JEE Advanced 2023

Analysis of the given sets:
Domain \( S = (0, 1) \cup (1, 2) \cup (3, 4) \). This set is a union of three disjoint open intervals. It contains infinitely many elements (cardinality of the continuum).
Codomain \( T = \{0, 1, 2, 3\} \). This set is finite with 4 elements.

Evaluating Statement (A):
The number of functions from a set \( S \) to a set \( T \) is given by \( |T|^{|S|} \). Here, \( |S| = \infty \) and \( |T| = 4 \).
Total functions = \( 4^{\infty} \), which is infinite.
Statement (A) is True.

Evaluating Statement (B):
A strictly increasing function \( f: S \to T \) requires that for any distinct \( x_1, x_2 \in S \) with \( x_1 < x_2 \), \( f(x_1) < f(x_2) \). This implies \( f \) must be injective (one-to-one).
By the Pigeonhole Principle, it is impossible to map an infinite set injectively into a finite set of 4 elements. Therefore, no strictly increasing function exists.
Statement (B) is False.

Evaluating Statement (C):
The domain \( S \) consists of 3 connected components (the intervals). The codomain \( T \) is a discrete set (in the standard topology). A continuous function mapping a connected space to a discrete space must be constant.
Thus, any continuous function \( f \) must be constant on each of the three intervals:
- \( f((0,1)) = c_1 \in T \)
- \( f((1,2)) = c_2 \in T \)
- \( f((3,4)) = c_3 \in T \)
For each interval, there are 4 choices for the constant value. Total continuous functions = \( 4 \times 4 \times 4 = 64 \).
Since \( 64 \le 120 \), this statement is correct.
Statement (C) is True.

Evaluating Statement (D):
As derived in (C), every continuous function from \( S \) to \( T \) is piecewise constant (locally constant). The derivative of a constant is 0. Thus, \( f'(x) = 0 \) for all \( x \in S \). Since the derivative exists everywhere in the domain, the function is differentiable.
Statement (D) is True.

Correct Options: A, C, D

Step 1: Determine Slope of Common Tangents
Equation of Parabola \( P: y^2 = 12x \implies 4a=12 \implies a=3 \).
Tangent to Parabola: \( y = mx + \frac{3}{m} \) ... (i)
Equation of Ellipse \( E: \frac{x^2}{6} + \frac{y^2}{3} = 1 \).
Tangent to Ellipse: \( y = mx \pm \sqrt{6m^2 + 3} \) ... (ii)
For common tangents, compare constants:
\( \frac{3}{m} = \pm \sqrt{6m^2 + 3} \implies \frac{9}{m^2} = 6m^2 + 3 \)
\( 6m^4 + 3m^2 - 9 = 0 \implies 2m^4 + m^2 - 3 = 0 \)
\( (2m^2 + 3)(m^2 - 1) = 0 \)
Since \( m^2 \ge 0 \), we have \( m^2 = 1 \implies m = \pm 1 \).

Step 2: Find Intersection of Tangents
Tangents are \( y = x + 3 \) (for \( m=1 \)) and \( y = -x - 3 \) (for \( m=-1 \)).
Intersection: \( x + 3 = -x - 3 \implies 2x = -6 \implies x = -3, y = 0 \).
Intersection Point is \( (-3, 0) \). Thus, (C) is True.

Step 3: Find Points of Contact & Area
On Parabola \( (a=3) \): Point \( (\frac{a}{m^2}, \frac{2a}{m}) \).
For \( m=1 \), \( A_1 = (3, 6) \). For \( m=-1 \), \( A_4 = (3, -6) \).
On Ellipse: Tangent \( x - y + 3 = 0 \). Tangent at \( (x_1, y_1) \) is \( \frac{x x_1}{6} + \frac{y y_1}{3} - 1 = 0 \).
Comparing coeffs: \( \frac{x_1/6}{1} = \frac{y_1/3}{-1} = \frac{-1}{3} \implies x_1 = -2, y_1 = 1 \). So \( A_2 = (-2, 1) \).
Similarly for \( m=-1 \), \( A_3 = (-2, -1) \).

Area of Quadrilateral \( A_1A_2A_3A_4 \):
Vertices: \( (3,6), (-2,1), (-2,-1), (3,-6) \).
This is a trapezoid with parallel vertical sides of lengths \( 6 - (-6) = 12 \) and \( 1 - (-1) = 2 \).
Height (horizontal distance) = \( 3 - (-2) = 5 \).
Area = \( \frac{1}{2} (12 + 2) \times 5 = 35 \) sq. units.
Thus, (A) is True.

Correct Options: A, C

Step 1: Calculate Base Areas
The function is \( f(x) = \frac{x^3}{3} - x^2 + \frac{5x}{9} + \frac{17}{36} \).
Area of the Red region (\( A_R \)) is the area under the curve in \( [0,1] \):
\( A_R = \int_0^1 f(x) dx = [\frac{x^4}{12} - \frac{x^3}{3} + \frac{5x^2}{18} + \frac{17x}{36}]_0^1 = \frac{1}{12} - \frac{1}{3} + \frac{5}{18} + \frac{17}{36} = 0.5 \).
Since the total area of the square is \( 1 \times 1 = 1 \), the Green area \( A_G = 1 - 0.5 = 0.5 \).

Step 2: Analyze Area Relations with Line \( L_h \)
Let \( A_R(h) \) and \( A_G(h) \) be the red and green areas below the line \( y=h \).
The total area below \( y=h \) in the square is \( h \). So, \( A_R(h) + A_G(h) = h \).
Let \( A_R^{above} \) and \( A_G^{above} \) be areas above the line.

Checking Statement (B): \( A_R^{above} = A_R(h) \)
This implies \( A_R(h) = 0.5 A_R = 0.25 \).
At \( h=0.25 \), since \( \min f(x) \approx 0.36 > 0.25 \), the line is entirely below the curve. So \( A_R(0.25) = 0.25 \times 1 = 0.25 \). Since \( 0.25 \in [1/4, 2/3] \), this is possible. (True)

Checking Statement (C): \( A_G^{above} = A_R(h) \)
\( 0.5 - A_G(h) = A_R(h) \implies A_G(h) + A_R(h) = 0.5 \).
Since \( A_G(h) + A_R(h) = h \), this implies \( h=0.5 \). Since \( 0.5 \in [1/4, 2/3] \), this is possible. (True)

Checking Statement (D): \( A_R^{above} = A_G(h) \)
\( 0.5 - A_R(h) = A_G(h) \implies A_G(h) + A_R(h) = 0.5 \).
This also implies \( h=0.5 \). (True)

Checking Statement (A): \( A_G^{above} = A_G(h) \)
This implies \( A_G(h) = 0.25 \).
\( A_G(h) = h - A_R(h) \). At max \( h=2/3 \), the line is above the curve, so \( A_R(2/3)=0.5 \).
\( A_G(2/3) = 2/3 - 0.5 \approx 0.167 \). Since \( A_G(h) \) increases with \( h \) and its max is \( 0.167 < 0.25 \), it never reaches 0.25. (False)

Correct Options: B, C, D

Step 1: Understand the definition of \(f(x)\)
The function \(f(x)\) is defined piecewise on intervals \(\left[\frac{1}{n+1}, \frac{1}{n}\right)\). For \(x\) in this interval, \(f(x) = \sqrt{n}\). As \(x \to 0\), \(n \to \infty\). Specifically, \(n \approx \frac{1}{x}\).

Step 2: Analyze the bounds for \(g(x)\)
We are given the inequality: \[ \int_{x^2}^x \sqrt{\frac{1-t}{t}} dt < g(x) < 2\sqrt{x} \] Let 's evaluate the integral on the left side, denoted as \(I(x)\), as \(x \to 0\). Substitute \(t = u^2 \implies dt = 2u du\). The limits change from \(x^2 \to x\) and \(x \to \sqrt{x}\). \[ I(x) = \int_{x}^{\sqrt{x}} \frac{\sqrt{1-u^2}}{u} \cdot 2u du = 2 \int_{x}^{\sqrt{x}} \sqrt{1-u^2} du \] For small \(x\) (and thus small \(u\)), \(\sqrt{1-u^2} \approx 1\). So, \(I(x) \approx 2 \int_{x}^{\sqrt{x}} 1 du = 2(\sqrt{x} - x) = 2\sqrt{x} - 2x\).

Step 3: Apply the Squeeze Theorem to the limit
We need to find \(L = \lim_{x \to 0} f(x)g(x)\). Let \(x \in \left[\frac{1}{n+1}, \frac{1}{n}\right)\). Then \(f(x) = \sqrt{n}\). Multiply the inequality for \(g(x)\) by \(f(x) = \sqrt{n}\): \[ \sqrt{n} (2\sqrt{x} - 2x) < f(x)g(x) < \sqrt{n}(2\sqrt{x}) \] Since \(x < \frac{1}{n}\), we have \(\sqrt{x} < \frac{1}{\sqrt{n}}\). Also, as \(n \to \infty\), \(x \approx \frac{1}{n}\), so \(\sqrt{x} \approx \frac{1}{\sqrt{n}}\). Evaluating the upper bound limit: \[ \lim_{n \to \infty} \sqrt{n} \cdot 2\left(\frac{1}{\sqrt{n}}\right) = 2 \] Evaluating the lower bound limit: \[ \lim_{n \to \infty} \sqrt{n} \cdot 2\left(\frac{1}{\sqrt{n}} - \frac{1}{n}\right) = \lim_{n \to \infty} \left(2 - \frac{2}{\sqrt{n}}\right) = 2 \] Since both bounds converge to 2, by the Squeeze Theorem, \(\lim_{x \to 0} f(x)g(x) = 2\).

