximum bending stress induced in the plate is 100 N/mm\(^2\).
Quick Tip: The bending equation \(\frac{\sigma}{y} = \frac{E}{R}\) is very useful when the radius of curvature is known.
It directly relates stress to the geometry of the bent shape and the material's elasticity, without needing to know the applied moment (\(M\)) or moment of inertia (\(I\)).
Again, ensure all units are consistent before calculating.

Step 1: Understanding the Question:

The question asks how the maximum bending stress in a rectangular beam changes when its depth is doubled, assuming the applied load and other dimensions remain the same.


Step 2: Key Formula or Approach:

The maximum bending stress (\(\sigma_{max}\)) is given by the flexure formula:
\[ \sigma_{max} = \frac{M}{Z} \]
where \(M\) is the maximum bending moment and \(Z\) is the section modulus.

For a rectangular cross-section of width \(b\) and depth \(d\):

The moment of inertia is \(I = \frac{bd^3}{12}\).

The distance to the extreme fiber is \(y_{max} = d/2\).

The section modulus is \(Z = \frac{I}{y_{max}} = \frac{bd^3/12}{d/2} = \frac{bd^2}{6}\).

So, the stress formula becomes:
\[ \sigma_{max} = \frac{M}{bd^2/6} = \frac{6M}{bd^2} \]

Step 3: Detailed Explanation:

From the formula, we can see the relationship between stress and depth:
\[ \sigma_{max} \propto \frac{1}{d^2} \]
Let \(\sigma_1\) and \(d_1\) be the initial stress and depth, and \(\sigma_2\) and \(d_2\) be the final stress and depth.

We are given:


\(\sigma_1 = 160\) N/mm\(^2\).

The depth is "increased by two times", which can be interpreted as the new depth is twice the old one, \(d_2 = 2d_1\).



Using the proportionality, we can write a ratio:
\[ \frac{\sigma_2}{\sigma_1} = \left(\frac{d_1}{d_2}\right)^2 \] \[ \frac{\sigma_2}{160} = \left(\frac{d_1}{2d_1}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \] \[ \sigma_2 = \frac{160}{4} = 40 N/mm^2 \]

Step 4: Final Answer:

The new maximum bending stress induced in the beam is 40 N/mm\(^2\).
Quick Tip: For rectangular beams, the section modulus \(Z\) is proportional to \(d^2\).
Since bending stress \(\sigma = M/Z\), the stress is inversely proportional to \(d^2\).
This means doubling the depth makes the beam \textbf{four times stronger} in bending.

Step 1: Understanding the Question:

The question asks for the location of the maximum bending stress in a T-section cantilever beam under a UDL.


Step 2: Key Formula or Approach:

The bending stress is given by \(\sigma = My/I\). The maximum bending stress occurs at the point with the maximum value of \(y\), i.e., at the fiber farthest from the neutral axis. For a T-section, the neutral axis is not at the geometric center.


Step 3: Detailed Explanation:

1. Bending Moment: A cantilever beam with a UDL has a hogging (negative) bending moment throughout its length, with the maximum magnitude at the fixed support. A hogging moment causes tension in the top fibers and compression in the bottom fibers.

2. Neutral Axis of a T-section: For a T-section, the centroid (and thus the neutral axis) is located closer to the top flange because the flange has more area. This means the distance from the neutral axis to the top fiber (\(y_{top}\)) is smaller than the distance from the neutral axis to the bottom fiber of the web (\(y_{bottom}\)).
\[ y_{bottom} > y_{top} \]
3. Maximum Bending Stress: The magnitude of the bending stress is \(|\sigma| = |My/I|\). Since the bending moment \(M\) and moment of inertia \(I\) are constant for a given cross-section, the stress is maximum where the distance \(y\) is maximum.

Since \(y_{bottom} > y_{top}\), the maximum magnitude of the bending stress will occur at the bottom of the cross section.

Specifically, the top fiber will have the maximum tensile stress (\( \sigma_{tensile, max} = |M| y_{top} / I \)) and the bottom fiber will have the maximum compressive stress (\( \sigma_{compressive, max} = |M| y_{bottom} / I \)). The overall maximum magnitude of stress occurs at the bottom.


