Step 1: Understanding the Question:

The question asks for the "order" of a given matrix.

The order of a matrix is a way to describe its dimensions, given as the number of rows by the number of columns.


Step 2: Detailed Explanation:

First, we count the number of horizontal lines, which are the rows of the matrix.

The given matrix is: \[ \begin{bmatrix} 1 & 6
2 & 0
7 & -1 \end{bmatrix} \]
Row 1 is [1 6].

Row 2 is [2 0].

Row 3 is [7 -1].

So, there are 3 rows.




Next, we count the number of vertical lines, which are the columns of the matrix.

Column 1 is \( \begin{bmatrix} 1
2
7 \end{bmatrix} \).

Column 2 is \( \begin{bmatrix} 6
0
-1 \end{bmatrix} \).

So, there are 2 columns.




The order is always written in the format (number of rows) \( \times \) (number of columns).

Therefore, the order of this matrix is \( 3 \times 2 \).


Step 3: Final Answer:

The matrix has 3 rows and 2 columns, so its order is \( 3 \times 2 \). This corresponds to option (B).
Quick Tip: To easily remember the format for the order of a matrix, think of "RC" - Rows first, then Columns. This simple mnemonic can help avoid confusion during exams.

Step 1: Understanding the Question:

The question asks for the value of a determinant when two of its rows or two of its columns are identical. This is a question about a fundamental property of determinants.


Step 2: Detailed Explanation:

One of the key properties of determinants is as follows:

Property: If any two rows or any two columns of a determinant are identical, then the value of the determinant is zero.




Reasoning:

Let's consider a determinant \( \Delta \).

Another property of determinants states that if we interchange any two rows (or columns), the sign of the determinant changes.

Suppose rows \( R_i \) and \( R_j \) of the determinant are identical.

If we interchange these two identical rows, the determinant itself does not change.

So, the new determinant is still \( \Delta \).

However, according to the interchange property, the new determinant should be \( -\Delta \).

Therefore, we have the equation:
\[ \Delta = -\Delta \] \[ 2\Delta = 0 \] \[ \Delta = 0 \]
This proves that the value of the determinant must be zero.


Step 3: Final Answer:

Based on the properties of determinants, if two rows or columns are identical, the determinant's value is 0. This corresponds to option (A).
Quick Tip: This is a crucial property to memorize. In an exam, if you are asked to evaluate a determinant, first quickly scan the rows and columns. If you spot any identical or proportional rows/columns, you can immediately state the answer is 0 without performing any calculations.

Step 1: Understanding the Question:

The question asks for the cofactor of a specific element, -4, in a given 3x3 determinant.


Step 2: Key Formula or Approach:

The cofactor of an element \( a_{ij} \) (located in the i-th row and j-th column) is given by the formula:
\[ C_{ij} = (-1)^{i+j} M_{ij} \]
where \( M_{ij} \) is the minor of the element \( a_{ij} \). The minor is the determinant of the submatrix formed by removing the i-th row and j-th column.


Step 3: Detailed Explanation:

First, we locate the element -4 in the determinant. It is in the 2nd row and 1st column. So, i = 2 and j = 1.

The element is \( a_{21} = -4 \).




Next, we calculate the minor \( M_{21} \). We do this by removing the 2nd row and 1st column from the original matrix:

Original determinant: \( \begin{vmatrix} 1 & 2 & 3
-4 & 3 & 6
2 & -7 & 9 \end{vmatrix} \)

Submatrix for \( M_{21} \): \( \begin{vmatrix} 2 & 3
-7 & 9 \end{vmatrix} \)

Now, we calculate the determinant of this 2x2 submatrix:
\[ M_{21} = (2)(9) - (3)(-7) = 18 - (-21) = 18 + 21 = 39 \]



Finally, we calculate the cofactor \( C_{21} \) using the formula:
\[ C_{21} = (-1)^{2+1} M_{21} \] \[ C_{21} = (-1)^3 \times 39 \] \[ C_{21} = -1 \times 39 = -39 \]

Step 4: Final Answer:

The cofactor of the element -4 is -39. This corresponds to option (D).
Quick Tip: The sign of the cofactor, given by \( (-1)^{i+j} \), follows a checkerboard pattern: \( \begin{pmatrix} + & - & +
- & + & -
+ & - & + \end{pmatrix} \). For the element at position (2,1), the sign is negative. You can quickly determine the sign and then just calculate the minor.

Step 1: Understanding the Question:

The question asks to classify the given matrix based on its properties. The options suggest we need to check for symmetry or skew-symmetry.


Step 2: Key Formula or Approach:

A square matrix A is called symmetric if it is equal to its transpose, i.e., \( A = A^T \). This means that the element in the i-th row and j-th column is equal to the element in the j-th row and i-th column, i.e., \( a_{ij} = a_{ji} \) for all i and j.

A square matrix A is called skew-symmetric if \( A = -A^T \), which means \( a_{ij} = -a_{ji} \). This implies all diagonal elements must be zero.


Step 3: Detailed Explanation:

Let the given matrix be A: \[ A = \begin{bmatrix} a & h & g
h & b & f
g & f & c \end{bmatrix} \]
Let's check the condition for a symmetric matrix, \( a_{ij} = a_{ji} \).

- Element at (1,2) is \( a_{12} = h \). Element at (2,1) is \( a_{21} = h \). So, \( a_{12} = a_{21} \).

- Element at (1,3) is \( a_{13} = g \). Element at (3,1) is \( a_{31} = g \). So, \( a_{13} = a_{31} \).

- Element at (2,3) is \( a_{23} = f \). Element at (3,2) is \( a_{32} = f \). So, \( a_{23} = a_{32} \).

Since the condition \( a_{ij} = a_{ji} \) holds for all elements, the matrix is symmetric.


Alternatively, we can find the transpose of A, \( A^T \), by interchanging rows and columns: \[ A^T = \begin{bmatrix} a & h & g
h & b & f
g & f & c \end{bmatrix} \]
We can see that \( A = A^T \). Therefore, the matrix is symmetric.


Step 4: Final Answer:

The given matrix satisfies the condition for a symmetric matrix. This corresponds to option (B).
Quick Tip: A symmetric matrix is visually symmetric about its main diagonal (from top-left to bottom-right). You can quickly check if the elements are mirrored across this diagonal. This matrix is a standard representation of a symmetric matrix used in various fields like physics and geometry.

Step 1: Understanding the Question:

The question asks for the inverse of the given matrix A.


Step 2: Key Formula or Approach:

The inverse of a matrix A, denoted \( A^{-1} \), is a matrix such that \( A A^{-1} = A^{-1} A = I \), where I is the identity matrix.

A simple way to check if a matrix is its own inverse is to multiply the matrix by itself. If the result is the identity matrix (\( A^2 = I \)), then it means \( A = A^{-1} \). Such a matrix is called an involutory matrix.


Step 3: Detailed Explanation:

Let's compute the product \( A \times A \): \[ A^2 = A \times A = \begin{bmatrix} 0 & 0 & 1
0 & 1 & 0
1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 1
0 & 1 & 0
1 & 0 & 0 \end{bmatrix} \]
We perform the matrix multiplication row by column:

The element in the 1st row, 1st column is \( (0)(0) + (0)(0) + (1)(1) = 1 \).

The element in the 1st row, 2nd column is \( (0)(0) + (0)(1) + (1)(0) = 0 \).

The element in the 1st row, 3rd column is \( (0)(1) + (0)(0) + (1)(0) = 0 \).


The element in the 2nd row, 1st column is \( (0)(0) + (1)(0) + (0)(1) = 0 \).

The element in the 2nd row, 2nd column is \( (0)(0) + (1)(1) + (0)(0) = 1 \).

The element in the 2nd row, 3rd column is \( (0)(1) + (1)(0) + (0)(0) = 0 \).


The element in the 3rd row, 1st column is \( (1)(0) + (0)(0) + (0)(1) = 0 \).

The element in the 3rd row, 2nd column is \( (1)(0) + (0)(1) + (0)(0) = 0 \).

The element in the 3rd row, 3rd column is \( (1)(1) + (0)(0) + (0)(0) = 1 \).


So, the resulting matrix is: \[ A^2 = \begin{bmatrix} 1 & 0 & 0
0 & 1 & 0
0 & 0 & 1 \end{bmatrix} = I \]
Since \( A^2 = I \), we can pre-multiply by \( A^{-1} \): \[ A^{-1}(A A) = A^{-1} I \] \[ (A^{-1}A) A = A^{-1} \] \[ I A = A^{-1} \] \[ A = A^{-1} \]

Step 4: Final Answer:

The inverse of matrix A is the matrix A itself. This corresponds to option (A).
Quick Tip: The given matrix is a permutation matrix. Specifically, it swaps the first and third elements. Applying the same swap twice returns the original arrangement. Recognizing special types of matrices like permutation matrices, identity matrices, or zero matrices can often lead to a much faster solution.

Step 1: Understanding the Question:

This is a definition-based question from algebra concerning rational fractions (expressions involving polynomials). We need to identify the correct term for a rational fraction where the degree of the numerator is greater than or equal to the degree of the denominator.


Step 2: Detailed Explanation:

A rational fraction is an expression of the form \( \frac{P(x)}{Q(x)} \), where P(x) and Q(x) are polynomials and \( Q(x) \neq 0 \).


There are two main types of rational fractions based on the degrees of the polynomials:


1. Proper Fraction: A rational fraction \( \frac{f(x)}{g(x)} \) is called a proper fraction if the degree of the numerator polynomial f(x) is less than the degree of the denominator polynomial g(x).

Example: \( \frac{x+1}{x^2+2} \). (Degree of numerator is 1, degree of denominator is 2).


2. Improper Fraction: A rational fraction \( \frac{f(x)}{g(x)} \) is called an improper fraction if the degree of the numerator polynomial f(x) is greater than or equal to the degree of the denominator polynomial g(x).

Example: \( \frac{x^3+1}{x^2+2} \) or \( \frac{x^2+1}{x^2+2} \).


The question states that deg f(x) \( \geq \) deg g(x), which directly matches the definition of an improper fraction.


Step 3: Final Answer:

The rational fraction is called an improper fraction. This corresponds to option (C).
Quick Tip: This terminology is analogous to numerical fractions. A fraction like 5/3 is "improper" because the numerator is larger than the denominator. A fraction like 2/3 is "proper". The same concept applies to the degrees of polynomials in rational expressions.

Step 1: Understanding the Question:

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


Step 2: Key Formula or Approach:

First, factor the denominator of the left-hand side: \( x^2+x-2 = (x+2)(x-1) \). The setup is correct.

The equation is: \[ \frac{3x}{(x+2)(x-1)} = \frac{A}{x+2} + \frac{B}{x-1} \]
To solve for A and B, we can combine the right side and equate the numerators.
\[ \frac{3x}{(x+2)(x-1)} = \frac{A(x-1) + B(x+2)}{(x+2)(x-1)} \]
This gives the identity: \[ 3x = A(x-1) + B(x+2) \]
We can find A and B by substituting strategic values for x (the roots of the denominator).


Step 3: Detailed Explanation:

To find B:

Let \( x = 1 \). This will make the term with A zero.

Substitute \( x=1 \) into the identity: \[ 3(1) = A(1-1) + B(1+2) \] \[ 3 = A(0) + B(3) \] \[ 3 = 3B \] \[ B = 1 \]



To find A:

Let \( x = -2 \). This will make the term with B zero.

Substitute \( x=-2 \) into the identity: \[ 3(-2) = A(-2-1) + B(-2+2) \] \[ -6 = A(-3) + B(0) \] \[ -6 = -3A \] \[ A = \frac{-6}{-3} = 2 \]

So, we have A = 2 and B = 1.


Step 4: Final Answer:

The ordered pair (A, B) is (2, 1). This corresponds to option (D).
Quick Tip: This method of substituting the roots of the denominator is called the "Heaviside cover-up method" and is the fastest way to find coefficients for distinct linear factors. To find A, "cover up" the (x+2) term on the left and substitute x = -2 into the rest: \( \frac{3x}{x-1} \rightarrow \frac{3(-2)}{-2-1} = \frac{-6}{-3} = 2 \).

Step 1: Understanding the Question:

Given the value of \( \tan A \), we need to find the value of \( \cos 2A \). This requires using a double-angle trigonometric identity.


Step 2: Key Formula or Approach:

There are several formulas for \( \cos 2A \), but the most direct one to use when \( \tan A \) is given is: \[ \cos 2A = \frac{1 - \tan^2 A}{1 + \tan^2 A} \]

Step 3: Detailed Explanation:

We are given \( \tan A = \frac{4}{3} \).

Substitute this value into the formula: \[ \cos 2A = \frac{1 - \left(\frac{4}{3}\right)^2}{1 + \left(\frac{4}{3}\right)^2} \]
First, calculate \( \tan^2 A \): \[ \left(\frac{4}{3}\right)^2 = \frac{16}{9} \]
Now, substitute this back into the expression: \[ \cos 2A = \frac{1 - \frac{16}{9}}{1 + \frac{16}{9}} \]
Simplify the numerator and the denominator by finding a common denominator (which is 9): \[ \cos 2A = \frac{\frac{9}{9} - \frac{16}{9}}{\frac{9}{9} + \frac{16}{9}} = \frac{\frac{9 - 16}{9}}{\frac{9 + 16}{9}} \] \[ \cos 2A = \frac{\frac{-7}{9}}{\frac{25}{9}} \]
We can cancel the 9 in the denominator of both the numerator and the main denominator: \[ \cos 2A = -\frac{7}{25} \]

Step 4: Final Answer:

The value of \( \cos 2A \) is \( -\frac{7}{25} \). This corresponds to option (A).
Quick Tip: Alternatively, you can visualize a right-angled triangle where \( \tan A = \frac{Opposite}{Adjacent} = \frac{4}{3} \). By Pythagoras' theorem, the hypotenuse is \( \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5 \). From this triangle, \( \cos A = \frac{3}{5} \). Then use the identity \( \cos 2A = 2\cos^2 A - 1 = 2(\frac{3}{5})^2 - 1 = 2(\frac{9}{25}) - 1 = \frac{18}{25} - \frac{25}{25} = -\frac{7}{25} \). Choose the method you find quickest.

Step 1: Understanding the Question:

The question asks for the value of the sum of \( \cos^{-1} x \) and \( \sin^{-1} x \). This is a standard identity in inverse trigonometric functions.


Step 2: Key Formula or Approach:

The key identity for inverse sine and cosine functions is: \[ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \]
This identity is valid for all \( x \) in the domain \( [-1, 1] \).


Step 3: Detailed Explanation:

This is a direct application of a fundamental property of inverse trigonometric functions.

Proof of the identity:

Let \( \theta = \sin^{-1} x \). By definition, this means \( \sin \theta = x \) and \( -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \).

We know the trigonometric co-function identity: \( \cos\left(\frac{\pi}{2} - \theta\right) = \sin \theta \).

Substituting \( \sin \theta = x \), we get \( \cos\left(\frac{\pi}{2} - \theta\right) = x \).

Now, we can take the inverse cosine of both sides: \[ \cos^{-1}\left(\cos\left(\frac{\pi}{2} - \theta\right)\right) = \cos^{-1} x \] \[ \frac{\pi}{2} - \theta = \cos^{-1} x \]
(This step is valid because \( 0 \leq \frac{\pi}{2} - \theta \leq \pi \), which is the range of \( \cos^{-1} \)).

Finally, substitute back \( \theta = \sin^{-1} x \): \[ \frac{\pi}{2} - \sin^{-1} x = \cos^{-1} x \]
Rearranging the terms gives the identity: \[ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \]

Step 4: Final Answer:

The value of \( \cos^{-1} x + \sin^{-1} x \) is \( \frac{\pi}{2} \). This corresponds to option (C).
Quick Tip: This is one of three important "complementary" identities for inverse trig functions that you should memorize: 1. \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \) 2. \( \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} \) 3. \( \sec^{-1} x + \csc^{-1} x = \frac{\pi}{2} \) They are frequently tested.

Step 1: Understanding the Question:

The question asks for the exact value of \( \sin 15^\circ \). We can find this by expressing 15° as a difference of two standard angles (like 45°, 30°, 60°).


Step 2: Key Formula or Approach:

We will use the angle subtraction formula for sine: \[ \sin(A - B) = \sin A \cos B - \cos A \sin B \]
We can write \( 15^\circ \) as \( 45^\circ - 30^\circ \).


Step 3: Detailed Explanation:

Let \( A = 45^\circ \) and \( B = 30^\circ \).

Applying the formula: \[ \sin(15^\circ) = \sin(45^\circ - 30^\circ) = \sin(45^\circ)\cos(30^\circ) - \cos(45^\circ)\sin(30^\circ) \]
We know the standard values for these angles: \[ \sin(45^\circ) = \frac{1}{\sqrt{2}}, \quad \cos(45^\circ) = \frac{1}{\sqrt{2}} \] \[ \sin(30^\circ) = \frac{1}{2}, \quad \cos(30^\circ) = \frac{\sqrt{3}}{2} \]
Substitute these values into the equation: \[ \sin(15^\circ) = \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right) \]
Combine the terms: \[ \sin(15^\circ) = \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} = \frac{\sqrt{3}-1}{2\sqrt{2}} \]
To match the format of the options, we rationalize the denominator by multiplying the numerator and denominator by \( \sqrt{2} \): \[ \sin(15^\circ) = \frac{(\sqrt{3}-1) \times \sqrt{2}}{(2\sqrt{2}) \times \sqrt{2}} = \frac{\sqrt{3}\sqrt{2} - 1\sqrt{2}}{2 \times 2} = \frac{\sqrt{6}-\sqrt{2}}{4} \]

Step 4: Final Answer:

The value of \( \sin 15^\circ \) is \( \frac{\sqrt{6}-\sqrt{2}}{4} \). This corresponds to option (B).
Quick Tip: The values for 15° and 75° are very common in competitive exams. It's highly beneficial to memorize them directly: - \( \sin(15^\circ) = \cos(75^\circ) = \frac{\sqrt{6}-\sqrt{2}}{4} \) - \( \cos(15^\circ) = \sin(75^\circ) = \frac{\sqrt{6}+\sqrt{2}}{4} \) - \( \tan(15^\circ) = \cot(75^\circ) = 2 - \sqrt{3} \)

Step 1: Understanding the Question:

We are given a relationship between \( \cos \theta \) and a variable \( x \). We need to find the corresponding relationship for \( \cos 3\theta \).


Step 2: Key Formula or Approach:

This problem has a standard result derived from De Moivre's theorem. If \( x = \cos\theta + i\sin\theta \), then \( \frac{1}{x} = \cos\theta - i\sin\theta \).
This leads to: \[ x + \frac{1}{x} = 2\cos\theta \] \[ x^n + \frac{1}{x^n} = 2\cos(n\theta) \]
We can use this general result directly.


Step 3: Detailed Explanation:

Method 1: Using the Standard Result

We are given \( x + \frac{1}{x} = 2 \cos \theta \).

Using the general identity \( x^n + \frac{1}{x^n} = 2\cos(n\theta) \), we can set \( n=3 \).

This immediately gives: \[ x^3 + \frac{1}{x^3} = 2\cos(3\theta) \]



Method 2: Algebraic Manipulation

We are given \( 2 \cos \theta = x + \frac{1}{x} \).

Let's cube both sides of the equation: \[ (2 \cos \theta)^3 = \left(x + \frac{1}{x}\right)^3 \] \[ 8 \cos^3 \theta = x^3 + 3 \cdot x^2 \cdot \left(\frac{1}{x}\right) + 3 \cdot x \cdot \left(\frac{1}{x}\right)^2 + \left(\frac{1}{x}\right)^3 \] \[ 8 \cos^3 \theta = x^3 + 3x + \frac{3}{x} + \frac{1}{x^3} \]
Group the terms: \[ 8 \cos^3 \theta = \left(x^3 + \frac{1}{x^3}\right) + 3\left(x + \frac{1}{x}\right) \]
We know that \( x + \frac{1}{x} = 2 \cos \theta \), so we substitute this back in: \[ 8 \cos^3 \theta = \left(x^3 + \frac{1}{x^3}\right) + 3(2 \cos \theta) \] \[ 8 \cos^3 \theta = \left(x^3 + \frac{1}{x^3}\right) + 6 \cos \theta \]
Now, rearrange to solve for \( x^3 + \frac{1}{x^3} \): \[ x^3 + \frac{1}{x^3} = 8 \cos^3 \theta - 6 \cos \theta \]
Factor out a 2 from the right-hand side: \[ x^3 + \frac{1}{x^3} = 2(4 \cos^3 \theta - 3 \cos \theta) \]
Using the triple angle identity for cosine, \( \cos 3\theta = 4 \cos^3 \theta - 3 \cos \theta \), we get: \[ x^3 + \frac{1}{x^3} = 2 \cos 3\theta \]

Step 4: Final Answer:

The expression for \( 2 \cos 3\theta \) is \( x^3 + \frac{1}{x^3} \). This corresponds to option (C).
Quick Tip: Recognizing the connection to complex numbers is the key to solving this type of problem quickly. The relations \( x^n + \frac{1}{x^n} = 2\cos(n\theta) \) and \( x^n - \frac{1}{x^n} = 2i\sin(n\theta) \) are powerful shortcuts.

Step 1: Understanding the Question:

The question asks to express \( \tan\left(\frac{B+C}{2}\right) \) in terms of an angle involving A, given that A, B, and C are angles of a triangle.


Step 2: Key Formula or Approach:

The fundamental property of a triangle is that the sum of its interior angles is 180° (or \( \pi \) radians). \[ A + B + C = 180^\circ \]
We will use this relation along with the trigonometric co-function identity \( \tan(90^\circ - \theta) = \cot \theta \).


Step 3: Detailed Explanation:

Start with the angle sum property of a triangle: \[ A + B + C = 180^\circ \]
We need an expression for \( B+C \), so we isolate these terms: \[ B + C = 180^\circ - A \]
The expression in the question involves \( \frac{B+C}{2} \), so we divide the entire equation by 2: \[ \frac{B+C}{2} = \frac{180^\circ - A}{2} \] \[ \frac{B+C}{2} = \frac{180^\circ}{2} - \frac{A}{2} \] \[ \frac{B+C}{2} = 90^\circ - \frac{A}{2} \]
Now, we take the tangent of both sides: \[ \tan\left(\frac{B+C}{2}\right) = \tan\left(90^\circ - \frac{A}{2}\right) \]
Using the co-function identity \( \tan(90^\circ - \theta) = \cot \theta \), with \( \theta = \frac{A}{2} \): \[ \tan\left(\frac{B+C}{2}\right) = \cot\left(\frac{A}{2}\right) \]

Step 4: Final Answer:

The expression \( \tan\left(\frac{B+C}{2}\right) \) is equal to \( \cot\frac{A}{2} \). This corresponds to option (B).
Quick Tip: For any problem involving the angles of a triangle, the first step is almost always to use the property A + B + C = 180°. This allows you to relate the sum of any two angles to the third angle.

Step 1: Understanding the Question:

This question is similar to the previous one. It asks to express \( \cos\left(\frac{B+C}{2}\right) \) in terms of a function of angle A.


Step 2: Key Formula or Approach:

We will again use the angle sum property of a triangle, \( A + B + C = 180^\circ \).
We will also use the trigonometric co-function identity \( \cos(90^\circ - \theta) = \sin \theta \).


Step 3: Detailed Explanation:

From the angle sum property of a triangle: \[ A + B + C = 180^\circ \]
Isolate the term \( B+C \): \[ B + C = 180^\circ - A \]
Divide by 2 to match the expression in the question: \[ \frac{B+C}{2} = \frac{180^\circ - A}{2} = 90^\circ - \frac{A}{2} \]
Now, take the cosine of both sides: \[ \cos\left(\frac{B+C}{2}\right) = \cos\left(90^\circ - \frac{A}{2}\right) \]
Using the co-function identity \( \cos(90^\circ - \theta) = \sin \theta \), with \( \theta = \frac{A}{2} \): \[ \cos\left(\frac{B+C}{2}\right) = \sin\left(\frac{A}{2}\right) \]

Step 4: Final Answer:

The value of \( \cos\left(\frac{B+C}{2}\right) \) is \( \sin\frac{A}{2} \). This corresponds to option (C).
Quick Tip: Remember the set of co-function identities for \( 90^\circ - \theta \): \( \sin(90^\circ - \theta) = \cos \theta \)
\( \cos(90^\circ - \theta) = \sin \theta \)
\( \tan(90^\circ - \theta) = \cot \theta \)
These are essential for solving problems involving angles in a triangle.

Step 1: Understanding the Question:

The question asks for the value of the sine function for the standard angle of 45°.


Step 2: Detailed Explanation:

We can determine this value by considering an isosceles right-angled triangle.

In such a triangle, the two angles other than the right angle are equal, so each is 45°.

Let the two equal sides (adjacent and opposite to the 45° angles) have a length of 1 unit.

By the Pythagorean theorem, the hypotenuse \( h \) is: \[ h^2 = 1^2 + 1^2 = 1 + 1 = 2 \] \[ h = \sqrt{2} \]
The definition of the sine of an angle in a right-angled triangle is: \[ \sin(\theta) = \frac{Length of the Opposite Side}{Length of the Hypotenuse} \]
For \( \theta = 45^\circ \): \[ \sin(45^\circ) = \frac{1}{\sqrt{2}} \]
This can also be written by rationalizing the denominator as \( \frac{\sqrt{2}}{2} \).


Step 3: Final Answer:

The value of \( \sin 45^\circ \) is \( 1/\sqrt{2} \). This corresponds to option (D).
Quick Tip: The trigonometric values for standard angles (0°, 30°, 45°, 60°, 90°) are fundamental and must be memorized for any competitive exam. Creating a table and practicing it can be very helpful.

Step 1: Understanding the Question:

Given the lengths of the three sides of a triangle, we need to find the value of the tangent of a half-angle, specifically \( \tan(A/2) \).


Step 2: Key Formula or Approach:

The half-angle formula for \( \tan(A/2) \) in terms of the sides of a triangle is: \[ \tan\left(\frac{A}{2}\right) = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \]
where \( s \) is the semi-perimeter of the triangle, calculated as \( s = \frac{a+b+c}{2} \).


Step 3: Detailed Explanation:

First, calculate the semi-perimeter \( s \): \[ s = \frac{13 + 14 + 15}{2} = \frac{42}{2} = 21 \]
Next, calculate the values of \( s-a \), \( s-b \), and \( s-c \): \[ s-a = 21 - 13 = 8 \] \[ s-b = 21 - 14 = 7 \] \[ s-c = 21 - 15 = 6 \]
Now, substitute these values into the half-angle formula: \[ \tan\left(\frac{A}{2}\right) = \sqrt{\frac{(7)(6)}{(21)(8)}} \]
Simplify the expression inside the square root: \[ \tan\left(\frac{A}{2}\right) = \sqrt{\frac{42}{168}} \]
The fraction \( \frac{42}{168} \) simplifies to \( \frac{1}{4} \) (since \( 4 \times 42 = 168 \)). \[ \tan\left(\frac{A}{2}\right) = \sqrt{\frac{1}{4}} \] \[ \tan\left(\frac{A}{2}\right) = \frac{1}{2} \]

Step 4: Final Answer:

The value of \( \tan\left(\frac{A}{2}\right) \) is \( 1/2 \). This corresponds to option (C).
Quick Tip: Another useful formula is \( \tan(A/2) = \frac{\Delta}{s(s-a)} \), where \( \Delta \) is the area of the triangle. Using Heron's formula, \( \Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = \sqrt{7056} = 84 \). Then \( \tan(A/2) = \frac{84}{21(8)} = \frac{84}{168} = \frac{1}{2} \). Both methods work well.

Step 1: Understanding the Question:

The question asks for the simplified value of the cyclic sum \( a^3 \cos(B - C) + b^3 \cos(C - A) + c^3 \cos(A - B) \) in a triangle ABC. This is a standard identity related to the properties of triangles.


Step 2: Key Formula or Approach:

This is a known, albeit complex, identity in trigonometry. The most straightforward approach in an exam setting is to recognize the identity or verify it with a special case.

Let's consider the term \( a \cos(B-C) \).

Using the sine rule, \( a = 2R \sin A \), and the angle sum property \( A = 180^\circ - (B+C) \), so \( \sin A = \sin(B+C) \). \[ a \cos(B-C) = 2R \sin(B+C) \cos(B-C) \]
Using the identity \( 2 \sin X \cos Y = \sin(X+Y) + \sin(X-Y) \): \[ a \cos(B-C) = R[\sin(2B) + \sin(2C)] \]
The full expression \( a^3 \cos(B-C) = a^2 \cdot [a \cos(B-C)] = (2R \sin A)^2 \cdot R(\sin 2B + \sin 2C) \). Summing this up is very complicated.

Therefore, we treat this as a standard result.


Step 3: Detailed Explanation:

The expression is a standard result in the study of properties of triangles: \[ \sum a^3 \cos(B - C) = 3abc \]
We can verify this result for a simple case, like an equilateral triangle.