Correct Option: C

Step 1: Visualize the Cube and Diagonal Sets
Let the cube vertices be \((x,y,z)\) where coordinates are 0 or 1. Set \(S\) consists of the 4 main diagonals (e.g., connecting \((0,0,0)\) to \((1,1,1)\)). Set \(F\) consists of the 12 face diagonals (e.g., connecting \((1,0,0)\) to \((0,1,0)\) on the \(z=0\) face). We need to find the maximum shortest distance between a line in \(S\) and a line in \(F\).

Step 2: Choose Representative Lines
Let's pick a main diagonal \(L_1 \in S\): The line \(OG\) passing through \(O(0,0,0)\) and \(G(1,1,1)\). Vector direction \(\vec{b_1}=( 1, 1, 1)\). We want to maximize the distance to a face diagonal. Face diagonals that share a vertex with \(OG\) intersect it, so the distance is 0. We must choose a face diagonal skew to \(OG\). Consider the face diagonal \(AC\) on the bottom face (\(z=0\)), connecting \(A(1,0,0)\) and \(C(0,1,0)\). Vector direction \(\vec{b_2}=( 0-1, 1-0, 0-0)=( -1, 1, 0)\). Point on \(L_1\): \(\vec{a_1}=( 0,0,0)\). Point on \(L_2\): \(\vec{a_2}=( 1,0,0)\).

Step 3: Calculate Shortest Distance
The shortest distance \(d\) between skew lines is given by: \[ d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right| \] Calculate \(\vec{b_1} \times \vec{b_2}\): \[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ -1 & 1 & 0 \end{vmatrix} = \hat{i}(0-1) - \hat{j}(0-(-1)) + \hat{k}(1-(-1)) = -\hat{i} - \hat{j} + 2\hat{k} \] Magnitude: \(|\vec{n}| = \sqrt{(-1)^2 + (-1)^2 + 2^2} = \sqrt{1+1+4} = \sqrt{6}\). Calculate dot product with \(\vec{a_2} - \vec{a_1} = (1, 0, 0)\): \[ (1, 0, 0) \cdot (-1, -1, 2) = -1 \] Shortest distance \(d = \frac{|-1|}{\sqrt{6}} = \frac{1}{\sqrt{6}}\). By symmetry, any non-intersecting pair of main and face diagonals will yield this distance. The intersecting ones yield 0. Thus, the maximum distance is \(\frac{1}{\sqrt{6}}\).

Correct Option: A

Step 1: Determine the set of points \(X\)
We need integer coordinates \((x,y)\) satisfying: 1. \(\frac{x^2}{8} + \frac{y^2}{20} < 1\) 2. \(y^2 < 5x\)
From (2), \(x\) must be positive. From (1), \(x^2 < 8 \implies x \in \{1, 2\}\). For \(x=1\): (1) \(\frac{1}{8} + \frac{y^2}{20} < 1 \implies \frac{y^2}{20} < \frac{7}{8} \implies y^2 < 17.5\). (2) \(y^2 < 5\). Integers satisfying \(y^2 < 5\) are \(y \in \{-2, -1, 0, 1, 2\}\). (5 points). For \(x=2\): (1) \(\frac{4}{8} + \frac{y^2}{20} < 1 \implies \frac{y^2}{20} < 0.5 \implies y^2 < 10\). (2) \(y^2 < 10\). Integers satisfying \(y^2 < 10\) are \(y \in \{-3, -2, -1, 0, 1, 2, 3\}\). (7 points). Total points \(n(X)=5 + 7=1 2\). Total ways to choose 3 points: \(^{12}C_3=\ frac{12 \times 11 \times 10}{6}=2 20\).

Step 2: Condition for Area to be a Positive Integer
Let vertices be \((x_1, y_1), (x_2, y_2), (x_3, y_3)\). Area \(\Delta = \frac{1}{2} | \sum (x_i y_{i+1} - x_{i+1} y_i) |\). Since all \(x_i \in \{1, 2\}\), for \(\Delta\) to be an integer, \(2\Delta\) (which is always an integer) must be even. Also, points must not be collinear (Area > 0). Since points lie on two vertical lines \(x=1\) and \(x=2\), any 3 points chosen such that they don't all lie on one line form a triangle with positive area. We must choose 2 points from one line and 1 from the other. Formula for \(2\Delta\) simplifies to \(|y_i - y_j|\) where \(y_i, y_j\) are the y-coordinates of the two points on the same vertical line. Thus, we need \(|y_i - y_j|\) to be an even number. This implies \(y_i\) and \(y_j\) must have the same parity (both odd or both even).

Step 3: Count Favorable Cases
Case A: 2 points from \(x=1\) (5 pts) and 1 point from \(x=2\) (7 pts). Points on \(x=1\): Evens \(\{-2, 0, 2\}\) (3), Odds \(\{-1, 1\}\) (2). Ways to choose pair with same parity: \(^3C_2 + ^2C_2 = 3 + 1 = 4\). Total ways = \(4 \times 7 = 28\). Case B: 2 points from \(x=2\) (7 pts) and 1 point from \(x=1\) (5 pts). Points on \(x=2\): Evens \(\{-2, 0, 2\}\) (3), Odds \(\{-3, -1, 1, 3\}\) (4). Ways to choose pair with same parity: \(^3C_2 + ^4C_2 = 3 + 6 = 9\). Total ways = \(9 \times 5 = 45\).

Total favorable outcomes = \(28 + 45 = 73\). Probability = \(\frac{73}{220}\).

Correct Option: B

Step 1: Identify Coordinates and Normal Equation
Let the point \( P \) on the parabola \( y^2 = 4ax \) be \( (am^2, -2am) \). The slope of the normal at this point is \( m \). The equation of the normal to the parabola \( y^2 = 4ax \) with slope \( m \) is given by: \[ y = mx - 2am - am^3 \] The focus \( F \) of the parabola is at \( (a, 0) \).

Step 2: Find Coordinates of Point Q
The normal meets the x-axis at point \( Q \). To find the coordinates of \( Q \), set \( y = 0 \) in the normal equation: \[ 0 = mx - 2am - am^3 \] \[ mx = 2am + am^3 \] \[ x = 2a + am^2 \] So, the coordinates of \( Q \) are \( (2a + am^2, 0) \).

Step 3: Calculate the Area of Triangle PFQ
The triangle \( PFQ \) has vertices \( P(am^2, -2am) \), \( F(a, 0) \), and \( Q(2a + am^2, 0) \). The base of the triangle lies on the x-axis and is the length of segment \( FQ \): \[ \text{Base } FQ = |x_Q - x_F| = |(2a + am^2) - a| = |a + am^2| = a(1+m^2) \] The height of the triangle is the absolute y-coordinate of \( P \): \[ \text{Height} = |-2am| = 2am \] The area of \( \Delta PFQ \) is: \[ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times a(1+m^2) \times 2am = a^2 m(1+m^2) \] Given that the area is 120, we have: \[ a^2 m(1+m^2) = 120 \]

Step 4: Solve for Integers a and m
We are given that \( a \) and \( m \) are positive integers. We can test integer values for \( m \):

• If \( m=1 \): \( a^2(1)(1^2+1) = 2a^2 = 120 \implies a^2 = 60 \) (not a perfect square).
• If \( m=2 \): \( a^2(2)(2^2+1) = 10a^2 = 120 \implies a^2 = 12 \) (not a perfect square).
• If \( m=3 \): \( a^2(3)(3^2+1) = 30a^2 = 120 \implies a^2 = 4 \implies a = 2 \). This is a valid integer solution.

Thus, the pair is \( (a, m) = (2, 3) \).

Correct Option: A

Step 1: Simplify the Given Equation
The equation is \( \sqrt{1 + \cos(2x)} = \sqrt{2} \tan^{-1}(\tan x) \). Using the identity \( 1 + \cos(2x) = 2\cos^2 x \), we get: \[ \sqrt{2\cos^2 x} = \sqrt{2} \tan^{-1}(\tan x) \] \[ \sqrt{2}|\cos x| = \sqrt{2} \tan^{-1}(\tan x) \] \[ |\cos x| = \tan^{-1}(\tan x) \] Since \( |\cos x| \ge 0 \), we must look for solutions where \( \tan^{-1}(\tan x) \ge 0 \).

Step 2: Analyze the Behavior of \( y = \tan^{-1}(\tan x) \)
The function \( f(x) = \tan^{-1}(\tan x) \) is periodic with period \( \pi \) and is defined as: \[ f(x) = x - k\pi \quad \text{for } x \in \left(k\pi - \frac{\pi}{2}, k\pi + \frac{\pi}{2}\right) \]

Step 3: Analyze Intersections in Each Interval
We examine the given set \( S = \left(-\frac{3\pi}{2}, -\frac{\pi}{2}\right) \cup \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \frac{3\pi}{2}\right) \).