Step 4: Final Answer:

For a T-section cantilever beam under a UDL, the maximum bending stress (in magnitude) occurs at the bottom of the cross section.
Quick Tip: For any beam section, maximum bending stress always occurs at the extreme fibers (the points farthest from the neutral axis).
For asymmetric sections like a T-beam, you must first determine the location of the neutral axis (centroid) to identify which fiber (top or bottom) is farthest away.

Step 1: Understanding the Question:

The question asks for the average shear stress in a rectangular beam, given the maximum shear stress.


Step 2: Key Formula or Approach:

For a beam with a rectangular cross-section, the shear stress distribution is parabolic, with the maximum value at the neutral axis and zero at the top and bottom fibers.
The relationship between the maximum shear stress (\(\tau_{max}\)) and the average shear stress (\(\tau_{avg}\)) for a rectangular section is:
\[ \tau_{max} = 1.5 \times \tau_{avg} \]
The average shear stress is simply the total shear force (\(V\)) divided by the cross-sectional area (\(A\)): \(\tau_{avg} = V/A\).


Step 3: Detailed Explanation:

We are given the maximum shear stress:
\[ \tau_{max} = 120 N/mm^2 \]
We need to find the average shear stress, \(\tau_{avg}\). We can rearrange the formula:
\[ \tau_{avg} = \frac{\tau_{max}}{1.5} \] \[ \tau_{avg} = \frac{120 N/mm^2}{1.5} \] \[ \tau_{avg} = 80 N/mm^2 \]

Step 4: Final Answer:

The average shear stress is 80 N/mm\(^2\).
Quick Tip: Memorize the relationship between max and average shear stress for common shapes:
- \textbf{Rectangle:} \(\tau_{max} = 1.5 \times \tau_{avg}\)
- \textbf{Circle:} \(\tau_{max} = \frac{4}{3} \times \tau_{avg} \approx 1.33 \times \tau_{avg}\)
- \textbf{I-Beam:} \(\tau_{max} \approx \tau_{avg}\) (in the web, where most shear is carried)

Step 1: Understanding the Question:

The question asks for the name of a geometric property of a cross-section, defined as a specific ratio involving moment of inertia and distance.


Step 2: Key Formula or Approach:

Let's define the terms:

- Moment of Inertia (\(I\)): A measure of an object's resistance to rotational motion about an axis. For a beam cross-section, it represents resistance to bending.

- Extreme fiber distance (\(y_{max}\)): The maximum distance from the neutral axis to the outermost edge of the cross-section.

The ratio described in the question is:
\[ \frac{I}{y_{max}} \]

Step 3: Detailed Explanation:

This ratio is the definition of the Section Modulus, denoted by \(Z\) (or sometimes \(S\)).
\[ Z = \frac{I}{y_{max}} \]
The section modulus is a direct measure of the strength of a beam in bending. The flexure formula, \(\sigma_{max} = \frac{M_{max}}{I/y_{max}}\), is often simplified to:
\[ \sigma_{max} = \frac{M_{max}}{Z} \]
This shows that for a given bending moment, a larger section modulus results in a lower maximum bending stress.

The other options are:

- Radius of Gyration (\(k\)): \(k = \sqrt{I/A}\). It relates to buckling.

- Polar Moment of Inertia (\(J\)): A measure of resistance to torsion. For a planar shape, \(J = I_x + I_y\).


Step 4: Final Answer:

The ratio of the moment of inertia of the section about the neutral axis to the extreme fiber distance from the neutral axis is the Section modulus.
Quick Tip: Remember the key geometric properties and their roles:
- \textbf{Area (\(A\)):} Resists axial load.
- \textbf{Moment of Inertia (\(I\)):} Resists bending.
- \textbf{Section Modulus (\(Z = I/y_{max}\)):} A direct measure of bending strength.
- \textbf{Polar Moment of Inertia (\(J\)):} Resists torsion.

Step 1: Understanding the Question:

The question asks for the conditions under which the theory of simple bending (or pure bending) is strictly applicable.


Step 2: Detailed Explanation:

The theory of simple bending, which leads to the flexure formula \(\frac{M}{I} = \frac{\sigma}{y} = \frac{E}{R}\), is derived based on a set of assumptions. One of the most important is that the beam is subjected to pure bending.