Verification for an equilateral triangle:

Let \( a = b = c \) and \( A = B = C = 60^\circ \).

The sum becomes: \[ a^3 \cos(60^\circ - 60^\circ) + b^3 \cos(60^\circ - 60^\circ) + c^3 \cos(60^\circ - 60^\circ) \]
Since \( a=b=c \), this is: \[ a^3 \cos(0^\circ) + a^3 \cos(0^\circ) + a^3 \cos(0^\circ) \]
As \( \cos(0^\circ) = 1 \): \[ a^3(1) + a^3(1) + a^3(1) = 3a^3 \]
Now, let's check the right-hand side from the options. The correct option is 3abc. \[ 3abc = 3(a)(a)(a) = 3a^3 \]
Since the Left Hand Side (LHS) equals the Right Hand Side (RHS), the identity holds for an equilateral triangle, giving us confidence in the result.


Step 4: Final Answer:

The value of the given summation is \( 3abc \). This corresponds to option (B).
Quick Tip: For complex cyclic sum identities in trigonometry, it's often not feasible to derive them during a timed exam. It's better to be familiar with common results. If you are unsure, testing a simple case like an equilateral triangle (a=b=c, A=B=C=60°) or an isosceles right triangle can help you confirm the correct option.

Step 1: Understanding the Question:

We need to find the principal value of the inverse cotangent of -1. The principal value is the value that lies within the defined principal value branch (range) of the function.


Step 2: Key Formula or Approach:

Let \( y = \cot^{-1}(-1) \). By definition, this means \( \cot(y) = -1 \).

The principal value range for the inverse cotangent function, \( y = \cot^{-1}(x) \), is \( 0 < y < \pi \).

This means the angle y must be in the first or second quadrant.


Step 3: Detailed Explanation:

First, we find the reference angle \( \alpha \) for which the cotangent value is positive 1. \[ \cot(\alpha) = 1 \]
This gives \( \alpha = \frac{\pi}{4} \) (or 45°).


Since \( \cot(y) = -1 \) is negative, the angle \( y \) must lie in a quadrant where cotangent is negative. Within the range \( (0, \pi) \), this is the second quadrant.


To find the angle in the second quadrant that has a reference angle of \( \alpha = \frac{\pi}{4} \), we use the relation: \[ y = \pi - \alpha \] \[ y = \pi - \frac{\pi}{4} \] \[ y = \frac{4\pi - \pi}{4} = \frac{3\pi}{4} \]
This value, \( \frac{3\pi}{4} \), lies within the principal value range \( (0, \pi) \).


Step 4: Final Answer:

The principle value of \( \cot^{-1}(-1) \) is \( \frac{3\pi}{4} \). This corresponds to option (D).
Quick Tip: Remember the principal value ranges for all inverse trigonometric functions. For negative inputs: - \( \sin^{-1}, \tan^{-1}, \csc^{-1} \) give angles in \( (-\pi/2, 0) \). - \( \cos^{-1}, \cot^{-1}, \sec^{-1} \) give angles in \( (\pi/2, \pi) \). For \( \cot^{-1}(-x) \), a useful identity is \( \cot^{-1}(-x) = \pi - \cot^{-1}(x) \). So, \( \cot^{-1}(-1) = \pi - \cot^{-1}(1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \).

Step 1: Understanding the Question:

The question asks us to perform the addition of two complex numbers.


Step 2: Key Formula or Approach:

To add two complex numbers of the form \( (a+bi) \) and \( (c+di) \), we add the real parts together and the imaginary parts together: \[ (a+bi) + (c+di) = (a+c) + (b+d)i \]

Step 3: Detailed Explanation:

The two complex numbers are \( -1 + 2i \) and \( \frac{1}{2} - i \).

First, identify and group the real parts: \( -1 \) and \( \frac{1}{2} \).

Add the real parts: \[ -1 + \frac{1}{2} = -\frac{2}{2} + \frac{1}{2} = -\frac{1}{2} \]
Next, identify and group the imaginary parts: \( 2i \) and \( -i \).

Add the imaginary parts: \[ 2i - i = (2 - 1)i = 1i = i \]
Combine the new real and imaginary parts to get the result: \[ -\frac{1}{2} + i \]

Step 4: Final Answer:

The sum of the complex numbers is \( -\frac{1}{2} + i \). This corresponds to option (C).
Quick Tip: Adding and subtracting complex numbers is just like combining like terms in algebra. Treat the imaginary unit 'i' as a variable, and simply combine the constant terms (real parts) and the 'i' terms (imaginary parts) separately.

Step 1: Understanding the Question:

The question asks for the product of two complex numbers. We should recognize that the two numbers are complex conjugates of each other.


Step 2: Key Formula or Approach:

The product of a complex number \( z = a + bi \) and its conjugate \( \bar{z} = a - bi \) is given by the formula: \[ z\bar{z} = (a+bi)(a-bi) = a^2 + b^2 \]
Alternatively, we can use the algebraic identity \( (x+y)(x-y) = x^2 - y^2 \).


Step 3: Detailed Explanation:

Method 1: Using the Conjugate Property

Let \( z = \cos\theta + i\sin\theta \). Then its conjugate is \( \bar{z} = \cos\theta - i\sin\theta \).

Here, \( a = \cos\theta \) and \( b = \sin\theta \).

Using the formula \( z\bar{z} = a^2 + b^2 \): \[ (\cos\theta + i\sin\theta)(\cos\theta - i\sin\theta) = (\cos\theta)^2 + (\sin\theta)^2 = \cos^2\theta + \sin^2\theta \]
Using the fundamental Pythagorean identity in trigonometry, \( \cos^2\theta + \sin^2\theta = 1 \).




Method 2: Using Algebraic Expansion

We expand the product using the difference of squares formula, \( (x+y)(x-y) = x^2 - y^2 \).
Let \( x = \cos\theta \) and \( y = i\sin\theta \). \[ (\cos\theta)^2 - (i\sin\theta)^2 \] \[ = \cos^2\theta - (i^2 \sin^2\theta) \]
Since \( i^2 = -1 \): \[ = \cos^2\theta - (-1 \cdot \sin^2\theta) \] \[ = \cos^2\theta + \sin^2\theta = 1 \]



Method 3: Using Euler's Formula

Euler's formula states \( e^{i\theta} = \cos\theta + i\sin\theta \).

This means \( \cos\theta - i\sin\theta = \cos(-\theta) + i\sin(-\theta) = e^{-i\theta} \).

The product is: \[ e^{i\theta} \cdot e^{-i\theta} = e^{i\theta - i\theta} = e^0 = 1 \]

Step 4: Final Answer:

The result of the multiplication is 1. This corresponds to option (A).
Quick Tip: The product of a complex number and its conjugate always results in a real number equal to the square of the modulus of the complex number (\( |z|^2 \)). For \( z = \cos\theta + i\sin\theta \), the modulus \( |z| = \sqrt{\cos^2\theta + \sin^2\theta} = \sqrt{1} = 1 \). Therefore, \( z\bar{z} = |z|^2 = 1^2 = 1 \).

Step 1: Understanding the Question:

We are given the general equation of a circle and asked to find its center and radius.


Step 2: Key Formula or Approach:

The general equation of a circle is given by \( x^2 + y^2 + 2gx + 2fy + c = 0 \).

The center of the circle is at the point \( (-g, -f) \).

The radius of the circle is given by the formula \( r = \sqrt{g^2 + f^2 - c} \).


Step 3: Detailed Explanation:

First, we compare the given equation \( x^2 + y^2 - 4x - 8y - 41 = 0 \) with the general form \( x^2 + y^2 + 2gx + 2fy + c = 0 \).

Comparing the coefficients of x: \[ 2g = -4 \implies g = -2 \]
Comparing the coefficients of y: \[ 2f = -8 \implies f = -4 \]
Comparing the constant term: \[ c = -41 \]
Now, we can find the center of the circle using the formula \( (-g, -f) \):

Center = \( (-(-2), -(-4)) = (2, 4) \).




Next, we calculate the radius using the formula \( r = \sqrt{g^2 + f^2 - c} \):
\[ r = \sqrt{(-2)^2 + (-4)^2 - (-41)} \] \[ r = \sqrt{4 + 16 + 41} \] \[ r = \sqrt{20 + 41} \] \[ r = \sqrt{61} \]

Step 4: Final Answer:

The center of the circle is (2, 4) and the radius is \( \sqrt{61} \). This corresponds to option (C).
Quick Tip: To find the center from the general equation, simply take half the coefficients of x and y and change their signs. For the radius, square these halved coefficients, subtract the constant term 'c', and then take the square root. Be careful with the sign of 'c'.

Step 1: Understanding the Question:

We need to find the number of common tangents to two given circles. This depends on the relative positions of the two circles.


Step 2: Key Formula or Approach:

First, find the center and radius of each circle. Let the centers be \( C_1, C_2 \) and radii be \( r_1, r_2 \).

Then, calculate the distance \( d \) between the centers, \( d = |C_1C_2| \).

Compare \( d \) with \( r_1 + r_2 \) and \( |r_1 - r_2| \) to determine the relative positions:

- If \( d > r_1 + r_2 \), circles are separate (4 common tangents).

- If \( d = r_1 + r_2 \), circles touch externally (3 common tangents).

- If \( |r_1 - r_2| < d < r_1 + r_2 \), circles intersect (2 common tangents).

- If \( d = |r_1 - r_2| \), circles touch internally (1 common tangent).

- If \( d < |r_1 - r_2| \), one circle is inside another (0 common tangents).


Step 3: Detailed Explanation:

For the first circle, \( C_1 \): \( x^2 + y^2 - x = 0 \)

Comparing with \( x^2 + y^2 + 2gx + 2fy + c = 0 \):
\( 2g_1 = -1 \implies g_1 = -1/2 \).
\( 2f_1 = 0 \implies f_1 = 0 \).
\( c_1 = 0 \).

Center \( C_1 = (-g_1, -f_1) = (1/2, 0) \).

Radius \( r_1 = \sqrt{g_1^2 + f_1^2 - c_1} = \sqrt{(-1/2)^2 + 0^2 - 0} = \sqrt{1/4} = 1/2 \).




For the second circle, \( C_2 \): \( x^2 + y^2 + x = 0 \)
\( 2g_2 = 1 \implies g_2 = 1/2 \).
\( 2f_2 = 0 \implies f_2 = 0 \).
\( c_2 = 0 \).

Center \( C_2 = (-g_2, -f_2) = (-1/2, 0) \).

Radius \( r_2 = \sqrt{g_2^2 + f_2^2 - c_2} = \sqrt{(1/2)^2 + 0^2 - 0} = \sqrt{1/4} = 1/2 \).




Distance between centers:

The distance \( d \) between \( C_1(1/2, 0) \) and \( C_2(-1/2, 0) \) is: \[ d = \sqrt{\left(\frac{1}{2} - \left(-\frac{1}{2}\right)\right)^2 + (0-0)^2} = \sqrt{\left(\frac{1}{2} + \frac{1}{2}\right)^2} = \sqrt{1^2} = 1 \]



Compare distance with sum of radii:

Sum of radii: \( r_1 + r_2 = 1/2 + 1/2 = 1 \).

Since \( d = r_1 + r_2 \) (as \( 1 = 1 \)), the two circles touch each other externally.


Step 4: Final Answer:

When two circles touch each other externally, they have exactly 3 common tangents (two direct and one transverse). This corresponds to option (D).
Quick Tip: Before doing any calculations, notice that the second circle's equation is obtained by replacing 'x' with '-x' in the first. This implies the second circle is a reflection of the first across the y-axis. They both pass through the origin (0,0), which must be their point of contact. This visual understanding confirms they touch externally.

Step 1: Understanding the Question:

We are given the center and radius of a circle and need to find its standard equation.


Step 2: Key Formula or Approach:

The standard equation of a circle with center at \( (h, k) \) and radius \( r \) is: \[ (x - h)^2 + (y - k)^2 = r^2 \]

Step 3: Detailed Explanation:

We are given:

Center \( (h, k) = (-3, 2) \).

Radius \( r = 4 \).


Substitute these values into the standard formula: \[ (x - (-3))^2 + (y - 2)^2 = 4^2 \]
Simplify the equation: \[ (x + 3)^2 + (y - 2)^2 = 16 \]

Step 4: Final Answer:

The equation of the circle is \( (x + 3)^2 + (y - 2)^2 = 16 \). This corresponds to option (C).
Quick Tip: Be very careful with the signs when substituting the center coordinates (h, k) into the formula \( (x - h)^2 + (y - k)^2 = r^2 \). The signs inside the brackets will be the opposite of the signs of the coordinates of the center.

Step 1: Understanding the Question:

This is a two-part question. We need to find:

1. The length of the latus rectum for the parabola \( y^2 = 12x \).

2. The focal distance of a specific point on that parabola.


Step 2: Key Formula or Approach:

For a parabola in the standard form \( y^2 = 4ax \):

- The length of the latus rectum is \( 4a \).

- The focal distance of any point \( (x_1, y_1) \) on the parabola is given by the formula \( x_1 + a \).


Step 3: Detailed Explanation:

Part 1: Length of the Latus Rectum

First, compare the given equation \( y^2 = 12x \) with the standard form \( y^2 = 4ax \).
\[ 4a = 12 \] \[ a = \frac{12}{4} = 3 \]
The length of the latus rectum is equal to \( 4a \). \[ Length of latus rectum = 12 \]



Part 2: Focal Distance

We need to find the focal distance of the point \( (x_1, y_1) = (3, -6) \).

(First, let's verify that the point lies on the parabola: \( (-6)^2 = 36 \) and \( 12(3) = 36 \). The point is on the parabola).

The formula for the focal distance is \( x_1 + a \).

We have \( x_1 = 3 \) and we found \( a = 3 \). \[ Focal distance = 3 + 3 = 6 \]

Step 4: Final Answer:

The length of the latus rectum is 12, and the focal distance of the point (3, -6) is 6. This corresponds to the pair (12, 6), which is option (D).
Quick Tip: The length of the latus rectum is simply the coefficient of x (or y) in the standard form \(y^2 = 4ax\) or \(x^2 = 4ay\). You can read it directly from the equation. The focal distance is the distance from a point on the parabola to the focus, which by definition is also its distance to the directrix. The distance to the directrix (\(x = -a\)) is \(|x_1 - (-a)| = x_1 + a\).

Step 1: Understanding the Question:

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


Step 2: Key Formula or Approach:

1. Identify the axis of symmetry: Since the x-coordinates of the vertex (0,0) and focus (0,-2) are the same, the axis of symmetry is the y-axis.

2. Determine the direction: The vertex is at the origin, and the focus is below the vertex (at y=-2). This means the parabola opens downwards.

3. Use the standard equation: The standard equation for a parabola with vertex at the origin opening downwards is \( x^2 = -4ay \), where \( a \) is the distance from the vertex to the focus.

The focus for this type of parabola is at \( (0, -a) \).


Step 3: Detailed Explanation:

The vertex is \( V(0,0) \) and the focus is \( S(0, -2) \).

The distance from the vertex to the focus is \( a \): \[ a = \sqrt{(0-0)^2 + (-2-0)^2} = \sqrt{(-2)^2} = 2 \]
Alternatively, by comparing the focus \( (0, -2) \) with the standard form \( (0, -a) \), we directly get \( a = 2 \).


Now, substitute the value of \( a = 2 \) into the standard equation for a downward-opening parabola, \( x^2 = -4ay \): \[ x^2 = -4(2)y \] \[ x^2 = -8y \]

Step 4: Final Answer:

The equation of the parabola is \( x^2 = -8y \). This corresponds to option (B).
Quick Tip: Quickly determine the orientation and direction. Vertex at origin and focus on an axis means a standard parabola. If the focus is on the y-axis, the equation is of the form \(x^2 = \dots\). If the y-coordinate of the focus is negative, it opens down, so use a minus sign: \(x^2 = -4ay\).

Step 1: Understanding the Question:

We need to find the eccentricity of the conic section represented by the given equation. The form of the equation suggests it is an ellipse.


Step 2: Key Formula or Approach:

First, convert the given equation to the standard form of an ellipse, \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).

The formula for eccentricity \( e \) is \( e = \sqrt{1 - \frac{minor axis squared}{major axis squared}} \).

- If \( a^2 > b^2 \), the formula is \( e = \sqrt{1 - \frac{b^2}{a^2}} \).

- If \( b^2 > a^2 \), the formula is \( e = \sqrt{1 - \frac{a^2}{b^2}} \).


Step 3: Detailed Explanation:

The given equation is \( x^2 + 2y^2 = 3 \).

To get it into standard form, we divide the entire equation by 3: \[ \frac{x^2}{3} + \frac{2y^2}{3} = 1 \] \[ \frac{x^2}{3} + \frac{y^2}{3/2} = 1 \]
Now, we compare this with the standard form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).

We have \( a^2 = 3 \) and \( b^2 = 3/2 \).

Since \( 3 > 3/2 \), we have \( a^2 > b^2 \). This means the major axis is horizontal.


Now we use the eccentricity formula for a horizontal ellipse: \[ e = \sqrt{1 - \frac{b^2}{a^2}} \] \[ e = \sqrt{1 - \frac{3/2}{3}} \] \[ e = \sqrt{1 - \frac{1}{2}} \] \[ e = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \]

Note on Answer Key:

The provided image indicates option (D) is correct, while our calculation yields option (A). There is a high probability of a typo in the question or the answer key. For the answer to be \( \frac{\sqrt{3}}{2} \), the equation would need to be \( x^2 + 4y^2 = const \) or \( 3x^2 + 4y^2 = const \). Based on the given equation, the calculation is correct. We will proceed with the mathematically derived answer.


Step 4: Final Answer:

The eccentricity of the ellipse \( x^2 + 2y^2 = 3 \) is \( \frac{1}{\sqrt{2}} \). This corresponds to option (A).
Quick Tip: Eccentricity 'e' of an ellipse is always between 0 and 1. This helps eliminate options like (B) and (C) immediately. To find \( e^2 \), you only need the ratio of the denominators in the standard form: \( e^2 = 1 - (smaller denominator)/(larger denominator) \).

Step 1: Understanding the Question:

We are asked to find the derivative of a product of two functions: \( e^x \) and \( x^2 + 1 \).


Step 2: Key Formula or Approach:

We must use the Product Rule for differentiation, which states: \[ \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx} \]

Step 3: Detailed Explanation:

Let \( u = e^x \) and \( v = x^2 + 1 \).

First, find the derivatives of \( u \) and \( v \) separately: \[ \frac{du}{dx} = \frac{d}{dx}(e^x) = e^x \] \[ \frac{dv}{dx} = \frac{d}{dx}(x^2 + 1) = 2x \]
Now, apply the product rule formula: \[ \frac{d}{dx}[e^x(x^2 + 1)] = (e^x)(2x) + (x^2 + 1)(e^x) \]
Factor out the common term \( e^x \): \[ = e^x [2x + (x^2 + 1)] \] \[ = e^x (x^2 + 2x + 1) \]
The expression in the parenthesis can also be written as \( (x+1)^2 \). Rearranging the terms to match the option: \[ = e^x (2x + x^2 + 1) \]

Step 4: Final Answer:

The derivative is \( e^x(2x + x^2 + 1) \). This corresponds to option (A).
Quick Tip: A useful shortcut for the product rule when one function is \(e^x\) is: \( \frac{d}{dx}(e^x f(x)) = e^x(f(x) + f'(x)) \). Here, \(f(x) = x^2+1\) and \(f'(x) = 2x\). The result is \(e^x((x^2+1) + 2x)\), which is the same.

Step 1: Understanding the Question:

We need to evaluate a standard limit involving an exponential function.


Step 2: Key Formula or Approach:

This is a standard limit formula. However, we can derive it using L'Hôpital's Rule because direct substitution leads to an indeterminate form.

When we substitute \( x=0 \):
Numerator: \( a^0 - 1 = 1 - 1 = 0 \).

Denominator: \( 0 \).

Since we have the indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that \( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \).


Step 3: Detailed Explanation:

Let \( f(x) = a^x - 1 \) and \( g(x) = x \).

Find the derivatives of the numerator and the denominator: \[ f'(x) = \frac{d}{dx}(a^x - 1) = a^x \ln(a) \] \[ g'(x) = \frac{d}{dx}(x) = 1 \]
Now, apply L'Hôpital's Rule: \[ \lim_{x \to 0} \frac{a^x - 1}{x} = \lim_{x \to 0} \frac{a^x \ln(a)}{1} \]
Now, we can substitute \( x=0 \) into the new expression: \[ = \frac{a^0 \ln(a)}{1} = \frac{1 \cdot \ln(a)}{1} = \ln(a) \]
Note: In many contexts, log a is understood to mean the natural logarithm, ln a.


Step 4: Final Answer:

The value of the limit is log a. This corresponds to option (A).
Quick Tip: This is a fundamental limit that is very useful to memorize for competitive exams: \( \lim_{x \to 0} \frac{a^x - 1}{x} = \ln a \). A special case of this is when \( a = e \), which gives \( \lim_{x \to 0} \frac{e^x - 1}{x} = \ln e = 1 \).

Step 1: Understanding the Question:

The question asks for the derivative of the inverse tangent function, \( \tan^{-1}(x) \).


Step 2: Key Formula or Approach:

This is a standard derivative formula in calculus. The derivation involves implicit differentiation.

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

Differentiate both sides with respect to \( x \): \[ \frac{d}{dx}(\tan(y)) = \frac{d}{dx}(x) \] \[ \sec^2(y) \cdot \frac{dy}{dx} = 1 \] \[ \frac{dy}{dx} = \frac{1}{\sec^2(y)} \]
Using the trigonometric identity \( \sec^2(y) = 1 + \tan^2(y) \): \[ \frac{dy}{dx} = \frac{1}{1 + \tan^2(y)} \]
Since \( \tan(y) = x \), we substitute this back: \[ \frac{dy}{dx} = \frac{1}{1 + x^2} \]

Step 3: Final Answer:

The derivative of \( \tan^{-1}(x) \) with respect to \( x \) is \( \frac{1}{1 + x^2} \). This corresponds to option (A).
Quick Tip: Memorizing the derivatives of the six inverse trigonometric functions is essential for speed in calculus problems. - \( \frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1-x^2}} \) - \( \frac{d}{dx}(\cos^{-1}x) = -\frac{1}{\sqrt{1-x^2}} \) - \( \frac{d}{dx}(\tan^{-1}x) = \frac{1}{1+x^2} \) Note the relationship between the derivatives of co-functions (e.g., sin/cos, tan/cot).

Step 1: Understanding the Question:

We are given a linear equation relating x and y and asked to find the derivative of y with respect to x, which is \( \frac{dy}{dx} \).


Step 2: Key Formula or Approach:

We can use two methods: implicit differentiation or rearranging the equation to make y the subject.


Step 3: Detailed Explanation:

Method 1: Implicit Differentiation

Differentiate each term of the equation \( 4x - 7y + 15 = 0 \) with respect to \( x \): \[ \frac{d}{dx}(4x) - \frac{d}{dx}(7y) + \frac{d}{dx}(15) = \frac{d}{dx}(0) \] \[ 4 - 7\frac{dy}{dx} + 0 = 0 \]
Now, solve for \( \frac{dy}{dx} \): \[ 4 = 7\frac{dy}{dx} \] \[ \frac{dy}{dx} = \frac{4}{7} \]



Method 2: Rearranging the Equation

First, solve the equation for \( y \): \[ 4x - 7y + 15 = 0 \] \[ 7y = 4x + 15 \] \[ y = \frac{4x + 15}{7} \] \[ y = \frac{4}{7}x + \frac{15}{7} \]
This equation is now in the slope-intercept form \( y = mx + c \), where the slope \( m \) is the derivative \( \frac{dy}{dx} \).

By comparing, we can see that \( m = \frac{4}{7} \).
Therefore, \( \frac{dy}{dx} = \frac{4}{7} \).


Step 4: Final Answer:

The derivative of y with respect to x is 4/7. This corresponds to option (D).
Quick Tip: For any linear equation in the form \( Ax + By + C = 0 \), the derivative \( \frac{dy}{dx} \) (which represents the slope) is always equal to \( -A/B \). In this case, \( A=4 \) and \( B=-7 \), so the slope is \( -4/(-7) = 4/7 \).

Step 1: Understanding the Question:

We are asked to find the second derivative of the function \( y = \cos x \). This means we need to differentiate the function twice.


Step 2: Key Formula or Approach:

We will use the standard derivatives of trigonometric functions. \[ \frac{d}{dx}(\cos x) = -\sin x \] \[ \frac{d}{dx}(\sin x) = \cos x \]

Step 3: Detailed Explanation:

First, find the first derivative of \( y = \cos x \): \[ \frac{dy}{dx} = \frac{d}{dx}(\cos x) = -\sin x \]
Now, differentiate the first derivative to find the second derivative: \[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}(-\sin x) \] \[ = -\frac{d}{dx}(\sin x) \] \[ = -(\cos x) = -\cos x \]

Step 4: Final Answer:

The second derivative of \( y = \cos x \) is \( -\cos x \). This corresponds to option (A).
Quick Tip: The derivatives of sine and cosine follow a cycle of four: \( \sin x \to \cos x \to -\sin x \to -\cos x \to \sin x \). This pattern can be useful for quickly finding higher-order derivatives. Since \( y = \cos x \), the first derivative is \( -\sin x \) and the second is \( -\cos x \).

Step 1: Understanding the Question:

We are given a function \( u \) of two variables, \( x \) and \( y \), and asked to find its partial derivative with respect to \( y \). This is denoted as \( \frac{\partial u}{\partial y} \).


Step 2: Key Formula or Approach:

To find the partial derivative of a function with respect to one variable (e.g., y), we treat all other variables (e.g., x) as constants and apply the standard rules of differentiation.


Step 3: Detailed Explanation:

The function is \( u(x, y) = e^x \sin y \).

We need to find \( \frac{\partial u}{\partial y} \).

In this process, we treat \( x \) and therefore \( e^x \) as a constant. \[ \frac{\partial u}{\partial y} = \frac{\partial}{\partial y}(e^x \sin y) \]
We can take the constant term \( e^x \) outside the derivative: \[ = e^x \frac{\partial}{\partial y}(\sin y) \]
The derivative of \( \sin y \) with respect to \( y \) is \( \cos y \). \[ = e^x \cos y \]

Step 4: Final Answer:

The first partial derivative of u with respect to y is \( e^x \cos y \). This corresponds to option (B).
Quick Tip: When performing partial differentiation, mentally "freeze" the variables you are not differentiating with respect to. Imagine \(e^x\) is just a number like 5. The problem then becomes finding the derivative of \(5 \sin y\), which is simply \(5 \cos y\). Then replace the 5 back with \(e^x\).

Step 1: Understanding the Question:

We need to find the derivative of the function \( e^{3\log x} \). The first step should always be to simplify the function if possible before differentiating.


Step 2: Key Formula or Approach:

We will use the following properties of logarithms and exponentials:

1. Logarithm power rule: \( n \log a = \log(a^n) \).

2. Inverse property: \( e^{\log a} = e^{\ln a} = a \) (assuming log denotes the natural logarithm).

3. Power rule for differentiation: \( \frac{d}{dx}(x^n) = nx^{n-1} \).


Step 3: Detailed Explanation:

First, simplify the expression \( e^{3\log x} \).

Using the logarithm power rule, we can rewrite the exponent: \[ 3\log x = \log(x^3) \]
So the original expression becomes: \[ e^{\log(x^3)} \]
Using the inverse property of \( e^x \) and \( \log x \), this simplifies to: \[ x^3 \]
Now, the problem is reduced to finding the derivative of \( x^3 \).

Using the power rule for differentiation: \[ \frac{d}{dx}(x^3) = 3x^{3-1} = 3x^2 \]

Step 4: Final Answer:

The derivative of \( e^{3\log x} \) is \( 3x^2 \). This corresponds to option (D).
Quick Tip: Always look for ways to simplify a function before differentiating or integrating. Expressions involving logarithms and exponentials, like \( e^{\log f(x)} \) or \( \log(e^{f(x)}) \), can almost always be simplified to just \( f(x) \), making the calculus much easier.

Step 1: Understanding the Question:

We are given a function \( u(x,y) \) and asked to compute the value of the expression \( x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} \). This expression is related to Euler's Theorem for homogeneous functions.


Step 2: Key Formula or Approach:

Euler's Theorem on Homogeneous Functions: If \( u(x, y) \) is a homogeneous function of degree \( n \), then \( x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = nu \).

A function \( u(x, y) \) is homogeneous of degree \( n \) if \( u(tx, ty) = t^n u(x, y) \) for any constant \( t \).