• Interval \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \):
  Here, \( \tan^{-1}(\tan x) = x \). The equation becomes \( |\cos x| = x \). Since LHS \( \ge 0 \), we need \( x \ge 0 \). The graph of \( y = \cos x \) starts at 1 and decreases to 0 at \( \pi/2 \approx 1.57 \). The line \( y = x \) goes from 0 to 1.57. They intersect exactly once in \( (0, \pi/2) \). (1 Solution)
• Interval \( \left(\frac{\pi}{2}, \frac{3\pi}{2}\right) \):
  Here, \( \tan^{-1}(\tan x) = x - \pi \). The equation becomes \( |\cos x| = x - \pi \). For RHS \( \ge 0 \), we need \( x \ge \pi \). We focus on \( [\pi, 3\pi/2) \). At \( x = \pi \), LHS = \( |\cos \pi| = 1 \), RHS = 0. At \( x = 3\pi/2 \), LHS = 0, RHS = \( \pi/2 \approx 1.57 \). Since LHS starts higher and ends lower than RHS, and both are continuous, they intersect exactly once in this interval. (1 Solution)
• Interval \( \left(-\frac{3\pi}{2}, -\frac{\pi}{2}\right) \):
  Here, \( \tan^{-1}(\tan x) = x + \pi \). The equation becomes \( |\cos x| = x + \pi \). For RHS \( \ge 0 \), we need \( x \ge -\pi \). We focus on \( [-\pi, -\pi/2) \). At \( x = -\pi \), LHS = 1, RHS = 0. At \( x = -\pi/2 \), LHS = 0, RHS = \( \pi/2 \). Similar to the previous case, they intersect exactly once. (1 Solution)

Conclusion:
Total number of real solutions is \( 1 + 1 + 1 = 3 \).

Correct Answer: 3

Step 1: Analyze the Function Segments
The function \( f(x) \) is defined piecewise on \( [0, 1] \). It describes a series of linear segments forming triangles. Let's calculate the area under the curve for each segment.

1. Interval \( [0, \frac{1}{2n}] \):
   \( f(x) = n(1-2nx) \). At \( x=0 \), \( f(0) = n \). At \( x=\frac{1}{2n} \), \( f(\frac{1}{2n}) = 0 \). This forms a triangle with base \( \frac{1}{2n} \) and height \( n \). Area \( A_1 = \frac{1}{2} \times \frac{1}{2n} \times n = \frac{1}{4} \).
2. Interval \( [\frac{1}{2n}, \frac{3}{4n}] \):
   \( f(x) = 2n(2nx-1) \). At \( x=\frac{1}{2n} \), \( f=0 \). At \( x=\frac{3}{4n} \), \( f = 2n(\frac{3}{2}-1) = n \). This is the left half of a triangle. Area \( A_2 \). Base width \( \frac{3}{4n} - \frac{1}{2n} = \frac{1}{4n} \). Area \( A_2 = \frac{1}{2} \times \frac{1}{4n} \times n = \frac{1}{8} \).
3. Interval \( [\frac{3}{4n}, \frac{1}{n}] \):
   \( f(x) = 4n(1-nx) \). At \( x=\frac{3}{4n} \), \( f = 4n(0.25) = n \). At \( x=\frac{1}{n} \), \( f=0 \). This is the right half of the triangle. Base width \( \frac{1}{n} - \frac{3}{4n} = \frac{1}{4n} \). Area \( A_3 = \frac{1}{2} \times \frac{1}{4n} \times n = \frac{1}{8} \).
4. Interval \( [\frac{1}{n}, 1] \):
   \( f(x) = \frac{n}{n-1}(nx-1) \). At \( x=\frac{1}{n} \), \( f=0 \). At \( x=1 \), \( f = \frac{n}{n-1}(n-1) = n \). This forms a large triangle (slope upwards). Base \( 1 - \frac{1}{n} = \frac{n-1}{n} \). Height \( n \). Area \( A_4 = \frac{1}{2} \times \frac{n-1}{n} \times n = \frac{n-1}{2} \).

Step 2: Calculate Total Area and Solve for n
Total Area = \( A_1 + A_2 + A_3 + A_4 \) \[ \text{Total Area} = \frac{1}{4} + \frac{1}{8} + \frac{1}{8} + \frac{n-1}{2} \] \[ = \frac{1}{4} + \frac{1}{4} + \frac{n-1}{2} = \frac{1}{2} + \frac{n-1}{2} = \frac{n}{2} \] Given that the Area is 4: \[ \frac{n}{2} = 4 \implies n = 8 \]

Step 3: Find Maximum Value of f
The function oscillates between 0 and \( n \). The peak values occur at \( x=0 \), \( x=\frac{3}{4n} \), and \( x=1 \). At all these points, \( f(x) = n \). Since \( n=8 \), the maximum value of the function is 8.

Correct Answer: 8

Step 1: Understand the Series Terms
Let \( T_k \) be the \( k \)-th term of the series \( S = 77 + 757 + 7557 + \dots \). The general term with \( r \) fives (where the total number of digits is \( r+2 \)) can be written as: \[ T = 7 \times 10^{r+1} + 5(10^r + 10^{r-1} + \dots + 10^1) + 7 \] Using the geometric series sum formula for the middle part: \[ T_r = 7 \times 10^{r+1} + 5 \times \frac{10(10^r - 1)}{10 - 1} + 7 \] \[ T_r = 70 \times 10^r + \frac{50}{9}(10^r - 1) + 7 \] The series sums terms corresponding to numbers with 0 fives up to 98 fives (based on the brace in the problem image indicating the last term has 98 fives). Let's index by the power of 10. The first term is 77, which corresponds to \( r=0 \) in the formula above relative to the number of 5s. Let's use the solution's indexing where \( r \) represents the number of digits. The sum is from \( r=2 \) to \( r=100 \) (since last term has 98 fives + 2 sevens = 100 digits). \[ T_r (\text{r digits}) = 7 \times 10^{r-1} + \frac{50}{9}(10^{r-2} - 1) + 7 \] Simplifying \( T_r \): \[ T_r = 7 \cdot 10^{r-1} + \frac{5}{9} \cdot 10^{r-1} - \frac{50}{9} + 7 = \frac{68}{9} 10^{r-1} + \frac{13}{9} \] Step 2: Sum the Series
We need to sum 𝑇 𝑟 T r ​ for 𝑟 = 2 r=2 to 100 100 . There are 100 − 2 + 1 = 99 100−2+1=99 terms. [ S = \sum_{r=2}^{100} \left( \frac{68}{9} 10^{r-1} + \frac{13}{9} \right) ] [ S = \frac{68}{9} (10^1 + 10^2 + \dots + 10^{99}) + 99 \times \frac{13}{9} ] [ S = \frac{68}{9} \times \frac{10(10^{99}-1)}{9} + 143 ] [ S = \frac{680}{81} (10^{99}-1) + 143 ] [ S = \frac{680 \cdot 10^{99} - 680 + 11583}{81} = \frac{680 \cdot 10^{99} + 10903}{81} ] Step 3: Relate to the Form (\frac{75\dots57 + m}{n})
Let 𝑁 = 75 … 57 N=75…57 be the number with 99 fives (as per the brace in the numerator in the question). This number has 99 + 2 = 101 99+2=101 digits. Using the derived formula for a number with 𝑟 r digits, for 𝑟 = 101 r=101 : [ N = \frac{68}{9} 10^{100} + \frac{13}{9} ] We can express the leading term of 𝑆 S using 𝑁 N : 𝑆 ≈ 680 ⋅ 10 99 81 = 68 ⋅ 10 100 81 = 1 9 ( 68 9 10 100 ) S≈ 81 680⋅10 99 ​ = 81 68⋅10 100 ​ = 9 1 ​ ( 9 68 ​ 10 100 ) . Substitute 68 9 10 100 = 𝑁 − 13 9 9 68 ​ 10 100 =N− 9 13 ​ : [ S = \frac{1}{9} \left[ \frac{9(N - 13/9) + 10903}{9} \right] ] Wait, let's substitute directly into 𝑆 = 680 ⋅ 10 99 + 10903 81 S= 81 680⋅10 99 +10903 ​ : Multiply numerator and denominator by something? No, notice the denominator is 81. We want denominator 𝑛 n . Let's rewrite 𝑆 S in terms of 𝑁 N : [ S = \frac{68 \cdot 10^{100} + 10903}{81} = \frac{\frac{68}{9} 10^{100} \cdot 9 + 10903}{81} ] Substitute 68 9 10 100 = 𝑁 − 13 9 9 68 ​ 10 100 =N− 9 13 ​ : [ S = \frac{9(N - \frac{13}{9}) + 10903}{81} = \frac{9N - 13 + 10903}{81} = \frac{9N + 10890}{81} ] [ S = \frac{9(N + 1210)}{81} = \frac{N + 1210}{9} ] Comparing this to 𝑆 = 𝑁 + 𝑚 𝑛 S= n N+m ​ : 𝑛 = 9 n=9 and 𝑚 = 1210 m=1210 . Step 4: Find 𝑚 + 𝑛 m+n
[ m + n = 1210 + 9 = 1219 ]

Answer: 1219

Step 1: Simplify the Complex Expression
We are given \( A = \left\{ \frac{1967 + 1686 i \sin \theta}{7 - 3 i \cos \theta} : \theta \in \mathbb{R} \right\} \). Notice the factors of 281 in the numerator: \( 1967 = 281 \times 7 \) \( 1686 = 281 \times 6 \) So, the expression becomes: \[ z = 281 \frac{7 + 6 i \sin \theta}{7 - 3 i \cos \theta} \] Step 2: Realize the Denominator
Multiply the numerator and denominator by the conjugate of the denominator, 7 + 3 𝑖 cos ⁡ 𝜃 7+3icosθ : [ z = 281 \frac{(7 + 6 i \sin \theta)(7 + 3 i \cos \theta)}{(7 - 3 i \cos \theta)(7 + 3 i \cos \theta)} ] The denominator becomes 7 2 + ( 3 cos ⁡ 𝜃 ) 2 = 49 + 9 cos ⁡ 2 𝜃 7 2 +(3cosθ) 2 =49+9cos 2 θ . The numerator expands to: [ 281 [ (49 - 18 \sin \theta \cos \theta) + i(21 \cos \theta + 42 \sin \theta) ] ] Step 3: Condition for Integer Value
For 𝑧 z to be a positive integer 𝑛 n , the imaginary part must be zero. [ 21 \cos \theta + 42 \sin \theta = 0 ] [ 21(\cos \theta + 2 \sin \theta) = 0 ] [ \cos \theta = -2 \sin \theta ] Step 4: Calculate the Integer Value
Substitute cos ⁡ 𝜃 = − 2 sin ⁡ 𝜃 cosθ=−2sinθ into the real part of the expression: Real Part = 281 49 − 18 sin ⁡ 𝜃 ( − 2 sin ⁡ 𝜃 ) 49 + 9 ( − 2 sin ⁡ 𝜃 ) 2 =281 49+9(−2sinθ) 2 49−18sinθ(−2sinθ) ​ [ = 281 \frac{49 + 36 \sin^2 \theta}{49 + 36 \sin^2 \theta} ] The fraction simplifies to 1. Thus, 𝑛 = 281 × 1 = 281 n=281×1=281 .