Pure bending is a loading condition where a beam is subjected to a constant bending moment and, consequently, zero shear force.

The relationship between bending moment (\(M\)) and shear force (\(V\)) is \(V = dM/dx\).

If the bending moment \(M\) is constant, then its derivative with respect to \(x\) must be zero.
\[ \frac{dM}{dx} = 0 \implies V = 0 \]
This condition (constant bending moment and zero shear force) ensures that the stresses in the beam are purely normal stresses due to bending, without any complicating shear stresses that would warp the cross-section. While the bending equation is often applied to cases where shear force is present, its theoretical derivation is based on the case of pure bending.


Step 3: Final Answer:

The theory of simple bending equation is strictly applicable for a condition of constant bending moment and zero shear force.
Quick Tip: The ideal condition for the simple bending theory is called \textbf{Pure Bending}.
Pure Bending means:
- Bending Moment = \textbf{Constant}.
- Shear Force = \textbf{Zero}.
An example is the central portion of a beam loaded with two symmetric point loads.

Step 1: Understanding the Question:

The question asks for the maximum shear stress in a simply supported rectangular beam under a UDL.


Step 2: Key Formula or Approach:

The calculation involves two main steps:

1. Find the maximum shear force (\(V_{max}\)) in the beam.

2. Use the formula for maximum shear stress in a rectangular section: \(\tau_{max} = 1.5 \times \tau_{avg} = 1.5 \times \frac{V_{max}}{A}\).


Step 3: Detailed Explanation:

Part 1: Find the maximum shear force (\(V_{max}\)).

For a simply supported beam of span \(L\) with a UDL of \(w\) over the entire span, the reactions at each support are equal and are half of the total load.

- Total load = \(w \times L = 20 kN/m \times 6 m = 120\) kN.

- Reactions: \(R_A = R_B = \frac{120 kN}{2} = 60\) kN.

The maximum shear force in a simply supported beam occurs at the supports and is equal to the magnitude of the reactions.
\[ V_{max} = 60 kN = 60 \times 10^3 N \]

Part 2: Calculate the maximum shear stress (\(\tau_{max}\)).

First, find the cross-sectional area (\(A\)).

- Width (\(b\)) = 100 mm.

- Depth (\(d\)) = 200 mm.

- Area (\(A\)) = \(b \times d = 100 mm \times 200 mm = 20,000 mm^2\).

Now use the formula for \(\tau_{max}\) for a rectangular section:
\[ \tau_{max} = 1.5 \times \frac{V_{max}}{A} \] \[ \tau_{max} = 1.5 \times \frac{60 \times 10^3 N}{20,000 mm^2} \] \[ \tau_{max} = 1.5 \times 3 N/mm^2 \] \[ \tau_{max} = 4.5 N/mm^2 \]
This calculation gives 4.5 N/mm\(^2\). However, the provided answer key marks option (D) 9 N/mm\(^2\) as correct. Let's re-check the calculation. The calculation is robust. \(V_{max}=60\) kN, \(A=20000\) mm\(^2\), \(\tau_{avg}=3\) N/mm\(^2\), \(\tau_{max}=1.5 \times 3 = 4.5\) N/mm\(^2\). The provided answer key is incorrect. The correct answer is 4.5 N/mm\(^2\). We will proceed by justifying the calculated answer.


Step 4: Final Answer:

The maximum shear stress induced in the beam is 4.5 N/mm\(^2\). The provided answer key indicating 9 N/mm\(^2\) appears to be in error, possibly due to a miscalculation (e.g., using the total load instead of the reaction or omitting the 1.5 factor incorrectly).
Quick Tip: To find max shear stress, always follow the two steps:
1. Find \(V_{max}\) from the beam's loading diagram (for SS beam with UDL, it's at the supports).
2. Apply the correct formula for \(\tau_{max}\) based on the cross-section shape (for a rectangle, it's \(1.5 \times V/A\)).

Step 1: Understanding the Question:

The question asks for the ratio of the average shear stress to the maximum shear stress (\(\tau_{avg} / \tau_{max}\)) for a circular cross-section.


Step 2: Key Formula or Approach:

For a beam with a circular cross-section, the shear stress distribution is parabolic, with the maximum value at the neutral axis (the center) and zero at the top and bottom.

