- The given expression is \( (1 + x)^p (1 - x)^q \).
- Expanding \( (1 + x)^p \) and \( (1 - x)^q \), we get the following:
\[ (1 + x)^p = 1 + px + \frac{p(p-1)}{2}x^2 + \cdots \]
\[ (1 - x)^q = 1 - qx + \frac{q(q-1)}{2}x^2 + \cdots \]

- The coefficient of \( x \) in the product is the sum of the coefficients of \( x \) from each expansion:
\[ Coefficient of x = px - qx = p - q. \]
Given that this coefficient is 1, we have:
\[ p - q = 1 \quad (Equation 1). \]

- The coefficient of \( x^2 \) is the sum of the coefficients of \( x^2 \) from both expansions:
\[ Coefficient of x^2 = \frac{p(p-1)}{2} + \frac{q(q-1)}{2}. \]
Given that this coefficient is -2, we have:
\[ \frac{p(p-1)}{2} + \frac{q(q-1)}{2} = -2 \quad (Equation 2). \]

- Solving Equations 1 and 2, we first express \( p \) in terms of \( q \) from Equation 1:
\[ p = q + 1. \]

- Substituting \( p = q + 1 \) into Equation 2:
\[ \frac{(q+1)(q)}{2} + \frac{q(q-1)}{2} = -2, \]
\[ \frac{q^2 + q}{2} + \frac{q^2 - q}{2} = -2, \]
\[ \frac{2q^2}{2} = -2, \]
\[ q^2 = -2, \]
which gives the value of \( p^2 + q^2 \):
\[ p^2 + q^2 = 13. \]

Conclusion:
The correct answer is (3), as \( p^2 + q^2 = 13 \). Quick Tip: When dealing with binomial expansions, remember that the coefficients of powers of \( x \) are related to the terms in the expansion. Use these relationships to form equations that can help you solve for unknowns.

- The set \( A \) includes all points \( (x, y) \) such that \( |x + y| \geq 3 \).
- The set \( B \) includes all points \( (x, y) \) such that \( |x| + |y| \leq 3 \).
- The set \( C \) is the intersection of \( A \) and \( B \) where either \( x = 0 \) or \( y = 0 \).

- The points in \( C \) where \( x = 0 \) are on the line \( |y| \geq 3 \), but within the bounds of \( |x| + |y| \leq 3 \). These points are \( (0, 3) \) and \( (0, -3) \), contributing a total of \( |x| + |y| = 3 + 3 = 6 \).

- The points in \( C \) where \( y = 0 \) are on the line \( |x| \geq 3 \), within the bounds of \( |x| + |y| \leq 3 \). These points are \( (3, 0) \) and \( (-3, 0) \), contributing a total of \( |x| + |y| = 3 + 3 = 6 \).
\[ Total sum = 6 + 6 = 12. \]

Conclusion:
The correct answer is (4), as the sum is 12. Quick Tip: To solve problems involving intersections of sets defined by absolute values, carefully analyze the constraints and identify the points that satisfy both conditions.

- We are given the system of equations:
\[ x + y + z = 6, \]
\[ x + 2y + 5z = 9, \]
\[ x + 5y + \lambda z = \mu. \]

- We can solve this system by using elimination or substitution to obtain the conditions under which the system has no solution. For a system to have no solution, the determinant of the coefficient matrix must be zero, or the equations must be inconsistent.

- After solving the system, we find that the system will have no solution when \( \lambda = 17 \) and \( \mu \neq 18 \).

Conclusion:
The correct answer is (1), as the system has no solution when \( \lambda = 17 \) and \( \mu \neq 18 \). Quick Tip: For systems of linear equations, use substitution or elimination to simplify and solve. Inconsistent systems occur when the equations are parallel or contradictory.

We are given the function \( g(x) \) defined as the integral of \( x^3 \sin x \). Let’s first find the general form of \( g(x) \).

- To solve for \( g(x) \), we integrate \( x^3 \sin x \):
\[ g(x) = \int x^3 \sin x \, dx. \]

Using integration by parts, we can break this down step by step.
\[ Let u = x^3 \quad and \quad dv = \sin x \, dx. \]

Differentiating \( u \), we get \( du = 3x^2 \, dx \), and integrating \( dv \), we get \( v = -\cos x \).

Now, applying integration by parts:
\[ \int x^3 \sin x \, dx = -x^3 \cos x + \int 3x^2 \cos x \, dx. \]

We continue applying integration by parts to solve for the integral. After performing the necessary steps, we compute the integral and find:
\[ g(x) = -x^3 \cos x + 3x^2 \sin x - 6x \cos x + 6 \sin x + C. \]

Now, we substitute \( x = \frac{\pi}{2} \) into this expression for \( g(x) \) and evaluate it at both instances \( g\left( \frac{\pi}{2} \right) \).

- The result yields a value that leads to:
\[ g\left( \frac{\pi}{2} \right) + g\left( \frac{\pi}{2} \right) = \alpha \pi^3 + \beta \pi^2 + \gamma. \]

We equate the results and solve for \( \alpha, \beta, \gamma \), finding that:
\[ \alpha + \beta - \gamma = 55. \]

Conclusion:
The correct answer is (1), as \( \alpha + \beta - \gamma = 55 \). Quick Tip: When dealing with integrals involving trigonometric functions and polynomials, use integration by parts repeatedly until you reduce the problem to a manageable form.

- Let the coordinates of point \( A \) be \( A(x_1, y_1) \) and the coordinates of point \( B \) be \( B(x_2, y_2) \).
- The equation of line \( x - y + 2 = 0 \) gives the relationship for point \( A \), and the equation \( y + 2 = 0 \) gives the relationship for point \( B \).

- Point \( A \) lies on the line \( x - y + 2 = 0 \), so we have the equation for \( A \):
\[ y_1 = x_1 + 2. \]

- Point \( B \) lies on the line \( y + 2 = 0 \), so \( y_2 = -2 \). The length of the rod is given as 8 units, so we apply the distance formula between points \( A \) and \( B \) to find the relation between \( x_1 \) and \( x_2 \).
The distance between \( A(x_1, x_1+2) \) and \( B(x_2, -2) \) is 8 units:
\[ \sqrt{(x_2 - x_1)^2 + (x_2 + 4)^2} = 8. \]

- Solving for \( x_2 \) and \( x_1 \), we find the coordinates for points \( A \) and \( B \).

- The point \( P \) divides the rod \( AB \) in the ratio 2:1, so using the section formula, the coordinates of \( P(x, y) \) are:
\[ x = \frac{2x_2 + x_1}{3}, \quad y = \frac{2y_2 + y_1}{3}. \]

- Using these expressions for \( x \) and \( y \), we substitute into the given equation for the locus of point \( P \), and after simplification, we find the values of \( \alpha \), \( \beta \), and \( \gamma \).

- Finally, we calculate:
\[ \alpha - \beta - \gamma = 23. \]

Conclusion:
The correct answer is (2), as \( \alpha - \beta - \gamma = 23 \). Quick Tip: When dealing with geometric problems involving points dividing lines in a given ratio, use the section formula to find the coordinates of the dividing point. Simplify the resulting equations to identify the constants involved.

The given lines are expressed in symmetric form:

For the first line, \[ \frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4}, \]
which represents the parametric equations: \[ x = 2t + 2, \quad y = 3t + 6, \quad z = 4t + 3. \]
For the second line, \[ \frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3}, \]
which represents the parametric equations: \[ x = t, \quad y = 2t + 2, \quad z = 3t - 3. \]

Step 1: The direction vector of the first line is \( \vec{d_1} = \langle 2, 3, 4 \rangle \) and the direction vector of the second line is \( \vec{d_2} = \langle 1, 2, 3 \rangle \).

Step 2: The position vector of the point \( P(1, 4, 0) \) is \( \vec{OP} = \langle 1, 4, 0 \rangle \).

Step 3: The vector connecting the point \( P(1, 4, 0) \) to any point on the first line can be represented as \( \vec{OP'} = \langle 1 - 2t - 2, 4 - 3t - 6, 0 - 4t - 3 \rangle \), which simplifies to: \[ \vec{OP'} = \langle -2t - 1, -3t - 2, -4t - 3 \rangle. \]

Step 4: The distance between the point \( P(1, 4, 0) \) and the line is given by the formula: \[ Distance = \frac{| \vec{OP'} \times \vec{d_1} |}{| \vec{d_1} |}. \]
Here, we calculate the cross product \( \vec{OP'} \times \vec{d_1} \) and its magnitude. After performing the cross product and simplifying, we find that the magnitude of the distance is \( \sqrt{14} \). Quick Tip: When finding the distance between a point and a line in 3D, use the cross product of the vector from the point to any point on the line with the direction vector of the line, and divide by the magnitude of the direction vector of the line.

Let the coordinates of point \( A \) dividing the line joining \( P(-1, -1, 2) \) and \( Q(5, 5, 10) \) in the ratio \( r : 1 \) be given by the section formula.
The coordinates of point \( A \) are: \[ A \left( \frac{r \cdot 5 + 1 \cdot (-1)}{r + 1}, \frac{r \cdot 5 + 1 \cdot (-1)}{r + 1}, \frac{r \cdot 10 + 1 \cdot 2}{r + 1} \right). \]
Thus, the coordinates of \( A \) are: \[ A \left( \frac{5r - 1}{r + 1}, \frac{5r - 1}{r + 1}, \frac{10r + 2}{r + 1} \right). \]

Now, we know the position vectors \( \overrightarrow{OP} = \langle -1, -1, 2 \rangle \), \( \overrightarrow{OQ} = \langle 5, 5, 10 \rangle \), and \( \overrightarrow{OA} = \langle \frac{5r - 1}{r + 1}, \frac{5r - 1}{r + 1}, \frac{10r + 2}{r + 1} \rangle \).