Answer: 281

Step 1: Analyze Plane and Sphere Geometry
The plane \( P \) is given by \( \sqrt{3}x + 2y + 3z = 16 \). Distance of this plane from the origin: \[ d_{origin} = \frac{|-16|}{\sqrt{(\sqrt{3})^2 + 2^2 + 3^2}} = \frac{16}{\sqrt{3+4+9}} = \frac{16}{4} = 4 \] The set \( S \) contains unit vectors (points on unit sphere \( |\vec{r}|=1 \)) whose distance from plane \( P \) is \( \frac{7}{2} = 3.5 \). The unit normal to the plane is \( \hat{n} = \frac{\sqrt{3}\hat{i} + 2\hat{j} + 3\hat{k}}{4} \). The distance of a point \( \vec{r} \) from the plane \( \vec{r} \cdot \hat{n} = 4 \) is \( |\vec{r} \cdot \hat{n} - 4| \). So, \( |\vec{r} \cdot \hat{n} - 4| = 3.5 \). This gives two possibilities: \( \vec{r} \cdot \hat{n} - 4 = -3.5 \implies \vec{r} \cdot \hat{n} = 0.5 \) or \( \vec{r} \cdot \hat{n} - 4 = 3.5 \implies \vec{r} \cdot \hat{n} = 7.5 \). Since \( \vec{r} \) is a unit vector, \( \vec{r} \cdot \hat{n} \le 1 \), so the second case is impossible. Thus, the points lie on the plane \( \vec{r} \cdot \hat{n} = 1/2 \). The intersection of the sphere \( |\vec{r}|=1 \) and the plane \( \vec{r} \cdot \hat{n} = 1/2 \) is a circle. Step 2: Find Circle Properties and Triangle Area
The circle is at distance ℎ = 1 / 2 h=1/2 from the origin. Radius of the circle 𝑅 𝑐 𝑖 𝑟 𝑐 𝑙 𝑒 = 𝑅 𝑠 𝑝 ℎ 𝑒 𝑟 𝑒 2 − ℎ 2 = 1 2 − ( 1 / 2 ) 2 = 3 2 R circle ​ = R sphere 2 ​ −h 2 ​ = 1 2 −(1/2) 2 ​ = 2 3 ​ ​ . The vectors 𝑢 ⃗ , 𝑣 ⃗ , 𝑤 ⃗ u , v , w satisfy ∣ 𝑢 ⃗ − 𝑣 ⃗ ∣ = ∣ 𝑣 ⃗ − 𝑤 ⃗ ∣ = ∣ 𝑤 ⃗ − 𝑢 ⃗ ∣ ∣ u − v ∣=∣ v − w ∣=∣ w − u ∣ , meaning they form an equilateral triangle inscribed in this circle. Side length 𝑎 a of an equilateral triangle inscribed in a circle of radius 𝑟 r is 𝑎 = 𝑟 3 a=r 3 ​ . 𝑎 = 3 2 × 3 = 3 2 a= 2 3 ​ ​ × 3 ​ = 2 3 ​ . Area of this triangle Δ = 3 4 𝑎 2 = 3 4 ( 9 4 ) = 9 3 16 Δ= 4 3 ​ ​ a 2 = 4 3 ​ ​ ( 4 9 ​ )= 16 9 3 ​ ​ . Step 3: Calculate Volume of Parallelepiped
The volume 𝑉 V of the parallelepiped formed by vectors 𝑢 ⃗ , 𝑣 ⃗ , 𝑤 ⃗ u , v , w from the origin is given by the scalar triple product [ 𝑢 ⃗ 𝑣 ⃗ 𝑤 ⃗ ] [ u v w ] . This is equal to 6 × 6× Volume of the tetrahedron 𝑂 𝑈 𝑉 𝑊 OUVW . Volume of tetrahedron = 1 3 × Base Area × Height = 3 1 ​ ×Base Area×Height . Here, base is the equilateral triangle on the circle plane, and height is the distance of the origin from that plane ( ℎ = 1 / 2 h=1/2 ). 𝑉 𝑡 𝑒 𝑡 = 1 3 × 9 3 16 × 1 2 = 3 3 32 V tet ​ = 3 1 ​ × 16 9 3 ​ ​ × 2 1 ​ = 32 3 3 ​ ​ . Volume 𝑉 = 6 × 3 3 32 = 9 3 16 V=6× 32 3 3 ​ ​ = 16 9 3 ​ ​ . Step 4: Compute Final Value
We need to find 80 3 𝑉 3 ​ 80 ​ V . [ \frac{80}{\sqrt{3}} \times \frac{9\sqrt{3}}{16} = \frac{80}{16} \times 9 = 5 \times 9 = 45 ]

Answer: 45

Step 1: Analyze the first expansion

Consider the expansion of \(\left( ax^2 + \frac{70}{27bx} \right)^4\).

The general term \( T_{r+1} \) is given by:

\[ T_{r+1} = \,^4C_r (ax^2)^{4-r} \left( \frac{70}{27bx} \right)^r \]

\[ T_{r+1} = \,^4C_r \cdot a^{4-r} \cdot x^{2(4-r)} \cdot \left( \frac{70}{27b} \right)^r \cdot x^{-r} \]

Grouping the powers of \( x \):

\[ T_{r+1} = \,^4C_r \cdot a^{4-r} \cdot \left( \frac{70}{27b} \right)^r \cdot x^{8-2r-r} = \,^4C_r \cdot a^{4-r} \cdot \left( \frac{70}{27b} \right)^r \cdot x^{8-3r} \]

We need the coefficient of \( x^5 \). Thus, equate the exponent of \( x \) to 5:

\[ 8 - 3r = 5 \implies 3r = 3 \implies r = 1 \]

Now, calculate the coefficient for \( r = 1 \):

\[ \text{Coeff}_1 = \,^4C_1 \cdot a^{4-1} \cdot \left( \frac{70}{27b} \right)^1 = 4 \cdot a^3 \cdot \frac{70}{27b} = \frac{280 a^3}{27b} \]

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Step 2: Analyze the second expansion

Consider the expansion of \(\left( ax - \frac{1}{bx^2} \right)^7\).

The general term \( T_{r+1} \) is given by:

\[ T_{r+1} = \,^7C_r (ax)^{7-r} \left( -\frac{1}{bx^2} \right)^r \]

\[ T_{r+1} = \,^7C_r \cdot a^{7-r} \cdot x^{7-r} \cdot (-1)^r \cdot b^{-r} \cdot x^{-2r} \]

Grouping the powers of \( x \):

\[ T_{r+1} = \,^7C_r \cdot a^{7-r} \cdot \frac{(-1)^r}{b^r} \cdot x^{7-r-2r} = \,^7C_r \cdot a^{7-r} \cdot \frac{(-1)^r}{b^r} \cdot x^{7-3r} \]

We need the coefficient of \( x^{-5} \). Thus, equate the exponent of \( x \) to -5:

\[ 7 - 3r = -5 \implies 3r = 12 \implies r = 4 \]

Now, calculate the coefficient for \( r = 4 \):

\[ \text{Coeff}_2 = \,^7C_4 \cdot a^{7-4} \cdot \frac{(-1)^4}{b^4} \]

Since \( \,^7C_4 = \,^7C_3 = \frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1} = 35 \), we have:

\[ \text{Coeff}_2 = 35 \cdot a^3 \cdot \frac{1}{b^4} = \frac{35 a^3}{b^4} \]

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Step 3: Equate the coefficients and solve for \( b \)

Given that the coefficients are equal:

\[ \frac{280 a^3}{27b} = \frac{35 a^3}{b^4} \]

Since \( a \) and \( b \) are nonzero, we can cancel \( a^3 \) and one \( b \) from the denominator:

\[ \frac{280}{27} = \frac{35}{b^3} \]

Rearranging to solve for \( b^3 \):

\[ b^3 = \frac{35 \cdot 27}{280} \]

\[ b^3 = \frac{27}{8} \quad (\text{Since } \frac{35}{280} = \frac{1}{8}) \]

\[ b = \sqrt[3]{\frac{27}{8}} = \frac{3}{2} \]

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Step 4: Find the value of \( 2b \)

\[ 2b = 2 \cdot \frac{3}{2} = 3 \]

Answer: 3

The system of linear equations is:

\[ x + 2y + z = 7 \]

\[ x + 0y + \alpha z = 11 \]

\[ 2x - 3y + \beta z = \gamma \]

Step 1: Calculate the determinant of the coefficient matrix (\( \Delta \))

\[ \Delta = \begin{vmatrix} 1 & 2 & 1 \\ 1 & 0 & \alpha \\ 2 & -3 & \beta \end{vmatrix} \]

Expanding along row 1:

\[ \Delta = 1(0 - (-3\alpha)) - 2(\beta - 2\alpha) + 1(-3 - 0) \]

\[ \Delta = 3\alpha - 2\beta + 4\alpha - 3 = 7\alpha - 2\beta - 3 \]

The condition for a unique solution is \( \Delta \neq 0 \), i.e., \( \beta \neq \frac{7\alpha - 3}{2} \).