Step 3: Detailed Explanation:

First, let's check if the given function \( u(x, y) \) is homogeneous. We substitute \( tx \) for \( x \) and \( ty \) for \( y \): \[ u(tx, ty) = \sin^{-1}\left(\frac{tx}{ty}\right) + \tan^{-1}\left(\frac{ty}{tx}\right) \]
The \( t \) in the numerator and denominator cancels out in both terms: \[ u(tx, ty) = \sin^{-1}\left(\frac{x}{y}\right) + \tan^{-1}\left(\frac{y}{x}\right) \]
This is the same as the original function \( u(x, y) \). \[ u(tx, ty) = u(x, y) \]
We can write this as: \[ u(tx, ty) = t^0 u(x, y) \]
This shows that \( u(x, y) \) is a homogeneous function of degree \( n = 0 \).


Now, we can apply Euler's Theorem: \[ x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = n \cdot u \]
Substituting \( n=0 \): \[ x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = 0 \cdot u = 0 \]

Step 4: Final Answer:

The value of the expression \( xu_x + yu_y \) is 0. This corresponds to option (B).
Quick Tip: Whenever you see the expression \( x u_x + y u_y \), your first thought should be Euler's Theorem. Quickly check if the function is homogeneous by replacing x with tx and y with ty. If all the t's cancel out, the degree is 0, and the answer is 0. This saves you from calculating the partial derivatives explicitly.

Step 1: Understanding the Question:

We are given an equation for S in terms of t, and we need to find its derivative with respect to t, \( \frac{ds}{dt} \).


Step 2: Key Formula or Approach:

We will use the power rule of differentiation, which states \( \frac{d}{dt}(t^n) = nt^{n-1} \). We also use the sum/difference rule and the constant multiple rule.


Step 3: Detailed Explanation:

The function is \( S = 12t - 3t^2 \).

We differentiate term by term: \[ \frac{ds}{dt} = \frac{d}{dt}(12t) - \frac{d}{dt}(3t^2) \]
For the first term, \( \frac{d}{dt}(12t) = 12 \cdot \frac{d}{dt}(t^1) = 12 \cdot 1 \cdot t^0 = 12 \).

For the second term, \( \frac{d}{dt}(3t^2) = 3 \cdot \frac{d}{dt}(t^2) = 3 \cdot (2t^1) = 6t \).

Combining the results: \[ \frac{ds}{dt} = 12 - 6t \]

Step 4: Final Answer:

The derivative of S with respect to t is \( 12 - 6t \). This corresponds to option (A).
Quick Tip: This is a standard polynomial differentiation. Remember that the derivative of a term \(at^n\) is \(ant^{n-1}\). The derivative of a linear term \(at\) is just the constant \(a\), and the derivative of a constant is zero.

Step 1: Understanding the Question:

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


Step 2: Key Formula or Approach:

There is no direct integration formula for \( \cot^2(x) \). We must first use a trigonometric identity to rewrite it in a form that can be integrated.

The relevant Pythagorean identity is: \( 1 + \cot^2(x) = \csc^2(x) \).

From this, we get \( \cot^2(x) = \csc^2(x) - 1 \).

We know the standard integral: \( \int \csc^2(x) \,dx = -\cot(x) + C \).


Step 3: Detailed Explanation:

Substitute the identity into the integral: \[ \int \cot^2(x) \,dx = \int (\csc^2(x) - 1) \,dx \]
Split the integral into two parts: \[ = \int \csc^2(x) \,dx - \int 1 \,dx \]
Now, integrate each part: \[ \int \csc^2(x) \,dx = -\cot(x) \] \[ \int 1 \,dx = x \]
Combine the results and add the constant of integration, C: \[ = -\cot(x) - x + C \]

Step 4: Final Answer:

The integral of \( \cot^2x \) is \( -\cot x - x + c \). This corresponds to option (D).
Quick Tip: Similarly, to integrate \( \tan^2(x) \), use the identity \( \tan^2(x) = \sec^2(x) - 1 \). This gives \( \int \tan^2(x) \,dx = \int (\sec^2(x) - 1) \,dx = \tan(x) - x + C \). These two integrals are common exam questions.

Step 1: Understanding the Question:

We are asked to find the indefinite integral of the function \( \frac{1}{\sqrt{a^2-x^2}} \).


Step 2: Key Formula or Approach:

This is a standard integration formula that results in an inverse trigonometric function.

The formula is: \[ \int \frac{1}{\sqrt{a^2-x^2}} \,dx = \sin^{-1}\left(\frac{x}{a}\right) + C \]
This formula is derived using a trigonometric substitution \( x = a\sin\theta \).


Step 3: Detailed Explanation:

This question requires the direct application of the standard integral formula.

By recalling the standard forms of integrals:

- \( \int \frac{1}{x^2+a^2} \,dx = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C \)

- \( \int \frac{1}{\sqrt{a^2-x^2}} \,dx = \sin^{-1}\left(\frac{x}{a}\right) + C \)

- \( \int \frac{1}{\sqrt{x^2+a^2}} \,dx = \ln|x + \sqrt{x^2+a^2}| + C \)

- \( \int \frac{1}{\sqrt{x^2-a^2}} \,dx = \ln|x + \sqrt{x^2-a^2}| + C \)

The given integral matches the second formula in this list.


Step 4: Final Answer:

The integral of \( \frac{1}{\sqrt{a^2-x^2}} \) is \( \sin^{-1}\frac{x}{a} + c \). This corresponds to option (C).
Quick Tip: It is crucial to memorize the standard integral formulas, especially those involving square roots of quadratic expressions. Pay close attention to the signs: \( \sqrt{a^2-x^2} \) leads to \( \sin^{-1} \), while \( \sqrt{x^2+a^2} \) and \( \sqrt{x^2-a^2} \) lead to logarithmic forms.

Step 1: Understanding the Question:

We need to find the indefinite integral of the product of \( e^x \) and \( \cos x \).


Step 2: Key Formula or Approach:

This integral is typically solved using integration by parts, often twice. The formula for integration by parts is \( \int u \,dv = uv - \int v \,du \).

There is also a standard formula for integrals of this type: \[ \int e^{ax}\cos(bx) \,dx = \frac{e^{ax}}{a^2+b^2}(a\cos(bx) + b\sin(bx)) + C \]

Step 3: Detailed Explanation:

Let \( I = \int e^x \cos x \,dx \).

We use integration by parts. Let \( u = \cos x \) and \( dv = e^x \,dx \).

Then \( du = -\sin x \,dx \) and \( v = e^x \).

Applying the formula: \[ I = (\cos x)(e^x) - \int (e^x)(-\sin x \,dx) \] \[ I = e^x \cos x + \int e^x \sin x \,dx \]
Now we must apply integration by parts to the new integral, \( \int e^x \sin x \,dx \).

Let \( u_2 = \sin x \) and \( dv_2 = e^x \,dx \).

Then \( du_2 = \cos x \,dx \) and \( v_2 = e^x \).
\[ \int e^x \sin x \,dx = (\sin x)(e^x) - \int (e^x)(\cos x \,dx) = e^x \sin x - I \]
Now, substitute this back into the equation for \( I \): \[ I = e^x \cos x + (e^x \sin x - I) \] \[ I = e^x(\cos x + \sin x) - I \]
Now, solve for \( I \): \[ 2I = e^x(\cos x + \sin x) \] \[ I = \frac{1}{2}e^x(\cos x + \sin x) + c \]

Step 4: Final Answer:

The integral is \( \frac{1}{2}e^x(\cos x + \sin x) + c \). This corresponds to option (A).
Quick Tip: For exams, memorizing the direct formulas for \( \int e^{ax}\cos(bx) \,dx \) and \( \int e^{ax}\sin(bx) \,dx \) can save a significant amount of time compared to performing integration by parts twice.

Step 1: Understanding the Question:

We need to find the indefinite integral of \( \frac{1}{\sqrt{x}} \).


Step 2: Key Formula or Approach:

First, rewrite the integrand using exponent notation: \( \frac{1}{\sqrt{x}} = \frac{1}{x^{1/2}} = x^{-1/2} \).

Then, use the power rule for integration: \[ \int x^n \,dx = \frac{x^{n+1}}{n+1} + C \quad (for n \neq -1) \]

Step 3: Detailed Explanation:

Rewrite the integral: \[ \int \frac{dx}{\sqrt{x}} = \int x^{-1/2} \,dx \]
Apply the power rule with \( n = -1/2 \): \[ = \frac{x^{-1/2 + 1}}{-1/2 + 1} + C \] \[ = \frac{x^{1/2}}{1/2} + C \]
Dividing by \( 1/2 \) is the same as multiplying by 2: \[ = 2x^{1/2} + C \]
Rewrite the result using radical notation: \[ = 2\sqrt{x} + C \]

Step 4: Final Answer:

The integral is \( 2\sqrt{x}+c \). This corresponds to option (C).
Quick Tip: The integral of \( 1/\sqrt{x} \) is a very common one. It's helpful to remember it directly. Differentiating the answer, \( \frac{d}{dx}(2\sqrt{x}) = 2 \cdot \frac{1}{2}x^{-1/2} = \frac{1}{\sqrt{x}} \), confirms the result quickly.

Step 1: Understanding the Question:

We need to find the indefinite integral of \( \sin(y/2) \).


Step 2: Key Formula or Approach:

We will use the standard integral \( \int \sin(u) \,du = -\cos(u) + C \) combined with the method of u-substitution.
A useful shortcut for linear arguments is: \( \int f(ay+b) \,dy = \frac{1}{a}F(ay+b) + C \), where \( F \) is the antiderivative of \( f \).


Step 3: Detailed Explanation:

Let's use u-substitution.
Let \( u = \frac{y}{2} \).

Then, differentiate \( u \) with respect to \( y \): \[ \frac{du}{dy} = \frac{1}{2} \implies dy = 2 \,du \]
Now, substitute \( u \) and \( dy \) into the original integral: \[ \int \sin\left(\frac{y}{2}\right) \,dy = \int \sin(u) \cdot (2 \,du) \] \[ = 2 \int \sin(u) \,du \]
Integrate with respect to \( u \): \[ = 2(-\cos(u)) + C = -2\cos(u) + C \]
Finally, substitute back \( u = \frac{y}{2} \): \[ = -2\cos\left(\frac{y}{2}\right) + C \]

Step 4: Final Answer:

The integral is \( -2 \cos \frac{y}{2} + c \). This corresponds to option (D).
Quick Tip: When integrating sin(ay) or cos(ay), the result is \(-\frac{1}{a}\cos(ay)\) or \( \frac{1}{a}\sin(ay) \) respectively. Just integrate the function as usual and then divide by the coefficient of the variable. Here the coefficient of y is 1/2, so we divide by 1/2 (which means multiplying by 2).

Step 1: Understanding the Question:

We need to evaluate the definite integral of the function f(x) = 1 over the interval from 0 to \( \pi \).


Step 2: Key Formula or Approach:

The fundamental theorem of calculus states that if \( F'(x) = f(x) \), then \( \int_{a}^{b} f(x) \,dx = F(b) - F(a) \).

First, we find the antiderivative of the integrand. The integrand is implicitly 1.
\[ \int 1 \,dx = x \]

Step 3: Detailed Explanation:

The antiderivative of the integrand (1) is \( F(x) = x \).

Now, we evaluate this antiderivative at the upper and lower limits of integration, which are \( b = \pi \) and \( a = 0 \). \[ \int_{0}^{\pi} 1 \,dx = [x]_{0}^{\pi} \] \[ = F(\pi) - F(0) \] \[ = (\pi) - (0) \] \[ = \pi \]
Geometrically, this integral represents the area of a rectangle with height 1 and width \( \pi \), so the area is \( 1 \times \pi = \pi \).


Step 4: Final Answer:

The value of the definite integral is \( \pi \). This corresponds to option (C).
Quick Tip: The definite integral \( \int_{a}^{b} C \,dx \) of any constant C is simply \( C \times (b-a) \). In this case, \( C=1 \), \( b=\pi \), and \( a=0 \), so the result is \( 1 \times (\pi - 0) = \pi \).

Step 1: Understanding the Question:

The question asks to simplify the definite integral of an even function over a symmetric interval \( [-a, a] \).


Step 2: Key Formula or Approach:

An even function is defined by the property \( f(-x) = f(x) \). Geometrically, its graph is symmetric with respect to the y-axis.

We use a property of definite integrals to split the interval: \[ \int_{-a}^{a} f(x)dx = \int_{-a}^{0} f(x)dx + \int_{0}^{a} f(x)dx \]

Step 3: Detailed Explanation:

Let's evaluate the first part of the split integral, \( \int_{-a}^{0} f(x)dx \).

We use a substitution. Let \( x = -u \). Then \( dx = -du \).

We must also change the limits of integration:
- When \( x = -a \), \( u = -(-a) = a \).

- When \( x = 0 \), \( u = -(0) = 0 \).

Substituting these into the integral gives: \[ \int_{a}^{0} f(-u)(-du) \]
Since \( f(x) \) is an even function, \( f(-u) = f(u) \). \[ \int_{a}^{0} f(u)(-du) = -\int_{a}^{0} f(u)du \]
We can reverse the limits of integration by changing the sign of the integral: \[ = \int_{0}^{a} f(u)du \]
Since \( u \) is a dummy variable, this is the same as \( \int_{0}^{a} f(x)dx \).




Now, substitute this result back into the original split equation: \[ \int_{-a}^{a} f(x)dx = \left(\int_{0}^{a} f(x)dx\right) + \int_{0}^{a} f(x)dx \] \[ = 2 \int_{0}^{a} f(x)dx \]

Step 4: Final Answer:

For an even function, the integral over a symmetric interval is twice the integral over the positive half of the interval. This corresponds to option (B).
Quick Tip: For an even function, the area under the curve from -a to 0 is identical to the area from 0 to a due to y-axis symmetry. Thus, the total area is simply twice the area of one side. For an odd function (\( f(-x) = -f(x) \)), the areas cancel out, and \( \int_{-a}^{a} f(x)dx = 0 \).

Step 1: Understanding the Question:

The question asks for the "area under the curve", which implies the total geometric area between the curve and the x-axis. This means we must treat any area below the x-axis as positive. The formula for area is \( A = \int_{a}^{b} |f(x)| \,dx \).


Step 2: Key Formula or Approach:

We need to analyze the sign of \( \sin x \) in the interval \( [0, 2\pi] \).

- In the interval \( [0, \pi] \), \( \sin x \geq 0 \), so \( |\sin x| = \sin x \).

- In the interval \( [\pi, 2\pi] \), \( \sin x \leq 0 \), so \( |\sin x| = -\sin x \).

Therefore, we must split the integral: \[ A = \int_{0}^{2\pi} |\sin x| \,dx = \int_{0}^{\pi} \sin x \,dx + \int_{\pi}^{2\pi} (-\sin x) \,dx \]

Step 3: Detailed Explanation:

First, evaluate the integral for the interval \( [0, \pi] \): \[ \int_{0}^{\pi} \sin x \,dx = [-\cos x]_{0}^{\pi} \] \[ = (-\cos(\pi)) - (-\cos(0)) = -(-1) - (-1) = 1 + 1 = 2 \]
Next, evaluate the integral for the interval \( [\pi, 2\pi] \): \[ \int_{\pi}^{2\pi} -\sin x \,dx = [\cos x]_{\pi}^{2\pi} \] \[ = (\cos(2\pi)) - (\cos(\pi)) = (1) - (-1) = 1 + 1 = 2 \]
The total area is the sum of the areas from the two intervals: \[ A = 2 + 2 = 4 \]

Step 4: Final Answer:

The total area under the curve \( f(x) = \sin x \) in the interval \( [0, 2\pi] \) is 4. This corresponds to option (D).
Quick Tip: Be careful with questions asking for "area". If you calculate \( \int_{0}^{2\pi} \sin x \,dx \) directly, you get 0, as the positive and negative areas cancel out. "Area under the curve" almost always implies the total absolute area. The area of one "hump" of the sine curve is 2. Over a full period, there are two humps, so the total area is 4.

Step 1: Understanding the Question:

We are asked to evaluate a definite integral where the upper and lower limits of integration are the same.


Step 2: Key Formula or Approach:

According to the Fundamental Theorem of Calculus, if F(x) is the antiderivative of f(x), then: \[ \int_{a}^{b} f(x)dx = F(b) - F(a) \]

Step 3: Detailed Explanation:

The question states that \( a = b \). We can substitute \( a \) for \( b \) in the formula: \[ \int_{a}^{a} f(x)dx = F(a) - F(a) \] \[ = 0 \]
Geometrically, a definite integral represents the area under a curve between two points. If the starting and ending points are the same, the width of the region is zero, and therefore its area is also zero.


Step 4: Final Answer:

When the limits of integration are equal, the value of the definite integral is 0. This corresponds to option (B).
Quick Tip: This is a fundamental property of definite integrals that you should know by heart. Any integral of the form \( \int_{a}^{a} f(x)dx \) is always zero, regardless of the function \( f(x) \).

Step 1: Understanding the Question:

We need to find the 'order' of the given differential equation.


Step 2: Detailed Explanation:

The order of a differential equation is defined as the order of the highest derivative that appears in the equation.

Let's examine the derivatives present in the equation: \[ \left[\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^3\right]^{6/5} = 6y \]
The derivatives involved are:
1. \( \frac{dy}{dx} \): This is the first derivative, so its order is 1.

2. \( \frac{d^2y}{dx^2} \): This is the second derivative, so its order is 2.




The highest order among all the derivatives in the equation is 2.

The powers and fractional exponents on the derivatives or the terms do not affect the order of the equation. They are related to the 'degree' of the equation, which is a different concept.


Step 3: Final Answer:

The order of the highest derivative is 2, so the order of the differential equation is 2. This corresponds to option (B).
Quick Tip: Don't confuse order with degree. - \textbf{Order}: The highest derivative (e.g., y'', y''', etc.). It's the most important classification. - \textbf{Degree}: The power of the highest-order derivative after the equation has been cleared of any radicals or fractional exponents involving the derivatives. In this case, raising both sides to the power of 5 gives a degree of 6.

Step 1: Understanding the Question:

We need to find the integrating factor for the given differential equation. First, we must arrange the equation into the standard linear form.


Step 2: Key Formula or Approach:

A first-order linear differential equation has the standard form: \[ \frac{dy}{dx} + P(x)y = Q(x) \]
The integrating factor (I.F.) for this form is given by the formula: \[ I.F. = e^{\int P(x)dx} \]

Step 3: Detailed Explanation:

First, rearrange the given equation \( \frac{dy}{dx} + 3x = 2y \) into the standard linear form. We need to move the term with 'y' to the left side. \[ \frac{dy}{dx} - 2y = -3x \]
Now, compare this with the standard form \( \frac{dy}{dx} + P(x)y = Q(x) \).
We can identify: \[ P(x) = -2 \] \[ Q(x) = -3x \]
Next, we calculate the integrating factor using its formula: \[ I.F. = e^{\int P(x)dx} = e^{\int (-2)dx} \]
Performing the integration in the exponent: \[ \int -2 \,dx = -2x \]
So, the integrating factor is: \[ I.F. = e^{-2x} \]

Step 4: Final Answer:

The integrating factor of the differential equation is \( e^{-2x} \). This corresponds to option (B).
Quick Tip: The most common mistake in these problems is incorrect identification of P(x). Always ensure the equation is in the exact standard form \( y' + P(x)y = Q(x) \) before you determine P(x). Pay close attention to the signs.

Step 1: Understanding the Question:

We are asked to rewrite the given differential equation into a standard linear form. The equation involves both dx and dy, so it could be linear in y (with x as the independent variable) or linear in x (with y as the independent variable).


Step 2: Key Formula or Approach:

The standard linear form with x as the dependent variable and y as the independent variable is: \[ \frac{dx}{dy} + P(y)x = Q(y) \]
We will try to arrange the given equation into this form.


Step 3: Detailed Explanation:

The given equation is: \[ dx + xdy = e^{-y}\sec^2y \, dy \]
To get \( \frac{dx}{dy} \), we should divide the entire equation by \( dy \). \[ \frac{dx}{dy} + \frac{xdy}{dy} = \frac{e^{-y}\sec^2y \, dy}{dy} \]
Simplifying each term: \[ \frac{dx}{dy} + x = e^{-y}\sec^2y \]
This equation is now in the standard linear form \( \frac{dx}{dy} + P(y)x = Q(y) \), where \( P(y) = 1 \) and \( Q(y) = e^{-y}\sec^2y \).


Step 4: Final Answer:

The linear form of the equation is \( \frac{dx}{dy} + x = e^{-y}\sec^2y \). This corresponds to option (D).
Quick Tip: If a differential equation is hard to arrange into the standard \( \frac{dy}{dx} \) linear form, always check if it can be arranged into the \( \frac{dx}{dy} \) linear form. The clue is often that the terms multiplying \( dy \) are functions of y only.

Step 1: Understanding the Question:

This is a theoretical question asking for the condition that defines an exact differential equation.


Step 2: Detailed Explanation:

A differential equation of the form \( M(x,y)dx + N(x,y)dy = 0 \) is said to be exact if its left-hand side is the total differential of some function \( u(x,y) \).

The total differential of \( u(x,y) \) is given by: \[ du = \frac{\partial u}{\partial x}dx + \frac{\partial u}{\partial y}dy \]
For the equation to be exact, we must have: \[ M = \frac{\partial u}{\partial x} \quad and \quad N = \frac{\partial u}{\partial y} \]
Now, we take the partial derivative of \( M \) with respect to \( y \) and \( N \) with respect to \( x \): \[ \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}\left(\frac{\partial u}{\partial x}\right) = \frac{\partial^2 u}{\partial y \partial x} \] \[ \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}\left(\frac{\partial u}{\partial y}\right) = \frac{\partial^2 u}{\partial x \partial y} \]
According to Clairaut's theorem, if the second partial derivatives are continuous, then the order of differentiation does not matter, meaning \( \frac{\partial^2 u}{\partial y \partial x} = \frac{\partial^2 u}{\partial x \partial y} \).

Therefore, the necessary and sufficient condition for the equation to be exact is: \[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \]

Step 3: Final Answer:

The condition for an exact differential equation is \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). This corresponds to option (B).
Quick Tip: A simple way to remember the condition is to think of it as "cross-differentiation": The coefficient of dx (M) is differentiated with respect to the "other" variable (y), and the coefficient of dy (N) is differentiated with respect to its "other" variable (x).

Step 1: Understanding the Question:

We need to find the complementary function (CF) for the given linear non-homogeneous differential equation. The CF is the general solution to the corresponding homogeneous equation.


Step 2: Key Formula or Approach:

The homogeneous equation is \( (D^3 - 8)y = 0 \). To solve this, we first form the auxiliary equation by replacing the operator \( D \) with a variable \( m \). \[ m^3 - 8 = 0 \]
We then find the roots of this auxiliary equation. The form of the CF depends on the nature of these roots (real, repeated, complex).


Step 3: Detailed Explanation:

The auxiliary equation is \( m^3 - 8 = 0 \), or \( m^3 = 8 \).

We can factor this as a difference of cubes, \( a^3 - b^3 = (a-b)(a^2+ab+b^2) \): \[ (m-2)(m^2 + 2m + 4) = 0 \]
One root is obtained from the first factor: \( m - 2 = 0 \implies m_1 = 2 \). This is a real, distinct root.


The other two roots come from the quadratic factor \( m^2 + 2m + 4 = 0 \). We use the quadratic formula \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ m = \frac{-2 \pm \sqrt{2^2 - 4(1)(4)}}{2(1)} = \frac{-2 \pm \sqrt{4 - 16}}{2} = \frac{-2 \pm \sqrt{-12}}{2} \] \[ m = \frac{-2 \pm \sqrt{4 \cdot 3 \cdot (-1)}}{2} = \frac{-2 \pm 2i\sqrt{3}}{2} = -1 \pm i\sqrt{3} \]
So, we have a pair of complex conjugate roots: \( m_{2,3} = -1 \pm i\sqrt{3} \). This is of the form \( \alpha \pm i\beta \) where \( \alpha = -1 \) and \( \beta = \sqrt{3} \).




Now we construct the complementary function from the roots:
- The real root \( m_1 = 2 \) gives the term \( C_1e^{2x} \).

- The complex pair \( \alpha \pm i\beta = -1 \pm i\sqrt{3} \) gives the term \( e^{\alpha x}(C_2\cos(\beta x) + C_3\sin(\beta x)) \), which is \( e^{-x}(C_2\cos(x\sqrt{3}) + C_3\sin(x\sqrt{3})) \).


Combining these parts gives the complete complementary function: \[ y_c = C_1e^{2x} + e^{-x}\{C_2\cos(x\sqrt{3}) + C_3\sin(x\sqrt{3})\} \]

Step 4: Final Answer:

The complementary function is \( C_1e^{2x} + e^{-x}\{C_2\cos(x\sqrt{3}) + C_3\sin(x\sqrt{3})\} \). This corresponds to option (D).
Quick Tip: Recognizing algebraic forms like the difference/sum of cubes is very helpful for solving auxiliary equations quickly. For \(m^3-a^3=0\), the roots are always \(a\) and \(a(-\frac{1}{2} \pm i\frac{\sqrt{3}}{2})\). Here \(a=2\).

Step 1: Understanding the Question:

This is a definition-based question that asks for the standard form of Bernoulli's differential equation.


Step 2: Detailed Explanation:

A Bernoulli differential equation is a first-order ordinary differential equation of the form: \[ \frac{dy}{dx} + P(x)y = Q(x)y^n \]
where \( n \) is any real number.


Let's analyze the options:
- (A) is a linear equation.
- (B) is non-linear due to \( (\frac{dy}{dx})^2 \) and is not the Bernoulli form.
- (C) matches the standard definition of a Bernoulli equation exactly, where P and Q are functions of x.
- (D) is a second-order differential equation.

Step 3: Final Answer:

The correct form for Bernoulli's equation is \( \frac{dy}{dx} + Py = Qy^n \). This corresponds to option (C).
Quick Tip: Bernoulli's equation is a generalization of the linear equation. If \(n=0\) or \(n=1\), the equation becomes a standard first-order linear equation. The key feature is the \( y^n \) term on the right-hand side.

Step 1: Understanding the Question:

This question asks for the standard rule to find the particular integral (PI) for a linear differential equation with constant coefficients when the right-hand side (the forcing function) is \( \cos(ax) \).


Step 2: Detailed Explanation:

For a differential equation of the form \( f(D)y = X(x) \), where \( f(D) \) is a polynomial in the differential operator \( D = \frac{d}{dx} \), the particular integral is formally written as: \[ y_{PI} = \frac{1}{f(D)} X(x) \]
When the forcing function is \( X(x) = \cos(ax) \) or \( \sin(ax) \), there is a specific rule for evaluating the operator \( \frac{1}{f(D)} \).


The rule states that you can replace every occurrence of \( D^2 \) in the polynomial \( f(D) \) with \( -a^2 \).
So, if \( X(x) = \cos(ax) \), then: \[ y_{PI} = \frac{1}{f(D^2)} \cos(ax) = \frac{1}{f(-a^2)} \cos(ax) \]
This rule is only valid provided that the denominator does not become zero, i.e., \( f(-a^2) \neq 0 \). If \( f(-a^2) = 0 \), it is a "case of failure" and a different procedure must be followed.


Step 3: Final Answer:

The formula for the particular integral is \( \frac{1}{f(-a^2)} \cos ax \), provided \( f(-a^2) \neq 0 \). This corresponds to option (A).
Quick Tip: The most crucial part of this rule is the substitution: replace \( D^2 \) with \( -a^2 \). Note the negative sign is outside the square. For example, if the function is \( \cos(3x) \), you replace \( D^2 \) with \( -(3^2) = -9 \), not \( (-3)^2 = 9 \).

Step 1: Understanding the Question:

We are asked to find the value of 1 unit of pressure in a new system of units and express it in the standard SI unit for pressure (Nm\(^{-2}\) or Pascal).


Step 2: Key Formula or Approach:

First, we need the dimensional formula for pressure.

Pressure (P) = \( \frac{Force}{Area} = \frac{Mass \times Acceleration}{Area} \)

The dimensional formula is: \[ [P] = \frac{[M][LT^{-2}]}{[L^2]} = [M L^{-1} T^{-2}] \]

Step 3: Detailed Explanation:

In the standard SI system:
- Unit of Mass ([M]): 1 kg
- Unit of Length ([L]): 1 m
- Unit of Time ([T]): 1 s
The SI unit of pressure is \( 1 \, kg \cdot m^{-1} \cdot s^{-2} \), which is 1 Nm\(^{-2}\).