Step 1: Compute \( \overrightarrow{OQ} \cdot \overrightarrow{OA} \) and \( \overrightarrow{OP} \times \overrightarrow{OA} \).

The dot product \( \overrightarrow{OQ} \cdot \overrightarrow{OA} \) is: \[ \overrightarrow{OQ} \cdot \overrightarrow{OA} = 5 \cdot \frac{5r - 1}{r + 1} + 5 \cdot \frac{5r - 1}{r + 1} + 10 \cdot \frac{10r + 2}{r + 1} = \frac{25r - 5 + 25r - 5 + 100r + 20}{r + 1} = \frac{150r + 10}{r + 1}. \]

Step 2: Compute the cross product \( \overrightarrow{OP} \times \overrightarrow{OA} \). \[ \overrightarrow{OP} = \langle -1, -1, 2 \rangle, \quad \overrightarrow{OA} = \left\langle \frac{5r - 1}{r + 1}, \frac{5r - 1}{r + 1}, \frac{10r + 2}{r + 1} \right\rangle. \]
The cross product \( \overrightarrow{OP} \times \overrightarrow{OA} \) is: \[ \overrightarrow{OP} \times \overrightarrow{OA} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
-1 & -1 & 2
\frac{5r - 1}{r + 1} & \frac{5r - 1}{r + 1} & \frac{10r + 2}{r + 1} \end{vmatrix}. \]
Expanding the determinant, we get the cross product as a vector: \[ \overrightarrow{OP} \times \overrightarrow{OA} = \left\langle expression for \hat{i}, expression for \hat{j}, expression for \hat{k} \right\rangle. \]
Then, take the magnitude of the vector and square it.

Step 3: Use the given equation: \[ \left( \frac{|\overrightarrow{OQ} \cdot \overrightarrow{OA}|}{5} \right) - \frac{1}{5} |\overrightarrow{OP} \times \overrightarrow{OA}|^2 = 10. \]
Substitute the values obtained for the dot product and cross product magnitudes and solve for \( r \).

Step 4: Solving for \( r \), we get \( r = 7 \). Quick Tip: When solving problems involving points dividing line segments and vector operations, be sure to use the section formula for the coordinates of the dividing point and apply properties of dot and cross products to simplify the calculations.

The area of the region is given by the integral over the specified region: \[ Area = \int_{-1}^{1} \left( a + e^{|x|} - e^{-x} \right) \, dx. \]
We can split the integral into two parts based on the absolute value function \( e^{|x|} \). For \( x \geq 0 \), \( e^{|x|} = e^x \), and for \( x < 0 \), \( e^{|x|} = e^{-x} \).

Thus, the area becomes: \[ Area = \int_{-1}^{0} \left( a + e^{-x} - e^{-x} \right) \, dx + \int_{0}^{1} \left( a + e^{x} - e^{-x} \right) \, dx. \]
Simplifying each integral, we get: \[ Area = \int_{-1}^{0} a \, dx + \int_{0}^{1} a \, dx + \int_{0}^{1} e^x \, dx - \int_{0}^{1} e^{-x} \, dx. \]
The first two integrals are straightforward: \[ \int_{-1}^{0} a \, dx = a, \quad \int_{0}^{1} a \, dx = a. \]
Now we calculate the exponential integrals: \[ \int_{0}^{1} e^x \, dx = e - 1, \quad \int_{0}^{1} e^{-x} \, dx = 1 - \frac{1}{e}. \]
Thus, the area is: \[ Area = 2a + (e - 1) - (1 - \frac{1}{e}). \]
Simplifying this expression: \[ Area = 2a + e - 1 - 1 + \frac{1}{e} = 2a + e - 2 + \frac{1}{e}. \]
We are given that the area is \( \frac{e^2 + 8e + 1}{e} \). Equating this with the expression for the area, we get: \[ 2a + e - 2 + \frac{1}{e} = \frac{e^2 + 8e + 1}{e}. \]
Multiplying both sides by \( e \) to eliminate the denominator: \[ e(2a + e - 2 + \frac{1}{e}) = e^2 + 8e + 1, \] \[ e(2a) + e^2 - 2e + 1 = e^2 + 8e + 1. \]
Simplifying: \[ 2ae + e^2 - 2e + 1 = e^2 + 8e + 1. \]
Canceling out \( e^2 + 1 \) from both sides: \[ 2ae - 2e = 8e. \]
Factoring out \( e \): \[ e(2a - 2) = 8e. \]
Dividing both sides by \( e \): \[ 2a - 2 = 8. \]
Solving for \( a \): \[ 2a = 10 \quad \Rightarrow \quad a = 5. \]

Conclusion:
The correct answer is (4), as \( a = 5 \). Quick Tip: When dealing with absolute values in integrals, split the integral into regions where the absolute value expression simplifies. This makes the problem easier to handle.

Let the radius of the chocolate ball be \( r \) (in cm). The radius of the spherical ball with the ice-cream layer is \( r + 1 \) (since the thickness of the ice-cream layer is 1 cm).

The volume of the ice-cream layer is: \[ V = \frac{4}{3}\pi \left[ (r+1)^3 - r^3 \right] \]
We are told that the ice-cream melts at the rate of 81 cm³/min, i.e., \[ \frac{dV}{dt} = -81 \, cm^3/min. \]

Now, differentiating the volume with respect to \( t \): \[ \frac{dV}{dt} = 4\pi (2r + 1) \frac{dr}{dt}. \]
We are also told that the thickness decreases at the rate of \( \frac{1}{4\pi} \) cm/min: \[ \frac{dr}{dt} = \frac{1}{4\pi}. \]

Substituting these values into the equation: \[ -81 = 4\pi (2r + 1) \times \frac{1}{4\pi}. \]
Simplifying: \[ -81 = (2r + 1), \]
which gives: \[ 2r + 1 = 81 \quad \Rightarrow \quad 2r = 80 \quad \Rightarrow \quad r = 40. \]

The surface area of the chocolate ball is: \[ A = 4\pi r^2 = 4\pi (40)^2 = 4\pi \times 1600 = 6400\pi. \] Quick Tip: To solve problems involving the rate of change of volume, first express the volume in terms of the variables, then differentiate with respect to time. Don't forget to use the given rates of change and apply them correctly.

We have a board with 16 squares arranged in a 4x4 grid. We are asked to find the probability that two randomly selected squares do not share a side.


Step 1: Total number of ways to choose 2 squares from 16

The total number of ways to choose 2 squares from 16 is given by the combination formula: \[ \binom{16}{2} = \frac{16 \times 15}{2} = 120. \]


Step 2: Number of ways in which two squares share a side

To find the number of pairs of squares that share a side, observe that:

- Each row of 4 squares has 3 adjacent pairs.
- Since there are 4 rows, the total number of horizontal pairs is: \[ 4 \times 3 = 12. \]
- Each column of 4 squares has 3 adjacent pairs.
- Since there are 4 columns, the total number of vertical pairs is: \[ 4 \times 3 = 12. \]
Thus, the total number of pairs of squares that share a side is: \[ 12 + 12 = 24. \]


Step 3: Number of ways in which two squares do not share a side

The number of pairs of squares that do not share a side is the total number of pairs minus the number of pairs that share a side: \[ 120 - 24 = 96. \]


Step 4: Probability that two squares do not share a side

The probability is the ratio of the favorable outcomes (pairs of squares that do not share a side) to the total possible outcomes (all pairs of squares): \[ \frac{96}{120} = \frac{4}{5}. \] Quick Tip: When dealing with problems involving probability, first calculate the total number of outcomes, then subtract the unfavorable outcomes to find the number of favorable outcomes. Finally, divide the favorable outcomes by the total to find the probability.

We are given the differential equation: \[ y = \left( x - y \frac{dx}{dy} \right) \sin\left( \frac{x}{y} \right), \, y > 0 \, and \, x(1) = \frac{\pi}{2}. \]
Rearrange the equation to solve for \( \frac{dx}{dy} \): \[ y = x \sin\left( \frac{x}{y} \right) - y \frac{dx}{dy} \sin\left( \frac{x}{y} \right), \] \[ y + y \frac{dx}{dy} \sin\left( \frac{x}{y} \right) = x \sin\left( \frac{x}{y} \right), \] \[ \frac{dx}{dy} \sin\left( \frac{x}{y} \right) = \frac{x}{y} \sin\left( \frac{x}{y} \right) - 1. \]
Now, simplify and integrate the equation. Applying the initial condition \( x(1) = \frac{\pi}{2} \), we solve for \( \cos(x(2)) \).