The condition for no solution or infinite solutions (when Cramer's rule fails) is \( \Delta = 0 \), i.e., \( \beta = \frac{7\alpha - 3}{2} \).

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Step 2: Calculate determinants \( \Delta_z \) for consistency checks

Using Cramer's Rule, let's find \( \Delta_z \):

\[ \Delta_z = \begin{vmatrix} 1 & 2 & 7 \\ 1 & 0 & 11 \\ 2 & -3 & \gamma \end{vmatrix} \]

Expanding along row 2 (or any row):

\[ \Delta_z = -1(2\gamma - (-21)) + 0 - 11(-3 - 4) \]

Let's stick to expanding along row 1 for consistency:

\[ \Delta_z = 1(0 - (-33)) - 2(\gamma - 22) + 7(-3 - 0) \]

\[ \Delta_z = 33 - 2\gamma + 44 - 21 = 56 - 2\gamma \]

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Step 3: Analyze each option in List-I

(P) If \( \beta = \frac{1}{2}(7\alpha - 3) \) and \( \gamma = 28 \)

• \( \beta = \frac{1}{2}(7\alpha - 3) \implies \Delta = 0 \).
• \( \gamma = 28 \implies \Delta_z = 56 - 2(28) = 0 \).
• When \( \Delta = 0 \) and all \( \Delta_x, \Delta_y, \Delta_z \) are 0, the system has infinitely many solutions.
• Checking \( \Delta_x \) with \( \gamma = 28 \) and \( 2\beta = 7\alpha - 3 \): \[ \Delta_x = 77\alpha - 22\beta - 33 = 77\alpha - 11(7\alpha - 3) - 33 = 0 \] Since \( \Delta = \Delta_x = \Delta_y = \Delta_z = 0 \), there are infinitely many solutions.
• Match: (P) \(\to\) (3)

(Q) If \( \beta = \frac{1}{2}(7\alpha - 3) \) and \( \gamma \neq 28 \)

• \( \beta = \frac{1}{2}(7\alpha - 3) \implies \Delta = 0 \).
• \( \gamma \neq 28 \implies \Delta_z = 56 - 2\gamma \neq 0 \).
• When \( \Delta = 0 \) and at least one of \( \Delta_x, \Delta_y, \Delta_z \) is non-zero, the system has no solution.
• Match: (Q) \(\to\) (2)

(R) If \( \beta \neq \frac{1}{2}(7\alpha - 3) \) where \( \alpha = 1 \) and \( \gamma \neq 28 \)

• \( \beta \neq \dots \implies \Delta \neq 0 \).
• \( \Delta \neq 0 \) guarantees a unique solution.
• The option "Unique solution" corresponds to (1).
• Match: (R) \(\to\) (1)

(S) If \( \beta \neq \frac{1}{2}(7\alpha - 3) \) where \( \alpha = 1 \) and \( \gamma = 28 \)

• \( \Delta \neq 0 \implies \) Unique solution.
• Let's find the specific solution for \( \alpha = 1 \) and \( \gamma = 28 \). Equations become: 1) \( x + 2y + z = 7 \) 2) \( x + z = 11 \implies z = 11 - x \) Substitute (2) into (1): \( x + 2y + (11 - x) = 7 \implies 2y = -4 \implies y = -2 \). Substitute into eq 3: \( 2x - 3y + \beta z = 28 \). This unique solution must satisfy the system regardless of \( \beta \) if it's consistent for all \( \beta \), but here we are finding the unique solution for the given parameters. Let's check if \( x=11, y=-2, z=0 \) (Option 4) works. Eq 1: \( 11 + 2(-2) + 0 = 7 \) (True) Eq 2: \( 11 + 1(0) = 11 \) (True) Eq 3: \( 2(11) - 3(-2) + \beta(0) = 22 + 6 = 28 \) (True) Since \( \gamma = 28 \), the unique solution is \( x=11, y=-2, z=0 \).
• Match: (S) \(\to\) (4)

Conclusion:

P \(\to\) 3, Q \(\to\) 2, R \(\to\) 1, S \(\to\) 4.

This corresponds to Option (A).

First, arrange the data in ascending order of \( x_i \) and create the frequency table:

\( x_i \)	3	4	5	8	10	11	Total
\( f_i \)	5	4	4	2	2	3	20

(P) Mean (\( \bar{x} \))

\[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \]

\[ \sum f_i x_i = (3 \times 5) + (4 \times 4) + (5 \times 4) + (8 \times 2) + (10 \times 2) + (11 \times 3) \]

\[ = 15 + 16 + 20 + 16 + 20 + 33 = 120 \]

\[ \bar{x} = \frac{120}{20} = 6 \]

Match: (P) \(\to\) (3)

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(Q) Median

Total frequency \( N = 20 \) (even).

The median is the average of the \( (\frac{N}{2}) \)-th and \( (\frac{N}{2} + 1) \)-th observations, i.e., the 10th and 11th terms.

Cumulative Frequencies (C.F.):

• \( x=3 \): 5
• \( x=4 \): 5 + 4 = 9
• \( x=5 \): 9 + 4 = 13 (This range includes the 10th to 13th terms)

Both the 10th and 11th observations are 5.

\[ \text{Median} = \frac{5 + 5}{2} = 5 \]

Match: (Q) \(\to\) (2)

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(R) Mean Deviation about the Mean

\[ \text{M.D.}(\bar{x}) = \frac{\sum f_i |x_i - \bar{x}|}{N} \]

Using \( \bar{x} = 6 \):

• \( |3-6|=3 \rightarrow f_i \cdot 3 = 5 \times 3 = 15 \)
• \( |4-6|=2 \rightarrow f_i \cdot 2 = 4 \times 2 = 8 \)
• \( |5-6|=1 \rightarrow f_i \cdot 1 = 4 \times 1 = 4 \)
• \( |8-6|=2 \rightarrow f_i \cdot 2 = 2 \times 2 = 4 \)
• \( |10-6|=4 \rightarrow f_i \cdot 4 = 2 \times 4 = 8 \)
• \( |11-6|=5 \rightarrow f_i \cdot 5 = 3 \times 5 = 15 \)

Sum = \( 15 + 8 + 4 + 4 + 8 + 15 = 54 \)

\[ \text{M.D.}(\bar{x}) = \frac{54}{20} = 2.7 \]

Match: (R) \(\to\) (4)

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(S) Mean Deviation about the Median

\[ \text{M.D.}(M) = \frac{\sum f_i |x_i - M|}{N} \]

Using \( M = 5 \):

• \( |3-5|=2 \rightarrow f_i \cdot 2 = 5 \times 2 = 10 \)
• \( |4-5|=1 \rightarrow f_i \cdot 1 = 4 \times 1 = 4 \)
• \( |5-5|=0 \rightarrow f_i \cdot 0 = 4 \times 0 = 0 \)
• \( |8-5|=3 \rightarrow f_i \cdot 3 = 2 \times 3 = 6 \)
• \( |10-5|=5 \rightarrow f_i \cdot 5 = 2 \times 5 = 10 \)
• \( |11-5|=6 \rightarrow f_i \cdot 6 = 3 \times 6 = 18 \)

Sum = \( 10 + 4 + 0 + 6 + 10 + 18 = 48 \)

\[ \text{M.D.}(M) = \frac{48}{20} = 2.4 \]

Match: (S) \(\to\) (5)

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Conclusion:

P \(\to\) 3, Q \(\to\) 2, R \(\to\) 4, S \(\to\) 5.

This corresponds to Option (A).

Step 1: Identify the lines and their direction vectors

Line \( \ell_1 \): \( \vec{r_1} = \lambda (\hat{i} + \hat{j} + \hat{k}) \). It passes through the origin \( O(0,0,0) \) and has direction vector \( \vec{d_1} = (1, 1, 1) \).

Line \( \ell_2 \): \( \vec{r_2} = (\hat{j} - \hat{k}) + \mu (\hat{i} + \hat{k}) \). It passes through point \( A(0, 1, -1) \) and has direction vector \( \vec{d_2} = (1, 0, 1) \).

Step 2: Determine the plane \( H_0 \)

Let \( H \) be a plane containing line \( \ell_1 \). The distance \( d(H) \) is the minimum distance between points of \( \ell_2 \) and \( H \). If \( \ell_2 \) intersects \( H \), this distance is zero. To maximize this distance, the plane \( H \) must be parallel to \( \ell_2 \). This unique plane \( H_0 \) contains \( \ell_1 \) and is parallel to \( \ell_2 \).

The normal vector \( \vec{n} \) to the plane \( H_0 \) must be perpendicular to both direction vectors \( \vec{d_1} \) and \( \vec{d_2} \).

\[ \vec{n} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 0 & 1 \end{vmatrix} \]

\[ \vec{n} = \hat{i}(1-0) - \hat{j}(1-1) + \hat{k}(0-1) = \hat{i} - \hat{k} \]

The normal vector is \( (1, 0, -1) \). Since the plane contains \( \ell_1 \), it passes through the origin \( (0,0,0) \).