In the new system:
- Unit of Mass ([M']): 1 kg
- Unit of Length ([L']): 1 m
- Unit of Time ([T']): 1 minute = 60 s




Let's express 1 unit of pressure in the new system using its fundamental units: \[ 1 unit of new pressure = 1 \, [M'] [L']^{-1} [T']^{-2} \]
Now, we substitute the SI equivalents for these new units: \[ = (1 \, kg) (1 \, m)^{-1} (1 \, minute)^{-2} \] \[ = (1 \, kg) (1 \, m)^{-1} (60 \, s)^{-2} \] \[ = 1 \, kg \cdot m^{-1} \cdot \frac{1}{60^2} \, s^{-2} \] \[ = \frac{1}{3600} \, (kg \cdot m^{-1} \cdot s^{-2}) \]
Since \( 1 \, kg \cdot m^{-1} \cdot s^{-2} = 1 \, Nm^{-2} \), we have: \[ = \frac{1}{3600} \, Nm^{-2} \]

Step 4: Final Answer:

One unit of pressure in the new system is equal to \( 1/3600 \) Nm\(^{-2}\). This corresponds to option (C).
Quick Tip: When converting units, write down the dimensional formula. Then, substitute the conversion factors for each fundamental unit into the formula. Pay close attention to the powers on each dimension.

Step 1: Understanding the Question:

We need to identify which of the given physical quantities has the dimensional formula \( [M L T^{-1}] \).


Step 2: Detailed Explanation:

Let's find the dimensional formula for each option:


(A) Speed:

Speed = \( \frac{Distance}{Time} \). Dimensions are \( \frac{[L]}{[T]} = [L T^{-1}] \). This is incorrect.


(B) Acceleration:

Acceleration = \( \frac{Velocity}{Time} \). Dimensions are \( \frac{[L T^{-1}]}{[T]} = [L T^{-2}] \). This is incorrect.


(C) Impulse:

Impulse = Force \( \times \) Time.

First, find the dimensions of Force: Force = Mass \( \times \) Acceleration = \( [M] \times [L T^{-2}] = [M L T^{-2}] \).

Now, dimensions of Impulse = \( [M L T^{-2}] \times [T] = [M L T^{-1}] \). This is correct.

(Alternatively, Impulse = Change in Momentum. Momentum = Mass \( \times \) Velocity = \( [M] \times [L T^{-1}] = [M L T^{-1}] \)).


(D) Force:

As calculated above, the dimensions of Force are \( [M L T^{-2}] \). This is incorrect.


Step 3: Final Answer:

The dimensional formula \( [M L T^{-1}] \) corresponds to Impulse. This is option (C).
Quick Tip: Remembering the dimensional formulas for Force (\(MLT^{-2}\)) and Energy/Work (\(ML^2T^{-2}\)) is very useful, as many other quantities can be quickly derived from them. For example, Impulse = Force \( \times \) Time, so you just multiply the dimensions of Force by T.

Step 1: Understanding the Question:

We are given a relationship between the magnitudes of the cross product and the dot product of two vectors. We need to find the magnitude of their sum.


Step 2: Key Formula or Approach:

1. Use the definitions of cross product and dot product to find the angle \( \theta \) between the vectors.

\( |\vec{A} \times \vec{B}| = AB \sin\theta \)

\( \vec{A} \cdot \vec{B} = AB \cos\theta \)

2. Use the formula for the magnitude of the sum of two vectors.

\( |\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + 2AB\cos\theta} \)

(Here, A and B represent the magnitudes \( |\vec{A}| \) and \( |\vec{B}| \)).


Step 3: Detailed Explanation:

Find the angle \( \theta \):

From the given condition: \[ AB \sin\theta = \sqrt{3} (AB \cos\theta) \]
Assuming \( A \neq 0 \) and \( B \neq 0 \), we can divide by \( AB \cos\theta \): \[ \frac{\sin\theta}{\cos\theta} = \tan\theta = \sqrt{3} \]
This means the angle between the vectors is \( \theta = 60^\circ \).




Find the magnitude of the sum:

Now substitute the value of \( \cos\theta \) into the magnitude formula. \[ \cos(60^\circ) = \frac{1}{2} \] \[ |\vec{A} + \vec{B}|^2 = A^2 + B^2 + 2AB\cos(60^\circ) \] \[ |\vec{A} + \vec{B}|^2 = A^2 + B^2 + 2AB\left(\frac{1}{2}\right) \] \[ |\vec{A} + \vec{B}|^2 = A^2 + B^2 + AB \]
Taking the square root of both sides: \[ |\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + AB} = (A^2 + B^2 + AB)^{1/2} \]

Step 4: Final Answer:

The value of \( |\vec{A} + \vec{B}| \) is \( (A^2 + B^2 + AB)^{1/2} \). This corresponds to option (A).
Quick Tip: The ratio \( \frac{|\vec{A} \times \vec{B}|}{\vec{A} \cdot \vec{B}} \) is a direct way to find the tangent of the angle between two vectors. This is a common pattern in competitive exam questions.

Step 1: Understanding the Question:

We need to identify which pair of vectors from the given list are parallel to each other.


Step 2: Key Formula or Approach:

Two non-zero vectors are parallel if one is a scalar multiple of the other. That is, \( \vec{V_1} = k \vec{V_2} \) for some scalar \( k \).

This implies that the ratio of their corresponding components must be constant. For two vectors \( \vec{V_1} = a_1\hat{i} + b_1\hat{j} + c_1\hat{k} \) and \( \vec{V_2} = a_2\hat{i} + b_2\hat{j} + c_2\hat{k} \), they are parallel if \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).


Step 3: Detailed Explanation:

Let's check the pairs:

\( \vec{A} \) and \( \vec{B} \): \( \vec{A} \) has no \( \hat{k} \) component, while \( \vec{B} \) has no \( \hat{j} \) component. They cannot be parallel.


\( \vec{A} \) and \( \vec{C} \): Both vectors only have \( \hat{i} \) and \( \hat{j} \) components. Let's check the ratio of their components. \[ \frac{C_x}{A_x} = \frac{5.1}{6} \quad and \quad \frac{C_y}{A_y} = \frac{6.8}{8} \]
Let's simplify the ratios: \[ \frac{5.1}{6} = \frac{51}{60} = \frac{17 \times 3}{20 \times 3} = \frac{17}{20} = 0.85 \] \[ \frac{6.8}{8} = \frac{68}{80} = \frac{17 \times 4}{20 \times 4} = \frac{17}{20} = 0.85 \]
Since the ratios are equal, \( \vec{A} \) and \( \vec{C} \) are parallel. Specifically, \( \vec{C} = 0.85 \vec{A} \).


\( \vec{A} \) and \( \vec{D} \): \( \vec{A} \) has no \( \hat{k} \) component, but \( \vec{D} \) does. They cannot be parallel.


\( \vec{C} \) and \( \vec{D} \): \( \vec{C} \) has no \( \hat{k} \) component, but \( \vec{D} \) does. They cannot be parallel.


Step 4: Final Answer:

The parallel vectors are \( \vec{A} \) and \( \vec{C} \). This corresponds to option (A).
Quick Tip: To quickly check for parallel vectors, first look for vectors that exist in the same plane (e.g., both are in the xy-plane). Then, check if the ratios of their components are the same. You can often see the relationship by factoring out common terms. Here, \( \vec{A} = 2(3\hat{i} + 4\hat{j}) \) and \( \vec{C} = 1.7(3\hat{i} + 4\hat{j}) \), making the parallel relationship obvious.

Step 1: Understanding the Question:

We are given the position of a particle as a function of time. We need to find its position at the instant its speed is maximum. Since motion is along the x-axis, speed is the magnitude of velocity.


Step 2: Key Formula or Approach:

1. Find the velocity function, \( v(t) \), by differentiating the position function \( x(t) \) with respect to time.

2. Find the acceleration function, \( a(t) \), by differentiating the velocity function \( v(t) \).

3. To find the time of maximum velocity, set the acceleration \( a(t) \) to zero and solve for \( t \).

4. Substitute this value of \( t \) back into the original position function \( x(t) \).


Step 3: Detailed Explanation:

1. Find velocity \( v(t) \):
\[ x(t) = 9t^2 - t^3 \] \[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(9t^2 - t^3) = 18t - 3t^2 \]
2. Find acceleration \( a(t) \):
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(18t - 3t^2) = 18 - 6t \]
3. Find time of maximum velocity:

Set \( a(t) = 0 \): \[ 18 - 6t = 0 \] \[ 6t = 18 \] \[ t = 3 \, s \]
(To verify it's a maximum, we can check the second derivative of velocity, which is \( \frac{da}{dt} = -6 \). Since it's negative, the velocity is indeed at a maximum).


4. Find position at t = 3 s:

Substitute \( t=3 \) into the position equation: \[ x(3) = 9(3)^2 - (3)^3 \] \[ x(3) = 9(9) - 27 \] \[ x(3) = 81 - 27 = 54 \, m \]

Step 4: Final Answer:

The position of the particle when it achieves maximum speed is 54 m. This corresponds to option (C).
Quick Tip: This is a standard maxima/minima problem from calculus applied to physics. Remember the sequence: position \( \xrightarrow{differentiate} \) velocity \( \xrightarrow{differentiate} \) acceleration. To find the maximum/minimum of a quantity, differentiate it and set the result to zero.

Step 1: Understanding the Question:

One ball goes up and another comes down. We need to find the time when their speeds are equal. Let's take the upward direction as positive and use \( g \approx 10 \, ms^{-2} \) as the options are integers.


Step 2: Key Formula or Approach:

We will use the first equation of motion, \( v = u + at \), for both balls. The acceleration for both is \( a = -g \).

- Ball A (projected up): \( u_A = +40 \, ms^{-1} \)

- Ball B (dropped): \( u_B = 0 \, ms^{-1} \)


Step 3: Detailed Explanation:

Velocity of Ball A at time t:
\[ v_A(t) = u_A + at = 40 - gt \]
Velocity of Ball B at time t:
\[ v_B(t) = u_B + at = 0 - gt = -gt \]
The negative sign indicates it's moving downward.




We are interested in when the magnitudes of their velocities are equal. The magnitude of velocity is speed. \[ |v_A(t)| = |v_B(t)| \]
The magnitude of \( v_B \) is always \( | -gt | = gt \).

The velocity of ball A is positive while it's going up and negative when it's coming down. It reaches its peak when \( v_A = 0 \), which occurs at \( t = 40/g = 40/10 = 4 \) s. We should assume the time we are looking for is \( t < 4 \) s.

In this case, \( v_A(t) = 40 - gt \) is positive, so \( |v_A(t)| = 40 - gt \).


Set the magnitudes equal: \[ 40 - gt = gt \] \[ 40 = 2gt \] \[ t = \frac{40}{2g} = \frac{20}{g} \]
Using \( g = 10 \, ms^{-2} \): \[ t = \frac{20}{10} = 2 \, s \]
Since \( t=2 \) s is less than 4 s, our assumption was correct.


Step 4: Final Answer:

The magnitudes of their velocities will be equal after 2 seconds. This corresponds to option (B).
Quick Tip: In problems involving magnitudes of velocities under gravity, be mindful of the direction of motion. The speed of the upwardly projected body decreases, while the speed of the dropped body increases. They will meet at a point in time.

Step 1: Understanding the Question:

We have two projectiles with the same initial speed but different angles, resulting in the same horizontal range. We are given the angle and maximum height for one projectile and asked to find the maximum height of the other.


Step 2: Key Formula or Approach:

1. For a given initial speed, the horizontal range is the same for two projection angles \( \theta_1 \) and \( \theta_2 \) if they are complementary, i.e., \( \theta_1 + \theta_2 = 90^\circ \).

2. The formula for maximum height is \( H = \frac{u^2 \sin^2\theta}{2g} \).


Step 3: Detailed Explanation:

1. Find the angle of the second stone:

The angle of the first stone is \( \theta_1 = \pi/3 \) radians, which is \( 60^\circ \).

Since the ranges are equal, the angle of the second stone must be complementary to the first. \[ \theta_2 = 90^\circ - \theta_1 = 90^\circ - 60^\circ = 30^\circ \]

2. Relate the maximum heights:

Let \( H_1 \) be the height for \( \theta_1 \) and \( H_2 \) be the height for \( \theta_2 \). \[ H_1 = \frac{u^2 \sin^2\theta_1}{2g} = 102 \, m \] \[ H_2 = \frac{u^2 \sin^2\theta_2}{2g} \]
To find \( H_2 \), we can take a ratio of the two heights to eliminate the common term \( \frac{u^2}{2g} \). \[ \frac{H_2}{H_1} = \frac{\sin^2\theta_2}{\sin^2\theta_1} \] \[ H_2 = H_1 \left( \frac{\sin\theta_2}{\sin\theta_1} \right)^2 \]
Substitute the known values: \[ H_2 = 102 \left( \frac{\sin 30^\circ}{\sin 60^\circ} \right)^2 \]
We know \( \sin 30^\circ = 1/2 \) and \( \sin 60^\circ = \sqrt{3}/2 \). \[ H_2 = 102 \left( \frac{1/2}{\sqrt{3}/2} \right)^2 = 102 \left( \frac{1}{\sqrt{3}} \right)^2 \] \[ H_2 = 102 \times \frac{1}{3} = 34 \, m \]

Step 4: Final Answer:

The maximum height reached by the other stone is 34 metres. This corresponds to option (D).
Quick Tip: For two complementary projection angles, the ratio of maximum heights is \( H_2/H_1 = \tan^2\theta_2 \). Since \( \theta_2 = 30^\circ \), \( \tan 30^\circ = 1/\sqrt{3} \). So, \( H_2 = H_1 \times (1/\sqrt{3})^2 = H_1/3 \). This is a useful shortcut.

Step 1: Understanding the Question:

We need to find the time \( t \) when the velocity vector of a projectile is perpendicular to its initial velocity vector.


Step 2: Key Formula or Approach:

Two vectors are perpendicular if their dot product is zero. We will write the initial velocity vector and the velocity vector at time \( t \), and then set their dot product to zero.


Step 3: Detailed Explanation:

The initial velocity vector is: \[ \vec{v}_0 = (u \cos\theta)\hat{i} + (u \sin\theta)\hat{j} \]
The velocity vector at time \( t \) is: \[ \vec{v}(t) = (u \cos\theta)\hat{i} + (u \sin\theta - gt)\hat{j} \]
For the vectors to be perpendicular, their dot product must be zero: \[ \vec{v}_0 \cdot \vec{v}(t) = 0 \] \[ ((u \cos\theta)\hat{i} + (u \sin\theta)\hat{j}) \cdot ((u \cos\theta)\hat{i} + (u \sin\theta - gt)\hat{j}) = 0 \]
Calculate the dot product: \[ (u \cos\theta)(u \cos\theta) + (u \sin\theta)(u \sin\theta - gt) = 0 \] \[ u^2 \cos^2\theta + u^2 \sin^2\theta - ugt \sin\theta = 0 \]
Factor out \( u^2 \) and use the identity \( \cos^2\theta + \sin^2\theta = 1 \): \[ u^2(\cos^2\theta + \sin^2\theta) - ugt \sin\theta = 0 \] \[ u^2(1) - ugt \sin\theta = 0 \] \[ u^2 = ugt \sin\theta \]
Assuming \( u \neq 0 \), we can divide by \( u \): \[ u = gt \sin\theta \]
Now, solve for \( t \): \[ t = \frac{u}{g \sin\theta} \]

Step 4: Final Answer:

The time is \( \frac{u}{g \sin \theta} \). This corresponds to option (A).
Quick Tip: This problem is a classic application of the dot product to check for orthogonality (perpendicularity). Whenever a question involves angles between vectors (especially 90 degrees), the dot product is usually the most efficient tool to use.

Step 1: Understanding the Question:

We need to determine the direction of the force of static friction on the front and rear wheels of a bicycle that is being pedalled and moving forward.


Step 2: Detailed Explanation:

Rear Wheel (Driving Wheel):

The cyclist applies a torque via the pedals and chain to the rear wheel. This torque makes the bottom of the rear wheel try to slide or push the ground backwards.

According to Newton's third law, the ground exerts an equal and opposite force on the wheel. This reaction force is the force of static friction, and it points in the forward direction. This forward frictional force is what propels the bicycle.


Front Wheel (Driven Wheel):

The front wheel is not connected to the pedals; it rolls freely. The bicycle's frame pushes the axle of the front wheel forward. Due to this push, the bottom of the front wheel has a tendency to slide forward over the ground.

The force of static friction from the ground opposes this tendency of sliding. Therefore, the friction on the front wheel acts in the backward direction. This backward friction is a form of rolling resistance.


Step 3: Final Answer:

The frictional force on the rear wheel is forward, and on the front wheel, it is backward. This corresponds to option (A).
Quick Tip: For any vehicle, the friction on a driven (powered) wheel is in the direction of motion, as it provides the propulsion. The friction on a non-driven (freely rolling) wheel opposes the motion and acts as a retarding force. The same logic applies to the front-wheel drive and rear-wheel drive cars.

Step 1: Understanding the Question:

We have a system of two blocks connected by a string on a rough surface. A force F pulls the 2 kg block. We need to find the value of F that ensures the tension in the string remains constant during motion.

The phrase "tension in the string is constant throughout" implies two things:
1. The tension is uniform along the length of the string. This would be true for a massless string, or for a massive ("heavy") string if it is not accelerating.
2. The tension does not change with time.
Since the string is described as "heavy", the only way for the tension to be uniform along its length during motion is if the acceleration of the system is zero. This means the blocks move at a constant velocity.


Step 2: Key Formula or Approach:

We will use the condition for motion with constant velocity (\( a = 0 \)), which means the net force on the system is zero.
The force of kinetic friction is given by \( f_k = \mu N \), where \( N \) is the normal force. On a horizontal surface, \( N = mg \). We will consider \( g = 10 \, m/s^2 \).


Step 3: Detailed Explanation:

Let's consider the two blocks and the string as a single system. For the system to move at a constant velocity, the applied force F must be exactly equal to the total frictional force acting on the system.


First, calculate the frictional force on the 4 kg block (\( m_1 \)): \[ f_1 = \mu \times N_1 = \mu \times m_1g \] \[ f_1 = 0.5 \times 4 \, kg \times 10 \, m/s^2 = 20 \, N \]

Next, calculate the frictional force on the 2 kg block (\( m_2 \)): \[ f_2 = \mu \times N_2 = \mu \times m_2g \] \[ f_2 = 0.5 \times 2 \, kg \times 10 \, m/s^2 = 10 \, N \]

The total frictional force on the system is the sum of the individual frictional forces: \[ f_{total} = f_1 + f_2 = 20 \, N + 10 \, N = 30 \, N \]

For the system to move with constant velocity (\( a = 0 \)), the net horizontal force must be zero. \[ F_{net} = F_{applied} - f_{total} = 0 \] \[ F - 30 \, N = 0 \] \[ F = 30 \, N \]

Step 4: Final Answer:

The value of the force F required to move the blocks at a constant velocity, thus ensuring constant tension, is 30 N. This corresponds to option (B).
Quick Tip: In problems with connected blocks, the condition "constant tension" usually implies constant acceleration. If the connecting string is described as "heavy" or "massive," constant tension along its length can only be achieved if the acceleration is zero (i.e., constant velocity).

Step 1: Understanding the Question:

The question asks for the mass flow rate of water (\( \frac{m}{t} \)) required to generate a specific amount of power from a given height. This is a problem of energy conversion.


Step 2: Key Formula or Approach:

The potential energy (PE) of a mass \(m\) of water at a height \(h\) is given by \( PE = mgh \).

Power (P) is the rate at which energy is converted, so \( P = \frac{PE}{t} \).

Substituting the formula for PE, we get: \[ P = \frac{mgh}{t} = \left(\frac{m}{t}\right)gh \]
We need to solve for the mass flow rate, \( \frac{m}{t} \).


Step 3: Detailed Explanation:

First, let's list the given values in SI units:
- Power, \( P = 1 \, MW = 1 \times 10^6 \, Watts \) (or J/s).

- Height, \( h = 10 \, m \).

- Acceleration due to gravity, \( g = 10 \, m/s^2 \).




Rearrange the power formula to solve for the mass flow rate \( \frac{m}{t} \): \[ \frac{m}{t} = \frac{P}{gh} \]
Substitute the given values into the formula: \[ \frac{m}{t} = \frac{1 \times 10^6}{10 \times 10} \] \[ \frac{m}{t} = \frac{10^6}{100} = \frac{10^6}{10^2} \] \[ \frac{m}{t} = 10^{6-2} = 10^4 \, Kg/s \]

Step 4: Final Answer:

A mass of \( 10^4 \) kilograms of water must fall per second. This corresponds to option (B).
Quick Tip: This calculation assumes 100% efficiency in converting potential energy to electrical energy. In real-world scenarios, the efficiency would be less than 100%, and a larger mass of water would be required. However, for exam problems, assume 100% efficiency unless stated otherwise.

Step 1: Understanding the Question:

We are given the kinetic energy at the highest point and the maximum height of a projectile. We need to find its initial launch velocity using the principle of conservation of energy.


Step 2: Key Formula or Approach:

According to the principle of conservation of mechanical energy, the total energy of the projectile remains constant throughout its flight (ignoring air resistance).

Total Energy at launch = Total Energy at highest point
\[ E_{initial} = E_{top} \] \[ KE_{initial} + PE_{initial} = KE_{top} + PE_{top} \]

Step 3: Detailed Explanation:

Let's define the energy at each point:
- At the launch point (ground):
- Height is 0, so \( PE_{initial} = 0 \).
- \( KE_{initial} = \frac{1}{2}mu^2 \), where \( u \) is the initial velocity.

- At the highest point:
- We are given \( KE_{top} = 200 \, J \).
- The potential energy is \( PE_{top} = mgh_{max} \).




Now, let's calculate \( PE_{top} \):
- Mass, \( m = 1 \, Kg \).
- Max height, \( h_{max} = 20 \, m \).
- Let's use \( g = 10 \, m/s^2 \). \[ PE_{top} = (1 \, Kg)(10 \, m/s^2)(20 \, m) = 200 \, J \]



Now, apply the conservation of energy principle: \[ KE_{initial} + 0 = KE_{top} + PE_{top} \] \[ \frac{1}{2}mu^2 = 200 \, J + 200 \, J = 400 \, J \]
Substitute the mass \( m = 1 \, Kg \) and solve for \( u \): \[ \frac{1}{2}(1)u^2 = 400 \] \[ u^2 = 800 \] \[ u = \sqrt{800} = \sqrt{400 \times 2} = 20\sqrt{2} \, m/s \]

Step 4: Final Answer:

The initial velocity of the projectile from the ground is \( 20\sqrt{2} \) m/s. This corresponds to option (C).
Quick Tip: At the highest point, the projectile's velocity is not zero; only its vertical component is zero. The kinetic energy at the top corresponds purely to the horizontal component of velocity. The total energy is the sum of this kinetic energy and the potential energy at that height.

Step 1: Understanding the Question:

We need to calculate the power of an engine given the mass of the train, the change in its velocity, and the time taken for this change.


Step 2: Key Formula or Approach:

Power is the rate at which work is done (\( P = W/t \)).

According to the Work-Energy Theorem, the work done on an object is equal to the change in its kinetic energy (\( W = \Delta KE \)).

Combining these, we get: \[ P = \frac{\Delta KE}{t} = \frac{KE_{final} - KE_{initial}}{t} = \frac{\frac{1}{2}mv^2 - \frac{1}{2}mu^2}{t} \]

Step 3: Detailed Explanation:

First, list the given values in SI units:
- Mass, \( m = 2.05 \times 10^6 \, Kg \).
- Initial velocity, \( u = 5 \, m/s \).
- Final velocity, \( v = 25 \, m/s \).
- Time, \( t = 5 \, minutes = 5 \times 60 = 300 \, s \).




Calculate the change in kinetic energy (\( \Delta KE \)): \[ \Delta KE = \frac{1}{2}m(v^2 - u^2) \] \[ \Delta KE = \frac{1}{2}(2.05 \times 10^6)((25)^2 - (5)^2) \] \[ \Delta KE = \frac{1}{2}(2.05 \times 10^6)(625 - 25) \] \[ \Delta KE = \frac{1}{2}(2.05 \times 10^6)(600) \] \[ \Delta KE = 300 \times 2.05 \times 10^6 = 615 \times 10^6 \, J \]



Now, calculate the power: \[ P = \frac{\Delta KE}{t} = \frac{615 \times 10^6 \, J}{300 \, s} \] \[ P = 2.05 \times 10^6 \, W \]
Since \( 1 \, MW = 10^6 \, W \), the power is \( 2.05 \, MW \).


Step 4: Final Answer:

The power of the engine is 2.05 MW. This corresponds to option (B).
Quick Tip: This question asks for the average power over the 5-minute interval. Be careful to convert all units to standard SI units (mass in kg, velocity in m/s, time in seconds) before performing calculations.

Step 1: Understanding the Question:

Two identical wires have the same initial frequency. The tension in one is changed, which changes its frequency. We need to find the resulting beat frequency.


Step 2: Key Formula or Approach:

1. The fundamental frequency (\(f\)) of a stretched string is given by \( f = \frac{1}{2L}\sqrt{\frac{T}{\mu}} \), where \(L\) is length, \(T\) is tension, and \(\mu\) is linear mass density. For identical wires, \(L\) and \(\mu\) are constant, so \( f \propto \sqrt{T} \).

2. The beat frequency is the difference between the two frequencies: \( f_{beat} = |f_2 - f_1| \).


Step 3: Detailed Explanation:

Let the initial frequency be \( f_1 = 100 \, Hz \) and the initial tension be \( T_1 \).

The tension in the second wire is increased by 21%. The new tension \( T_2 \) is: \[ T_2 = T_1 + 0.21 T_1 = 1.21 T_1 \]
Since \( f \propto \sqrt{T} \), we can write the relationship between the new frequency \( f_2 \) and the old frequency \( f_1 \) as: \[ \frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1}} \]
Substitute the values: \[ \frac{f_2}{100} = \sqrt{\frac{1.21 T_1}{T_1}} = \sqrt{1.21} = 1.1 \]
Solve for the new frequency \( f_2 \): \[ f_2 = 100 \times 1.1 = 110 \, Hz \]
The frequency of the first wire remains \( f_1 = 100 \, Hz \).

Now, calculate the number of beats produced per second (beat frequency): \[ f_{beat} = |f_2 - f_1| = |110 - 100| = 10 \, Hz \]

Step 4: Final Answer:

The number of beats produced is 10. This corresponds to option (B).
Quick Tip: For small percentage changes in tension (\(x%\)), the percentage change in frequency is approximately \( (x/2)% \). Here, the change is large (21%), so the approximation is not accurate. It's better to use the direct square root relationship \( f' = f \sqrt{1 + fractional change} \).

Step 1: Understanding the Question:

We are given the maximum velocity and maximum acceleration of a body in Simple Harmonic Motion (S.H.M.) and need to find its amplitude.


Step 2: Key Formula or Approach:

For a body in S.H.M. with amplitude \(A\) and angular frequency \(\omega\):
- Maximum velocity is given by \( v_{max} = A\omega \).

- Maximum acceleration is given by \( a_{max} = A\omega^2 \).

We can solve this system of two equations to find \(A\).


Step 3: Detailed Explanation:

We are given:
1. \( v_{max} = A\omega = 1 \)
2. \( a_{max} = A\omega^2 = 4 \)




We can express \( a_{max} \) in terms of \( v_{max} \): \[ a_{max} = (A\omega)\omega = v_{max} \cdot \omega \]
Substitute the given values to find \(\omega\): \[ 4 = (1) \cdot \omega \] \[ \omega = 4 \, rad/s \]



Now that we have the angular frequency \(\omega\), we can use the equation for maximum velocity to find the amplitude \(A\): \[ v_{max} = A\omega \] \[ 1 = A \cdot (4) \] \[ A = \frac{1}{4} = 0.25 \, m \]

Step 4: Final Answer:

The amplitude of the S.H.M. is 0.25 metres. This corresponds to option (D).
Quick Tip: A very quick way to solve this is to take the ratio of the given quantities. \( a_{max} / v_{max} = (A\omega^2) / (A\omega) = \omega \). And amplitude \( A = v_{max}^2 / a_{max} = (A\omega)^2 / (A\omega^2) = A^2\omega^2 / A\omega^2 = A \). So, \( A = (1)^2 / 4 = 0.25 \).

Step 1: Understanding the Question:

We are given the frequencies of two simple pendulums and asked to find the time period of a third pendulum whose length is the difference of the lengths of the first two.


Step 2: Key Formula or Approach:

The time period \( T \) of a simple pendulum of length \( l \) is given by \( T = 2\pi\sqrt{\frac{l}{g}} \).

Frequency \( f \) is the reciprocal of the time period, \( f = \frac{1}{T} \).

From these, we can relate length and time period: \( T^2 = \frac{4\pi^2}{g}l \), which means \( l \propto T^2 \).


Step 3: Detailed Explanation:

For pendulum 1:

Frequency \( f_1 = \frac{1}{4} \) Hz.

Time period \( T_1 = \frac{1}{f_1} = 4 \) s.

Since \( l \propto T^2 \), we can write \( l_1 = k T_1^2 = k(4^2) = 16k \) for some constant \( k = \frac{g}{4\pi^2} \).


For pendulum 2:

Frequency \( f_2 = \frac{1}{3} \) Hz.

Time period \( T_2 = \frac{1}{f_2} = 3 \) s.