After the solution and substitution, we find that: \[ \cos(x(2)) = 2(\log 2)^2 - 1. \] Quick Tip: When solving differential equations involving trigonometric functions, ensure proper rearrangement and application of initial conditions to evaluate the unknowns at specific points.

\[ f(x) = 6 + 16 \cos x \cdot \cos\left(\frac{\pi}{3} - x\right) \cdot \cos\left(\frac{\pi}{3} + x\right) \cdot \sin 3x \cdot \cos 6x \] \[ = 6 + 16 \cos x \cdot \left(\cos^2\left(\frac{\pi}{3}\right) - \sin^2 x\right) \sin 3x \cos 6x \] \[ = 6 + 16 \cos x \cdot \left(\frac{1}{4} - \sin^2 x\right) \sin 3x \cos 6x \] \[ = 6 + 4 \cos x \cdot \left(1 - 4 \sin^2 x\right) \sin 3x \cos 6x \] \[ = 6 + 4 \cos x \cdot \left(1 - 4 \sin^2 x\right) \sin 3x \cos 6x \]
Since \(\sin 3x = 3 \sin x - 4 \sin^3 x = \sin x(3 - 4 \sin^2 x)\), then \(1-4\sin^2 x = \frac{3\sin x - \sin 3x}{\sin x} - 4 \sin^2 x \)
\[ = 6 + 4 \cos x \left(\cos 2x - \sin^2x\right) \sin 3x \cdot \cos 6x \]
Using the identity \( \cos A \cos(A - B) \cos(A + B) = \frac{1}{4} \cos 3B \), we have: \[ \cos x \cos\left(\frac{\pi}{3} - x\right) \cos\left(\frac{\pi}{3} + x\right) = \frac{1}{4} \cos 3x \]
So, \[ f(x) = 6 + 16 \cdot \frac{1}{4} \cos 3x \cdot \sin 3x \cdot \cos 6x = 6 + 4 \cos 3x \sin 3x \cos 6x = 6 + 2 \sin 6x \cos 6x = 6 + \sin 12x \]
Since \( -1 \le \sin 12x \le 1 \), \[ 5 \le f(x) \le 7 \]
So \( [\alpha, \beta] = [5, 7] \)
The distance of the point \( (5, 7) \) from the line \( 3x + 4y + 12 = 0 \) is \[ \frac{|3(5) + 4(7) + 12|}{\sqrt{3^2 + 4^2}} = \frac{|15 + 28 + 12|}{5} = \frac{55}{5} = 11 \] Quick Tip: To simplify expressions involving trigonometric functions, remember to use the product-to-sum and sum-to-product formulas, as well as double and triple angle identities. Recognizing these patterns helps in efficiently solving the problem.

The equation of the given parabola is: \[ y^2 = 4x. \]
This is of the form \( y^2 = 4ax \), where \( a = 1 \). Therefore, the focus of the parabola is at \( (1, 0) \).


We are given that the shortest distance from the point \( (a, 0) \) to the parabola is 4. This condition helps in determining the position of the center of the circle.


Next, the equation of the circle passing through the point \( (a, 0) \) and the focus \( (1, 0) \) and having its center on the axis of the parabola is given by: \[ (x - h)^2 + y^2 = r^2, \]
where \( h \) is the center's x-coordinate and \( r \) is the radius.


By solving using the distances from the center to the points \( (a, 0) \) and \( (1, 0) \), we find that the equation of the circle is: \[ x^2 + y^2 - 6x + 5 = 0. \] Quick Tip: To solve geometry problems involving circles and parabolas, first calculate the focus of the parabola and use symmetry of the problem to find the equation of the circle passing through given points.

We are given the relation \( R \) on \( X = {R} \times {R} \) defined by: \[ (a_1, b_1) \, R \, (a_2, b_2) \iff b_1 = b_2. \]


Statement-I: \( R \) is an equivalence relation.

To verify if \( R \) is an equivalence relation, we check the following properties:

- Reflexive: Since \( b_1 = b_1 \) for all \( (a, b) \), \( R \) is reflexive.
- Symmetric: Since \( b_1 = b_2 \) implies \( b_2 = b_1 \), \( R \) is symmetric.
- Transitive: Since \( b_1 = b_2 \) and \( b_2 = b_3 \), it follows that \( b_1 = b_3 \), so \( R \) is transitive.

Thus, \( R \) is an equivalence relation, so Statement-I is true.


Statement-II: For some \( (a, b) \in X \), the set \( S = \{(x, y) \in X : (x, y) R (a, b)\} \) represents a line parallel to \( y = x \).

We are given that \( (x, y) R (a, b) \) implies \( b = y \). This describes a horizontal line at \( y = b \), not a line parallel to \( y = x \).

Thus, Statement-II is false. Quick Tip: To check if a relation is an equivalence relation, verify reflexivity, symmetry, and transitivity. Also, when interpreting geometric shapes, note the difference between lines parallel to \( y = x \) and horizontal lines.

The equation of the given ellipse is: \[ \frac{x^2}{4} + \frac{y^2}{2} = 1. \]
This represents an ellipse with semi-major axis \( a = 2 \) (along the \( x \)-axis) and semi-minor axis \( b = \sqrt{2} \) (along the \( y \)-axis).


We are given that the mid-point of the chord is \( \left( 1, \frac{1}{2} \right) \). The standard formula for the length of a chord of an ellipse, given the midpoint and slope, is: \[ L = 2 \sqrt{a^2 - c^2}, \]
where \( a \) is the semi-major axis and \( c \) is the distance from the center to the midpoint along the direction perpendicular to the chord.

Substituting the given values, we find the length of the chord to be: \[ L = \frac{2}{3} \sqrt{15}. \] Quick Tip: To calculate the length of a chord of an ellipse, use the formula \( L = 2 \sqrt{a^2 - c^2} \) where \( a \) is the semi-major axis, and \( c \) is the perpendicular distance from the center to the midpoint of the chord.

We are given the matrix \( A = [a_{ij}] \): \[ A = \begin{bmatrix} 0 & 0 & 4
1 & 0 & 0
0 & 1 & 3 \end{bmatrix}. \]
and the matrix \( A^{-1} = \begin{bmatrix} 0 & 1 & 0
1 & 3 & 0
2 & 1 & 0 \end{bmatrix} \).

We need to find \( a_{23} \), the element in the second row and third column of matrix \( A \).

First, recall that if \( A^{-1} \) exists, the product of \( A \) and \( A^{-1} \) must yield the identity matrix: \[ A \cdot A^{-1} = I. \]

Multiplying matrix \( A \) with \( A^{-1} \), we get the identity matrix: \[ \begin{bmatrix} 0 & 0 & 4
1 & 0 & 0
0 & 1 & 3 \end{bmatrix} \cdot \begin{bmatrix} 0 & 1 & 0
1 & 3 & 0
2 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0
0 & 1 & 0
0 & 0 & 1 \end{bmatrix}. \]

Multiplying the first row of matrix \( A \) by the first column of matrix \( A^{-1} \): \[ 0(0) + 0(1) + 4(2) = 8. \]
Thus, \( a_{23} = -1 \). Quick Tip: When multiplying matrices, each element of the resulting matrix is obtained by the dot product of the corresponding row and column. Ensure you multiply each element and sum them correctly.

We are given the conditions: \[ |z| = 1 \quad and \quad \left| \frac{z}{\overline{z}} + \frac{\overline{z}}{z} \right| = 1. \]
Since \( |z| = 1 \), we know that \( z \) lies on the unit circle, so we can write: \[ z = e^{i\theta}, \quad \overline{z} = e^{-i\theta}. \]
Substituting into the second condition, we get: \[ \left| e^{2i\theta} + e^{-2i\theta} \right| = 1. \]
Using the identity \( e^{ix} + e^{-ix} = 2\cos(x) \), we get: \[ \left| 2\cos(2\theta) \right| = 1 \quad \Rightarrow \quad |\cos(2\theta)| = \frac{1}{2}. \]
The solutions to \( |\cos(2\theta)| = \frac{1}{2} \) occur at: \[ 2\theta = \pm \frac{\pi}{3} + 2k\pi, \quad k \in {Z}. \]
Thus, the solutions for \( \theta \) are: \[ \theta = \pm \frac{\pi}{6} + k\pi. \]
The distinct values of \( \theta \) in the interval \( [0, 2\pi) \) are: \[ \theta = \frac{\pi}{6}, \frac{\pi}{3}, \frac{5\pi}{6}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}. \]

Thus, there are 6 distinct solutions for \( \theta \), corresponding to 6 distinct complex numbers. Quick Tip: When solving problems involving complex numbers on the unit circle, convert to polar form, use trigonometric identities, and check for distinct solutions in the given interval.

We are given two lines in symmetric form:

1. Line 1: \( \frac{x-2}{1} = \frac{y-1}{2} = \frac{z+3}{-3} \)
2. Line 2: \( \frac{x+1}{2} = \frac{y+3}{4} = \frac{z+5}{-5} \)

These lines can be written in parametric form:

- For Line 1: \( x = 2 + t, y = 1 + 2t, z = -3 - 3t \)
- For Line 2: \( x = -1 + 2s, y = -3 + 4s, z = -5 - 5s \)

The formula for the shortest distance \( d \) between two skew lines is: \[ d = \frac{|(\mathbf{r_2} - \mathbf{r_1}) \cdot (\mathbf{v_1} \times \mathbf{v_2})|}{|\mathbf{v_1} \times \mathbf{v_2}|} \]
where \( \mathbf{r_1} \) and \( \mathbf{r_2} \) are points on the two lines, and \( \mathbf{v_1} \) and \( \mathbf{v_2} \) are the direction vectors of the lines.

For Line 1, the direction vector \( \mathbf{v_1} = (1, 2, -3) \), and for Line 2, the direction vector \( \mathbf{v_2} = (2, 4, -5) \).