Equation of plane \( H_0 \):

\[ 1(x - 0) + 0(y - 0) - 1(z - 0) = 0 \implies x - z = 0 \]

Step 3: Evaluate List-I entries

(P) The value of \( d(H_0) \)

This is the perpendicular distance of any point on the parallel line \( \ell_2 \) from the plane \( H_0 \). Using point \( A(0, 1, -1) \) on \( \ell_2 \):

\[ d(H_0) = \left| \frac{1(0) + 0(1) - 1(-1)}{\sqrt{1^2 + 0^2 + (-1)^2}} \right| = \frac{|1|}{\sqrt{2}} = \frac{1}{\sqrt{2}} \]

This corresponds to option (5) in List-II, which is \( \frac{1}{\sqrt{2}} \).

(Q) The distance of the point \( (0, 1, 2) \) from \( H_0 \)

Using the distance formula for point \( (0, 1, 2) \) and plane \( x - z = 0 \):

\[ D = \left| \frac{0 - 2}{\sqrt{1^2 + 0^2 + (-1)^2}} \right| = \frac{|-2|}{\sqrt{2}} = \sqrt{2} \]

This corresponds to option (4) in List-II, which is \( \sqrt{2} \).

(R) The distance of origin from \( H_0 \)

Since the plane \( H_0 \) passes through the origin, the distance is 0.

This corresponds to option (3) in List-II, which is 0.

(S) Distance of origin from the point of intersection of planes \( y=z \), \( x=1 \), and \( H_0 \)

Find the intersection point:

1. \( x = 1 \)
2. \( H_0: x - z = 0 \implies z = x \implies z = 1 \)
3. \( y = z \implies y = 1 \)

The intersection point is \( P(1, 1, 1) \). The distance from origin is:

\[ OP = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \]

This corresponds to option (1) in List-II, which is \( \sqrt{3} \).

Conclusion:

\( (P) \to (5), (Q) \to (4), (R) \to (3), (S) \to (1) \)

Correct Option: (B)

Step 1: Simplify the given complex equation

Let \( z = x + iy \), where \( x, y \in \mathbb{R} \) and \( y \neq 0 \) (since the imaginary part is non-zero).

The given equation is:

\[ |z|^3 + 2z^2 + 4\bar{z} - 8 = 0 \]

Substituting \( z = x + iy \) and \( \bar{z} = x - iy \):

\[ |z|^3 + 2(x + iy)^2 + 4(x - iy) - 8 = 0 \]

\[ |z|^3 + 2(x^2 - y^2 + 2ixy) + 4x - 4iy - 8 = 0 \]

Grouping real and imaginary parts:

\[ [|z|^3 + 2(x^2 - y^2) + 4x - 8] + i[4xy - 4y] = 0 \]

For a complex number to be zero, both real and imaginary parts must be zero.

Step 2: Solve for coordinates \( x \) and \( y \)

From the imaginary part:

\[ 4xy - 4y = 0 \implies 4y(x - 1) = 0 \]

Since \( y \neq 0 \), we must have \( x = 1 \).

Now substitute \( x = 1 \) into the real part equation:

\[ |z|^3 + 2(1^2 - y^2) + 4(1) - 8 = 0 \]

Note that \( |z| = \sqrt{x^2 + y^2} = \sqrt{1 + y^2} \). So \( |z|^3 = (1 + y^2)^{3/2} \).

\[ (1 + y^2)^{3/2} + 2 - 2y^2 + 4 - 8 = 0 \]

\[ (1 + y^2)^{3/2} - 2y^2 - 2 = 0 \]

\[ (1 + y^2)^{3/2} = 2(y^2 + 1) \]

Let \( u = 1 + y^2 \). Since \( y \neq 0 \), \( u > 1 \). The equation becomes:

\[ u^{3/2} = 2u \]

Dividing by \( u \) (since \( u \neq 0 \)):

\[ u^{1/2} = 2 \implies \sqrt{u} = 2 \implies u = 4 \]

So, \( 1 + y^2 = 4 \implies y^2 = 3 \).

Thus, \( x = 1 \) and \( y^2 = 3 \).

Step 3: Evaluate List-I entries

(P) \( |z|^2 \)

\[ |z|^2 = x^2 + y^2 = 1 + 3 = 4 \]

Match: (P) \(\to\) (2)

(Q) \( |z - \bar{z}|^2 \)

\[ z - \bar{z} = (x + iy) - (x - iy) = 2iy \]

\[ |z - \bar{z}|^2 = |2iy|^2 = (2|y|)^2 = 4y^2 = 4(3) = 12 \]

Match: (Q) \(\to\) (1)

(R) \( |z|^2 + |z + \bar{z}|^2 \)

\[ |z|^2 = 4 \]

\[ z + \bar{z} = (x + iy) + (x - iy) = 2x = 2(1) = 2 \]

\[ |z + \bar{z}|^2 = 2^2 = 4 \]

\[ \text{Sum} = 4 + 4 = 8 \]

Match: (R) \(\to\) (3)

(S) \( |z + 1|^2 \)

\[ z + 1 = (1 + iy) + 1 = 2 + iy \]

\[ |z + 1|^2 = 2^2 + y^2 = 4 + 3 = 7 \]

Match: (S) \(\to\) (5)

Conclusion:

\( (P) \to (2), (Q) \to (1), (R) \to (3), (S) \to (5) \)

Correct Option: (B)

1. Velocity at the bottom of the slide ($\vec{u}_0$):
Using conservation of energy for the ball sliding down height $h$: $$mgh = \frac{1}{2} m u_0^2 \implies u_0 = \sqrt{2gh}$$ Since the slide becomes horizontal, the velocity vector is $\vec{u}_0 = \sqrt{2gh}\hat{x}$. Thus, statement (A) is correct.

2. Projectile motion from the terrace:
The ball leaves the terrace at height $3h$ with horizontal velocity $u_x = \sqrt{2gh}$ and $u_z = 0$. The vertical velocity $v_z$ upon reaching the ground is found using: $$v_z^2 = u_z^2 + 2g(3h) \implies v_z = \sqrt{6gh}$$ The velocity just before impact is $\vec{v}_{impact} = \sqrt{2gh}\hat{x} - \sqrt{6gh}\hat{z}$. The angle $\theta$ with horizontal is given by: $$\tan \theta = \frac{|v_z|}{|v_x|} = \frac{\sqrt{6gh}}{\sqrt{2gh}} = \sqrt{3} \implies \theta = 60^{\circ}$$ Thus, statement (C) is correct.

3. Velocity after bounce ($\vec{v}$):
The coefficient of restitution is $e = 1/\sqrt{3}$. Horizontal component remains unchanged: $v'_x = \sqrt{2gh}$. Vertical component reverses direction and scales by $e$: $$v'_z = e |v_z| = \frac{1}{\sqrt{3}} (\sqrt{6gh}) = \sqrt{2gh}$$ So, $\vec{v} = \sqrt{2gh}\hat{x} + \sqrt{2gh}\hat{z} = \sqrt{2gh}(\hat{x} + \hat{z})$. Statement (B) suggests $\vec{v} = \sqrt{2gh}(\hat{x} - \hat{z})$, which has the wrong vertical direction. Thus, (B) is incorrect.

4. Distance ratio ($d/h_1$):
Time to fall $3h$: $t = \sqrt{2(3h)/g} = \sqrt{6h/g}$. Horizontal distance $d = u_x t = \sqrt{2gh} \cdot \sqrt{6h/g} = \sqrt{12h^2} = 2\sqrt{3}h$. Max height after bounce $h_1 = (v'_z)^2 / (2g) = (2gh)/(2g) = h$. Ratio $d/h_1 = 2\sqrt{3}h / h = 2\sqrt{3}$. Thus, statement (D) is correct.

1. Blue Light Analysis (Brewster's Angle):
The problem states there is no reflection for the incident blue light. This occurs when light is polarized in the plane of incidence and incident at the Brewster angle ($i_p$). Thus, Statement (A) is correct.
The refractive index for blue light is given as $\mu = \sqrt{3}$. $$i = i_p = \tan^{-1}(\mu) = \tan^{-1}(\sqrt{3}) = 60^{\circ}$$ Using Snell's law at the first surface: $$1 \cdot \sin(60^{\circ}) = \sqrt{3} \sin r_1 \implies \frac{\sqrt{3}}{2} = \sqrt{3} \sin r_1 \implies r_1 = 30^{\circ}$$ For the prism, deviation $\delta = i + e - A$. Given $\delta = 60^{\circ}$: $$60^{\circ} = 60^{\circ} + e - A \implies e = A$$ Also, $r_1 + r_2 = A \implies 30^{\circ} + r_2 = A \implies r_2 = A - 30^{\circ}$. At the emergent surface: $$\mu \sin r_2 = \sin e \implies \sqrt{3} \sin(A - 30^{\circ}) = \sin A$$ Expanding $\sin(A - 30^{\circ})$: $$\sqrt{3}(\sin A \cos 30^{\circ} - \cos A \sin 30^{\circ}) = \sin A$$ $$\sqrt{3}(\frac{\sqrt{3}}{2}\sin A - \frac{1}{2}\cos A) = \sin A$$ $$\frac{3}{2}\sin A - \frac{\sqrt{3}}{2}\cos A = \sin A \implies \frac{1}{2}\sin A = \frac{\sqrt{3}}{2}\cos A \implies \tan A = \sqrt{3}$$ So, Prism Angle $A = 60^{\circ}$. Consequently, emergent angle $e = A = 60^{\circ}$. Statement (D) is correct ($e=60^{\circ}$). Statement (B) is incorrect (calculates $A=45^{\circ}$).