Similarly, \( l_2 = k T_2^2 = k(3^2) = 9k \).

From this we see that \( l_1 > l_2 \), so the difference is positive.


For the new pendulum:

The length of the new pendulum is \( l_{new} = l_1 - l_2 \). \[ l_{new} = 16k - 9k = 7k \]
The time period of this new pendulum, \( T_{new} \), is related to its length by \( l_{new} = k T_{new}^2 \). \[ 7k = k T_{new}^2 \] \[ T_{new}^2 = 7 \] \[ T_{new} = \sqrt{7} \, s \]

Step 4: Final Answer:

The time period of the pendulum of length (\( l_1 - l_2 \)) is \( \sqrt{7} \) s. This corresponds to option (C).
Quick Tip: For simple pendulums, the length is directly proportional to the square of the time period (\( l \propto T^2 \)). This relationship is very useful for comparing pendulums without needing to calculate the constant \( g/4\pi^2 \).

Step 1: Understanding the Question:

This is a problem about the Doppler effect for sound waves. A sound source is moving away from a stationary observer, and we need to find the apparent wavelength detected by the observer.


Step 2: Key Formula or Approach:

When a source moves away from a stationary observer, the observed frequency (\(f_o\)) is lower than the source frequency (\(f_s\)). The formula is: \[ f_o = f_s \left( \frac{v}{v + v_s} \right) \]
where \(v\) is the speed of sound and \(v_s\) is the speed of the source.

Wavelength is related to frequency and speed by \( \lambda = v/f \).
The source wavelength is \( \lambda_s = v/f_s \).
The observed wavelength is \( \lambda_o = v/f_o \).


Step 3: Detailed Explanation:

We can derive a formula for the apparent wavelength.
From the frequency formula, \( \frac{f_s}{f_o} = \frac{v+v_s}{v} \).

Substituting \( f = v/\lambda \): \[ \frac{v/\lambda_s}{v/\lambda_o} = \frac{\lambda_o}{\lambda_s} = \frac{v+v_s}{v} \] \[ \lambda_o = \lambda_s \left( \frac{v+v_s}{v} \right) \]
This shows that when the source moves away, the observed wavelength increases (a "stretch").




Now, substitute the given values:
- Source wavelength, \( \lambda_s = 50 \, cm \).
- Source speed, \( v_s = \frac{1}{5}v \).
\[ \lambda_o = 50 \left( \frac{v + \frac{1}{5}v}{v} \right) \] \[ \lambda_o = 50 \left( \frac{\frac{6}{5}v}{v} \right) \] \[ \lambda_o = 50 \left( \frac{6}{5} \right) = 10 \times 6 = 60 \, cm \]

Step 4: Final Answer:

The wavelength of the sound heard by the observer is 60 cm. This corresponds to option (D).
Quick Tip: Remember the physical effect: when a source moves away, the waves are "stretched out" behind it, leading to a longer wavelength and a lower frequency (lower pitch). When it moves towards, the waves are "compressed," leading to a shorter wavelength and higher frequency.

Step 1: Understanding the Question:

The question asks about the requirement for good acoustics in a hall, specifically regarding reverberation time.


Step 2: Detailed Explanation:

Reverberation is the persistence of sound in a particular space after the original sound is produced. Reverberation time is the time it takes for the sound to "fade away" or decay.


- Option (A): If there are no sound-absorbing materials (like carpets, curtains, or acoustic panels), the sound waves will reflect excessively off the hard surfaces (walls, ceiling, floor). This leads to a very long reverberation time, causing echoes and making speech or music sound muddled and unclear. This is undesirable.


- Option (B): A maximum reverberation time is the same issue as in option (A) and is not good for sound clarity.


- Option (C): A reverberation time of zero would mean that sound dies out instantly. This is characteristic of an anechoic chamber. Such an environment sounds unnatural and "dead," and is not suitable for performances like concerts, where some resonance is needed to enrich the sound.


- Option (D): An optimum reverberation time provides a balance. It allows the sound to persist long enough to blend and create a sense of fullness and space (liveness) but is short enough to ensure that individual sounds and words remain clear and distinct. The ideal reverberation time varies depending on the purpose of the hall (e.g., longer for orchestral music, shorter for lectures).


Step 3: Final Answer:

For good acoustics, the reverberation time must be optimized for the intended use of the hall. This corresponds to option (D).
Quick Tip: Think of the difference between singing in a tiled bathroom (long reverberation) versus a room full of soft furniture (short reverberation). Neither is perfect for a concert. Good concert hall acoustics are a carefully engineered compromise between these extremes.

Step 1: Understanding the Question:

For an ideal gas in a closed vessel (constant volume), a small percentage change in pressure results in a specific absolute change in temperature. We need to find the initial temperature.


Step 2: Key Formula or Approach:

For an ideal gas at constant volume (isochoric process), Gay-Lussac's Law applies: \[ \frac{P}{T} = constant \]
where T must be in Kelvin.
This implies \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \).

For small changes, we can use the differential form: \( \frac{dP}{P} = \frac{dT}{T} \). This can be approximated as \( \frac{\Delta P}{P} \approx \frac{\Delta T}{T} \).


Step 3: Detailed Explanation:

Let the initial pressure be \( P_1 \) and initial temperature be \( T_1 \).

The pressure is increased by 0.5%, so the change in pressure is \( \Delta P = 0.005 P_1 \).

The fractional change in pressure is \( \frac{\Delta P}{P_1} = 0.005 \).

The increase in temperature is \( \Delta T = 2^\circC \). A change of 2\(^\circ\)C is equal to a change of 2 Kelvin, so \( \Delta T = 2 \, K \).




Using the approximation for small changes: \[ \frac{\Delta P}{P_1} = \frac{\Delta T}{T_1} \] \[ 0.005 = \frac{2}{T_1} \]
Solve for the initial temperature \( T_1 \) in Kelvin: \[ T_1 = \frac{2}{0.005} = \frac{2}{5/1000} = \frac{2000}{5} = 400 \, K \]



The options are given in degrees Celsius. We need to convert the initial temperature from Kelvin to Celsius: \[ T(^\circC) = T(K) - 273.15 \] \[ T_1(^\circC) = 400 - 273 = 127^\circC \]

Step 4: Final Answer:

The initial temperature of the gas is 127\(^{\circ}\)C. This corresponds to option (B).
Quick Tip: Always remember to use absolute temperature (Kelvin) in gas law calculations. Also, a change in temperature (\(\Delta T\)) has the same numerical value in Celsius and Kelvin, but the absolute temperatures are different.

Step 1: Understanding the Question:

We need to identify the physical quantity that is conserved during the free expansion of an ideal gas.


Step 2: Key Formula or Approach:

Free expansion (also known as Joule expansion) is a process where a gas expands into a vacuum. The key characteristics are:
1. No work is done (\(W=0\)) because the gas expands against zero external pressure.

2. The system is typically considered to be thermally isolated, so there is no heat exchange (\(Q=0\)).

We analyze the consequences using the First Law of Thermodynamics: \( \Delta U = Q - W \).


Step 3: Detailed Explanation:

Applying the First Law of Thermodynamics to the free expansion process: \[ \Delta U = 0 - 0 = 0 \]
This means the change in the internal energy of the gas is zero. The internal energy remains constant.


For an ideal gas, the internal energy is a function of temperature only (\( U = f(T) \)).

Therefore, if \( \Delta U = 0 \), it implies that the change in temperature is also zero (\( \Delta T = 0 \)).

So, the temperature of the ideal gas remains constant during free expansion.


- Pressure decreases because the gas occupies a larger volume.
- Volume increases by definition of expansion.
- The ratio of pressure to volume (\(P/V\)) also changes.


Step 4: Final Answer:

During the free expansion of an ideal gas, the temperature remains constant. This corresponds to option (A).
Quick Tip: Remember the three key conditions for free expansion of an ideal gas: \( W=0 \), \( Q=0 \), and therefore \( \Delta U=0 \). For an ideal gas, \( \Delta U=0 \) directly implies \( \Delta T=0 \).

Step 1: Understanding the Question:

We are given the specific heat per unit mass (\(c_V\)) and the specific heat per mole (\(C_V\)) for a gas. We need to find the mass of a single atom.


Step 2: Key Formula or Approach:

1. Relate molar specific heat (\(C_V\)) and specific heat (\(c_V\)) using the molar mass (\(M\)): \( C_V = M \cdot c_V \).

2. The molar mass (\(M\)) is the mass of one mole of the substance.

3. One mole contains Avogadro's number (\( N_A \approx 6.022 \times 10^{23} \)) of atoms.

4. The mass of a single atom (\(m_{atom}\)) is the molar mass divided by Avogadro's number: \( m_{atom} = \frac{M}{N_A} \).


Step 3: Detailed Explanation:

The given values are:
- Molar specific heat, \( C_V = 3 \, cal/mol/K \).
- Specific heat, \( c_V = 0.075 \, cal/kg/K \).


There seems to be a unit inconsistency in the problem statement, as a molar mass calculated from these values would be extremely large. It is highly probable that the specific heat was intended to be given in units of cal/g/K. Let's assume this correction.
Corrected specific heat, \( c_V = 0.075 \, cal/g/K \).


Now, calculate the molar mass \( M \) in grams per mole: \[ M = \frac{C_V}{c_V} = \frac{3 \, cal/mol/K}{0.075 \, cal/g/K} \] \[ M = \frac{3}{75/1000} = \frac{3000}{75} = 40 \, g/mol \]
This molar mass (40 g/mol) corresponds to the element Argon, which is a monoatomic gas, so the values are consistent.


Finally, calculate the mass of a single atom: \[ m_{atom} = \frac{M}{N_A} = \frac{40 \, g/mol}{6.022 \times 10^{23} \, atoms/mol} \] \[ m_{atom} \approx \frac{40}{6} \times 10^{-23} \, g \approx 6.67 \times 10^{-23} \, g \]

Step 4: Final Answer:

The mass of one atom of the gas is approximately \( 6.67 \times 10^{-23} \) gm. This corresponds to option (A).
Quick Tip: In physics and chemistry problems, always perform a unit analysis. If the initial calculation leads to an unreasonable result (like a molar mass of 40 kg/mol), re-check the units for a likely typo. Grams are often used instead of kilograms in chemistry contexts.

Step 1: Understanding the Question:

We need to find the exponent \(x\) in the temperature-volume relationship for an adiabatic process involving a rigid diatomic ideal gas.


Step 2: Key Formula or Approach:

1. The primary equation for an adiabatic process is \( PV^\gamma = constant \), where \( \gamma \) is the adiabatic index (ratio of specific heats, \( C_P/C_V \)).

2. For an ideal gas, \( P = \frac{nRT}{V} \).

3. The adiabatic index \( \gamma = 1 + \frac{2}{f} \), where \( f \) is the number of degrees of freedom.


Step 3: Detailed Explanation:

First, determine the degrees of freedom for a rigid diatomic gas at room temperature. It has 3 translational and 2 rotational degrees of freedom, so \( f = 3 + 2 = 5 \). Vibrational modes are not excited at room temperature.

Next, calculate the adiabatic index \( \gamma \): \[ \gamma = 1 + \frac{2}{f} = 1 + \frac{2}{5} = \frac{7}{5} \]
Now, convert the \( PV \) relationship to a \( TV \) relationship. Start with \( PV^\gamma = K_1 \) (constant).
Substitute \( P = \frac{nRT}{V} \) from the ideal gas law: \[ \left(\frac{nRT}{V}\right)V^\gamma = K_1 \] \[ nRT V^{\gamma-1} = K_1 \]
Since \( n \) and \( R \) are constants, we can combine them into a new constant \( K_2 = K_1/(nR) \). \[ T V^{\gamma-1} = K_2 \]
This is the required relationship. By comparing \( T V^{\gamma-1} = constant \) with the given form \( TV^x = constant \), we can see that: \[ x = \gamma - 1 \]
Substitute the value of \( \gamma \): \[ x = \frac{7}{5} - 1 = \frac{7-5}{5} = \frac{2}{5} \]

Step 4: Final Answer:

The value of x is \( 2/5 \). This corresponds to option (B).
Quick Tip: Memorize the three forms of the adiabatic equation for an ideal gas: 1. \( PV^\gamma = constant \) 2. \( TV^{\gamma-1} = constant \) 3. \( P^{1-\gamma}T^\gamma = constant \) Knowing these directly saves time on derivations during an exam.

Step 1: Understanding the Question:

A device that can operate as a Carnot engine is reversed to work as a refrigerator. We are given its efficiency as an engine and the work input as a refrigerator, and we need to find the heat extracted from the cold reservoir.


Step 2: Key Formula or Approach:

1. The efficiency of a Carnot heat engine is \( \eta = 1 - \frac{T_C}{T_H} \), where \( T_C \) and \( T_H \) are the temperatures of the cold and hot reservoirs, respectively.
2. The Coefficient of Performance (COP) of a refrigerator is \( \beta = \frac{Q_C}{W} \), where \( Q_C \) is the heat absorbed from the cold reservoir and \( W \) is the work done on the system.
3. For a Carnot cycle, the COP is also given by \( \beta = \frac{T_C}{T_H - T_C} \).
We can relate \( \eta \) and \( \beta \).


Step 3: Detailed Explanation:

First, let's establish the relationship between \( \eta \) and \( \beta \).
From the efficiency formula, \( \eta = \frac{T_H - T_C}{T_H} \).

From the COP formula, \( \beta = \frac{T_C}{T_H - T_C} \).

Let's manipulate the efficiency formula: \( \frac{1}{\eta} = \frac{T_H}{T_H - T_C} \). \[ \frac{1}{\eta} - 1 = \frac{T_H}{T_H - T_C} - \frac{T_H - T_C}{T_H - T_C} = \frac{T_H - (T_H - T_C)}{T_H - T_C} = \frac{T_C}{T_H - T_C} = \beta \]
So, the relationship is \( \beta = \frac{1}{\eta} - 1 = \frac{1-\eta}{\eta} \).




Now, use the given values:
- Efficiency, \( \eta = \frac{1}{10} \).
- Work done, \( W = 10 \, J \).

Calculate the COP (\( \beta \)): \[ \beta = \frac{1 - 1/10}{1/10} = \frac{9/10}{1/10} = 9 \]
Now, use the definition of COP to find the heat absorbed, \( Q_C \): \[ \beta = \frac{Q_C}{W} \] \[ 9 = \frac{Q_C}{10 \, J} \] \[ Q_C = 9 \times 10 = 90 \, J \]

Step 4: Final Answer:

The amount of energy absorbed from the lower temperature reservoir is 90 J. This corresponds to option (C).
Quick Tip: For a Carnot cycle, the relationship between engine efficiency (\(\eta\)) and refrigerator COP (\(\beta\)) is \( \beta = \frac{1-\eta}{\eta} \). Memorizing this can be a useful shortcut.

Step 1: Understanding the Question:

We are given the energies of two incident photons and the work function of a metal. We need to find the ratio of the maximum velocities of the emitted photoelectrons.


Step 2: Key Formula or Approach:

We use Einstein's photoelectric equation: \[ KE_{max} = E_{photon} - \phi \]
where \( KE_{max} = \frac{1}{2}mv_{max}^2 \), \( E_{photon} \) is the energy of the incident photon, and \( \phi \) is the work function of the metal.


Step 3: Detailed Explanation:

Let's calculate the maximum kinetic energy for each case.


For the first photon (\( E_1 = 2.5 \, eV \)):
\[ KE_1 = E_1 - \phi = 2.5 \, eV - 1.5 \, eV = 1.0 \, eV \]
So, \( \frac{1}{2}mv_1^2 = 1.0 \, eV \).


For the second photon (\( E_2 = 3.5 \, eV \)):
\[ KE_2 = E_2 - \phi = 3.5 \, eV - 1.5 \, eV = 2.0 \, eV \]
So, \( \frac{1}{2}mv_2^2 = 2.0 \, eV \).




Now, we find the ratio of the velocities. We can take the ratio of the two kinetic energy equations: \[ \frac{\frac{1}{2}mv_1^2}{\frac{1}{2}mv_2^2} = \frac{1.0 \, eV}{2.0 \, eV} \]
The \( \frac{1}{2}m \) terms cancel out: \[ \frac{v_1^2}{v_2^2} = \frac{1}{2} \]
Take the square root of both sides to find the ratio of the velocities: \[ \frac{v_1}{v_2} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \]
So, the ratio \( v_1 : v_2 \) is \( 1 : \sqrt{2} \).


Step 4: Final Answer:

The ratio of the maximum velocities is 1 : \( \sqrt{2} \). This corresponds to option (D).
Quick Tip: When finding ratios, you often don't need to convert units (like eV to Joules) as long as they are consistent, because they will cancel out. The velocity is proportional to the square root of the maximum kinetic energy.

Step 1: Understanding the Question:

This is a definition-based question from optics concerning the phenomenon of total internal reflection.


Step 2: Detailed Explanation:

Total internal reflection can occur when light travels from a denser medium to a rarer (less dense) medium.

According to Snell's Law: \( n_1 \sin\theta_1 = n_2 \sin\theta_2 \), where \( n_1 > n_2 \).

Here, \( \theta_1 \) is the angle of incidence in the denser medium, and \( \theta_2 \) is the angle of refraction in the rarer medium.


The critical angle (\( \theta_c \)) is defined as the specific angle of incidence \( \theta_1 \) for which the angle of refraction \( \theta_2 \) is exactly 90 degrees.

At this angle, the refracted ray travels exactly along the boundary surface between the two media.

If the angle of incidence is greater than the critical angle, the light does not refract into the rarer medium at all; instead, it is completely reflected back into the denser medium.


Therefore, by definition, when the angle of incidence is the critical angle, the angle of refraction is 90\(^{\circ}\).


Step 3: Final Answer:

At the critical angle, the angle of refraction is 90\(^{\circ}\). This corresponds to option (B).
Quick Tip: Remember the condition for total internal reflection: light must travel from a denser to a rarer medium, and the angle of incidence must be greater than the critical angle. The critical angle itself is the boundary case where refraction happens at 90\(^{\circ}\).

Step 1: Understanding the Question:

The question asks to identify the quantum number that determines the fundamental shape of an electron's orbital in an atom.


Step 2: Detailed Explanation:

There are four primary quantum numbers that describe an electron in an atom:


- n (Principal Quantum Number): This number describes the main energy level or shell of the electron. It primarily determines the size and energy of the orbital. Its values are positive integers (1, 2, 3, ...).


- l (Azimuthal or Angular Momentum Quantum Number): This number describes the shape of the orbital and corresponds to the subshell. Its values range from 0 to n-1.
- l = 0 corresponds to an s-orbital, which is spherical.
- l = 1 corresponds to a p-orbital, which is dumbbell-shaped.
- l = 2 corresponds to a d-orbital, with more complex shapes (e.g., cloverleaf).
- l = 3 corresponds to an f-orbital, with even more complex shapes.


- m\(_l\) (Magnetic Quantum Number): This number describes the orientation of the orbital in three-dimensional space. Its values range from -l to +l, including 0.


- m\(_s\) (Spin Quantum Number): This number describes the intrinsic angular momentum of the electron, which has two possible states: +1/2 or -1/2.


From these definitions, the quantum number that describes the shape of the orbital is the azimuthal quantum number, symbolized by 'l'.


Step 3: Final Answer:

The symbol for the quantum number that describes the shape of an atomic orbital is l. This corresponds to option (A).
Quick Tip: A simple mnemonic to remember the roles: - \textbf{n} = Size/Energy (Shell) - \textbf{l} = Shape (Subshell) - \textbf{m\(_l\)} = Orientation (Orbital) - \textbf{m\(_s\)} = Spin

Step 1: Understanding the Question:

The question provides a statement and asks for the scientific principle or rule it defines.


Step 2: Detailed Explanation:

Let's analyze the given principles:


- Pauli's Principle (Pauli Exclusion Principle): This fundamental principle of quantum mechanics states exactly what is quoted in the question: no two electrons in a single atom can have the same four quantum numbers (n, l, m\(_l\), and m\(_s\)). An implication of this is that an atomic orbital can hold a maximum of two electrons, and these two electrons must have opposite spins.


- Hund's Rule: This rule deals with the filling of degenerate (equal-energy) orbitals within a subshell. It states that electrons will fill empty orbitals singly before pairing up, and that electrons in singly occupied orbitals will have the same spin.


- Aufbau Principle: This principle states that electrons fill atomic orbitals of the lowest available energy levels before occupying higher levels.


- Lewis Rule (Octet Rule): This is a concept in chemical bonding stating that atoms tend to bond in such a way that they each have eight electrons in their valence shell, giving them the same electronic configuration as a noble gas.


The statement in the question is the precise definition of the Pauli Exclusion Principle.


Note on Answer Key: The provided image marks Hund's Rule as the correct answer. This is incorrect. The statement is the definition of the Pauli Exclusion Principle. For the purpose of this solution, we will proceed with the factually correct answer.


Step 3: Final Answer:

The statement is known as the Pauli Exclusion Principle. This corresponds to option (A).
Quick Tip: Remember the key ideas: - \textbf{Aufbau:} Build up from lowest energy. - \textbf{Pauli Exclusion:} No two electrons are identical. - \textbf{Hund's Rule:} Maximize spin (spread out before pairing up).

Step 1: Understanding the Question:

We need to identify all elements from atomic number 1 to 20 whose ground-state electron configurations have all their electrons paired. This means every occupied orbital is completely filled.


Step 2: Detailed Explanation:

An element has no unpaired electrons if all of its occupied atomic subshells are completely filled with electrons. Let's list the electron configurations for Z = 1 to 20 and identify these elements.


- Z = 1 (H): 1s\(^1\) - 1 unpaired electron.
- Z = 2 (He): 1s\(^2\) - No unpaired electrons (1s subshell is full).

- Z = 3 (Li): [He] 2s\(^1\) - 1 unpaired electron.

- Z = 4 (Be): [He] 2s\(^2\) - No unpaired electrons (2s subshell is full).

- Z = 5 (B): [He] 2s\(^2\) 2p\(^1\) - 1 unpaired electron.

- Z = 6 (C): [He] 2s\(^2\) 2p\(^2\) - 2 unpaired electrons (Hund's rule).

- Z = 7 (N): [He] 2s\(^2\) 2p\(^3\) - 3 unpaired electrons.

- Z = 8 (O): [He] 2s\(^2\) 2p\(^4\) - 2 unpaired electrons.

- Z = 9 (F): [He] 2s\(^2\) 2p\(^5\) - 1 unpaired electron.

- Z = 10 (Ne): [He] 2s\(^2\) 2p\(^6\) - No unpaired electrons (2p subshell is full).

- Z = 11 (Na): [Ne] 3s\(^1\) - 1 unpaired electron.

- Z = 12 (Mg): [Ne] 3s\(^2\) - No unpaired electrons (3s subshell is full).

- Z = 13 (Al): [Ne] 3s\(^2\) 3p\(^1\) - 1 unpaired electron.

- Z = 14 (Si): [Ne] 3s\(^2\) 3p\(^2\) - 2 unpaired electrons.

- Z = 15 (P): [Ne] 3s\(^2\) 3p\(^3\) - 3 unpaired electrons.

- Z = 16 (S): [Ne] 3s\(^2\) 3p\(^4\) - 2 unpaired electrons.

- Z = 17 (Cl): [Ne] 3s\(^2\) 3p\(^5\) - 1 unpaired electron.

- Z = 18 (Ar): [Ne] 3s\(^2\) 3p\(^6\) - No unpaired electrons (3p subshell is full).

- Z = 19 (K): [Ar] 4s\(^1\) - 1 unpaired electron.

- Z = 20 (Ca): [Ar] 4s\(^2\) - No unpaired electrons (4s subshell is full).


The elements with no unpaired electrons are Helium (He), Beryllium (Be), Neon (Ne), Magnesium (Mg), Argon (Ar), and Calcium (Ca).


Step 3: Final Answer:

Counting these elements, we find there are 6 of them. This corresponds to option (D).
Quick Tip: Elements with no unpaired electrons are those with completely filled subshells. These are the noble gases (He, Ne, Ar) and the alkaline earth metals (Be, Mg, Ca) within the first 20 elements.

Step 1: Understanding the Question:

We need to identify the statement that does not accurately describe the general properties of covalent compounds.


Step 2: Detailed Explanation:

Let's analyze the properties of covalent compounds based on their bonding and structure. Covalent compounds consist of discrete molecules held together by relatively weak intermolecular forces (like van der Waals forces or hydrogen bonds).


- (A) They are generally insoluble in water: This is a typical property. Water is a polar solvent. Nonpolar covalent compounds (like oil, CCl\(_4\)) do not dissolve in water. While some polar covalent compounds (like sugar, ethanol) do dissolve, the general rule "like dissolves like" makes this statement largely true for the broad class of covalent compounds.


- (B) They consist of molecules: This is the defining characteristic of covalent compounds (excluding network covalent solids). The atoms are bonded into discrete, neutral units called molecules.


- (C) They exist as solids, liquids or gases: Due to the wide range of strengths of intermolecular forces, covalent compounds can exist in all three states at room temperature. For example, methane (gas), water (liquid), and iodine (solid). This is a valid property.


- (D) The reactions between them are fast: This is generally not a property of covalent compounds. Reactions involving covalent compounds require the breaking of strong covalent bonds and the formation of new ones. This process often requires a significant amount of activation energy, making the reactions relatively slow. In contrast, reactions between ionic compounds in solution are often instantaneous because they involve the interaction of free-moving ions.


Step 3: Final Answer:

The statement that is not a property of covalent compounds is that their reactions are fast. This corresponds to option (D).
Quick Tip: Contrast covalent properties with ionic properties. Ionic compounds are crystalline solids, soluble in water, conduct electricity when molten or dissolved, and their reactions in solution are very fast. Covalent compounds are often the opposite on these points.

Step 1: Understanding the Question:

The question requires us to determine the number of covalent bonds in each of the three given molecules (H\(_2\), N\(_2\), HCl) and then calculate their sum.


Step 2: Detailed Explanation:

We will find the number of covalent bonds in each molecule by considering the valence electrons and the need to achieve a stable electron configuration (a duet for hydrogen, an octet for nitrogen and chlorine).


1. Hydrogen molecule (H\(_2\)):

A hydrogen atom (H) has 1 valence electron. To achieve a stable configuration like Helium, it needs one more electron. Two hydrogen atoms share their single electrons to form one pair, resulting in a single covalent bond (H-H).

Number of bonds in H\(_2\) = 1.


2. Nitrogen molecule (N\(_2\)):

A nitrogen atom (N) has 5 valence electrons. To achieve a stable octet, it needs three more electrons. Two nitrogen atoms share three pairs of electrons, resulting in a triple covalent bond (N\(\equiv\)N).

Number of bonds in N\(_2\) = 3.


3. Hydrogen Chloride molecule (HCl):

A hydrogen atom (H) has 1 valence electron, and a chlorine atom (Cl) has 7 valence electrons. Hydrogen needs one more electron, and chlorine needs one more electron to complete their respective stable configurations. They share one pair of electrons, resulting in a single covalent bond (H-Cl).

Number of bonds in HCl = 1.


4. Sum of the bonds:

Total number of covalent bonds = (Bonds in H\(_2\)) + (Bonds in N\(_2\)) + (Bonds in HCl)

Total = 1 + 3 + 1 = 5.


Step 3: Final Answer:

The sum of covalent bonds in the three molecules is 5. This corresponds to option (B).
Quick Tip: To quickly determine the number of bonds between atoms in simple diatomic molecules, consider how many electrons each atom needs to gain to achieve a noble gas configuration. This number is often the number of covalent bonds it will form.

Step 1: Understanding the Question:

The question asks for the mass of Sodium Hydroxide (NaOH) needed to prepare a specific volume of a solution with a given normality.


Step 2: Key Formula or Approach:

1. Calculate the Equivalent Weight (EW) of NaOH.


EW = Molecular Weight (MW) / n-factor.


2. Calculate the number of gram equivalents required.


Gram Equivalents = Normality (N) \( \times \) Volume (V in litres).


3. Calculate the required mass.


Mass = Gram Equivalents \( \times \) Equivalent Weight.


Step 3: Detailed Explanation:

1. Calculate Molecular and Equivalent Weight of NaOH:
\[ MW of NaOH = 23 (Na) + 16 (O) + 1 (H) = 40 \, g/mol \]
Since the n-factor for NaOH is 1:

\[ EW of NaOH = \frac{40}{1} = 40 \, g/equivalent \]

2. Calculate Gram Equivalents:
\[ Normality (N) = 0.1 \, N = 0.1 \, equivalents/litre \] \[ Volume (V) = 5.0 \, litres \] \[ Gram Equivalents = 0.1 \frac{equivalents}{litre} \times 5.0 \, litres = 0.5 \, equivalents \]

3. Calculate Required Mass:
\[ Mass of NaOH = Gram Equivalents \times EW \] \[ Mass of NaOH = 0.5 \, equivalents \times 40 \frac{g}{equivalent} = 20 \, g \]

Step 4: Final Answer:

20 grams of NaOH are required.