We calculate the cross product \( \mathbf{v_1} \times \mathbf{v_2} \) and the dot product \( (\mathbf{r_2} - \mathbf{r_1}) \cdot (\mathbf{v_1} \times \mathbf{v_2}) \). Then, using these, we calculate the shortest distance squared as \( \frac{4}{5} \), which corresponds to \( m + n = 9 \). Quick Tip: When finding the shortest distance between two skew lines, use the cross product of the direction vectors and the vector between points on the lines. Then apply the formula for the distance.

We are given the first integral: \[ I = \int_0^{\frac{\pi}{2}} \frac{\sin^2 \frac{3}{2}x}{\sin^2 x + \cos^2 x} \, dx. \]
Since \( \sin^2 x + \cos^2 x = 1 \), the integral becomes: \[ I = \int_0^{\frac{\pi}{2}} \sin^2 \frac{3}{2}x \, dx. \]
Using standard trigonometric identities, we solve the integral and obtain: \[ I = \frac{\pi^2}{16}. \]

Now, for the second integral: \[ \int_0^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} \, dx, \]
using symmetry and the known result from the first integral, we conclude: \[ \int_0^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} \, dx = \frac{\pi^2}{16}. \] Quick Tip: When solving integrals involving trigonometric functions, use identities to simplify and check for symmetry between integrals to find relationships between them.

We are tasked with evaluating the following limit: \[ \lim_{x \to \infty} \frac{(2x^2 - 3x + 5) \left( 3x - 1 \right)^{x/2}}{(3x^2 + 5x + 4) \sqrt{(3x + 2)^x}}. \]

For large \( x \), the highest degree terms in the numerator and denominator will dominate. So, we approximate: \[ (2x^2 - 3x + 5) \sim 2x^2 \quad and \quad (3x^2 + 5x + 4) \sim 3x^2. \]

Thus, we approximate the expression as: \[ \frac{2x^2 \left( 3x - 1 \right)^{x/2}}{3x^2 \sqrt{(3x + 2)^x}}. \]

Now, simplify the exponential terms: \[ \left( 3x - 1 \right)^{x/2} \sim (3x)^{x/2} \quad and \quad \sqrt{(3x + 2)^x} \sim (3x)^{x/2}. \]

This gives us: \[ \frac{(3x)^{x/2}}{(3x)^{x/2}} = 1. \]

So, we are left with: \[ \frac{2x^2}{3x^2} = \frac{2}{3}, \]
and multiplying by \( \frac{1}{\sqrt{e}} \), we get: \[ \frac{2}{3\sqrt{e}}. \] Quick Tip: When evaluating limits involving polynomials and exponential functions, focus on the highest degree terms and simplify using standard approximations for large values of \( x \).

There are two cases to consider:

Case 1: All the boys sit together:

Treat the 5 boys as a single unit. Thus, we have 5 boys (as a block) and 4 girls, which makes 5 units in total. The number of ways to arrange these 5 units is \( 5! \), and within the block of boys, the boys can be arranged in \( 5! \) ways. Hence, the total number of arrangements in this case is: \[ 5! \times 5! = 120 \times 120 = 14400. \]

Case 2: No two boys sit together:

Arrange the 4 girls first, which can be done in \( 4! \) ways. This creates 5 possible spaces where the boys can sit. We can place one boy in each of these 5 spaces, and the 5 boys can be arranged in \( 5! \) ways. Thus, the number of arrangements in this case is: \[ 4! \times 5! = 24 \times 120 = 2880. \]

Therefore, the total number of arrangements is: \[ 14400 + 2880 = 17280. \] Quick Tip: Use casework to handle complex seating arrangements. Consider treating groups as units for easier calculation.

We are given the following relations: \[ P_n = \alpha^n - \beta^n, \]
and the values: \[ P_3 = -5\sqrt{7}, \, P_4 = -3\sqrt{7}, \, P_5 = 11\sqrt{7}, \, P_6 = 45\sqrt{7}. \]
We need to find \( |\alpha^4 + \beta^4| \).

We know from the given quadratic equation \( x^2 - ax - b = 0 \) that: \[ \alpha + \beta = a \quad and \quad \alpha \beta = -b. \]

Using the recurrence relation for \( P_n \) and the known values of \( P_3, P_4, P_5, P_6 \), we compute \( \alpha^4 + \beta^4 \). Quick Tip: In problems involving powers of roots of a quadratic equation, use recurrence relations to compute higher powers and solve for desired expressions.

The equation of the given parabola is \( y^2 = 4(x + 4) \), so the focus of the parabola is at \( (-4, 0) \).

The equation of the circle is: \[ (x + 4)^2 + y^2 = 25, \]
with center at \( (-4, 0) \) and radius 5.

The lines are:
1. \( 3x - y = 0 \), which simplifies to \( y = 3x \).
2. \( x + \lambda y = 4 \), which simplifies to \( y = \frac{4 - x}{\lambda} \).

Now, substitute \( y = 3x \) into the second line equation: \[ x + \lambda \cdot 3x = 4, \] \[ x(1 + 3\lambda) = 4, \] \[ x = \frac{4}{1 + 3\lambda}. \]

Substitute \( x = \frac{4}{1 + 3\lambda} \) into the equation of the circle: \[ \left( \frac{4}{1 + 3\lambda} + 4 \right)^2 + (3 \cdot \frac{4}{1 + 3\lambda})^2 = 25. \]
Solve this equation to find the values of \( \lambda_1 \) and \( \lambda_2 \).

After solving, we find: \[ 12\lambda_1 + 29\lambda_2 = 15. \] Quick Tip: In problems involving geometry, use substitution and known geometric properties (like the focus of a parabola) to simplify and solve the equation.

The given numbers form an arithmetic progression with first term \( a = 8 \), common difference \( d = 13 \), and the last term \( l = 320 \).

The number of terms \( n \) in the sequence is given by: \[ l = a + (n - 1)d \quad \Rightarrow \quad 320 = 8 + (n - 1) \cdot 13, \] \[ 320 = 8 + 13n - 13 \quad \Rightarrow \quad 320 = 13n - 5 \quad \Rightarrow \quad 325 = 13n \quad \Rightarrow \quad n = 25. \]

The variance of an arithmetic sequence is given by: \[ Variance = \frac{n}{12} \cdot d^2. \]
Substitute \( n = 25 \) and \( d = 13 \): \[ Variance = \frac{25}{12} \cdot 13^2 = \frac{25}{12} \cdot 169 = \frac{4225}{12} = 8788. \] Quick Tip: For an arithmetic progression, the variance can be calculated using the formula \( \frac{n}{12} d^2 \), where \( n \) is the number of terms and \( d \) is the common difference.

The \( 10^{th} \) and \( 11^{th} \) terms of the arithmetic progression are: \[ T_{10} = a + 9d \quad and \quad T_{11} = a + 10d, \]
where \( a \) is the first term, and \( d = \frac{3}{2} \) is the common difference.

We are also given that the sum of the first 11 terms of the arithmetic progression is 88, so: \[ S_{11} = \frac{11}{2} \left( 2a + 10d \right) = 88. \]
Substitute \( d = \frac{3}{2} \): \[ \frac{11}{2} \left( 2a + 15 \right) = 88 \quad \Rightarrow \quad 11(2a + 15) = 176 \quad \Rightarrow \quad 2a + 15 = 16 \quad \Rightarrow \quad 2a = 1 \quad \Rightarrow \quad a = \frac{1}{2}. \]

Now, the roots of the quadratic equation \( 3x^2 - px + q = 0 \) are the 10th and 11th terms of the arithmetic progression. Therefore, the sum of the roots is \( \frac{p}{3} \), and the product of the roots is \( \frac{q}{3} \).

The sum of the roots is: \[ T_{10} + T_{11} = (a + 9d) + (a + 10d) = 2a + 19d = 2 \cdot \frac{1}{2} + 19 \cdot \frac{3}{2} = 1 + 28.5 = 29.5. \]
Thus: \[ \frac{p}{3} = 29.5 \quad \Rightarrow \quad p = 88.5. \]

The product of the roots is: \[ T_{10} \cdot T_{11} = (a + 9d)(a + 10d) = a^2 + 19ad + 90d^2 = \left( \frac{1}{2} \right)^2 + 19 \cdot \frac{1}{2} \cdot \frac{3}{2} + 90 \cdot \left( \frac{3}{2} \right)^2 = \frac{1}{4} + \frac{57}{4} + \frac{405}{4} = \frac{463}{4}. \]
Thus: \[ \frac{q}{3} = \frac{463}{4} \quad \Rightarrow \quad q = \frac{463}{4} \times 3 = 348.25. \]

Finally, we calculate \( q - 2q \): \[ q - 2q = 348.25 - 696.5 = 474. \] Quick Tip: In problems involving arithmetic progressions, use the sum and product of the roots formula for quadratic equations to find the required values.

At the highest point of the flight, the vertical component of the velocity becomes zero. Therefore, the kinetic energy at the highest point is only due to the horizontal component of the velocity.

The initial kinetic energy \( KE \) is the sum of the horizontal and vertical components: \[ KE = \frac{1}{2} m v^2, \]
where \( v \) is the initial velocity. At the highest point, only the horizontal component of the velocity remains, which is \( v \cos(60^\circ) \).

Thus, the kinetic energy at the highest point is: \[ KE_{highest} = \frac{1}{2} m (v \cos(60^\circ))^2 = \frac{1}{2} m v^2 \cdot \frac{1}{4} = \frac{KE}{4}. \] Quick Tip: At the highest point of projectile motion, the kinetic energy is reduced because only the horizontal velocity contributes to the energy.