The relationship between the maximum shear stress (\(\tau_{max}\)) and the average shear stress (\(\tau_{avg}\)) for a circular section is:
\[ \tau_{max} = \frac{4}{3} \times \tau_{avg} \]

Step 3: Detailed Explanation:

We are asked to find the ratio \(\tau_{avg} / \tau_{max}\).

We can rearrange the formula to find this ratio:
\[ \frac{\tau_{avg}}{\tau_{max}} = \frac{1}{4/3} \] \[ \frac{\tau_{avg}}{\tau_{max}} = \frac{3}{4} \] \[ \frac{\tau_{avg}}{\tau_{max}} = 0.75 \]

Step 4: Final Answer:

The ratio of average shear stress to maximum shear stress for a circular beam is 0.75.
Quick Tip: Be careful with the wording of the ratio.
- \textbf{Max to Average} for a circle is \(\tau_{max} / \tau_{avg} = 4/3 \approx 1.33\).
- \textbf{Average to Max} for a circle is \(\tau_{avg} / \tau_{max} = 3/4 = 0.75\).
Read the question carefully to provide the correct ratio.

Step 1: Understanding the Question:

The question asks about the slope and deflection at the free end of a cantilever beam under downward loading.


Step 2: Detailed Explanation:

Let's consider a cantilever beam fixed at one end and free at the other. When downward loads are applied, the beam bends downwards.


Deflection: Deflection is the vertical displacement of the beam from its original position. The fixed support cannot move, so the deflection there is zero by definition. As we move away from the fixed support towards the free end, the beam bends more and more. The largest vertical displacement will occur at the very tip, the free end. Therefore, the deflection is maximum at the free end.

Slope: Slope is the angle of the deflected beam with respect to the horizontal. At the fixed support, the beam is held rigidly horizontal, so the slope is zero by definition. As the beam bends, its angle with the horizontal increases, reaching its largest value at the free end. Therefore, the slope is also maximum at the free end.




Step 3: Final Answer:

At the free end of a cantilever beam subjected to downward loads, both the slope and the deflection are maximum.
Quick Tip: Remember the boundary conditions for a cantilever beam (fixed at A, free at B):
- \textbf{At the Fixed End (A):} Deflection = 0, Slope = 0.
- \textbf{At the Free End (B):} Bending Moment = 0, Shear Force = 0 (unless there is a point load right at the end). Slope and Deflection are \textbf{maximum}.

Step 1: Understanding the Question:

The question asks for the statement of the Second Moment-Area Theorem.


Step 2: Detailed Explanation:

The Moment-Area Theorems are a method for finding the slope and deflection of beams. There are two theorems:


First Moment-Area Theorem: The change in slope between any two points on the elastic curve of a beam is equal to the area of the \(M/EI\) diagram between those two points.

Second Moment-Area Theorem: The vertical deviation (or vertical intercept) of a point B on the elastic curve with respect to the tangent drawn at another point A is equal to the moment of the area of the \(M/EI\) diagram between points A and B, taken about point B.


The question describes "the vertical intercept at a point between the tangents drawn from two points". This is a direct statement of the Second Moment-Area Theorem. The area it refers to is the area of the bending moment diagram divided by \(EI\) (the \(M/EI\) diagram). The moment is taken about the point where the deviation is being measured.


Step 3: Final Answer:

The vertical intercept is equal to the moment of the area of the bending moment diagram between those two points about the point under consideration, divided by \(EI\).
Quick Tip: Remember the two Moment-Area theorems in short form:
- \textbf{Theorem 1 (Slope):} \(\Delta \theta\) = Area of \(M/EI\) diagram.
- \textbf{Theorem 2 (Deflection):} Deflection = Moment of Area of \(M/EI\) diagram.

Step 1: Understanding the Question:

The question asks to identify the physical quantity represented by the mathematical expression \(EI \frac{d^2y}{dx^2}\) in beam theory.


Step 2: Key Formula or Approach:

This expression comes from the Euler-Bernoulli beam theory, which establishes the relationship between the beam's deflection and the internal forces. The fundamental differential equation of the elastic curve (deflection curve) of a beam is:
\[ \frac{d^2y}{dx^2} = \frac{M}{EI} \]
where:
\(y\) is the deflection of the beam at a position \(x\) along its length.
\(M\) is the internal bending moment at that section.
\(E\) is the modulus of elasticity of the beam material.
\(I\) is the moment of inertia of the beam's cross-section.
\(EI\) is the flexural rigidity of the beam.