2. Red Light Analysis (Minimum Deviation):
For red light, minimum deviation $\delta_{min} = 30^{\circ}$. $$\mu_R = \frac{\sin(\frac{A + \delta_{min}}{2})}{\sin(\frac{A}{2})} = \frac{\sin(\frac{60^{\circ} + 30^{\circ}}{2})}{\sin(\frac{60^{\circ}}{2})} = \frac{\sin 45^{\circ}}{\sin 30^{\circ}} = \frac{1/\sqrt{2}}{1/2} = \sqrt{2}$$ Statement (C) is correct.

1. Steady State Analysis ($t < t_0$, Switch K Open):
When K is open, the capacitor branch is disconnected. The circuit consists of the 15V and 5V sources and the resistor network. Let's find the resistance $R$. Given that the current through the $1\Omega$ resistor is $1\text{A}$, we can analyze the potentials. Using nodal analysis or equivalent circuit reduction (as shown in the solution image), the equation for the current involves $R$. The solution derived leads to: $$1 = \frac{30 - 5R}{4R + 3} \implies 4R + 3 = 30 - 5R \implies 9R = 27 \implies R = 3\Omega$$ Thus, Statement (A) is correct.
With $R=3\Omega$, the current $I_1$ in the middle branch ($3\Omega$ resistor) is calculated as: $$I_1 = \frac{V_{effective}}{R_{branch}} = \frac{6V}{3\Omega} = 2\text{A}$$ Thus, Statement (B) is correct.

2. Transient Analysis ($t \geq t_0$, Switch K Closed):
Closing switch K completes the circuit for the capacitor $C = 2\mu\text{F}$. Note that the capacitor is in series with a $3\Omega$ resistor in its specific branch. We find the Thevenin equivalent circuit seen by this branch:

• $V_{th} = 6\text{V}$
• $R_{th} = 0.6\Omega$ (or $3/5 \Omega$)

The total resistance in the charging loop is $R_{total} = R_{th} + R_{series} = 0.6 + 3 = 3.6\Omega$. The time constant is $\tau = R_{total} C = 3.6\Omega \times 2\mu\text{F} = 7.2\mu\text{s}$.

Checking Statement (D): As $t \to \infty$, the capacitor fully charges to $V_{th}$. $$Q_{max} = C \cdot V_{th} = 2\mu\text{F} \cdot 6\text{V} = 12\mu\text{C}$$ Thus, Statement (D) is correct.

Checking Statement (C): The charging current is given by $i(t) = \frac{V_{th}}{R_{total}} e^{-(t-t_0)/\tau}$. At $t = t_0 + 7.2\mu\text{s}$, the time elapsed is exactly one time constant ($t-t_0 = \tau$). $$i(\tau) = \frac{6}{3.6} e^{-1} \approx 1.667 \times 0.36 = 0.6\text{A}$$ Thus, Statement (C) is correct.

Conclusion: All statements A, B, C, and D are correct.

1. Conservation of Angular Momentum:
The system consists of the pivoted bar and the moving mass. Since the pivot force exerts no torque about the pivot point, angular momentum is conserved about the hinge during the collision. Let the initial velocity of the mass be $u = 5.00 \, \text{m/s}$ and its final rebound speed be $v$ (so velocity is $-v$). The bar is initially at rest and acquires angular velocity $\omega$. The mass strikes at distance $r = L/2$. $$L_{initial} = L_{final}$$ $$m u \left(\frac{L}{2}\right) = I_{rod} \omega - m v \left(\frac{L}{2}\right)$$ Substitute $I_{rod} = \frac{ML^2}{3}$ and given values ($M=1, L=0.2, m=0.1, u=5$): $$0.1 \times 5 \times \frac{0.2}{2} = \frac{1 \times (0.2)^2}{3} \omega - 0.1 \times v \times \frac{0.2}{2}$$ $$0.05 = \frac{0.04}{3} \omega - 0.01 v$$ Multiplying by 300 to clear decimals and denominators: $$15 = 4\omega - 3v \quad \dots (1)$$

2. Coefficient of Restitution (Elastic Collision):
For an elastic collision, $e=1$. The relative velocity of separation equals the relative velocity of approach at the point of impact. Velocity of impact point on rod after collision = $v_{rod} = \omega(L/2)$. Velocity of mass after collision = $-v$ (backward). $$v_{separation} = v_{rod} - v_{mass} = \omega(L/2) - (-v) = \frac{L}{2}\omega + v$$ $$v_{approach} = u - 0 = 5$$ Using $e = \frac{v_{separation}}{v_{approach}} = 1$: $$\frac{0.2}{2}\omega + v = 5 \implies 0.1\omega + v = 5 \implies v = 5 - 0.1\omega \quad \dots (2)$$

3. Solving the equations:
Substitute (2) into (1): $$15 = 4\omega - 3(5 - 0.1\omega)$$ $$15 = 4\omega - 15 + 0.3\omega$$ $$30 = 4.3\omega \implies \omega = \frac{300}{43} \approx 6.98 \, \text{rad/s}$$ Now find $v$: $$v = 5 - 0.1(6.98) = 5 - 0.698 = 4.302 \approx 4.30 \, \text{m/s}$$ Thus, $\omega = 6.98 \, \text{rad s}^{-1}$ and $v = 4.30 \, \text{m s}^{-1}$.

1. Determine Liquid Height:
Rate of filling $R = 250 \, \text{cm}^3/\text{s}$. Time elapsed $t = 10 \, \text{s}$. Total Volume filled $V = R \times t = 2500 \, \text{cm}^3$. The base area of the container is $50 \, \text{cm} \times 5 \, \text{cm} = 250 \, \text{cm}^2$. Height of liquid column $x = \frac{\text{Volume}}{\text{Base Area}} = \frac{2500}{250} = 10 \, \text{cm}$.

2. Equivalent Capacitance Model:
The conducting walls are the $50 \, \text{cm} \times 50 \, \text{cm}$ faces, separated by distance $d = 5 \, \text{cm}$. The filling of the liquid creates two capacitors in parallel (since the potential difference is the same across the height, but the dielectric varies along the area): 1. Liquid filled capacitor ($C_1$): Height 10 cm. 2. Air filled capacitor ($C_2$): Remaining height $50 - 10 = 40$ cm.

3. Calculation:
Width of plates $w = 50 \, \text{cm} = 0.5 \, \text{m}$. Plate separation $d = 5 \, \text{cm} = 0.05 \, \text{m}$. Liquid height $h_1 = 10 \, \text{cm} = 0.1 \, \text{m}$. Air height $h_2 = 40 \, \text{cm} = 0.4 \, \text{m}$. Area $A_1 = w \times h_1 = 0.5 \times 0.1 = 0.05 \, \text{m}^2$. Area $A_2 = w \times h_2 = 0.5 \times 0.4 = 0.20 \, \text{m}^2$. Dielectric constants: $K_{liq} = 3, K_{air} = 1$.

$$C_1 = \frac{K_{liq} \epsilon_0 A_1}{d} = \frac{3 \epsilon_0 (0.05)}{0.05} = 3\epsilon_0$$ $$C_2 = \frac{K_{air} \epsilon_0 A_2}{d} = \frac{1 \epsilon_0 (0.20)}{0.05} = 4\epsilon_0$$ $$C_{total} = C_1 + C_2 = 3\epsilon_0 + 4\epsilon_0 = 7\epsilon_0$$ Given $\epsilon_0 = 9 \times 10^{-12} \, \text{F/m}$: $$C_{total} = 7 \times (9 \times 10^{-12}) = 63 \times 10^{-12} \, \text{F} = 63 \, \text{pF}$$

1. Adiabatic Process ($A \to f$):
For an adiabatic expansion of an ideal gas from state $(T_A, V_0)$ to $(T_f, 5V_0)$, the relation is $TV^{\gamma-1} = \text{constant}$. $$T_A V_0^{\gamma-1} = T_f (5V_0)^{\gamma-1}$$ $$T_A = T_f \left( \frac{5V_0}{V_0} \right)^{\gamma-1} = T_f (5)^{\gamma-1} \quad \dots (1)$$

2. Isothermal Process ($B \to f$):
For an isothermal expansion from state $(T_B, V_0)$ to $(T_f, 5V_0)$, the temperature remains constant. $$T_{initial} = T_{final} \implies T_B = T_f \quad \dots (2)$$

3. Ratio Calculation:
We need the ratio $T_A / T_B$. Substituting (2) into (1): $$T_A = T_B (5)^{\gamma-1}$$ $$\frac{T_A}{T_B} = 5^{\gamma-1}$$

1. Formula for Gravitational Acceleration:
The acceleration due to gravity at a height $h$ from the Earth's surface is given by: $$g(h) = \frac{GM}{(R+h)^2}$$ where $R$ is the radius of the Earth.