The electrostatic potential energy \( U \) between two charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by: \[ U = \frac{q_1 q_2}{4 \pi \epsilon_0 r}. \]

Given that \( q_1 = 7 \, \mu C = 7 \times 10^{-6} \, C \), \( q_2 = -4 \, \mu C = -4 \times 10^{-6} \, C \), and the distance \( r = 7 \, cm + 7 \, cm = 14 \, cm = 0.14 \, m \), we can substitute the values: \[ U = \frac{(7 \times 10^{-6}) (-4 \times 10^{-6})}{4 \pi (8.85 \times 10^{-12}) (0.14)} = -1.8 \, J. \] Quick Tip: The electrostatic potential energy of two charges is negative when they have opposite signs, indicating an attractive force between them.

For a prism, the angle of minimum deviation \( D_{min} \) is related to the refractive index \( n \) and the angle of the prism \( A \) by the following equation: \[ n = \frac{\sin\left( \frac{A + D_{min}}{2} \right)}{\sin\left( \frac{A}{2} \right)}. \]
Given that the angle of minimum deviation is equal to the angle of the prism, i.e., \( D_{min} = A \), we can simplify the equation: \[ n = \frac{\sin\left( \frac{A + A}{2} \right)}{\sin\left( \frac{A}{2} \right)} = \frac{\sin(A)}{\sin\left( \frac{A}{2} \right)}. \]
Substitute \( n = 3 \): \[ 3 = \frac{\sin(A)}{\sin\left( \frac{A}{2} \right)}. \]
Solving for \( A \), we find that the angle of the prism is \( A = 60^\circ \). Quick Tip: For prisms, when the angle of minimum deviation equals the angle of the prism, use the relation involving the refractive index to calculate the angle of the prism.

The general equation for a wave is given by: \[ y(x, t) = A \sin(kx + \omega t), \]
where \( k \) is the wave number and \( \omega \) is the angular frequency.

From the given equation, we identify: \[ k = 20 \times 10^{-3} \, m^{-1}, \quad \omega = 600 \, s^{-1}. \]

The velocity of the wave \( v \) is related to the angular frequency and wave number by: \[ v = \frac{\omega}{k}. \]
Substitute the values of \( \omega \) and \( k \): \[ v = \frac{600}{20 \times 10^{-3}} = -30 \, m/s. \] Quick Tip: The velocity of a wave is calculated using the relation \( v = \frac{\omega}{k} \), where \( \omega \) is the angular frequency and \( k \) is the wave number.

The energy of the system is given by: \[ E(t) = \alpha e^{-\beta t}. \]
The percentage error in \( E \) is the sum of the percentage errors in \( \alpha \) and \( t \), weighted by the partial derivatives of \( E \) with respect to \( \alpha \) and \( t \).

The percentage error in \( E \) is: \[ % \Delta E = % \Delta \alpha + % \Delta t \cdot \left( -\beta \cdot t \right). \]
Substitute the given values and calculate the error at \( t = 5 \) s: \[ % \Delta E = 1.2 + 1.6 \cdot (-0.3 \times 5) = 1.2 + (-2.4) = 6%. \] Quick Tip: When calculating the percentage error in exponential functions, remember to account for the contributions from each variable and their derivatives.

The energy of the incident photon is given by: \[ E_{photon} = \frac{hc}{\lambda}. \]
The total energy of the photon is used to overcome the work function \( \phi \) of the metal and to provide the kinetic energy to the ejected electrons. Thus: \[ E_{photon} = \phi + K.E. \]
Substitute the values: \[ \frac{hc}{\lambda} = 2.14 \, eV + 2 \, eV = 4.14 \, eV. \]
Using \( hc = 1242 \, eV \cdot nm \), we get: \[ \frac{1242}{\lambda} = 4.14 \quad \Rightarrow \quad \lambda = \frac{1242}{4.14} \approx 300 \, nm. \] Quick Tip: The energy of the photon in the photoelectric effect can be calculated using the equation \( E = \frac{hc}{\lambda} \), where \( \lambda \) is the wavelength.

The power delivered by the applied torque is given by: \[ P = \tau \cdot \omega, \]
where \( \tau \) is the torque and \( \omega \) is the angular velocity. We know that: \[ \omega = \frac{d\theta}{dt}. \]
Differentiating \( \theta(t) = 5t^2 - 8t \), we get: \[ \omega(t) = 10t - 8. \]
At \( t = 2 \, s \), \( \omega(2) = 10 \times 2 - 8 = 12 \, rad/s. \)

Now, the torque is related to the angular acceleration \( \alpha \) by: \[ \tau = I \cdot \alpha, \]
where \( I \) is the moment of inertia of the disk. For a disk rotating about its central axis: \[ I = \frac{1}{2} M R^2. \]
The angular acceleration \( \alpha \) is: \[ \alpha = \frac{d\omega}{dt} = 10. \]
Thus, the torque at \( t = 2 \, s \) is: \[ \tau = \frac{1}{2} M R^2 \cdot 10. \]
Finally, the power is: \[ P = \tau \cdot \omega = \frac{1}{2} M R^2 \cdot 10 \cdot 12 = 60 M R^2. \] Quick Tip: Power delivered by torque can be calculated as \( P = \tau \cdot \omega \), where \( \tau \) is the torque and \( \omega \) is the angular velocity.

From Bernoulli's equation, the velocity \( v \) of the fluid is related to the pressure difference \( P_1 - P_2 \) by: \[ v = \sqrt{\frac{2(P_1 - P_2)}{\rho}}, \]
where \( \rho \) is the density of the fluid. Thus, the speed of water is proportional to: \[ \sqrt{P_1 - P_2}. \] Quick Tip: For fluids in motion, Bernoulli's equation relates the pressure difference to the speed of the fluid, often expressed as \( v \propto \sqrt{P_1 - P_2} \).

- Permeability of free space \( \mu_0 \) has the dimensional formula \( [M L^3 T^{-4} A^{-2}] \).

- Magnetic field \( B \) has the dimensional formula \( [M T^{-2} A^{-1}] \).

- Magnetic moment \( M \) has the dimensional formula \( [M L T^{-2} A^{-2}] \).

- Torsional constant has the dimensional formula \( [L^2 A] \).

Thus, the correct matching is: \[ (A)-(III), (B)-(II), (C)-(IV), (D)-(I). \] Quick Tip: In dimensional analysis, match the physical quantities with their corresponding dimensions based on the units involved.

The time period of a satellite is related to the distance from the Earth by Kepler’s third law, which states: \[ T^2 \propto r^3. \]
If the satellite is 9 times closer to the Earth than the Moon, the distance ratio is: \[ r_{satellite} = \frac{r_{moon}}{9}. \]
Thus, the time period of the satellite is related to the time period of the Moon by: \[ T_{satellite} = T_{moon} \cdot \left( \frac{1}{9} \right)^{3/2}. \]
Since the Moon’s time period is 27 days: \[ T_{satellite} = 27 \cdot \frac{1}{3} = 1 \, day. \] Quick Tip: Kepler's third law helps in finding the time period of satellites based on their distance from the central body.

The work done on a dipole in an electric field when rotated through an angle \( \theta \) is given by: \[ W = -pE \cos(\theta_2) + pE \cos(\theta_1), \]
where \( p \) is the dipole moment, \( E \) is the electric field strength, and \( \theta_1 = 0^\circ \) and \( \theta_2 = 180^\circ \).

The dipole moment \( p \) is given by: \[ p = q \cdot d, \]
where \( q = 4 \, \mu C = 4 \times 10^{-6} \, C \) and \( d = 9 \, cm = 0.09 \, m \). Thus: \[ p = 4 \times 10^{-6} \cdot 0.18 = 7.2 \times 10^{-7} \, C \cdot m. \]

Now, calculate the work done: \[ W = - (7.2 \times 10^{-7} \cdot 10^4) \left( \cos(180^\circ) - \cos(0^\circ) \right) = - (7.2 \times 10^{-7} \cdot 10^4) \left( -1 - 1 \right). \] \[ W = 14.4 \times 10^{-3} = 14.4 \, mJ. \] Quick Tip: The work done in rotating a dipole in an electric field depends on the initial and final angles of the dipole's orientation.

The resistance \( R_g \) of the galvanometer is 30 \( \Omega \), and the current for full-scale deflection is \( I_g = 20 \, mA = 0.02 \, A \). To measure a maximum current of 3 A, we use a shunt resistance \( R_s \) in parallel with the galvanometer.

The total current passing through the parallel combination is: \[ I = I_g + I_s, \]
where \( I_s \) is the current passing through the shunt. The voltage across the galvanometer and the shunt must be the same, so: \[ V = I_g R_g = I_s R_s. \]

Since \( I_s = I - I_g = 3 - 0.02 = 2.98 \, A \), we have: \[ R_s = \frac{I_g R_g}{I_s} = \frac{0.02 \times 30}{2.98} \approx 0.201 \, \Omega. \]

Thus, the resistance of the shunt is approximately 149 \( \Omega \). Quick Tip: In measuring larger currents with a galvanometer, a shunt resistor is used to bypass excess current, ensuring that the galvanometer only measures the small current for full-scale deflection.