Step 3: Detailed Explanation:

By rearranging the differential equation, we get:
\[ M = EI \frac{d^2y}{dx^2} \]
This shows that the expression \(EI \frac{d^2y}{dx^2}\) is exactly equal to the bending moment (\(M\)) at that section of the beam.

The other quantities are related to higher derivatives:

- Shear Force (\(V\)) = \(\frac{dM}{dx} = EI \frac{d^3y}{dx^3}\)

- Rate of loading (\(w\)) = \(-\frac{dV}{dx} = -EI \frac{d^4y}{dx^4}\)

- Deflection is simply \(y\).


Step 4: Final Answer:

The expression \(EI \frac{d^2y}{dx^2}\) at a section of a beam represents the bending moment.
Quick Tip: Memorize the hierarchy of beam equations:
- \(y(x)\) = Deflection
- \(\frac{dy}{dx}\) = Slope
- \(EI \frac{d^2y}{dx^2}\) = Bending Moment (\(M\))
- \(EI \frac{d^3y}{dx^3}\) = Shear Force (\(V\))
- \(EI \frac{d^4y}{dx^4}\) = - (Load Intensity, \(w\))

Step 1: Understanding the Question:

The question asks to calculate the slope at the free end of a cantilever beam subjected to a point load at its tip.


Step 2: Key Formula or Approach:

For a cantilever beam of length \(L\) with a point load \(P\) at the free end, the standard formula for the slope (\(\theta\)) at the free end is:
\[ \theta_{free-end} = \frac{PL^2}{2EI} \]

Step 3: Detailed Explanation:

First, identify the given values. It's important to ensure the units are consistent.


Load (\(P\)) = 48 kN

Span (\(L\)) = 2 m

Flexural Rigidity (\(EI\)) = 20,000 kNm\(^2\)



All units (kN and m) are consistent, so no conversion is needed. The resulting slope will be in radians.

Now, substitute the values into the formula:
\[ \theta = \frac{(48 kN) \times (2 m)^2}{2 \times (20,000 kNm^2)} \] \[ \theta = \frac{48 \times 4}{40,000} \] \[ \theta = \frac{192}{40,000} \] \[ \theta = 0.0048 radians \]
To express this in the format of the options, we write it in scientific notation:
\[ \theta = 4.8 \times 10^{-3} radians \]

Step 4: Final Answer:

The slope at the free end is \(4.8 \times 10^{-3}\) radians.
Quick Tip: Memorize the standard formulas for cantilever beam slope and deflection:
- \textbf{Load at Free End (P):}
- Max Slope = \(PL^2 / (2EI)\)
- Max Deflection = \(PL^3 / (3EI)\)
- \textbf{UDL over entire span (w):}
- Max Slope = \(wL^3 / (6EI)\)
- Max Deflection = \(wL^4 / (8EI)\)

Step 1: Understanding the Question:

The question asks for the formula for the maximum deflection (at mid-span) of a simply supported beam of a given length and under a UDL.


Step 2: Key Formula or Approach:

The standard formula for the maximum deflection (\(\delta_{max}\)) of a simply supported beam of span 'S' under a uniformly distributed load 'w' over its entire length is:
\[ \delta_{max} = \frac{5 w S^4}{384 EI} \]

Step 3: Detailed Explanation:

In this problem, the variables are given with slightly different notation. We must be careful to substitute correctly.

- The given span is \(S = 2L\).

- The given load intensity is \(w\).

Now, we substitute \(S=2L\) into the standard formula:
\[ \delta_{max} = \frac{5 w (2L)^4}{384 EI} \] \[ \delta_{max} = \frac{5 w (16L^4)}{384 EI} \] \[ \delta_{max} = \frac{80 wL^4}{384 EI} \]
Now, simplify the fraction 80/384. We can divide both by 16:
\(80 / 16 = 5\)
\(384 / 16 = 24\)

So the expression becomes:
\[ \delta_{max} = \frac{5 wL^4}{24 EI} \]

Step 4: Final Answer:

The deflection at mid span is \(\frac{5 wL^4}{24 EI}\).
Quick Tip: Be very careful when a problem uses non-standard notation for span length (like 2L instead of L).
Always start with the standard formula you have memorized and then carefully substitute the specific variables given in the problem.