This corresponds to option (A).
Quick Tip: For monobasic bases (like NaOH, KOH) and monoacidic acids (like HCl, HNO\(_3\)), Normality is equal to Molarity.
You could calculate moles (Molarity \( \times \) Volume) and then multiply by Molar Mass to get the same result.

Step 1: Understanding the Question:

We are given the masses of three different gases in a mixture and need to calculate the total number of molecules.


Step 2: Key Formula or Approach:

1. Calculate the number of moles for each gas using the formula: Moles = \( \frac{Given Mass}{Molar Mass} \).


Remember that oxygen, nitrogen, and hydrogen are diatomic (O\(_2\), N\(_2\), H\(_2\)).


2. Find the total number of moles by summing the moles of each gas.


3. Calculate the total number of molecules using Avogadro's law: Total Molecules = Total Moles \( \times \) Avogadro's Number (N\(_A\)).


Step 3: Detailed Explanation:

1. Molar Masses:

- Oxygen (O\(_2\)): \( 2 \times 16 = 32 \) g/mol


- Nitrogen (N\(_2\)): \( 2 \times 14 = 28 \) g/mol


- Hydrogen (H\(_2\)): \( 2 \times 1 = 2 \) g/mol


2. Number of Moles:

- Moles of O\(_2\): \( \frac{8 \, g}{32 \, g/mol} = 0.25 \, mol \)


- Moles of N\(_2\): \( \frac{14 \, g}{28 \, g/mol} = 0.5 \, mol \)


- Moles of H\(_2\): \( \frac{8 \, g}{2 \, g/mol} = 4.0 \, mol \)


3. Total Moles:

- Total Moles = 0.25 + 0.5 + 4.0 = 4.75 mol


4. Total Molecules:

- Total Molecules = Total Moles \( \times \) N\(_A\)


- Total Molecules = \( 4.75 \times (6 \times 10^{23}) \)


- Total Molecules = \( 28.5 \times 10^{23} = 2.85 \times 10^{24} \)


Step 4: Final Answer:

The total number of molecules in the mixture is \( 2.85 \times 10^{24} \).


This corresponds to option (C).
Quick Tip: A common mistake is forgetting that common gaseous elements like oxygen, nitrogen, and hydrogen are diatomic.
Always use the molar mass of the molecule (e.g., O\(_2\)), not the atom (O), when calculating moles from mass.

Step 1: Understanding the Question:

We need to calculate the equivalent weight for each of the given compounds and identify which one has the largest value.


Step 2: Key Formula or Approach:

Equivalent Weight (EW) is defined as:

\[ EW = \frac{Molecular Weight (MW)}{n-factor} \]
The n-factor depends on the type of substance and the reaction context.


Assuming standard acid-base or salt reactions:

- For an acid, n-factor is its basicity (number of replaceable H\(^+\) ions).


- For a salt, n-factor is the total magnitude of positive or negative charge on the ions.


Step 3: Detailed Explanation:

(A) Na\(_2\)CO\(_3\):

This is a salt.


It dissociates into 2 Na\(^+\) ions and one CO\(_3\)\(^{2-}\) ion.


The total positive charge is 2. So, n-factor = 2.

\[ EW = \frac{106}{2} = 53 \]

(B) H\(_3\)PO\(_4\):

This is phosphoric acid, a tribasic acid.


It can donate up to 3 protons. So, n-factor = 3.

\[ EW = \frac{98}{3} \approx 32.67 \]

(C) H\(_2\)C\(_2\)O\(_4\).2H\(_2\)O:

This is oxalic acid dihydrate, a dibasic acid.


It has two replaceable protons. So, n-factor = 2.

\[ EW = \frac{126}{2} = 63 \]

(D) AlCl\(_3\):

This is a salt.


It dissociates into one Al\(^{3+}\) ion and three Cl\(^-\) ions.


The total positive charge is 3. So, n-factor = 3.

\[ EW = \frac{133.5}{3} = 44.5 \]

Comparison:

Comparing the calculated equivalent weights: 53, 32.67, 63, 44.5.


The highest value is 63, which corresponds to oxalic acid dihydrate.


Step 4: Final Answer:

The equivalent weight is highest for H\(_2\)C\(_2\)O\(_4\).2H\(_2\)O.


This corresponds to option (C).
Quick Tip: Remember that the n-factor can change depending on the reaction.
For example, in redox reactions, the n-factor for oxalic acid is also 2 (as C goes from +3 to +4), but for H\(_3\)PO\(_4\), it can react to form salts like NaH\(_2\)PO\(_4\) (n=1) or Na\(_2\)HPO\(_4\) (n=2).
However, unless specified, assume complete reaction.

Step 1: Understanding the Question:

The question asks for the value of the ionic product of water (K\(_w\)) in a specific acidic solution at a standard temperature.


Step 2: Key Formula or Approach:

The ionic product of water, K\(_w\), is the equilibrium constant for the autoionization of water (H\(_2\)O \( \rightleftharpoons \) H\(^+\) + OH\(^-\)).


It is defined as:

\[ K_w = [H^+][OH^-] \]
A key property of K\(_w\) is that its value depends only on temperature.


For any dilute aqueous solution at a given temperature, the value of K\(_w\) is constant.


Step 3: Detailed Explanation:

The temperature is given as 25\(^{\circ}\)C.


At 25\(^{\circ}\)C, the standard value for the ionic product of water is a well-known constant:

\[ K_w = 1.0 \times 10^{-14} \, mol^2/L^2 \]
The presence of 0.01M HCl in the solution does not change the value of K\(_w\).


The HCl will increase the [H\(^+\)] concentration to 0.01 M, and according to Le Chatelier's principle, the [OH\(^-\)] concentration will decrease to maintain the constant product K\(_w\).


[OH\(^-\)] = \( K_w / [H^+] = (1.0 \times 10^{-14}) / (0.01) = 1.0 \times 10^{-12} \) M.


However, the value of K\(_w\) itself remains \( 1.0 \times 10^{-14} \).


Step 4: Final Answer:

The ionic product (K\(_w\)) of the solution at 25\(^{\circ}\)C is \( 1.0 \times 10^{-14} \) mol\(^2\)/L\(^2\).


This corresponds to option (C).
Quick Tip: This is a conceptual question.
Remember that equilibrium constants (like K\(_w\), K\(_a\), K\(_b\)) are only dependent on temperature.
The addition of acids or bases changes the concentrations of the species involved but not the value of the equilibrium constant itself.

Step 1: Understanding the Question:

We need to identify which pair of chemicals, when mixed, will form a buffer solution.


Step 2: Key Formula or Approach:

A buffer solution is a solution that resists changes in pH upon the addition of small amounts of an acid or a base.


There are two main types:

1. Acidic Buffer: A mixture of a weak acid and its salt with a strong base (which provides the conjugate base).


Example: Acetic acid (CH\(_3\)COOH) and sodium acetate (CH\(_3\)COONa).


2. Basic Buffer: A mixture of a weak base and its salt with a strong acid (which provides the conjugate acid).


Example: Ammonium hydroxide (NH\(_4\)OH) and ammonium chloride (NH\(_4\)Cl).


Step 3: Detailed Explanation:

Let's analyze each option:


(A) HCl + NaCl:

HCl is a strong acid.


NaCl is the salt of a strong acid (HCl) and a strong base (NaOH).


A mixture of a strong acid and its salt is not a buffer.


(B) CH\(_3\)COOH + CH\(_3\)COONa:

CH\(_3\)COOH (acetic acid) is a weak acid.


CH\(_3\)COONa (sodium acetate) is its salt with a strong base (NaOH).


This salt provides the conjugate base, CH\(_3\)COO\(^-\).


This combination of a weak acid and its conjugate base forms an acidic buffer solution.


(C) CH\(_3\)COOH + NaCl:

CH\(_3\)COOH is a weak acid.


NaCl is a salt that is not related to acetic acid (it doesn't provide the conjugate base).


This mixture is not a buffer.


(D) NH\(_4\)OH + NaOH:

NH\(_4\)OH (ammonium hydroxide) is a weak base.


NaOH is a strong base.


A mixture of a weak base and a strong base is not a buffer.


The strong base would suppress the dissociation of the weak base.


Step 4: Final Answer:

The combination that forms a buffer solution is CH\(_3\)COOH + CH\(_3\)COONa.


This corresponds to option (B).
Quick Tip: The key to identifying a buffer is to look for a "weak conjugate pair".
This means a weak acid and its conjugate base (e.g., HF and F\(^-\)), or a weak base and its conjugate acid (e.g., NH\(_3\) and NH\(_4\)\(^+\)).
The conjugate partner is usually supplied by a salt.

Step 1: Understanding the Question:

We are performing electrolysis of molten AlCl\(_3\) and need to calculate the volume of chlorine gas produced at the anode under standard conditions (STP).


Step 2: Key Formula or Approach:

1. Calculate the total electric charge (Q) passed through the electrolyte: \( Q = I \times t \).


2. Determine the half-reaction at the anode (oxidation).


3. Use Faraday's laws of electrolysis to find the moles of product.


The number of moles of electrons is \( \frac{Q}{F} \), where F is the Faraday constant (96500 C/mol).


4. Use the stoichiometry of the half-reaction to relate moles of electrons to moles of Cl\(_2\).


5. Calculate the volume of gas at STP using the molar volume: 1 mole of any ideal gas at STP occupies 22.4 L or 22400 mL.


Step 3: Detailed Explanation:

1. Calculate Charge (Q):
\[ Q = I \times t = 0.5 \, A \times 96.5 \, s = 48.25 \, C \]

2. Anode Half-Reaction:

At the anode, chloride ions are oxidized to chlorine gas:

\[ 2Cl^- \rightarrow Cl_2(g) + 2e^- \]

3. Calculate Moles of Electrons:
\[ Moles of e^- = \frac{Q}{F} = \frac{48.25 \, C}{96500 \, C/mol} = 0.0005 \, mol \]

4. Calculate Moles of Cl\(_2\):

From the half-reaction, 2 moles of electrons produce 1 mole of Cl\(_2\) gas.


Therefore:

\[ Moles of Cl_2 = \frac{Moles of e^-}{2} = \frac{0.0005}{2} = 0.00025 \, mol \]

5. Calculate Volume at STP:
\[ Volume of Cl_2 = Moles of Cl_2 \times Molar Volume at STP \] \[ Volume of Cl_2 = 0.00025 \, mol \times 22400 \, mL/mol \] \[ Volume of Cl_2 = 5.6 \, mL \]

Step 4: Final Answer:

The volume of Cl\(_2\) gas liberated is 5.6 ml.


This corresponds to option (C).
Quick Tip: Notice the numbers given: 96.5 seconds.
This is a strong hint to use the Faraday constant, as 96.5 is 96500 / 1000.
This often simplifies calculations.
\( Q = 0.5 \times 96.5 = 48.25 = 96500/2000 \).
So, moles of electrons = 1/2000 = 0.0005.

Step 1: Understanding the Question:

This is a definition-based question from the topic of electrolysis.


We need to identify the term for the mass of a substance deposited by one Faraday of charge.


Step 2: Detailed Explanation:

Let's define the terms:


- Faraday (F): One Faraday is the magnitude of the electric charge per mole of electrons.

\( 1 F \approx 96500 \, C/mol \).


- Electrochemical Equivalent (ECE): This is the mass of a substance deposited or liberated by the passage of 1 Coulomb of charge.


- Equivalent Weight (EW): This is the mass of a substance that combines with or displaces 1 mole of hydrogen atoms, or 8 grams of oxygen, or 1 mole of electrons.


In electrochemistry, it's the molar mass divided by the number of electrons transferred per formula unit in the redox reaction (n-factor).


According to Faraday's laws, the passage of 1 Faraday (1 mole of electrons) will deposit \( \frac{1}{n} \) moles of a substance, where n is the charge of the ion.


The mass deposited is \( \frac{1}{n} \times Molar Mass \).


This quantity, \( \frac{Molar Mass}{n} \), is precisely the definition of the Equivalent Weight.


Step 3: Final Answer:

The amount of substance deposited by the passage of 1F of electricity is one gram equivalent weight.


This corresponds to option (B).
Quick Tip: Remember the key distinction:
- 1 Coulomb deposits the Electrochemical Equivalent (ECE).
- 1 Faraday (96500 C) deposits the Equivalent Weight (EW).
The relationship is: EW = ECE \( \times \) 96500.

Step 1: Understanding the Question:

We are asked to calculate the electromotive force (emf) of a galvanic cell given in standard cell notation and with standard reduction potentials for the half-cells.


Step 2: Key Formula or Approach:

1. Identify the anode and cathode from the cell notation.


The convention is Anode \( || \) Cathode.


2. The anode is where oxidation occurs, and the cathode is where reduction occurs.


3. The standard emf of the cell (E\(^{\circ}_{cell}\)) is calculated using the standard reduction potentials of the cathode and anode:

\[ E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} \]
4. Since the concentrations of the ions are 1M, the conditions are standard, and the cell emf is equal to the standard cell emf (E\(_cell\) = E\(^{\circ}_{cell}\)).


Step 3: Detailed Explanation:

From the cell notation Sn\(|\)Sn\(^{2+}\) (1M)\(||\)Ag\(^+\) (1M)\(||\)Ag:


- Anode (Oxidation): Sn \( \rightarrow \) Sn\(^{2+}\) + 2e\(^-\). The Sn electrode is the anode.


- Cathode (Reduction): Ag\(^+\) + e\(^-\) \( \rightarrow \) Ag. The Ag electrode is the cathode.


The given standard reduction potentials are:

\[ E^{\circ}_{Sn^{2+}|Sn} = E^{\circ}_{anode} = -0.14 \, V \] \[ E^{\circ}_{Ag^{+}|Ag} = E^{\circ}_{cathode} = +0.80 \, V \]

Now, we apply the formula for the standard cell emf:

\[ E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} \] \[ E^{\circ}_{cell} = (+0.80 \, V) - (-0.14 \, V) \] \[ E^{\circ}_{cell} = 0.80 \, V + 0.14 \, V = 0.94 \, V \]

Step 4: Final Answer:

The emf of the cell is 0.94 V.


This corresponds to option (D).
Quick Tip: A spontaneous galvanic cell must have a positive E\(_cell\).
The formula E\(_cell\) = E\(^{\circ}_{cathode}\) - E\(^{\circ}_{anode}\) ensures this.
An alternative is E\(_cell\) = E\(^{\circ}_{reduction}\) + E\(^{\circ}_{oxidation}\), but you must remember to reverse the sign of the potential for the oxidation half-reaction.
The first method is generally safer and less prone to sign errors.

Step 1: Understanding the Question:

We are given the standard reduction potentials for three species and asked to rank them in order of their strength as reducing agents (reducing power).


Step 2: Key Formula or Approach:

- Standard Reduction Potential (E\(^{\circ}\)): A measure of the tendency of a chemical species to be reduced (gain electrons).


A higher (more positive) E\(^{\circ}\) means a greater tendency to be reduced.


- Reducing Agent: A substance that causes another substance to be reduced, and in the process, it gets oxidized (loses electrons).


- Relationship: A strong reducing agent is a substance that is easily oxidized.


A substance is easily oxidized if its tendency to be reduced is very low.


Therefore, a stronger reducing agent has a lower (more negative) standard reduction potential.


Step 3: Detailed Explanation:

The given standard reduction potentials (E\(^{\circ}\)) are:

- E\(^{\circ}\)(A) = +0.68 V


- E\(^{\circ}\)(B) = -2.54 V


- E\(^{\circ}\)(C) = -0.50 V


To find the order of reducing power, we need to arrange these potentials in increasing order (from most negative to most positive).


The order of the E\(^{\circ}\) values is:
\[ E^{\circ}(B) < E^{\circ}(C) < E^{\circ}(A) \] \[ -2.54 \, V < -0.50 \, V < +0.68 \, V \]
Since a lower reduction potential implies a stronger reducing agent, the order of reducing power is:

\[ Reducing Power: B > C > A \]

Step 4: Final Answer:

The correct order of reducing power is B > C > A.


This corresponds to option (D).
Quick Tip: Remember this simple rule:
\textbf{L}ower \textbf{R}eduction \textbf{P}otential = \textbf{S}tronger \textbf{R}educing \textbf{P}ower.
Conversely, a higher reduction potential means a stronger oxidizing power.

Step 1: Understanding the Question:

The question asks to identify the anions that cause permanent hardness in water when they form salts with magnesium (Mg\(^{2+}\)) and calcium (Ca\(^{2+}\)) ions.


Step 2: Key Formula or Approach:

Water hardness is primarily caused by dissolved divalent cations, mainly Ca\(^{2+}\) and Mg\(^{2+}\).


The type of hardness depends on the accompanying anion.


- Temporary Hardness (Carbonate Hardness): Caused by the presence of bicarbonate (hydrogen carbonate, HCO\(_3\)\(^{-}\)) salts of calcium and magnesium.


It is called "temporary" because it can be removed by boiling.


- Permanent Hardness (Non-carbonate Hardness): Caused by the presence of chloride (Cl\(^-\)) and sulfate (SO\(_4\)\(^{2-}\)) salts of calcium and magnesium.


These salts are soluble and are not removed by boiling.


Step 3: Detailed Explanation:

Based on the definitions above, permanent hardness is due to salts like CaCl\(_2\), MgCl\(_2\), CaSO\(_4\), and MgSO\(_4\).


The anions responsible are therefore chloride (Cl\(^-\)) and sulfate (SO\(_4\)\(^{2-}\)).


Let's analyze the options:

- (A) Cl\(^-\), SO\(_4\)\(^{2-}\): These are the anions that cause permanent hardness.


- (B) Cl\(^-\), NO\(_3\)\(^{-}\): Nitrates can contribute to hardness but chlorides and sulfates are the primary causes.


- (C) HCO\(_3\)\(^{-}\), Cl\(^-\): Bicarbonate (HCO\(_3\)\(^{-}\)) causes temporary hardness.


- (D) CO\(_3\)\(^{2-}\), HCO\(_3\)\(^{-}\): Carbonates are largely insoluble and bicarbonate causes temporary hardness.


Step 4: Final Answer:

The anions responsible for permanent hardness are Cl\(^-\) and SO\(_4\)\(^{2-}\).


This corresponds to option (A).
Quick Tip: A simple way to remember:
Temporary hardness is caused by bicarbonates and can be removed by boiling.
Permanent hardness is caused by "everything else", primarily chlorides and sulfates, and cannot be removed by boiling.

Step 1: Understanding the Question:

The question asks about the chemical used to regenerate permutit after it has been used for water softening.


Step 2: Key Formula or Approach:

Permutit is a type of zeolite (sodium aluminium silicate, represented as Na\(_2\)Z) used in ion-exchange water softeners.


- Softening Process: Hard water containing Ca\(^{2+}\) and Mg\(^{2+}\) ions is passed through the permutit.


The Ca\(^{2+}\) and Mg\(^{2+}\) ions displace the Na\(^+\) ions from the permutit.

\[ Na_2Z + Ca^{2+}(aq) \rightarrow CaZ + 2Na^{+}(aq) \]
- Exhausted Permutit: After some time, the permutit becomes "exhausted" (e.g., CaZ or MgZ).


- Regeneration: To make the softener usable again, the process must be reversed by passing a concentrated solution of a sodium salt through it.


Step 3: Detailed Explanation:

The regeneration reaction is:

\[ CaZ + 2NaCl(concentrated) \rightarrow Na_2Z + CaCl_2(aq) \]
A concentrated solution of sodium chloride (NaCl), commonly known as brine, is used for this purpose because it is inexpensive.


Step 4: Final Answer:

Exhausted permutit is regenerated by washing with a solution of NaCl.


This corresponds to option (B).
Quick Tip: The permutit process is an ion-exchange process.
To regenerate the original material (sodium permutit), you need to flood it with a high concentration of the ion that was originally displaced (sodium ions).
Common salt (NaCl) is the cheapest source of sodium ions.

Step 1: Understanding the Question:

We are given the masses of two hardness-causing salts in a specific mass of water.


We need to calculate the total hardness in parts per million (ppm) expressed as calcium carbonate equivalents.


Step 2: Key Formula or Approach:

1. Calculate the CaCO\(_3\) equivalent mass for each salt.

\[ CaCO_3 equivalent = Mass of salt \times \frac{MW of CaCO_3}{MW of salt} \]
2. Sum the CaCO\(_3\) equivalents to get the total equivalent mass.


3. Calculate the hardness in ppm.

\[ Hardness (ppm) = \frac{Total mass of CaCO_3 equivalent (in mg)}{Mass of water (in kg)} \]

Step 3: Detailed Explanation:

Given:

- Mass of CaSO\(_4\) = 27.2 mg


- Mass of MgSO\(_4\) = 2.4 mg


- Mass of water = 2 kg


- MW of CaSO\(_4\) = 136


- MW of MgSO\(_4\) = 120


- MW of CaCO\(_3\) = 100


1. CaCO\(_3\) equivalent of CaSO\(_4\):
\[ = 27.2 \, mg \times \frac{100}{136} = 20 \, mg of CaCO_3 equivalent \]

2. CaCO\(_3\) equivalent of MgSO\(_4\):
\[ = 2.4 \, mg \times \frac{100}{120} = 2.0 \, mg of CaCO_3 equivalent \]

3. Total CaCO\(_3\) equivalent mass:
\[ Total equivalent mass = 20 \, mg + 2 \, mg = 22 \, mg \]

4. Calculate Hardness in ppm:

The total equivalent mass (22 mg) is present in 2 kg of water.


Hardness in ppm is the equivalent mass per 1 kg of water.

\[ Hardness (ppm) = \frac{22 \, mg}{2 \, kg} = 11 \, mg/kg = 11 \, ppm \]

Step 4: Final Answer:

The total hardness of the water is 11 ppm.


This corresponds to option (A).
Quick Tip: The molecular weight of CaCO\(_3\) (100) is a convenient standard.
The key is to convert every hardness-causing salt into its CaCO\(_3\) equivalent before summing them up.
Finally, remember that ppm for water hardness is mg of CaCO\(_3\) equivalent per litre (or kg) of water.

Step 1: Understanding the Question:

We need to evaluate the correctness of two separate statements related to corrosion.


Step 2: Detailed Explanation:

Statement I: The lower the pH greater is the corrosion.

A lower pH indicates a higher concentration of H\(^+\) ions (higher acidity).


In acidic solutions, the reduction of H\(^+\) ions is a common cathodic reaction that drives corrosion:

\[ 2H^+ + 2e^- \rightarrow H_2(g) \]
A higher concentration of H\(^+\) ions facilitates this reaction, which accelerates the rate of oxidation (corrosion) at the anode.


Therefore, this statement is correct.


Statement II: Electrochemical Corrosion always occurs at the anodic area.

Electrochemical corrosion is a process involving an anode and a cathode.


- At the anode, oxidation occurs, which for a metal is the process of corrosion.


Example: \( Fe \rightarrow Fe^{2+} + 2e^- \).


- At the cathode, reduction occurs, consuming the electrons produced at the anode.


Corrosion is the destructive oxidation of the metal, which happens at the anode by definition.


Therefore, this statement is correct.


Step 3: Final Answer:

Both Statement I and Statement II are correct.


This corresponds to option (A).
Quick Tip: Remember the mnemonic "An Ox, Red Cat" for electrochemistry:
\textbf{An}ode is for \textbf{Ox}idation, \textbf{Red}uction is at the \textbf{Cat}hode.
Corrosion is an oxidation process, so it must happen at the anode.

Step 1: Understanding the Question:

The question asks for the chemical name of rust.


Step 2: Detailed Explanation:

Rust is the common term for the reddish-brown substance that forms on the surface of iron or its alloys when exposed to oxygen and moisture.


The chemical process involves the oxidation of iron (Fe) to ferric ions (Fe\(^{3+}\)).


These ferric ions combine with oxygen and water to form a complex, hydrated compound.


The chemical formula for rust is generally written as Fe\(_2\)O\(_3\).\(n\)H\(_2\)O.


The chemical name for Fe\(_2\)O\(_3\) is iron(III) oxide or ferric oxide.


The ".nH\(_2\)O" part indicates that it is hydrated.


Therefore, the chemical name for rust is Hydrated Ferric Oxide.


Step 3: Final Answer:

Rust is chemically known as Hydrated Ferric Oxide.


This corresponds to option (A).
Quick Tip: Remember the difference between "ferrous" and "ferric".
Ferrous refers to the iron(II) ion (Fe\(^{2+}\)).
Ferric refers to the iron(III) ion (Fe\(^{3+}\)).
Rust contains the more oxidized ferric ion.

Step 1: Understanding the Question:

We need to identify the monomer of the polymer Teflon and then count the number of fluorine atoms in that monomer molecule.


Step 2: Detailed Explanation:

1. Identify the Polymer and Monomer:

The polymer is Teflon.


The chemical name for Teflon is Polytetrafluoroethylene (PTFE).


The name "Polytetrafluoroethylene" indicates that the repeating monomer unit is "tetrafluoroethylene".


2. Determine the Structure and Formula of the Monomer:

"Ethylene" has the formula C\(_2\)H\(_4\) (CH\(_2\)=CH\(_2\)).


"Tetrafluoro" means that the four hydrogen atoms in ethylene are replaced by four fluorine atoms.


So, the structure of the monomer, tetrafluoroethylene (X), is CF\(_2\)=CF\(_2\).


The chemical formula is C\(_2\)F\(_4\).


3. Count the Fluorine Atoms:

In the molecule C\(_2\)F\(_4\), there are four fluorine atoms.


Step 3: Final Answer:

The number of fluorine atoms in the monomer of Teflon is 4.


This corresponds to option (C).
Quick Tip: Polymer names often give a direct clue to the monomer.
"Poly-" means many, so what follows is usually the name of the monomer.
For example, Poly(ethylene) comes from ethylene, Poly(styrene) from styrene, and Poly(tetrafluoroethylene) from tetrafluoroethylene.

Step 1: Understanding the Question:

We need to classify the polymer Bakelite based on its properties.


Step 2: Detailed Explanation:

Let's define the types of polymers listed:


- Thermoplastic Polymer: Polymers that become soft and moldable upon heating and solidify upon cooling. The process is reversible.


- Elastomer: Polymers with high elasticity that return to their original shape after being stretched.


- Fibre: Thread-forming polymers with high tensile strength.


- Thermosetting Polymer: Polymers that undergo irreversible chemical changes upon heating, forming a hard, rigid, three-dimensional network structure that cannot be remolded.


Bakelite is a condensation polymer of phenol and formaldehyde.


During its formation, extensive cross-linking creates a rigid 3D network.


Once it is cured by heating, it becomes permanently hard and cannot be melted.


This property is the definition of a thermosetting polymer.


Step 3: Final Answer:

Bakelite is an example of a thermosetting polymer.


This corresponds to option (D).
Quick Tip: Remember the key difference:
\textbf{Thermo-plastic} is like plastic that can be melted and reformed.
\textbf{Thermo-set} is "set" permanently by heat, like baking a cake – you can't un-bake it.
Bakelite's rigidity and heat resistance are classic traits of a thermoset.

Step 1: Understanding the Question:

We need to identify the correct chemical structure for the repeating unit of neoprene rubber from the given options.


Step 2: Detailed Explanation:

Neoprene is a synthetic rubber.


It is the polymer of the monomer chloroprene.


The chemical name for chloroprene is 2-chloro-1,3-butadiene.


The repeating unit of neoprene (polychloroprene) is:

\[ \left[ - CH_2 - \underset{Cl}{\underset{|}{C}} = CH - CH_2 - \right]_n \]
Let's examine the options provided in the image:


- Option 2 shows a Cl (chlorine) atom attached to the second carbon of the four-carbon chain.


This matches the structure of the neoprene repeating unit.


- The other options show incorrect substituent groups (phenyl, fluorine, methyl).


Step 3: Final Answer:

The correct structure is the one containing the chlorine atom.


This corresponds to option (B).
Quick Tip: Associate the names of common synthetic rubbers with their key monomer or feature:
- \textbf{Neoprene} \(\rightarrow{}\) Monomer is \textbf{chloro}prene \(\rightarrow{}\) Contains Chlorine (Cl).
- Natural Rubber \(\rightarrow{}\) Monomer is isoprene (2-\textbf{methyl}-1,3-butadiene) \(\rightarrow{}\) Contains a Methyl group (CH\(_3\)).
- Buna-S (SBR) \(\rightarrow{}\) Monomers are \textbf{B}utadiene and \textbf{S}tyrene.

Step 1: Understanding the Question:

We need to distinguish between primary and secondary fuels and identify which of the options is a secondary fuel.


Step 2: Detailed Explanation:

- Primary Fuels: These are fuels found in nature and can be used directly or with minimal processing.


- Secondary Fuels: These are fuels that are manufactured from primary fuels through a conversion process.


Let's classify the options:


- Natural gas: Extracted directly from the earth. It is a primary fuel.


- Coal gas: Manufactured from coal by destructive distillation. It is a secondary fuel.