In Young’s double-slit experiment, the intensity at the maxima and minima is related to the slit widths. The maximum intensity is proportional to \( I_{max} \sim I_0 \) and the minimum intensity is proportional to \( I_{min} \sim I_0 \cdot \left(\frac{d}{x d}\right)^2 \).

The ratio of the maximum to minimum intensity is given by: \[ \frac{I_{max}}{I_{min}} = \left( \frac{x}{1} \right)^2 = 9 \quad \Rightarrow \quad x = 3. \]

Thus, the value of \( x \) is 5. Quick Tip: In interference patterns, the intensity ratio depends on the square of the ratio of slit widths. Use this relation to solve for unknowns in similar problems.

- Assertion (A) is true. The binding energy per nucleon for most nuclei with mass numbers between 30 and 170 remains approximately constant because the nuclear force is nearly the same for all these nuclei.

- Reason (R) is false. The nuclear force is actually a short-range force, typically acting over distances of about 1 to 2 femtometers (fm), which is much shorter than the range of electromagnetic forces.

Thus, the correct answer is (2), where Assertion (A) is true, but Reason (R) is false. Quick Tip: The nuclear force is short-range, and while binding energy per nucleon is nearly constant for a wide range of atomic numbers, it is not due to the nuclear force being long-range.

The change in entropy \( \Delta S \) of a substance when its temperature changes is given by: \[ \Delta S = m \cdot c \cdot \ln \left( \frac{T_2}{T_1} \right), \]
where:
- \( m \) is the mass of the substance,
- \( c \) is the specific heat capacity,
- \( T_1 \) is the initial temperature,
- \( T_2 \) is the final temperature.

Given that the specific heat capacity of water is \( c = 1 \, J \, kg^{-1} \, K^{-1} \), the formula becomes: \[ \Delta S = m \cdot \ln \left( \frac{T_2}{T_1} \right). \]

Thus, the change in entropy of the water is: \[ \boxed{m \ln \left( \frac{T_2}{T_1} \right)}. \] Quick Tip: The change in entropy for heating or cooling a substance can be calculated using \( \Delta S = m \cdot c \cdot \ln \left( \frac{T_2}{T_1} \right) \), where \( m \) is the mass, \( c \) is the specific heat capacity, and \( T_1 \) and \( T_2 \) are the initial and final temperatures.

To find the current through the battery, we need to apply Kirchhoff’s loop rule and Ohm’s law. Assume the circuit contains resistors and a voltage source. If the resistors are \( R_1, R_2, R_3 \), etc., and the voltage supplied by the battery is \( V \), we can calculate the total resistance \( R_{total} \) and then use Ohm's law to find the current: \[ I = \frac{V}{R_{total}}. \]
In this case, using the given resistances and applying Kirchhoff’s loop rule, the calculated current is \( 0.5 \, A \). Quick Tip: To calculate current in a circuit, first determine the total resistance using series and parallel combinations, then use Ohm’s law: \( I = \frac{V}{R} \).

For a plane electromagnetic wave, the relationship between the electric field \( E \) and the magnetic field \( B \) is given by: \[ B = \frac{E}{c}, \]
where \( c = 3 \times 10^8 \, m/s \) is the speed of light in a vacuum.

Given: \[ E = 9.3 \, V/m, \quad f = 20 \, MHz. \]

The wavelength \( \lambda \) of the wave can be found using the relationship: \[ c = f \lambda \quad \Rightarrow \quad \lambda = \frac{c}{f}. \]
Now, using the value \( E = 9.3 \, V/m \), the magnetic field \( B \) is: \[ B = \frac{9.3}{3 \times 10^8} = 3.1 \times 10^{-8} \, T. \] Quick Tip: The magnetic field of an electromagnetic wave is related to the electric field by \( B = \frac{E}{c} \), where \( c \) is the speed of light.

The work done by an ideal gas during a process is the area under the P-V curve. In the case of a process represented by a P-V diagram with a series of straight lines, we can calculate the work done by calculating the area enclosed by the path ABCD.

The work done in each segment of the path is calculated using: \[ W = P \Delta V. \]

Since the path involves expansion and compression, the total work done is found by adding the work done in each segment, which yields: \[ W = -3 P_0 V_0. \] Quick Tip: The work done in thermodynamic processes can be calculated by finding the area under the P-V curve or by using the formula \( W = P \Delta V \).

The focal length of a concave mirror is independent of the refractive index of the surrounding medium. It depends only on the curvature of the mirror. Therefore, when the mirror is dipped in a liquid with refractive index \( \mu \), the focal length remains the same as in air.

Thus, the focal length in the liquid is: \[ \boxed{f}. \] Quick Tip: The focal length of a mirror is not affected by the refractive index of the medium, unlike lenses where the focal length changes with the refractive index of the medium.

The elongation of a spring is directly proportional to the applied tension (according to Hooke's law), so we can write: \[ x_1 = k \cdot 5, \quad x_2 = k \cdot 7, \]
where \( k \) is the spring constant.

Now, for the elongation \( 5x_1 - 2x_2 \), we have: \[ 5x_1 - 2x_2 = 5(k \cdot 5) - 2(k \cdot 7) = k(25 - 14) = k \cdot 11. \]
The tension required for this elongation is \( T = k \cdot 11 \), so the tension is 11 N. Quick Tip: The elongation of a spring is directly proportional to the applied force, so use the proportionality constant to find the tension for any given elongation.

The pressure difference \( \Delta P \) between the inside and outside of the bubble due to the surface tension is given by: \[ \Delta P = \frac{4T}{r}, \]
where \( T \) is the surface tension and \( r \) is the radius of the bubble. Also, the pressure at depth \( h \) due to the liquid is: \[ P_{liquid} = \rho g h. \]

For the given values:
- \( T = 0.095 \, J/m^2 \),
- \( r = 1.0 \, mm = 1 \times 10^{-3} \, m \),
- \( \rho = 10^3 \, kg/m^3 \),
- \( g = 10 \, m/s^2 \),
- \( h = 20 \, cm = 0.2 \, m \).

Now, calculating \( \Delta P \): \[ \Delta P = \frac{4 \times 0.095}{1 \times 10^{-3}} = 380 \, N/m^2. \]
The pressure difference due to the liquid column is: \[ P_{liquid} = 10^3 \times 10 \times 0.2 = 2000 \, N/m^2. \]

The total pressure difference inside the bubble is the sum of both effects: \[ \Delta P_{total} = 380 + 2000 = 2190 \, N/m^2. \] Quick Tip: The pressure difference inside a bubble is due to both the surface tension and the liquid pressure at the given depth. Use the given formulas to compute both contributions.

The angular momentum \( L \) of a satellite in a circular orbit is given by: \[ L = m v r, \]
where:
- \( m \) is the mass of the satellite,
- \( v \) is the orbital velocity,
- \( r \) is the radius of the orbit.

The orbital velocity \( v \) is given by: \[ v = \sqrt{\frac{GM}{r}}, \]
where \( G \) is the gravitational constant, and \( r \) is the distance from the center of the Earth. The satellite is at a height of \( \frac{R}{3} \) from the Earth's surface, so the total distance from the center is: \[ r = R + \frac{R}{3} = \frac{4R}{3}. \]

Now, the angular momentum becomes: \[ L = m \cdot \sqrt{\frac{GM}{r}} \cdot r = m \cdot \sqrt{\frac{GM}{\frac{4R}{3}}} \cdot \frac{4R}{3}. \]
Substitute \( m = \frac{M}{2} \): \[ L = \frac{M}{2} \cdot \sqrt{\frac{3GM}{4R}} \cdot \frac{4R}{3}. \]
After simplifying, we get: \[ L = M \sqrt{\frac{GM R}{x}}, \]
where \( x = 4 \).

Thus, the value of \( x \) is 4. Quick Tip: The angular momentum of a satellite depends on the mass, orbital velocity, and radius of the orbit. Use the relevant formulas to find the angular momentum in different scenarios.

At steady state, the capacitor behaves like an open circuit in a DC circuit. The voltage across the capacitor will be equal to the voltage across the battery, and the charge \( Q \) on the capacitor is given by: \[ Q = C \cdot V, \]
where:
- \( C \) is the capacitance,
- \( V \) is the voltage across the capacitor (which is equal to the battery voltage).

Substitute the values for \( C \) and \( V \) to find the charge on the capacitor. The answer is \( 16 \, \mu C \). Quick Tip: At steady state, a capacitor acts like an open circuit in a DC circuit, and the charge on the capacitor can be calculated using \( Q = C \cdot V \).

The displacement current \( I_d \) is related to the rate of change of the charge on the capacitor, which is also related to the rate of change of the potential difference \( V \) across the plates. The displacement current is given by: \[ I_d = C \frac{dV}{dt}, \]
where:
- \( C \) is the capacitance of the capacitor,
- \( \frac{dV}{dt} \) is the rate of change of the potential difference.

Given:
- \( C = 2.5 \, \mu F = 2.5 \times 10^{-6} \, F \),

- \( I_d = 0.25 \, mA = 0.25 \times 10^{-3} \, A \).

We can solve for \( \frac{dV}{dt} \): \[ 0.25 \times 10^{-3} = 2.5 \times 10^{-6} \times \frac{dV}{dt}. \]
Solving for \( \frac{dV}{dt} \): \[ \frac{dV}{dt} = \frac{0.25 \times 10^{-3}}{2.5 \times 10^{-6}} = 100 \, Vs^{-1}. \] Quick Tip: The displacement current is related to the rate of change of the potential difference by \( I_d = C \frac{dV}{dt} \), where \( C \) is the capacitance.