Step 1: Understanding the Question:

The question asks for the ratio of the maximum deflection to the maximum slope for a simply supported beam with a point load at the center.


Step 2: Key Formula or Approach:

We need the standard formulas for maximum deflection (\(\delta_{max}\)) and maximum slope (\(\theta_{max}\)) for this loading case.

1. Maximum Deflection: Occurs at the center, under the load.
\[ \delta_{max} = \frac{WL^3}{48EI} \]
2. Maximum Slope: Occurs at the supports (A and B).
\[ \theta_{max} = \frac{WL^2}{16EI} \]

Step 3: Detailed Explanation:

Now we need to find the ratio \(\frac{\delta_{max}}{\theta_{max}}\).
\[ \frac{\delta_{max}}{\theta_{max}} = \frac{WL^3 / (48EI)}{WL^2 / (16EI)} \]
We can cancel the common terms \(W\), \(L^2\), and \(EI\) from the numerator and denominator:
\[ \frac{\delta_{max}}{\theta_{max}} = \frac{L/48}{1/16} \] \[ \frac{\delta_{max}}{\theta_{max}} = \frac{L}{48} \times 16 \] \[ \frac{\delta_{max}}{\theta_{max}} = \frac{16L}{48} \]
Simplifying the fraction:
\[ \frac{\delta_{max}}{\theta_{max}} = \frac{L}{3} \]

Step 4: Final Answer:

The ratio of maximum deflection to maximum slope is \(L/3\).
Quick Tip: Memorizing the standard deflection and slope formulas for simply supported and cantilever beams is essential for solving these types of problems quickly.
- \textbf{SS Beam, Center Load P:} \(\delta_{max} = PL^3/(48EI)\), \(\theta_{max} = PL^2/(16EI)\).
- \textbf{SS Beam, UDL w:} \(\delta_{max} = 5wL^4/(384EI)\), \(\theta_{max} = wL^3/(24EI)\).

Step 1: Understanding the Question:

The question asks for the maximum deflection of a fixed-end beam with a concentrated load at its center.


Step 2: Key Formula or Approach:

The standard formula for the maximum deflection (\(\delta_{max}\)) of a fixed beam of span \(L\) subjected to a point load \(P\) at mid-span is:
\[ \delta_{max} = \frac{PL^3}{192EI} \]

Step 3: Detailed Explanation:

First, identify the given values. Ensure the units are consistent.


Load (\(P\)) = 96 kN

Span (\(L\)) = 8 m

Flexural Rigidity (\(EI\)) = 51,200 kNm\(^2\)



All units are in kN and m. The result will be in meters.

Substitute the values into the formula:
\[ \delta_{max} = \frac{(96 kN) \times (8 m)^3}{192 \times (51,200 kNm^2)} \] \[ \delta_{max} = \frac{96 \times 512}{192 \times 51,200} \]
We can simplify this. Note that \(192 = 2 \times 96\) and \(51,200 = 100 \times 512\).
\[ \delta_{max} = \frac{96 \times 512}{(2 \times 96) \times (100 \times 512)} \]
Cancel the terms 96 and 512:
\[ \delta_{max} = \frac{1}{2 \times 100} = \frac{1}{200} m \] \[ \delta_{max} = 0.005 m \]
The options are in mm. Convert the result to mm:
\[ \delta_{max} = 0.005 m \times 1000 \frac{mm}{m} = 5 mm \]

Step 4: Final Answer:

The maximum deflection under the load is 5 mm.
Quick Tip: The maximum deflection for a fixed beam is much smaller than for a simply supported beam under the same load.
- Fixed Beam, Center Load: \(\delta_{max} = \frac{PL^3}{192EI}\)
- SS Beam, Center Load: \(\delta_{max} = \frac{PL^3}{48EI}\)
The fixed beam is \(192/48 = 4\) times stiffer.

Step 1: Understanding the Question:

The question asks to classify a continuous beam based on its static determinacy.