- Lignite: A type of coal mined directly from the ground. It is a primary fuel.


- Crude oil: Extracted from underground reservoirs. It is a primary fuel.


Step 3: Final Answer:

Coal gas is not a primary fuel; it is a secondary fuel.


This corresponds to option (B).
Quick Tip: Ask yourself: "Is this fuel found in nature as-is, or is it made from something else?"
If it's found in nature (like coal, oil, wood), it's primary.
If it's made in a factory (like gasoline, coke, coal gas), it's secondary.

Step 1: Understanding the Question:

We need to identify the typical pH range for rain that is classified as acid rain.


Step 2: Detailed Explanation:

- Normal Rain: Normal, unpolluted rainwater is naturally acidic due to dissolved atmospheric CO\(_2\), which forms weak carbonic acid.


This brings the pH of normal rain down to about 5.6.


- Acid Rain: Rain with a pH lower than 5.6 is considered acid rain.


It is caused by pollutants like sulfur dioxide and nitrogen oxides forming sulfuric and nitric acid.


The typical pH of acid rain in affected areas is generally between 4.0 and 5.5, but can be lower.


The range 3.5 - 5.6 best represents the general definition and observed values for acid rain.


Step 3: Final Answer:

The pH of acid rain is generally in the range of 3.5 - 5.6.


This corresponds to option (B).
Quick Tip: Remember that the cutoff for acid rain is a pH of 5.6 (the pH of normal rain).
Any value below this is technically acid rain.
Choose the option that best represents the acidic conditions below 5.6.

Step 1: Understanding the Question:

The question asks to identify the layer of the Earth's atmosphere that contains the ozone layer.


Step 2: Detailed Explanation:

The Earth's atmosphere is divided into several layers. From the ground up, they are:


1. Troposphere (0 to ~12 km): Where weather occurs.


2. Stratosphere (~12 to ~50 km): This layer contains the vast majority (about 90%) of atmospheric ozone, forming the protective "ozone layer".


This layer absorbs harmful UV radiation from the Sun, causing the temperature to increase with altitude.


3. Mesosphere (~50 to ~85 km): Where meteors burn up.


4. Thermosphere (~85 to ~600 km): Where the aurora occurs.


Step 3: Final Answer:

The ozone layer is present in the Stratosphere.


This corresponds to option (C).
Quick Tip: A simple mnemonic for the atmospheric layers from the ground up is "The Straight Man's Testicles":
\textbf{T}roposphere, \textbf{S}tratosphere, \textbf{M}esosphere, \textbf{T}hermosphere.
The ozone layer is in the second layer up, the Stratosphere.

Step 1: Understanding the Question:

We need to identify the correct mining term for an access tunnel to a mineral deposit located inside a hill.


Step 2: Detailed Explanation:

Let's define the different types of mine access:


- Shaft: A vertical tunnel from the surface to a deep deposit. Best for flat terrain.


- Incline / Decline: Sloping tunnels. A decline goes down, an incline goes up.


- Adit: A horizontal or nearly horizontal tunnel driven into the side of a hill or mountain.


This method uses the topography to provide easy, level access for transport and drainage.


Given the "hilly terrain", an adit is the specific term for a horizontal entry into the hillside.


Step 3: Final Answer:

The mode of access to a deposit in a hilly terrain is an adit.


This corresponds to option (C).
Quick Tip: Associate the type of access with the terrain:
- Flat terrain, deep deposit \(\rightarrow{}\) \textbf{Shaft} (vertical).
- Flat terrain, shallow deposit \(\rightarrow{}\) \textbf{Decline/Incline} (sloped).
- Hilly/Mountainous terrain \(\rightarrow{}\) \textbf{Adit} (horizontal).

Step 1: Understanding the Question:

The question asks for the definition of the term "borehole survey" in the context of mining or geology.


Step 2: Detailed Explanation:

When a borehole is drilled, especially a deep one, it rarely follows a perfectly straight path.


It can deviate from its intended course due to various factors like the geology of the rock, drilling parameters, and the drill string's alignment.


A borehole survey is the specific process of measuring the borehole's actual trajectory or path in three-dimensional space.


This involves measuring its deviation from the vertical (inclination) and its direction in the horizontal plane (azimuth) at various points along its depth.


While measuring depth (A) and identifying properties (D, known as logging) are also parts of borehole analysis, the term "survey" specifically refers to mapping the path and its deviation.


Step 3: Final Answer:

Borehole survey means the measurement of the deviation of the borehole from its planned trajectory.


This corresponds to option (C).
Quick Tip: In surveying, the primary goal is to determine positions and paths.
Think of a borehole survey as creating a map of the hole's journey underground, not just measuring its length or what it passed through.
This is crucial for accurately locating the position of a mineral deposit intersected by the borehole.

Step 1: Understanding the Question:

This question asks for the name of a specific chemical pre-treatment process used in the cementation method of shaft sinking.


Step 2: Detailed Explanation:

The Cementation method is a ground treatment technique used to seal water-bearing fissures and stabilize weak ground, particularly during shaft sinking.


It involves injecting cement grout under pressure into boreholes drilled around the proposed shaft area.


In certain types of rock, especially those with very fine fissures, the penetration of the cement grout can be hindered by high friction and the rock's chemical properties.


To overcome this, a preliminary chemical treatment is applied.


This treatment, known as Pre-Silicatisation, involves injecting a solution of sodium silicate.


This chemical reacts with the rock mass in a way that lubricates the fissures, reducing the friction and making it easier for the subsequent cement grout to penetrate deeply and effectively.


Step 3: Final Answer:

The process of chemically treating the holes to aid cement passage is called Pre-Silicatisation.


This corresponds to option (B).
Quick Tip: The prefix "Pre-" indicates that this is a preparatory step done before the main process.
Here, it's a pre-treatment before the main cement injection (grouting).
This helps distinguish it from "Silicatisation," which might be part of the main grouting mix itself.

Step 1: Understanding the Question:

The question asks for the specific name of holes designed to relieve water pressure behind the lining of a mine shaft.


Step 2: Detailed Explanation:

When a shaft is sunk through water-bearing strata, water can seep into the space behind the permanent concrete or brick lining.


If this water is not drained, it will accumulate and exert a significant hydrostatic pressure on the back of the lining.


This pressure can be strong enough to damage or even cause the failure of the shaft lining.


To prevent this, small-diameter holes or pipes, known as weep holes, are intentionally left in the lining, typically at regular intervals.


These holes allow any trapped water to "weep" or drain out from behind the lining into the shaft, where it can be collected and pumped to the surface.


This relieves the hydrostatic pressure and ensures the stability of the shaft lining.


Step 3: Final Answer:

The holes designed for pressure relief behind a shaft lining are called weep holes.


This corresponds to option (B).
Quick Tip: The term "weep hole" is used widely in civil engineering for any opening that allows water to drain from behind a retaining structure, like a retaining wall, bridge abutment, or, in this case, a shaft lining.
The name itself suggests its function: allowing the structure to "weep" out trapped water.

Step 1: Understanding the Question:

The question asks for the property of an explosive that describes its ability to withstand storage conditions (heat, humidity, time) without degrading.


Step 2: Detailed Explanation:

Let's define the properties of explosives:


- Stability: This refers to the chemical stability of an explosive.


It is the ability of the explosive to be stored for long periods under varying environmental conditions (like heat, humidity, sunlight) without decomposing, changing its properties, or becoming dangerously unstable.


This directly matches the description in the question.


- Sensitivity: This is a measure of the ease with which an explosive can be initiated or detonated by an external stimulus like impact, friction, heat, or shock.


High sensitivity means it detonates easily; low sensitivity means it is safer to handle.


- Density: This is the mass per unit volume of the explosive.


It affects the explosive's velocity of detonation and the amount of energy that can be packed into a borehole.


- Resistance: This is a general term and could refer to water resistance or electrical resistance, but it is not the primary term for describing storage life.


Step 3: Final Answer:

The characteristic that describes the storage life of an explosive under various conditions is its stability.


This corresponds to option (A).
Quick Tip: Think of the difference between stability and sensitivity:
- \textbf{Stability} is about the explosive's "shelf life" and its ability to resist decomposing on its own.
- \textbf{Sensitivity} is about its reaction to being "provoked" by an external shock or spark.
A good explosive is stable (safe to store) but sensitive enough to be detonated when intended.

Step 1: Understanding the Question:

We need to identify the specific class of permitted explosives that is suitable for "solid blasting" in underground coal mines.


Step 2: Detailed Explanation:

In underground coal mining, special "permitted" explosives are used to minimize the risk of igniting flammable methane gas or coal dust.


These explosives are classified into different categories (P1, P2, P3, P4, P5) based on their safety characteristics and intended use.


Solid Blasting is a technique where blasting is carried out in a coal face "off the solid," meaning without a pre-existing free face created by cutting or shearing.


This is a particularly hazardous operation because the explosive energy is highly confined, increasing the risk of a gas or dust ignition.


Therefore, it requires the safest category of explosives.


- P1 Explosives: General-purpose explosives for use in coal where there is minimal gas.


- P3 Explosives: For use in stone drifts or seams with higher gas content.


- P4 Explosives: This category is obsolete.


- P5 Explosives: These are "sheathed" or equivalent-to-sheathed (EQS) explosives designed with the highest degree of safety.


They produce a low-temperature, short-duration flame and are specifically approved for the most hazardous applications, including solid blasting and blasting in fractured ground.


Step 3: Final Answer:

The type of explosive used for solid blasting is the P5 category.


This corresponds to option (D).
Quick Tip: The "P" number in permitted explosives indicates the level of safety, with a higher number generally signifying a safer explosive for gassier conditions.
P5 is the "top tier" of safety, required for the most dangerous blasting scenario, which is solid blasting.

Step 1: Understanding the Question:

The question asks to identify the chemical compound or mixture typically used as the primary or priming charge in a detonator.


Step 2: Detailed Explanation:

A detonator is a device used to trigger a larger, less sensitive main explosive charge.


It works as a multi-stage explosive train:


1. Initiation: An external stimulus (like an electric current heating a fuse head or a shock from a fuse) initiates the first charge.


2. Prime Charge (Primary Explosive): This is a highly sensitive explosive that detonates from the initial stimulus.


Its purpose is to generate enough shock to detonate the next, more powerful charge.


A very common prime charge composition is ASA, which is a mixture of Azide (Lead Azide), Styphnate (Lead Styphnate), and powdered Aluminium.


3. Base Charge (Secondary Explosive): This is a less sensitive but more powerful explosive that makes up the bulk of the detonator's power.


It is detonated by the shockwave from the prime charge.


PETN (Pentaerythritol tetranitrate) and TNT (Trinitrotoluene) are powerful secondary explosives and are commonly used as the base charge in detonators.


Ammonium Nitrate is the main component of ANFO, a bulk blasting agent, not a primary charge.


Step 3: Final Answer:

The prime charge in a detonator is commonly ASA.


This corresponds to option (C).
Quick Tip: Think of a detonator like lighting a fire.
You use a match (the initiation) to light the kindling (the prime charge, e.g., ASA), which then provides a strong enough flame to ignite the main log (the base charge, e.g., PETN).

Step 1: Understanding the Question:

We need to arrange the listed mining activities into the correct chronological order for a standard blasting cycle.


Step 2: Detailed Explanation:

Let's analyze the logical flow of a drill-and-blast operation:


1. Drilling: The first step is to create holes in the rock or coal face where the explosives will be placed.


2. Charging: The drilled holes are then "charged" or loaded with the explosives and detonators.


3. Stemming: After charging, the mouth of the hole is plugged with an inert material like clay or sand stemming.


This confines the explosive energy within the hole, ensuring it is used to break the rock rather than blowing out of the hole (a "blowout").


4. Blasting: Once the area is clear of personnel, the explosive charges are detonated.


5. Mucking: After the blast and a safety period for fumes to clear, the broken rock or coal (the "muck") is loaded and hauled away.


This sequence is Drilling \(\rightarrow{}\) Charging \(\rightarrow{}\) Stemming \(\rightarrow{}\) Blasting \(\rightarrow{}\) Mucking.


Step 3: Final Answer:

The correct sequence of operations is Drilling, Charging, Stemming, Blasting, Mucking.


This corresponds to option (A).
Quick Tip: Think of the process logically, like baking a cake.
You have to prepare the pan (\textbf{Drilling}), put the batter in (\textbf{Charging}), put it in the oven (\textbf{Stemming/Blasting}), and then take it out to serve (\textbf{Mucking}).
You can't muck before you blast, and you can't blast before you drill and charge the holes.

Step 1: Understanding the Question:

The question asks about the texture of igneous rocks that form from magma cooling near the Earth's surface.


Step 2: Detailed Explanation:

The texture of an igneous rock refers to the size, shape, and arrangement of its mineral crystals.


The most important factor controlling crystal size is the cooling rate of the magma or lava.


- Slow Cooling: When magma cools slowly, deep within the Earth's crust (intrusive or plutonic rocks), the atoms have a long time to migrate and arrange themselves into large, well-formed crystals.


This results in a coarse-grained texture (also called phaneritic), where individual crystals are visible to the naked eye (e.g., Granite).


- Rapid Cooling: When magma or lava cools quickly, at or near the Earth's surface (extrusive or volcanic rocks), the atoms have very little time to form large crystals.


This results in a fine-grained texture (also called aphanitic), where the individual crystals are too small to be seen without a microscope (e.g., Basalt, Rhyolite).


Since the question specifies consolidation "nearer to the surface," this implies rapid cooling.


Conglomerate (D) is a sedimentary rock, not an igneous rock.


Step 3: Final Answer:

Rocks formed from magma cooling near the surface have a fine texture.


This corresponds to option (C).
Quick Tip: Remember the simple relationship:
Slow cooling = Large crystals = Coarse texture (Intrusive).
Fast cooling = Small crystals = Fine texture (Extrusive).
The closer to the surface, the faster the cooling.

Step 1: Understanding the Question:

The question asks for the correct sequence of coal formation, known as coalification, which represents the increasing rank of coal.


Step 2: Detailed Explanation:

Coalification is the process by which plant matter is converted into coal under the influence of heat and pressure over millions of years.


The process involves a progressive increase in carbon content and a decrease in moisture and volatile matter.


The sequence of coal ranks, from lowest to highest, is as follows:


1. Peat: This is the initial stage.


It is an accumulation of partially decayed vegetation or organic matter.


It has high moisture content and low carbon content.


2. Lignite: With increased pressure and temperature, peat is transformed into lignite (also called brown coal).


It is the lowest rank of coal.


3. Bituminous Coal: Further burial, heat, and pressure convert lignite into bituminous coal (also called soft coal).


This is a large and important category of coal, often subdivided into sub-bituminous and bituminous.


4. Anthracite: This is the highest rank of coal (also called hard coal).


It forms from bituminous coal under the highest heat and pressure.


It has the highest carbon content, the lowest moisture, and burns with the cleanest flame.


The correct sequence of formation is therefore Peat \(\rightarrow{}\) Lignite \(\rightarrow{}\) Bituminous \(\rightarrow{}\) Anthracite.


Step 3: Final Answer:

The correct sequence of coal formation is Peat \(\rightarrow{}\) Lignite \(\rightarrow{}\) Bituminous \(\rightarrow{}\) Anthracite.


This corresponds to option (B).
Quick Tip: Think of it as squeezing water out and concentrating carbon.
\textbf{P}eat is like a wet sponge of plant material.
Squeeze it a bit, you get \textbf{L}ignite.
Squeeze harder, you get \textbf{B}ituminous.
Squeeze the hardest, you get \textbf{A}nthracite, which is almost pure carbon.
The sequence is alphabetical if you remember the four main types: A, B, L, P. The process goes in reverse alphabetical order of rank: P \(\rightarrow{}\) L \(\rightarrow{}\) B \(\rightarrow{}\) A.

Step 1: Understanding the Question:

The question asks for the geological term for the vertical component of displacement across a fault.


Step 2: Detailed Explanation:

In structural geology, the displacement of rock layers or a seam across a fault is described by several components:


- Throw: This is the measure of the vertical displacement of the fault.


It is the vertical distance between a point on one side of the fault and the corresponding point on the other side.


This directly matches the question's description.


- Heave: This is the measure of the horizontal displacement of the fault.


- Hade: This is the angle that the fault plane makes with the vertical.


It is the complement of the dip angle (Hade = 90\(^{\circ}\) - Dip).


- Want: This is a mining term for a barren area where the coal seam is missing, which could be caused by a fault.


- Inclination: This is a general term for a slope or angle, often used for the dip of strata.


Step 3: Final Answer:

The vertical displacement of a seam in a normal fault is known as the throw.


This corresponds to option (A).
Quick Tip: Remember the two main components of fault displacement:
- \textbf{Throw} is the vertical part (what you "throw" up or down).
- \textbf{Heave} is the horizontal part (what you "heave" sideways).

Step 1: Understanding the Question:

We need to identify the sedimentary rock from the given list of four rock types.


Step 2: Detailed Explanation:

Let's classify each rock:


- Granite: This is an igneous rock.


Specifically, it is an intrusive (or plutonic) igneous rock, formed from the slow cooling of magma beneath the Earth's surface.


- Basalt: This is an igneous rock.


Specifically, it is an extrusive (or volcanic) igneous rock, formed from the rapid cooling of lava on the Earth's surface.


- Sandstone: This is a classic sedimentary rock.


It is formed from the compaction and cementation (lithification) of sand-sized grains of minerals, rock, or organic material.


- Schist: This is a metamorphic rock.


It is formed when other rocks (like shale) are subjected to intense heat and pressure, causing the mineral grains to recrystallize and align in layers or bands (a property called foliation).


Step 3: Final Answer:

Sandstone is the example of a sedimentary rock in the list.


This corresponds to option (C).
Quick Tip: Associate the rock types with their formation process:
- \textbf{Igneous} = from "fire" (molten magma/lava).
- \textbf{Sedimentary} = from "sediment" (sand, mud, shells).
- \textbf{Metamorphic} = "morphed" or changed by heat and pressure.

Step 1: Understanding the Question:

The question asks for the specific term for the point of origin of an earthquake.


Step 2: Detailed Explanation:

Let's define the given terms related to earthquakes:


- Focus (or Hypocenter): This is the point within the Earth along a fault where the initial rupture and release of energy occurs.


It is the true origin point of the earthquake.


- Epicenter: This is the point on the Earth's surface directly above the focus.


It is the location where the earthquake's effects are often the most intense.


- Iso-seismal line: A line on a map connecting points of equal earthquake intensity.


- Co-seismal line: A line on a map connecting points where the seismic waves arrive at the same time.


The question asks for the point where the earthquake originates, which is the focus.


Step 3: Final Answer:

The point where an earthquake originates is known as the focus.


This corresponds to option (D).
Quick Tip: Remember the difference:
- \textbf{Focus is the underground origin.
- \textbf{Epicenter} is the surface point directly above the focus.
Think of "Epi-" as a prefix meaning "upon" or "over," so the epicenter is upon the center.

Step 1: Understanding the Question:

We need to identify the scientific hypothesis that specifically explains the formation of the Earth and the solar system.


Step 2: Detailed Explanation:

Let's analyze the given hypotheses:


- Nebular Hypothesis: This is the most widely accepted model that explains the formation and evolution of the Solar System.


It proposes that the Sun, the Earth, and all other planets formed from a giant, rotating cloud of interstellar gas and dust called a solar nebula.


- Plate Tectonic Hypothesis (or Theory): This theory explains the large-scale movements of the Earth's lithosphere (the crust and upper mantle).


It explains phenomena like continental drift, earthquakes, and volcanic activity, but not the origin of the Earth itself.


- Big Bang Hypothesis (or Theory): This is the leading cosmological model for the observable universe.


It describes how the universe expanded from an initial state of extremely high density and temperature, but it does not specifically detail the formation of the Earth.


- Null Hypothesis: This is a term used in statistics and scientific testing.


It is a default statement that there is no relationship between two measured phenomena. It has nothing to do with cosmology.


Step 3: Final Answer:

The Nebular Hypothesis explains the origin of the Earth.


This corresponds to option (A).
Quick Tip: Distinguish the scales of these theories:
- \textbf{Big Bang} = Origin of the entire Universe.
- \textbf{Nebular Hypothesis} = Origin of our Solar System (including Earth).
- \textbf{Plate Tectonics} = How the surface of the Earth moves and changes.

Step 1: Understanding the Question:

The question asks for the best definition of the geological process of erosion from the given options.


Step 2: Detailed Explanation:

Let's define the key geological terms:


- Weathering: This is the in-situ (in one place) breakdown of rocks, soil, and minerals.


It includes physical breakdown (disintegration, e.g., by frost) and chemical breakdown

(decomposition, e.g., by acid rain).


Option (B) describes weathering.


- Erosion: This is the process where weathered material (like soil, rock fragments) is moved from one location to another.


It involves both the initial picking-up or breakdown of the material and its subsequent transportation by agents like water, wind, ice, or gravity.


Therefore, erosion encompasses both the disintegration (or detachment) of material and its transportation.


- Abrasion and Deflation (Option A) are specific mechanisms of erosion, primarily by wind, but they don't define the entire process.


Abrasion is the scraping/wearing away, and deflation is the lifting of loose material.


Option (D) provides the most complete definition by including both the breakdown/detachment aspect (Disintegration) and the crucial movement aspect (Transportation).


Step 3: Final Answer:

The process of erosion includes disintegration and transportation.


This corresponds to option (D).
Quick Tip: A simple way to remember the difference:
\textbf{Weathering} makes the mess (breaks the rock down).
\textbf{Erosion} cleans it up (carries the pieces away).

Step 1: Understanding the Question:

We need to arrange the given four minerals in order from hardest to softest, according to the Mohs scale of mineral hardness.


Step 2: Detailed Explanation:

The Mohs scale is a qualitative ordinal scale that characterizes the scratch resistance of various minerals through the ability of a harder material to scratch a softer material.


Let's recall the hardness values for the minerals listed:


- Topaz: Hardness = 8
- Quartz: Hardness = 7
- Orthoclase (a type of feldspar): Hardness = 6
- Fluorite: Hardness = 4


Arranging these in decreasing order of hardness (from hardest to softest) gives:


Topaz (8) \(>\) Quartz (7) \(>\) Orthoclase (6) \(>\) Fluorite (4)


Step 3: Final Answer:

The correct decreasing order of hardness is Topaz \(>\) Quartz \(>\) Orthoclase \(>\) Fluorite.


This corresponds to option (A).
Quick Tip: It is very helpful to memorize the 10 minerals of the Mohs scale for geology questions.
A common mnemonic is: "\textbf{T}all \textbf{G}irls \textbf{C}an \textbf{F}ight \textbf{A}nd \textbf{O}ften \textbf{Q}uit, \textbf{T}hey \textbf{C}ry \textbf{D}iamonds."
1. \textbf{T}alc 2. \textbf{G}ypsum 3. \textbf{C}alcite 4. \textbf{F}luorite 5. \textbf{A}patite 6. \textbf{O}rthoclase 7. \textbf{Q}uartz 8. \textbf{T}opaz 9. \textbf{C}orundum 10. \textbf{D}iamond

Step 1: Understanding the Question:

The question asks to identify the type of geological fold based on the relative age of the rock layers within it.


Step 2: Detailed Explanation:

Geological folds are wavelike structures formed when rock layers are bent or buckled.


The two most basic types of folds are anticlines and synclines:


- Anticline: This is an arch-like fold where the rock layers are bent upwards.


Due to this arch shape, the oldest rock layers are found in the core (the center) of the fold.


As you move outwards from the center, the rock layers become progressively younger.


The description "older rock formation is present inside the fold" and "younger rock formation is present outside" perfectly matches the definition of an anticline.


- Syncline: This is a trough-like fold where the rock layers are bent downwards.


In a syncline, the youngest rock layers are in the core, and the layers get older as you move outwards.


- Chevron fold and Recumbent fold describe the shape and orientation of the fold, not the age relationship of the strata.


A chevron fold has sharp, angular hinges, and a recumbent fold is an anticline or syncline that has been overturned onto its side.


Step 3: Final Answer:

A fold with older rocks in the core and younger rocks on the outside is an anticline.


This corresponds to option (B).
Quick Tip: A simple way to remember:
\textbf{A}nticline = \textbf{A}-shaped arch, with \textbf{A}ncient rocks in the middle.
\textbf{S}yncline = \textbf{S}inks down like a trough, with younger rocks in the middle.

Step 1: Understanding the Question:

The question asks to identify the metamorphic rock that has limestone as its parent rock (protolith).


Step 2: Detailed Explanation:

Metamorphic rocks are formed when existing rocks are changed by heat, pressure, or chemical reactions.


Let's look at the parent rocks for the given options:


- Marble: This is a non-foliated metamorphic rock composed primarily of recrystallized carbonate minerals, most commonly calcite or dolomite.


The parent rock of marble is limestone (composed of calcite) or dolostone (composed of dolomite).


This matches the question.


- Slate: This is a fine-grained, foliated metamorphic rock formed from the low-grade metamorphism of shale or mudstone.


- Schist: This is a medium- to coarse-grained, foliated metamorphic rock.


It forms at a higher grade of metamorphism than slate, often from shale or mudstone as the original parent rock.


- Gneiss: This is a high-grade, foliated metamorphic rock characterized by distinct bands of different minerals.


It can form from various parent rocks, including granite or sedimentary rocks like shale.


Step 3: Final Answer:

The metamorphic rock formed from the recrystallization of limestone is marble.


This corresponds to option (C).
Quick Tip: It's useful to remember some common parent rock \(\rightarrow{}\) metamorphic rock pairs:
- Limestone \(\rightarrow{}\) Marble
- Sandstone \(\rightarrow{}\) Quartzite
- Shale \(\rightarrow{}\) Slate \(\rightarrow{}\) Phyllite \(\rightarrow{}\) Schist \(\rightarrow{}\) Gneiss (this shows increasing metamorphic grade)
- Granite \(\rightarrow{}\) Gneiss

Step 1: Understanding the Question:

The question asks for the specific term for a roof collapse that occurs shortly after supports are removed in a mining operation, like in the goaf of a longwall or depillaring panel.


Step 2: Detailed Explanation:

In underground mining methods where pillars are extracted or longwall panels advance, an area called the "goaf" or "gove" is created where the roof is intentionally allowed to collapse.


This collapse happens in stages:


- Local Fall (or Immediate Fall): This refers to the collapse of the immediate roof strata, which are the rock layers directly above the coal seam.


This fall happens relatively soon after the supports (like hydraulic chocks in a longwall) are advanced or withdrawn.


It involves the rock breaking and falling into the void under its own weight.


- Main Fall: This is a much larger and more significant collapse of the stronger, thicker main roof strata that lie above the immediate roof.


The main fall occurs periodically, after a significant area of the goaf has been created, as the massive rock beds break under the immense stress.


An Air blast is a violent rush of air caused by a large, sudden roof fall (like a main fall) that displaces a huge volume of air.


A Rock burst is a violent, explosive failure of rock under high stress, different from a gravity-induced fall.


The question specifies a fall that takes place "soon after withdrawal of supports," which is the definition of a local fall.


Step 3: Final Answer:

The roof fall that occurs soon after support withdrawal is called a local fall.


This corresponds to option (A).
Quick Tip: Think of the roof as having two parts: a weak lower layer and a strong upper layer.
The \textbf{Local fall} is the weak layer collapsing right away.
The \textbf{Main fall} is the strong layer breaking later, after a large area is exposed.

Step 1: Understanding the Question:

The question asks for the specific term for the violent failure of coal pillars due to the sudden release of stored strain energy.


Step 2: Detailed Explanation:

In deep mines, the immense weight of the overlying rock (overburden) creates very high stresses in the rock mass and in the coal pillars left for support.


Coal, like other rock materials, is elastic and can store a significant amount of strain energy when compressed.


- Coal Bump (or Bump): This is the specific term used in coal mining to describe the sudden, violent failure of a coal pillar or face.


It occurs when the stored elastic strain energy is released explosively, causing the coal to be ejected with great force.


This phenomenon is a major hazard in deep coal mines.


- Rock Burst: This is a more general term used in hard rock mining for the same phenomenon of violent rock failure due to the sudden release of stored strain energy.


While technically similar, "coal bump" is the more specific and appropriate term for the failure of coal pillars.


- Air Blast: This is a secondary effect, a powerful gust of wind caused by the sudden displacement of air from a large rock fall or a bump.


- Premature Collapse: This is a more general term for any structural failure that happens earlier than expected, but it doesn't specifically describe the violent, energy-driven nature of a bump.


Step 3: Final Answer:

The violent burst of coal pillars from the release of stored energy is called a coal bump.


This corresponds to option (C).
Quick Tip: Remember the distinction:
- \textbf{Rock Burst} is the general term for hard rock.
- \textbf{Coal Bump} is the specific term for coal.
Both describe the same physical principle: explosive release of stored strain energy.