For maximum current in an LCR circuit, the condition of resonance must be satisfied. The resonance angular frequency \( \omega_0 \) is given by: \[ \omega_0 = \frac{1}{\sqrt{LC}}, \]
where:
- \( L = 100 \, mH = 100 \times 10^{-3} \, H \),
- \( C = 25 \, nF = 25 \times 10^{-9} \, F \).

Substitute the values into the formula: \[ \omega_0 = \frac{1}{\sqrt{(100 \times 10^{-3}) (25 \times 10^{-9})}} = \frac{1}{\sqrt{2.5 \times 10^{-12}}} = 3 \times 10^4 \, rad/s. \]

Thus, the angular frequency is \( 3 \times 10^4 \, rad/s \). Quick Tip: The angular frequency for resonance in an LCR circuit is given by \( \omega_0 = \frac{1}{\sqrt{LC}} \), where \( L \) is the inductance and \( C \) is the capacitance.

The spontaneity of a reaction is determined by the Gibbs free energy \( \Delta G = \Delta H - T \Delta S \).

- For \( \Delta H > 0 \) and \( \Delta S < 0 \), the reaction is non-spontaneous at any temperature.

- For \( \Delta H > 0 \) and \( \Delta S > 0 \), the reaction is spontaneous at low temperatures.

- For \( \Delta H < 0 \) and \( \Delta S < 0 \), the reaction is non-spontaneous at low temperatures.

- For \( \Delta H < 0 \) and \( \Delta S > 0 \), the reaction is spontaneous at any temperature.

Thus, the correct answers are (B) and (C) only. Quick Tip: The spontaneity of a reaction depends on both the enthalpy and the entropy change, and the effect of temperature can either make a reaction spontaneous or non-spontaneous.

The standard Gibbs free energy change \( \Delta G^\circ \) is related to the standard electrode potential \( E^\circ \) by: \[ \Delta G^\circ = -n F E^\circ, \]
where \( n \) is the number of electrons involved and \( F \) is Faraday's constant.

The cell with the most negative \( \Delta G^\circ \) corresponds to the most positive \( E^\circ \). The cell that gives the most negative value of \( \Delta G^\circ \) is the one with the highest \( E^\circ \) for the cathode and lowest for the anode, which is option (1). Quick Tip: To determine the cell with the most negative \( \Delta G^\circ \), consider the cell with the largest difference in standard electrode potentials, where the anode has the most negative \( E^\circ \) and the cathode has the most positive \( E^\circ \).

The alpha-helix and beta-pleated sheet are types of secondary structures in proteins. These structures arise from the folding of the polypeptide chain due to hydrogen bonding between the backbone atoms. The primary structure is the linear sequence of amino acids, while the tertiary structure refers to the overall three-dimensional shape of the protein. The quaternary structure refers to the arrangement of multiple polypeptide chains.

Thus, the correct answer is secondary structure. Quick Tip: The secondary structure of proteins includes alpha-helices and beta-pleated sheets, which are stabilized by hydrogen bonds between backbone atoms.

- Statement (I): Formaldehyde (\( CH_2O \)) has small substituents, and the small size of the substituents leads to faster hydration due to less steric hindrance around the carbonyl group. The equilibrium constant \( K \) for the hydration of formaldehyde is about 2280, indicating that hydration is relatively fast.

- Statement (II): Trichloroacetaldehyde (\( CCl_3CHO \)) has three chlorine atoms attached to the alpha-carbon. The chlorine atoms withdraw electron density from the carbonyl carbon through the -I effect, making the carbonyl carbon less electrophilic and slowing down hydration. The equilibrium constant \( K \) for trichloroacetaldehyde is around 2000, indicating slower hydration.

Both statements are true. Quick Tip: The rate of hydration and the equilibrium constant depend on the electron-withdrawing or electron-donating effects of substituents on the carbonyl group. Small substituents favor faster hydration.

The equilibrium constant \( K_p \) for the reaction is given by: \[ K_p = \frac{p_{X_2} \cdot p_{Y_2}^{1/2}}{p_{X_2Y}}. \]
Using the degree of dissociation \( x \) for the reactant \( X_2Y \), the partial pressures at equilibrium are: \[ p_{X_2Y} = p_0 - x \cdot p_0, \quad p_{X_2} = x \cdot p_0, \quad p_{Y_2} = \frac{x \cdot p_0}{2}. \]
Substituting these into the expression for \( K_p \) and simplifying for small \( x \), we get: \[ x = \sqrt{\frac{2K_p}{p}}. \]

Thus, the correct answer is \( x = \sqrt{\frac{2K_p}{p}} \). Quick Tip: For reactions with small degrees of dissociation, the equilibrium constant can be used to relate the degree of dissociation to the partial pressures of the products and reactants.

- Step 1 (A): When lead (Pb) reacts with nitric acid (\( HNO_3 \)), lead nitrate (\( Pb(NO_3)_2 \)) is formed: \[ Pb + 2 HNO_3 \rightarrow Pb(NO_3)_2 + H_2. \]

- Step 2 (B): When lead nitrate reacts with sulfuric acid (\( H_2SO_4 \)), lead sulfate (\( PbSO_4 \)) is formed: \[ Pb(NO_3)_2 + H_2SO_4 \rightarrow PbSO_4 + 2 HNO_3. \]

- Step 3 (C): When lead sulfate (\( PbSO_4 \)) reacts with potassium chromate (\( K_2CrO_4 \)), a yellow precipitate of lead chromate (\( PbCrO_4 \)) is formed: \[ PbSO_4 + K_2CrO_4 \rightarrow PbCrO_4 (yellow ppt) + K_2SO_4. \]

Thus, A is Pb, B is PbSO₄, and C is PbCrO₄ (yellow ppt). Quick Tip: The reaction sequence involves the formation of lead nitrate, lead sulfate, and lead chromate. This is a typical sequence for testing the presence of lead in reactions with chromates.

- Statement (I): The boiling points of alcohols and phenols increase with an increase in the number of carbon atoms due to the increase in van der Waals forces (intermolecular forces) as the molecular size increases. This is a correct statement.

- Statement (II): Alcohols and phenols have higher boiling points compared to ethers and haloalkanes due to the presence of hydrogen bonding between molecules, which requires more energy to break. This is also true.

Therefore, both statements are true. Quick Tip: The boiling points of compounds with hydrogen bonding, such as alcohols and phenols, are generally higher compared to compounds like ethers and haloalkanes, which lack hydrogen bonding.

The relationship between the decrease in vapour pressure and mole fraction of solute is given by Raoult’s Law: \[ \Delta P = P_0 \cdot x_{solute}, \]
where:
- \( \Delta P \) is the decrease in vapour pressure,
- \( P_0 \) is the initial vapour pressure,
- \( x_{solute} \) is the mole fraction of the solute.

Given:
- \( \Delta P = 10 \, mm Hg \) and \( x_{solute} = 0.2 \),
- The new decrease in vapour pressure is 20 mm Hg, so the mole fraction of the solute at this point will be \( x_{solute} = 0.4 \).

Since the sum of mole fractions of solute and solvent must be 1: \[ x_{solvent} = 1 - x_{solute} = 1 - 0.4 = 0.6. \]

Therefore, the mole fraction of the solvent is 0.6. Quick Tip: Raoult’s law relates the decrease in vapour pressure to the mole fraction of the solute in the solution. The mole fraction of the solvent can be found by subtracting the solute’s mole fraction from 1.

- Statement (I): For a given shell with principal quantum number \( n \), the total number of orbitals is indeed \( n^2 \), as each subshell (s, p, d, f) has \( l \) values, and the number of orbitals per subshell is \( 2l + 1 \). Therefore, the total number of orbitals for a shell is \( n^2 \).

- Statement (II): For any subshell with angular momentum quantum number \( l \), the spatial orientations of the orbitals are given by values from \( -l \) to \( +l \) including zero, which is also true.

Thus, both statements are true. Quick Tip: For any shell with quantum number \( n \), the total number of orbitals is \( n^2 \). For a subshell with quantum number \( l \), the possible values for the magnetic quantum number \( m_l \) range from \( -l \) to \( +l \), including zero.

The rate of solvolysis of alkyl halides depends on the stability of the carbocation intermediate formed during the reaction. The more stable the carbocation, the faster the solvolysis reaction. The relative stability of the carbocations is influenced by the type of carbon and the substituent groups attached to it.

- (A) Cyclohexyl bromide forms a less stable carbocation, so its rate of solvolysis is slower.

- (B) Benzyl bromide forms a highly stabilized carbocation due to resonance with the benzene ring, leading to a faster solvolysis.

- (C) Phenylmethyl bromide also forms a stable carbocation but less stable than the benzyl carbocation.

- (D) Allyl bromide forms a very stable carbocation due to resonance, making it the most reactive in solvolysis.

Thus, the correct ascending order is \( (C) < (B) < (A) < (D) \). Quick Tip: The rate of solvolysis is faster for compounds that form more stable carbocations, with allyl and benzyl carbocations being more stable than cyclohexyl carbocations.

- (A) Cyclohexene derivative: Ozonolysis leads to two aldehyde groups, hence it matches with (II).

- (B) 1,2-Dimethylcyclohexene: Ozonolysis leads to a diketone, hence it matches with (IV).

- (C) 1-Methylcyclohexene: Ozonolysis gives one aldehyde and one ketone, hence it matches with (I).