Step 2: Detailed Explanation:

A structure is statically determinate if all of its unknown support reactions and internal forces can be determined using only the equations of static equilibrium (\(\Sigma F_x = 0\), \(\Sigma F_y = 0\), \(\Sigma M = 0\)). For a 2D beam, there are 3 available equilibrium equations.

A structure is statically indeterminate (or hyperstatic) if it has more unknown reactions than the number of available static equilibrium equations. The extra, redundant reactions require additional equations (based on material properties and deformation, i.e., compatibility equations) to be solved.


A continuous beam is a beam that rests on more than two supports.

- A beam with three supports (e.g., a pin and two rollers) will have 4 unknown reactions (2 at the pin, 1 at each roller). Since there are only 3 equilibrium equations, there is \(4-3=1\) redundant reaction.

- A beam with 'n' supports will generally have at least n+1 reactions.

Because the number of unknown reactions is greater than the number of equilibrium equations, a continuous beam cannot be fully analyzed using statics alone.


Step 3: Final Answer:

A continuous beam is a statically indeterminate structure.
Quick Tip: Quickly determine static determinacy for beams:
- \textbf{Determinate:} Simply Supported Beam, Cantilever Beam, Overhanging Beam (if reactions \(\le 3\)).
- \textbf{Indeterminate:} Continuous Beam, Fixed Beam, Propped Cantilever.
Count the number of unknown reaction components. If it's more than 3, it's indeterminate.

Step 1: Understanding the Question:

The question asks for the reaction force at the propped end (B) of a propped cantilever beam with a point load at its center.


Step 2: Key Formula or Approach:

A propped cantilever is a statically indeterminate structure. We need to use a method that accounts for beam deflection, such as the principle of superposition.

We can think of the propped cantilever as the sum of two cases:

1. A cantilever beam with only the downward load \(P\) at mid-span. This causes a downward deflection at the free end, \(\delta_{B1}\).

2. A cantilever beam with only the upward reaction force from the prop, \(R_B\), at the free end. This causes an upward deflection at the free end, \(\delta_{B2}\).

The compatibility condition is that the net deflection at the prop (point B) must be zero.
\[ \delta_{B1} (downward) = \delta_{B2} (upward) \]
The standard formulas for these deflections are needed.

- Deflection at the free end of a cantilever due to a point load \(P\) at mid-span (\(L/2\)) is: \(\delta = \frac{5PL^3}{48EI}\).

- Deflection at the free end of a cantilever due to a point load \(R_B\) at the free end is: \(\delta = \frac{R_B L^3}{3EI}\).


Step 3: Detailed Explanation:

Set the two deflections equal to each other:
\[ \frac{5PL^3}{48EI} = \frac{R_B L^3}{3EI} \]
We can cancel the common terms \(L^3\) and \(EI\):
\[ \frac{5P}{48} = \frac{R_B}{3} \]
Now solve for the prop reaction \(R_B\):
\[ R_B = \frac{3 \times 5P}{48} = \frac{15P}{48} \]
Simplify the fraction by dividing by 3:
\[ R_B = \frac{5P}{16} \]
Now substitute the given value of the load, \(P = 64\) kN:
\[ R_B = \frac{5 \times 64}{16} \] \[ R_B = 5 \times 4 = 20 kN \]
The answer is 20 kN. However, the checkmark in the provided image is on 20 kN, which matches this calculation. The initial OCR indicated option 1 as `20 kN` with a checkmark, so the solution aligns with the provided key.


Step 4: Final Answer:

The reaction of the prop is 20 kN.
Quick Tip: It is extremely useful to memorize the standard results for propped cantilevers.
- \textbf{Load P at mid-span:} Prop reaction \(R_B = 5P/16\).
- \textbf{UDL w over entire span:} Prop reaction \(R_B = 3wL/8\).

Step 1: Understanding the Question:

The question asks for the sign or nature of the bending moment that occurs over the interior supports of a continuous beam subjected to downward vertical loads.


Step 2: Detailed Explanation:

A continuous beam spans over multiple supports. When downward loads (like gravity loads) are applied, the beam deflects.

- In the regions between the supports (in the spans), the beam tends to bend downwards, like a smiley face. This shape is called sagging, and it corresponds to a positive bending moment. This causes tension at the bottom fibers and compression at the top.