Step 1: Understanding the Question:

We need to identify the specific mining term for a roadway that is driven through non-coal rock strata to connect different coal seams.


Step 2: Detailed Explanation:

Let's define the terms for underground roadways:


- Gallery (or Roadway, Heading): A general term for a horizontal or near-horizontal underground passage, typically driven within the coal seam itself.


- Cross-cut: A passage driven between two parallel galleries within the same coal seam, often for ventilation or transport.


- Tunnel: A general civil engineering term for an underground passage.


In mining, it's sometimes used for main access ways from the surface, but a more specific term is usually preferred.


- Cross Measure Drift (or Drivage): This is the specific term for a roadway driven through the "measures" or rock strata that lie between coal seams.


Its purpose is to "cross" these strata to connect workings in one seam to another, either at a different level or in a different area.


The key parts of the definition are "driven through stone" (cross measure) and "to connect two or more coal seams".


Step 3: Final Answer:

An underground roadway driven through rock to connect different coal seams is known as a cross measure drift.


This corresponds to option (D).
Quick Tip: Break down the term "Cross Measure Drift":
- \textbf{Drift} is a horizontal or near-horizontal mine opening.
- \textbf{Measure} refers to the rock strata or layers.
- \textbf{Cross} means it is driven across the strata, not along them (like a roadway in a seam).

Step 1: Understanding the Question:

The question asks for the preferred shape of coal pillars in a room-and-pillar mining system that uses trackless mobile equipment like shuttle cars.


Step 2: Detailed Explanation:

In traditional room-and-pillar mining, pillars were often square or rectangular.


This layout creates sharp, 90-degree turns at intersections.


While this is manageable for conveyor belts or rail-bound systems, it is inefficient and hazardous for large, rubber-tired vehicles like shuttle cars and load-haul-dump (LHD) machines.


These vehicles have a large turning radius and navigating sharp right-angle turns is slow, difficult, and can cause collisions.


To improve the efficiency and safety of these trackless mining systems, pillars are often developed in a rhombus or diamond shape.


This creates intersections where the turns are obtuse (greater than 90 degrees), allowing for smoother, faster, and safer maneuvering of the equipment.


Circular pillars are sometimes used but are much more difficult to create and are not standard for this purpose.


Step 3: Final Answer:

The preferred shape of pillars for shuttle car and locomotive haulage is a rhombus.


This corresponds to option (A).
Quick Tip: Think about driving a car.
It's much easier and faster to navigate a gentle curve or an intersection at an angle greater than 90 degrees than it is to make a sharp 90-degree turn.
The same principle applies to large mining machinery.
Rhombus-shaped pillars facilitate this smoother traffic flow.

Step 1: Understanding the Question:

We need to identify the mining method that is well-suited for extracting coal seams that contain one or more layers of non-coal material, known as "dirt bands" or "partings".


Step 2: Detailed Explanation:

- Board and Pillar method (or Room and Pillar): This method involves driving a series of entries (rooms) into the seam, leaving behind pillars of coal to support the roof.


While it is flexible, selectively mining around or excluding a dirt band within the extraction height can be difficult and inefficient, often leading to dilution (mixing waste rock with coal).


- Longwall method: This is a high-production method where a large block of coal (a "panel") is extracted in a single, continuous operation.


A machine called a shearer cuts coal from a long face (up to 400m).


Modern longwall shearers are very adaptable.


They can be programmed to cut at specific horizons, allowing them to selectively cut the coal sections while leaving the dirt band either in the roof or the floor, if possible.


Even if the band must be cut, it can be handled more systematically than in room and pillar.


This ability to manage in-seam waste makes longwall a suitable method.


- Slicing method: This refers to methods for extracting very thick seams by taking them out in horizontal or inclined "slices".


While it can handle thick seams, it doesn't inherently offer a better solution for thin dirt bands than other methods.


- Hydraulic Mining: This uses high-pressure water jets to break and transport coal.


It is not selective and would mix the coal and dirt band together, making it unsuitable.


Step 3: Final Answer:

The longwall method is highly suitable for extracting coal seams with dirt bands due to the selective cutting capabilities of modern shearers.


This corresponds to option (B).
Quick Tip: Longwall mining is a highly mechanized and systematic method.
This high degree of control allows for "selective mining," where the cutting machine can be precisely positioned to either avoid cutting a waste band or to handle it separately, thus improving the quality of the mined coal.

Step 1: Understanding the Question:

The question asks for the preferred pillar extraction (depillaring) pattern when hydraulic stowing is used to fill the void.


Step 2: Detailed Explanation:

Hydraulic stowing involves filling the goaf (the void left after extraction) with a sand-water slurry to control roof subsidence.


This requires the goaf to be systematically and securely barricaded to contain the slurry.


- Diagonal line of extraction is a common method, but it can create a long, continuous goaf line which is difficult to manage with stowing.


- Step diagonal line of extraction is a modification where pillars are extracted in a staggered or "stepped" pattern.


This method has significant advantages when combined with hydraulic stowing:


1. It breaks the long goaf line into smaller, more manageable sections.


2. It makes it easier to erect strong barricades to contain the sand.


3. It improves overall roof control and safety during the simultaneous process of extraction and filling.


For these reasons, the step diagonal method is the preferred technique when depillaring with hydraulic stowing.


Step 3: Final Answer:

The preferred method of extraction is the step diagonal line.


This corresponds to option (A).
Quick Tip: The choice of extraction pattern is often determined by the method of goaf management.
For caving, a long, straight or diagonal line might be preferred to promote roof fall.
For stowing, a method that creates contained, easily-barricaded sections, like the step diagonal line, is superior.

Step 1: Understanding the Question:

We need to calculate the area of the roof supported by a single coal pillar using the Tributary Area Theory.


Step 2: Key Formula or Approach:

The Tributary Area Theory (TAT) is a simple method for estimating the load on a pillar.


It assumes that each pillar supports the weight of the overlying rock in a "tributary area" that extends halfway to each adjacent pillar.


In a regular grid of pillars, this tributary area is simply the centre-to-centre distance between pillars multiplied in both directions.

\[ Tributary Area = (Centre-to-centre distance)_1 \times (Centre-to-centre distance)_2 \]

Step 3: Detailed Explanation:

The problem gives the following data:

- Centre-to-centre size of pillars = 40 m x 40 m


- Depth = 280 m (This is extra information, not needed for calculating the area).


- Gallery width = 4.0 m (This is also extra information).




According to the Tributary Area Theory, the area of roof supported by one pillar is equal to the square of the centre-to-centre distance.

\[ Area Supported = 40 \, m \times 40 \, m = 1600 \, m^2 \]

Step 4: Final Answer:

The area of roof a pillar supports according to the tributary area theory is 1600 sq.m.


This corresponds to option (C).
Quick Tip: Don't get confused by extra information.
The Tributary Area Theory is very straightforward: the supported area is simply the centre-to-centre pillar spacing squared (for a square grid).
The gallery width would be needed to find the actual size of the pillar itself (Pillar size = Centre distance - Gallery width).
The depth would be needed to calculate the stress on the pillar (Stress = Overburden Pressure * Tributary Area / Pillar Area).

Step 1: Understanding the Question:

The question asks about the consequences of an improperly designed pipeline profile in a hydraulic stowing system.


Step 2: Detailed Explanation:

Hydraulic stowing transports a sand-water slurry.


The hydraulic profile refers to the vertical layout of the pipeline.


An incorrect profile, particularly one with high points (summits) where the pipe rises and then falls, can lead to serious flow problems.


At these high points, if the pressure inside the pipe drops below the vapor pressure of the water, the water can start to boil even at ambient temperature.


This phenomenon is called cavitation, and it leads to the formation of vapor-filled cavities or bubbles in the flow.


These cavities can disrupt the continuous flow of the slurry.


When these bubbles travel to a region of higher pressure downstream, they collapse violently, which can cause severe damage to pipes, create noise and vibration (pulsation), and ultimately lead to flow instabilities that can contribute to the settling of sand and jamming of pipes.


However, the primary hydraulic phenomenon that occurs due to a negative pressure gradient in an incorrect profile is the formation of cavities.


The other options are consequences of this primary issue.


Step 3: Final Answer:

An incorrect hydraulic profile leads to the formation of cavities in the flow.


This corresponds to option (A).
Quick Tip: While pipe jamming is a major operational problem in stowing, its root cause from a fluid dynamics perspective in an incorrect profile is often cavitation.
The pressure dropping to near-vacuum at high points in the pipeline creates vapor pockets, disrupting the flow and leading to other problems.

Step 1: Understanding the Question:

We need to identify a key factor that determines the width of a gallery or heading in a modern Board and Pillar mine.


Step 2: Detailed Explanation:

While several factors influence the final design of a mine layout, the width of the headings is constrained by both geotechnical stability and operational requirements.


- Depth of working and rock strength determine the maximum \textit{safe width that can be excavated without excessive support.


- Size of pillars is related to the gallery width through the extraction ratio, but doesn't directly determine the width.


- Ventilation required can be met by adjusting both width and height to achieve a target cross-sectional area.


- Face machinery used: In modern mechanized Board and Pillar mining, the dimensions of the equipment are a critical and often primary constraint.


The heading must be wide enough to allow the cutting machine (e.g., a Continuous Miner) to operate effectively and to provide sufficient clearance for loading and hauling equipment (e.g., Shuttle Cars) to maneuver.


The size of the machinery often dictates the minimum practical width of the heading, which must then be checked for geotechnical stability.


Therefore, the choice of face machinery is a direct and major determinant of the heading width.


Step 3: Final Answer:

The width of the heading is heavily dependent on the face machinery used.


This corresponds to option (C).
Quick Tip: In modern mining, the system is designed around the equipment.
The mine layout (like gallery width and intersection angles) must accommodate the machinery that will be used.
This operational constraint is often considered first, and then the design is verified for safety and stability based on rock mechanics principles.

Step 1: Understanding the Question:

We need to calculate the powder factor for a blasting operation in a coal heading, given the dimensions of the heading, the advance per blast, and the amount of explosive used.


Step 2: Key Formula or Approach:

Powder Factor (PF) is a measure of the efficiency of explosive use.


It is defined as the mass of rock or coal broken per unit mass of explosive used.

\[ Powder Factor = \frac{Mass of coal broken (in tonnes)}{Mass of explosive used (in kg)} \]
First, we need to calculate the volume of coal broken, and then its mass.


Step 3: Detailed Explanation:

1. Calculate the volume of coal broken per blast:

Volume = Width \( \times \) Height \( \times \) Advance

\[ Volume = 4 \, m \times 2.5 \, m \times 1 \, m = 10 \, m^3 \]

2. Calculate the mass of coal broken:

The specific gravity of coal is given as 1.2 t/m\(^3\).


This is the density of the coal.


Mass = Volume \( \times \) Density

\[ Mass = 10 \, m^3 \times 1.2 \, t/m^3 = 12 \, tonnes (te) \]

3. Calculate the Powder Factor:

Mass of explosive used = 5 kg

\[ Powder Factor = \frac{12 \, te}{5 \, kg} = 2.4 \, te/kg \]

Step 4: Final Answer:

The powder factor is 2.40 te/kg.


This corresponds to option (B).
Quick Tip: Pay close attention to the units.
The desired unit for the powder factor is tonnes/kg.
Ensure your mass of coal is in tonnes and your mass of explosive is in kg before dividing.
Also, be aware that "specific gravity" given in units of t/m\(^3\) is effectively the density.

Step 1: Understanding the Question:

The question asks where the main development openings (like haulage drives) are typically located in relation to a dipping orebody in a metal mine.


Step 2: Detailed Explanation:

In mining steeply dipping orebodies, the rock mass above the ore is called the hanging wall, and the rock mass below it is called the footwall.


- The hanging wall becomes unsupported as the ore is mined out from beneath it.


It is subjected to tensile and shear stresses and is generally considered unstable and prone to failure or collapse into the mined-out area (stope).


- The footwall, on the other hand, remains as the base of the workings.


It is generally under compression and is much more stable and competent than the hanging wall.


For the long-term security and longevity of essential mine infrastructure like main haulage levels, shafts, and ramps, it is standard and safe practice to locate them within the stable footwall rock, away from the zone of instability created by mining.


Step 3: Final Answer:

Most of the main development in metal mining is carried out in the footwall.


This corresponds to option (B).
Quick Tip: Think of it like building a house on a hillside.
You would build the foundation on the solid ground below the slope (the footwall), not on the unstable ground above the slope that could slide down (the hanging wall).

Step 1: Understanding the Question:

The question asks for the specific name of a pillar that is left between the top of a stope and the mining level above it.


Step 2: Detailed Explanation:

In underground metal mining, different types of pillars are left for support:


- Crown Pillar: This is the pillar of rock or ore left above a stope to provide support to the floor of the level immediately above it.


It forms the "crown" of the stope.


This directly matches the description in the question.


- Sill Pillar: This is the pillar of rock or ore left at the \textit{bottom of a stope, forming its floor or "sill" above the level below.


- Rib Pillar: These are pillars left between adjacent stopes to support the side walls.


- Barrier Pillar: These are very large pillars left to separate major mining panels or to protect main roadways and shafts from the stresses of extraction areas.


Step 3: Final Answer:

The pillar left to protect the upper level is known as a crown pillar.


This corresponds to option (A).
Quick Tip: Remember the location relative to the stope (the excavated room):
- \textbf{Crown is at the top (like a crown on a head).
- \textbf{Sill} is at the bottom (like a window sill).
- \textbf{Rib} is on the side (like a rib cage).

Step 1: Understanding the Question:

We need to identify which mining method is unsafe to use in an orebody that is prone to spontaneous heating (self-ignition when exposed to air).


Step 2: Detailed Explanation:

Spontaneous heating is a major hazard, especially in sulphide ore bodies and some types of coal.


It requires a large surface area of broken material, a supply of oxygen (air), and sufficient time for the slow oxidation reaction to build up heat to the point of ignition.


A suitable mining method must prevent these conditions.


Let's analyze the methods:


- Sub level Caving, Cut and fill, Room and Pillar: In these methods, the broken ore is removed relatively quickly from the working area.


Ventilation can be controlled, and there is less long-term storage of broken ore in the stope.


- Shrinkage Stoping: This method is unique because it involves breaking the ore and leaving most of it inside the stope to act as a working platform for the miners and to support the stope walls.


The broken ore is only drawn out from the bottom after the entire stope has been mined out.


This means a large volume of broken ore (with a huge surface area) remains in the stope, permeable to air, for a very long period.


This creates the perfect conditions for spontaneous heating to occur and develop into an uncontrollable underground fire.


Therefore, shrinkage stoping is extremely dangerous and unsuitable for ores prone to spontaneous heating.


Step 3: Final Answer:

Shrinkage Stoping is not suitable for an ore body subjected to spontaneous heating.


This corresponds to option (C).
Quick Tip: The key feature of Shrinkage Stoping is using the broken ore itself as a stockpile and working platform inside the mine.
Anytime you have large amounts of broken, reactive material sitting for a long time with air access, you have a risk of spontaneous combustion.

Step 1: Understanding the Question:

The question asks for the mining term for a vertical or inclined connection between two levels that is excavated in a downward direction.


Step 2: Detailed Explanation:

In underground mining, connections between different horizontal levels are essential for ventilation, transport of ore, and access for workers.


The terminology depends on the direction of excavation:


- Raise: A vertical or inclined opening driven upwards from a lower level to connect to an upper level.


- Winze: A vertical or inclined opening driven downwards from an upper level to connect to a lower level.


The question explicitly states the connection is made "from upper level to lower level," which means it is being driven downwards.


Therefore, the correct term is a winze.


Cross cuts and cross measure drifts are horizontal openings.


Step 3: Final Answer:

The connection driven from an upper level to a lower level is known as a winze.


This corresponds to option (A).
Quick Tip: A simple way to remember:
- You \textbf{Raise} your hands \textbf{up}. A raise is driven upwards.
- You might \textbf{Wince} if you look \textbf{down} a deep hole. A winze is driven downwards.

Step 1: Understanding the Question:

We need to select the most appropriate mining method for an orebody with a specific set of characteristics: it is massive (very large in all three dimensions), the ore itself is weak, and the wall rock can be either weak or strong.


Step 2: Detailed Explanation:

Let's evaluate the suitability of each method based on the given conditions:


- Sublevel Stoping & Shrinkage Stoping: These are "open stoping" methods.


They require the ore and the wall rocks to be strong because large empty voids (stopes) are created.


They are completely unsuitable for weak ore that would collapse uncontrollably.


- Cut and fill method: This method is used for orebodies with weak wall rocks, because artificial fill is used to support the walls as mining progresses.


However, it requires the ore to be at least moderately strong to form a stable roof (the "back") under which miners can work.


It is not ideal for very weak, massive orebodies.


- Block Caving: This method is specifically designed for large, massive, low-grade orebodies where the ore is weak and prone to collapse.


The method works by undercutting a large "block" of ore, which removes its support.


The weak ore then collapses under its own weight in a controlled manner.


The broken ore is then extracted from drawpoints below the collapsed block.


This method perfectly matches the given conditions: massive orebody and weak ore.


Step 3: Final Answer:

The method suitable for a massive orebody with weak ore is Block Caving.


This corresponds to option (D).
Quick Tip: Caving methods are the only methods that take advantage of weak ore and rock.
If you see "weak ore" and "massive" in the description, your first thought should be a caving method like Block Caving or Sublevel Caving.
Since Block Caving is the primary method for weak ore, it's the correct choice.

Step 1: Understanding the Question:

We need to identify a mining method that is specifically suited for extracting narrow vein deposits.


Step 2: Detailed Explanation:

Narrow vein deposits pose a challenge because it's difficult to mine only the valuable ore without also mining the worthless wall rock, a problem called dilution.


- Sub level caving and Room and Pillar are generally used for much wider, more massive or tabular orebodies.


- Cut and fill method is very common for vein deposits of various widths and is known for being selective, but there is an even more specialized method for very narrow veins.


- Resuing Stoping: This is a highly selective mining method developed specifically for very narrow, often high-grade, veins.


The process involves two separate steps in each cycle:


1. First, a slice of the waste wall rock adjacent to the vein is blasted and removed (or used as fill).


2. Second, the now-exposed narrow vein of clean ore is blasted and collected separately.


This two-stage process minimizes dilution, which is critical when the vein is very narrow and valuable.


Step 3: Final Answer:

Resuing is a specialized method used for the extraction of narrow vein deposits.


This corresponds to option (B).
Quick Tip: The name "Resuing" is unique and specifically associated with narrow vein mining.
If you see it as an option for a question about narrow veins, it is very likely the correct answer.
It's the ultimate method for minimizing dilution when the ore is only a few inches or centimeters wide.

Step 1: Understanding the Question:

The question asks for the general classification of the rill stoping method.


Step 2: Detailed Explanation:

Mining methods can be classified based on the direction of extraction relative to the haulage level.


- Overhand (or Overhead) Stoping: In this class of methods, mining starts from a lower level and progresses upwards.


Miners typically work on top of broken ore or fill, and the ore is passed down to the lower level for transport.


- Underhand Stoping: In these methods, mining starts from an upper level and progresses downwards.


Miners work on solid ground, and the broken ore falls to the bottom of the stope.


- Breast Stoping: In this method, mining proceeds horizontally into a face, similar to room and pillar workings.


Rill stoping is a method where the stope is excavated in such a way that the face is maintained at the angle of repose of the broken ore.


The working face advances upwards and outwards from a central raise.


Since the overall direction of extraction is upwards from the haulage level, it is classified as a type of overhand stoping.


Step 3: Final Answer:

Rill stoping is a form of overhand stoping.


This corresponds to option (A).
Quick Tip: The key to classifying stoping methods is the vertical direction of progress.
If you're mining up, it's overhand.
If you're mining down, it's underhand.
Rill stoping involves working upwards, so it's overhand.

Step 1: Understanding the Question:

We need to select the best mining method for an orebody that is thin, flat, and has strong ore and strong wall rocks.


Step 2: Detailed Explanation:

Let's analyze the conditions:

- Flat Dip, Thin Orebody: This describes a tabular, sheet-like deposit.


- Strong Ore and Strong Walls: This means the rock mass is very competent and can support itself over significant spans without collapsing.


Now let's evaluate the methods:

- Sublevel stoping, Sub level Caving, Top slicing: These methods are generally used for thick and/or steeply dipping orebodies. They are not suitable for thin, flat deposits.


- Room and Pillar: This method is ideal for flat to gently dipping, tabular deposits.


It involves excavating "rooms" and leaving behind "pillars" of ore to support the roof.


The conditions of strong ore and strong walls are perfect for this method, as they allow for stable pillars and wide rooms to be created with minimal artificial support, leading to a safe and economical operation.


Step 3: Final Answer:

The preferred method for a thin, flat orebody with strong ore and walls is the Room and pillar method.


This corresponds to option (C).
Quick Tip: The combination of "flat dip" and "strong rock" is the classic indicator for the Room and Pillar method.
This method is used worldwide for mining coal, salt, potash, and limestone deposits that fit this description.

Step 1: Understanding the Question:

We need to classify the sub-level stoping method into one of the major categories of underground mining methods.


Step 2: Detailed Explanation:

Underground mining methods are broadly classified based on how the void created by ore extraction is supported:


1. Artificially Supported Methods: The walls of the stope are supported using materials brought into the mine, such as waste rock, sand, or cement (e.g., Cut-and-Fill stoping).


2. Supported by Ore (Pillars): The roof is supported by pillars of the ore itself, which are left unmined (e.g., Room and Pillar, Shrinkage Stoping).


3. Caving Methods: The ore and/or the surrounding wall rock is intentionally collapsed in a controlled manner (e.g., Block Caving, Sub-level Caving).


4. Open Stoping Methods: These methods are used in orebodies where both the ore and the surrounding wall rocks are strong and competent.


Large underground openings (stopes) are created and left empty (open) after the ore is removed, relying on the natural strength of the rock mass for stability.


Sub-level stoping is a classic example of an open stoping method.


It involves creating very large, open stopes, and it can only be applied in orebodies with strong ore and strong host rock that will not collapse once the ore is removed.


Step 3: Final Answer:

Sub-level stoping belongs to the open stoping type.


This corresponds to option (D).
Quick Tip: The key to open stoping methods is "strong rock."
If the description of a method involves creating large empty rooms without backfilling and without intentional collapse, it's an open stoping method.

Step 1: Understanding the Question:

We need to identify the specific mining term for a horizontal roadway that connects the main shaft to the orebody on a particular level.


Step 2: Detailed Explanation:

In underground mine development, a main vertical or inclined shaft provides access to different depths or "levels".


On each level, a network of horizontal roadways is created.


- A shaft station or plat is excavated around the shaft on each level to serve as a hub for operations.


- From the shaft station, a primary horizontal roadway is driven through the barren country rock to reach the mineral deposit or orebody.


This specific roadway is called a cross-cut because it "cuts across" the rock strata to get to the ore.


- Once the cross-cut intersects the orebody, other roadways called drifts or drives are then developed along the length of the orebody.


- A cross measure drift is a term more commonly used in coal mining to connect different seams.


In metal mining, cross-cut is the standard term.


Step 3: Final Answer:

The roadway driven to interconnect the shaft and the ore body is known as a cross-cut.


This corresponds to option (B).
Quick Tip: Think of the development sequence:
1. \textbf{Shaft} (vertical access).
2. \textbf{Cross-cut} (horizontal access from shaft to ore).
3. \textbf{Drift} (horizontal roadway along the ore).

Step 1: Understanding the Question:

We need to find the ratio of aerodynamic resistance of two mine roadways that are identical except for their lengths.


Step 2: Key Formula or Approach:

The aerodynamic resistance (R) of a mine airway is given by Atkinson's formula: \[ R = \frac{k L P}{A^3} \]
where:
- \(k\) = friction factor (depends on surface characteristics)
- \(L\) = length of the roadway
- \(P\) = perimeter of the roadway's cross-section
- \(A\) = cross-sectional area of the roadway


Step 3: Detailed Explanation:

The problem states that the two roadways have:
- Similar cross-sectional areas (\(A_A = A_B\)).
- Similar surface characteristics (\(k_A = k_B\)).
- Similar area implies a similar perimeter (\(P_A = P_B\)).


This means that all the terms in the resistance formula are constant except for the length (L).


Therefore, the resistance (R) is directly proportional to the length (L).

\[ R \propto L \]
We can find the ratio of the resistances by taking the ratio of the lengths: \[ \frac{R_A}{R_B} = \frac{L_A}{L_B} \]
Given \( L_A = 100 \, m \) and \( L_B = 200 \, m \).

\[ \frac{R_A}{R_B} = \frac{100}{200} = \frac{1}{2} \]
The ratio R\(_A\): R\(_B\) is 1 : 2.


Step 4: Final Answer:

The ratio of their resistances R\(_A\): R\(_B\) is 1 : 2.


This corresponds to option (A).
Quick Tip: For airways with the same shape, size, and lining, the resistance is simply proportional to the length.
If you double the length, you double the resistance.

Step 1: Understanding the Question:

We need to find the ratio of air quantities flowing through three parallel airways that are identical except for their lengths.


Step 2: Key Formula or Approach:

1. For airways in parallel, the pressure drop (H) across each branch is the same.


2. The fundamental ventilation equation is \( H = R Q^2 \), where H is pressure, R is resistance, and Q is quantity.


3. From the previous question, we know that for these airways, resistance is directly proportional to length (\( R \propto L \)).


Step 3: Detailed Explanation:

Since the pressure drop is the same for all three roadways: \[ H = R_A Q_A^2 = R_B Q_B^2 = R_C Q_C^2 \]
This means that \( Q^2 \) is inversely proportional to R: \[ Q^2 \propto \frac{1}{R} \]
Taking the square root, we find that the quantity Q is inversely proportional to the square root of the resistance: \[ Q \propto \frac{1}{\sqrt{R}} \]
Since resistance is proportional to length (\( R \propto L \)), we can say: \[ Q \propto \frac{1}{\sqrt{L}} \]
Now we can write the ratio of the quantities: \[ Q_A : Q_B : Q_C = \frac{1}{\sqrt{L_A}} : \frac{1}{\sqrt{L_B}} : \frac{1}{\sqrt{L_C}} \]
Substitute the given lengths \( L_A=100 \), \( L_B=200 \), \( L_C=300 \): \[ Q_A : Q_B : Q_C = \frac{1}{\sqrt{100}} : \frac{1}{\sqrt{200}} : \frac{1}{\sqrt{300}} \] \[ Q_A : Q_B : Q_C = \frac{1}{\sqrt{100 \times 1}} : \frac{1}{\sqrt{100 \times 2}} : \frac{1}{\sqrt{100 \times 3}} \] \[ Q_A : Q_B : Q_C = \frac{1}{10\sqrt{1}} : \frac{1}{10\sqrt{2}} : \frac{1}{10\sqrt{3}} \]
To simplify the ratio, we can multiply all parts by 10: \[ Q_A : Q_B : Q_C = 1 : \frac{1}{\sqrt{2}} : \frac{1}{\sqrt{3}} \]

Step 4: Final Answer:

The ratio of the quantities is 1 : \( 1/\sqrt{2} \) : \( 1/\sqrt{3} \).


This corresponds to option (D).
Quick Tip: Remember the key rules for parallel and series circuits in ventilation:
- \textbf{Series:} Quantities are equal (\(Q_T = Q_A = Q_B\)), Resistances add up (\(R_T = R_A + R_B\)).
- \textbf{Parallel:} Pressures are equal (\(H_A = H_B\)), Quantities add up (\(Q_T = Q_A + Q_B\)).
For parallel flow, air will always prefer the path of least resistance, so the shortest airway gets the most air.

Step 1: Understanding the Question:

We need to identify the type of centrifugal fan that has a "non-overloading" power characteristic.


Step 2: Detailed Explanation:

The power characteristic of a fan is the relationship between the power it consumes and the volume of air it moves.


- Overloading Characteristic: This means the power consumed by the fan motor continuously increases as the airflow quantity increases (which happens when the mine resistance decreases).


If the resistance becomes very low, the fan will try to move a huge amount of air, and the power demand can exceed the motor's rating, causing it to "overload" and potentially burn out.


Forward-bladed and Radial-bladed centrifugal fans typically exhibit this overloading characteristic.


- Non-overloading Characteristic: This means the power consumption curve reaches a peak at some airflow quantity and then decreases as the quantity increases further.


This is a very desirable safety feature.


It ensures that no matter how low the mine resistance becomes, the power required by the fan will never exceed a certain maximum value, which can be matched to the motor's rating.


Backward-bladed and backward-curved centrifugal fans are known for this self-limiting, non-overloading power characteristic.


Step 3: Final Answer:

The backward-bladed centrifugal fan has non-overloading power characteristics.


This corresponds to option (C).
Quick Tip: Remember the key safety feature:
- Forward blades = Power keeps rising = Overloading (bad).
- Backward blades = Power peaks and then drops = Non-overloading (good).
This is a major reason why backward-bladed fans are very common in main mine ventilation systems.