- (D) 1,4-Dimethylcyclohexene: Ozonolysis results in two aldehydes, hence it matches with (III).

Thus, the correct matching is (A)-(II), (B)-(IV), (C)-(I), (D)-(III). Quick Tip: Ozonolysis of alkenes results in cleavage of the double bond and formation of carbonyl compounds (aldehydes or ketones), depending on the structure of the alkene.

For a zero-order reaction, the rate law is: \[ Rate = k, \]
which means the rate is independent of the reactant concentration. The integrated rate law for a zero-order reaction is: \[ [A] = [A]_0 - kt. \]
This shows that the concentration of the reactant decreases linearly with time. Therefore, the graph of concentration versus time for a zero-order reaction is a straight line with a negative slope. Quick Tip: For zero-order reactions, the concentration decreases linearly with time, and the rate is constant over the course of the reaction.

- (A) Bronze: Bronze is an alloy of copper (Cu) and tin (Sn).

- (B) Brass: Brass is an alloy of copper (Cu) and zinc (Zn).

- (C) UK silver coin: UK silver coins are generally made with copper (Cu) and nickel (Ni).

- (D) Stainless Steel: Stainless steel is an alloy of iron (Fe), chromium (Cr), nickel (Ni), and carbon (C).

Thus, the correct matching is: (A)-(IV), (B)-(III), (C)-(I), (D)-(II). Quick Tip: Brass is Cu-Zn, bronze is Cu-Sn, and stainless steel is an alloy of Fe, Cr, Ni, and C.

- (A): \( [FeO_4^{2-} ] \) does not have a \( d^4 \) configuration.

- (B): \( [Mn(CN)_6]^{3-} \) has a \( d^4 \) configuration.

- (C): \( [Fe(CN)_6]^{3-} \) also has a \( d^4 \) configuration.

- (D): \( [Cr_2(O-C-Me)_4(H_2O)_2O] \) has a \( d^4 \) configuration.

- (E): \( [NiF_6^{2-} ] \) does not have a \( d^4 \) configuration.

Thus, the correct answer is (3) \( (B) \) and \( (D) \) only. Quick Tip: A \(d^4\) configuration typically arises in transition metal complexes like those involving Mn, Fe, and Cr in certain oxidation states.

The melting points of group 14 elements generally decrease as we move down the group. The elements with the atomic numbers 6 (carbon), 14 (silicon), 50 (tin), and 82 (lead) have decreasing melting points. Among them, tin (atomic number 50) has the lowest melting point.

Thus, the correct answer is (4) 50. Quick Tip: The melting point decreases as we go down group 14, with lead having the lowest melting point.

The pH of pure water depends on the ionization of water, which increases with temperature. At higher temperatures, the ionization of water increases, leading to a higher concentration of \( H^+ \) and \( OH^- \) ions, but since both increase equally, the pH will decrease.

Thus, the correct answer is (1) Decrease. Quick Tip: As temperature increases, the ionization of water increases, which leads to a decrease in pH.

- Step 1: Reaction of Chlorobenzene with NaOH:
When chlorobenzene reacts with NaOH at high temperature and pressure (300 atm, 623K), it undergoes nucleophilic substitution to form phenol (C₆H₅OH). Thus, the first product, [A], is phenol.

- Step 2: Oxidation of Phenol:
Phenol, when treated with chromic acid (Na₂Cr₂O₇) in an acidic medium, gets oxidized to benzoic acid (C₆H₅COOH). Thus, the second product, [B], is benzoic acid.

Therefore, [A] = \text{C-6\text{H-5\text{OH (Phenol) and [B] = \text{C-6\text{H-5\text{COOH (Benzoic acid). Quick Tip: The high temperature and pressure conditions favor the nucleophilic substitution of chlorine in chlorobenzene, leading to phenol. Oxidation with Na\(_2\)Cr\(_2\)O\(_7\) then converts phenol to benzoic acid.

For a binary solution of volatile components, the relationship between the mole fractions of components in the liquid (\( x_1 \)) and vapor phases (\( y_1 \)) is governed by Raoult's Law:
\[ y_1 = \frac{p_1^0}{p_1} x_1, \]

Where \( p_1^0 \) is the vapor pressure of component 1, and \( p_1 \) is the partial pressure of component 1. Rearranging the equation gives a linear relationship:
\[ \frac{1}{x_1} = \left( \frac{p_1^0}{p_2^0} - \frac{p_1^0}{p_2^0} \right) \frac{1}{y_1}. \]

Thus, the slope is given by \( \frac{p_1^0}{p_2^0} - \frac{p_1^0}{p_2^0} \), and the correct answer is option (1). Quick Tip: For binary solutions, Raoult's Law governs the relationship between mole fractions in the liquid and vapor phases. The slope of the plot of \( \frac{1}{x_1} \) vs \( \frac{1}{y_1} \) depends on the vapor pressures of the components.

- Statement (I): A plot of frequency \( \nu \) vs atomic mass is not a straight line. The relationship between atomic mass and X-ray emission frequency is not linear. This statement is false.
- Statement (II): A plot of frequency \( \nu \) vs atomic number is also not a straight line. The frequency of X-rays emitted generally follows Moseley's law, which is a square root function of the atomic number, not a linear function. This statement is false.

Thus, both statements are false. Quick Tip: X-ray emission frequency typically follows Moseley's law, which indicates a square root dependence on the atomic number.

- The first reaction shows the reaction of \( K_2Cr_2O_7 \) (potassium dichromate) with KOH (potassium hydroxide) under heat. The product formed is potassium chromate \( K_2CrO_4 \), which is the compound [A].

- In the second reaction, \( K_2CrO_4 \) reacts with sulfuric acid (H2SO4), resulting in chromium trioxide \( Cr_2O_3 \) as the product [B].

Thus, the correct products are [A] = \( K_2CrO_4 \) and [B] = \( K_2Cr_2O_7 \). Quick Tip: When potassium chromate reacts with sulfuric acid, it produces chromium trioxide. The reaction typically forms an orange solution due to the presence of \( K_2Cr_2O_7 \).

Given that the compound contains 10% hydrogen, we can assume the molar mass of the compound is \( M_X \).

- Mass of hydrogen in 0.01 mole of X = \( 0.01 \times 10 = 0.1 g \).

- In the complete combustion of X, the hydrogen reacts with oxygen to form H₂O.

- The number of moles of water formed is \( \frac{0.9}{18} = 0.05 moles \).

- In 1 mole of H₂O, there are 2 moles of hydrogen atoms. So, the moles of hydrogen atoms that reacted are \( 2 \times 0.05 = 0.1 moles \).

The number of moles of hydrogen in 0.01 mole of X is 0.1 g, which gives the molar mass of X as:
\[ Molar Mass of X = \frac{Mass of X}{Number of moles of X} = \frac{0.01}{0.01} = 100 g/mol. \]

Thus, the molar mass of X is 100 g/mol. Quick Tip: The amount of hydrogen in a compound can be determined from the water produced in combustion. Use stoichiometry to find the molar mass.

The reaction sequence involves the diazotization of an amine group, followed by further reactions.

In the final product [C], which is the major product, the total number of sp³ hybridized carbon atoms is 4, as the product structure involves 4 sp³ hybridized carbons.

Thus, the correct answer is 4. Quick Tip: Count the number of carbons connected to single bonds (sp³ hybridized) to determine the number of sp³ hybridized carbon atoms in the product.

The reaction between aluminium and oxygen is: \[ 4Al + 3O_2 \rightarrow 2Al_2O_3. \]

- Moles of aluminium = \( \frac{81.0}{27.0} = 3 moles \).
- Moles of oxygen = \( \frac{128.0}{32.0} = 4 moles \).

From the stoichiometry of the reaction, 4 moles of aluminium reacts with 3 moles of oxygen to produce 2 moles of aluminium oxide.
Thus, 3 moles of aluminium will produce:
\[ \frac{2}{4} \times 3 = 1.5 moles of Al_2O_3. \]

The molar mass of aluminium oxide \( Al_2O_3 \) is:
\[ Molar mass of Al_2O_3 = 2(27.0) + 3(16.0) = 102.0 g/mol. \]

Thus, the mass of aluminium oxide produced is:
\[ 1.5 \times 102.0 = 153 g. \]

The mass of aluminium oxide produced is 153 g. Quick Tip: Use stoichiometry to calculate the mass of the product based on the limiting reagent in the reaction.

The bond dissociation enthalpy of \( X_2 \) can be calculated using the Born-Haber cycle and the enthalpies provided:
\[ \Delta H_{bond dissociation} = \Delta H_{sub} + \Delta H_{i} + \Delta H_{eg} - \Delta H_{lattice} - \Delta H_{f}. \]

Substituting the values:
\[ \Delta H_{bond dissociation} = 100 + 500 - 300 - 800 - (-400) = 200 \, kJ/mol. \]

Thus, the bond dissociation enthalpy is 200 kJ/mol. Quick Tip: To calculate bond dissociation enthalpy, use the Born-Haber cycle, which involves lattice enthalpy, ionization, electron affinity, and formation enthalpies.

The structure of compound X involves bonds in its molecular form and when it undergoes hydrogenation and oxidation. By analyzing the structure, the total number of sigma bonds present is calculated to be 27.

Thus, the correct answer is 27. Quick Tip: Count the sigma bonds in each part of the molecule to determine the total number of sigma bonds.