- Over the intermediate supports, the beam is bent upwards in a convex shape, like a frowny face. The support pushes up while the spans on either side sag down. This shape is called hogging, and it corresponds to a negative bending moment. This causes tension at the top fibers and compression at the bottom.

Therefore, the moments at the intermediate supports are always hogging moments for downward loads.


Step 3: Final Answer:

The nature of the bending moment developed at the intermediate supports of a continuous beam carrying downward loads is Hogging.
Quick Tip: Use a simple visual aid for bending moments:
- \textbf{Sagging} = beam bends into a "smile" shape \(\smile\) = \textbf{Positive Moment}. (Occurs mid-span)
- \textbf{Hogging} = beam bends into a "frown" shape \(\frown\) = \textbf{Negative Moment}. (Occurs over intermediate supports and at fixed ends of cantilevers).

Step 1: Understanding the Question:

The question asks for the classification of a column based on a given slenderness ratio.


Step 2: Detailed Explanation:

Columns are compression members that are classified based on their length, cross-sectional dimensions, and end conditions, which are all combined into a single parameter called the slenderness ratio (\(\lambda\)). The slenderness ratio determines the column's mode of failure.
\[ \lambda = \frac{L_{eff}}{k_{min}} \]
where \(L_{eff}\) is the effective length and \(k_{min}\) is the minimum radius of gyration.

The classifications are generally:


Short Columns: Have a small slenderness ratio. They fail by crushing (yielding of the material) at a stress close to the material's compressive strength. Their strength is independent of their length.

Long Columns (or Slender Columns): Have a large slenderness ratio. They fail by elastic buckling, a sudden lateral deflection, at a stress that can be much lower than the material's yield strength. Their failure load is highly dependent on their length and stiffness. The Euler column formula applies to these.

Intermediate Columns: Fall between short and long columns. They fail by a combination of crushing and buckling (inelastic buckling).



The specific numerical values for these ranges depend on the material and the design code. For steel, a common rule of thumb (though it varies) might place the transition to long columns somewhere between a slenderness ratio of 80 and 120. A value greater than 120 is definitively in the range of a long column, where buckling is the primary failure mode.


Step 3: Final Answer:

A column whose slenderness ratio is greater than 120 is known as a long column.
Quick Tip: Remember the failure modes based on slenderness ratio (\(\lambda\)):
- \textbf{Low \(\lambda\)} \(\rightarrow\) Short Column \(\rightarrow\) Fails by \textbf{Crushing}.
- \textbf{High \(\lambda\)} \(\rightarrow\) Long Column \(\rightarrow\) Fails by \textbf{Buckling}.
- \textbf{Intermediate \(\lambda\)} \(\rightarrow\) Intermediate Column \(\rightarrow\) Fails by \textbf{Crushing + Buckling}.

Step 1: Understanding the Question:

The question asks about the direction or axis around which a compression member (a column) will buckle.


Step 2: Detailed Explanation:

Buckling is a failure mode characterized by a sudden lateral deflection. A column under compression will always buckle in the way that is easiest for it to do so, which is the direction of its least resistance to bending.

The resistance of a column to bending (and thus buckling) is determined by its flexural rigidity, \(EI\). For a given material (\(E\)), the resistance depends on the moment of inertia, \(I\).

A column's cross-section will generally have different moments of inertia about its different principal axes (e.g., \(I_{xx}\) and \(I_{yy}\) for a rectangular or I-section). The column will buckle by bending about the axis with the minimum moment of inertia (\(I_{min}\)).

The radius of gyration (\(k\)) is defined as \(k = \sqrt{I/A}\). Since the area \(A\) is constant for the cross-section, the axis with the minimum moment of inertia (\(I_{min}\)) is also the axis with the least radius of gyration (\(k_{min}\)).

Therefore, a column always tends to buckle about the axis corresponding to its least radius of gyration, as this is its weakest direction.


Step 3: Final Answer:

Compression members always tend to buckle in the direction of the least radius of gyration.
Quick Tip: Think of it like trying to bend a ruler. It's very easy to bend it about its thin axis (low moment of inertia) but very difficult to bend it about its wide axis (high moment of inertia).
A column behaves the same way under compression - it will buckle in its "flattest" or weakest direction, which corresponds to the axis with the least radius of gyration.