2. Setup Expressions for P and Q:
For satellite P, height $h_P = R/3$: $$g_P = \frac{GM}{(R + \frac{R}{3})^2} = \frac{GM}{(\frac{4R}{3})^2} = \frac{9GM}{16R^2}$$ For satellite Q, height $h_Q$: $$g_Q = \frac{GM}{(R + h_Q)^2}$$

3. Use the Given Ratio:
Given ratio $\frac{g_P}{g_Q} = \frac{36}{25}$. Substituting the expressions: $$\frac{g_P}{g_Q} = \frac{\frac{9GM}{16R^2}}{\frac{GM}{(R+h_Q)^2}} = \frac{9}{16R^2} \cdot (R+h_Q)^2 = \left( \frac{3(R+h_Q)}{4R} \right)^2$$ Equating to the given value: $$\left( \frac{3(R+h_Q)}{4R} \right)^2 = \frac{36}{25} = \left(\frac{6}{5}\right)^2$$ Taking the square root of both sides: $$\frac{3(R+h_Q)}{4R} = \frac{6}{5}$$

4. Solve for $h_Q$:
$$\frac{R+h_Q}{4R} = \frac{2}{5} \implies R+h_Q = \frac{8R}{5}$$ $$h_Q = \frac{8R}{5} - R = \frac{3R}{5}$$ Thus, option (A) is correct.

1. Work Function Calculation:
The work function $\phi$ of the metal is determined by the threshold wavelength $\lambda_{th} = 310$ nm. $$\phi = \frac{hc}{\lambda_{th}} = \frac{1240 \text{ eV}\cdot\text{nm}}{310 \text{ nm}} = 4 \text{ eV}$$

2. Energy of Incident Photon:
Using the photoelectric equation $K_{max} = E_{photon} - \phi$, where maximum kinetic energy $K_{max} = 1.95$ eV: $$E_{photon} = K_{max} + \phi = 1.95 \text{ eV} + 4 \text{ eV} = 5.95 \text{ eV}$$

3. Atomic Transition Analysis:
The photon is emitted from a transition $n=4 \to n=3$ in a Hydrogen-like atom with atomic number $Z$. The energy difference is: $$\Delta E = 13.6 Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$$ $$5.95 = 13.6 Z^2 \left( \frac{1}{3^2} - \frac{1}{4^2} \right)$$ $$5.95 = 13.6 Z^2 \left( \frac{1}{9} - \frac{1}{16} \right)$$ $$5.95 = 13.6 Z^2 \left( \frac{16 - 9}{144} \right) = 13.6 Z^2 \left( \frac{7}{144} \right)$$

4. Solve for Z:
$$Z^2 = \frac{5.95 \times 144}{13.6 \times 7}$$ $$Z^2 = \frac{856.8}{95.2} = 9$$ $$Z = 3$$ The atomic number is 3 (Lithium ion $Li^{2+}$).

Analysis of the System:
We have an object S placed at the midpoint between a Lens L ($f_L = 10$) and Mirror M2 ($f_2 = -12$). The distance L-M2 is 20 cm, so S is 10 cm from both. We investigate paths where the final image coincides with S. There are three valid configurations for the distance $d$ between L and M1 ($d = n/7$).

Path Setup: S $\to$ M2 $\to$ Lens $\to$ M1 $\to$ ...
1. **Reflection from M2:** Object distance $u = -10$ cm. $\frac{1}{v} + \frac{1}{-10} = \frac{1}{-12} \implies \frac{1}{v} = \frac{1}{10} - \frac{1}{12} = \frac{1}{60}$. $v = +60$ cm (Virtual image $I_1$ to the right of M2). 2. **Refraction through Lens:** Distance of $I_1$ from Lens = $20 (\text{L-M2}) + 60 = 80$ cm. Since light is traveling left towards the lens, $u_L = -80$ cm. $\frac{1}{v_L} - \frac{1}{-80} = \frac{1}{10} \implies \frac{1}{v_L} = \frac{1}{10} - \frac{1}{80} = \frac{7}{80}$. $v_L = \frac{80}{7}$ cm (Real image $I_2$ formed to the left of Lens).

Case 1: Retracing Path (Normal Incidence on M1)
For the image to form back at S via the exact same path, rays must hit M1 normally and retrace. This happens if $I_2$ forms at the Center of Curvature of M1 ($C_1$). Distance $d = v_L + R_1 = \frac{80}{7} + 20 = \frac{80 + 140}{7} = \frac{220}{7}$. Here, $n = 220$.

Case 2: Image on Pole of M1
If $I_2$ forms exactly at the pole of M1, rays reflect symmetrically. Distance $d = v_L = \frac{80}{7}$. So $n = 80$. Verification: Rays reflect from pole, travel back $80/7$ to Lens. Image forms at $v'=80$ cm (right of lens). This point acts as object for M2 ($u = 80-20=60$ cm). M2 forms image at $S$ (calculation: $1/v + 1/60 = 1/-12 \to v=-10$). Correct.

Case 3: Image at Focus of M1 (Parallel Return)
If $I_2$ forms at the Focus of M1 ($F_1$), reflected rays become parallel. Distance $d = v_L + f_1 = \frac{80}{7} + 10 = \frac{150}{7}$. So $n = 150$. Verification: Parallel rays from M1 hit Lens. Lens focuses them at its focus (10 cm). S is at 10 cm. Correct.

Answer: The possible values for $n$ are 80, 150, or 220.

1. Lens Formula:
The thin lens formula is $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$. Since it's a real image formed by a convex lens, the object distance $u$ is negative and image distance $v$ is positive. Let $|u| = 10 \text{ cm}$ and $|v| = 20 \text{ cm}$. The formula becomes $\frac{1}{f} = \frac{1}{20} + \frac{1}{10} = \frac{3}{20} \implies f = \frac{20}{3} \text{ cm}$.

2. Error Propagation Formula:
Differentiating the lens formula: $$-\frac{df}{f^2} = -\frac{dv}{v^2} - \frac{du}{u^2}$$ For maximum error calculation, we add the absolute errors: $$\frac{\Delta f}{f^2} = \frac{\Delta v}{v^2} + \frac{\Delta u}{u^2}$$ Rearranging to find relative error in focal length $\frac{\Delta f}{f}$: $$\frac{\Delta f}{f} = f \left( \frac{\Delta v}{v^2} + \frac{\Delta u}{u^2} \right)$$ $$= \frac{20}{3} \left( \frac{0.2}{20^2} + \frac{0.1}{10^2} \right)$$ $$= \frac{20}{3} \left( \frac{0.2}{400} + \frac{0.1}{100} \right)$$ $$= \frac{20}{3} \left( 0.0005 + 0.001 \right)$$ $$= \frac{20}{3} \times 0.0015 = \frac{20}{3} \times \frac{15}{10000} = \frac{20 \times 5}{10000} = \frac{100}{10000} = 0.01$$

3. Percentage Error:
Percentage error = $\frac{\Delta f}{f} \times 100\% = 0.01 \times 100\% = 1\%$. The value of $n$ is 1.

1. Work Done in Isobaric Process:
The total moles of gas $n_{total} = n_1 + n_2 = 2 + 1 = 3 \text{ moles}$. Work done at constant pressure is given by $W = n_{total} R \Delta T$. Given $W = 66 \text{ J}$. $$3 R \Delta T = 66 \implies R \Delta T = 22 \text{ J}$$

2. Change in Internal Energy ($\Delta U$):
The change in internal energy for a mixture is the sum of changes for individual components. $\Delta U = \Delta U_1 + \Delta U_2 = n_1 C_{V1} \Delta T + n_2 C_{V2} \Delta T$. For monatomic gas ($n_1=2, \gamma=5/3$): $f_1 = 3$, so $C_{V1} = \frac{3}{2}R$. For diatomic gas ($n_2=1, \gamma=7/5$): $f_2 = 5$, so $C_{V2} = \frac{5}{2}R$. Substituting values: $$\Delta U = 2 \times \frac{3}{2} R \Delta T + 1 \times \frac{5}{2} R \Delta T$$ $$\Delta U = 3 R \Delta T + 2.5 R \Delta T = 5.5 R \Delta T$$ $$\Delta U = \frac{11}{2} R \Delta T$$

3. Calculate Final Value:
Substitute $R \Delta T = 22$: $$\Delta U = \frac{11}{2} \times 22 = 11 \times 11 = 121 \text{ J}$$

1. Setup Geometry with Similar Triangles:
Let $H = 4 \text{ m}$ be the height of the lamp post and $h = 1.6 \text{ m}$ be the height of the person. Let $x_1$ be the distance of the person from the lamp post. Let $x_2$ be the distance of the tip of the shadow from the lamp post. From similar triangles formed by the light source, the top of the person's head, and the shadow tip: $$\frac{H}{x_2} = \frac{h}{x_2 - x_1}$$ Substitute values ($H=4, h=1.6$): $$\frac{4}{x_2} = \frac{1.6}{x_2 - x_1}$$ $$4(x_2 - x_1) = 1.6 x_2$$ $$4x_2 - 4x_1 = 1.6 x_2 \implies 2.4 x_2 = 4 x_1 \implies x_2 = \frac{4}{2.4} x_1 = \frac{5}{3} x_1$$

2. Differentiate with respect to time:
Given speed of person $v_{person} = \frac{dx_1}{dt} = 60 \text{ cm/s}$. We need to find the speed of the shadow tip relative to the person. First, find speed of shadow tip $\frac{dx_2}{dt}$: $$\frac{dx_2}{dt} = \frac{5}{3} \frac{dx_1}{dt} = \frac{5}{3} \times 60 = 100 \text{ cm/s}$$

3. Relative Velocity:
Speed of shadow tip with respect to the person ($v_{rel}$): $$v_{rel} = v_{shadow} - v_{person} = \frac{dx_2}{dt} - \frac{dx_1}{dt}$$ $$v_{rel} = 100 - 60 = 40 \text{ cm/s}$$

1. Locate Center of Mass (CM):
Let the rod lie along the x-axis with the 30g mass at $x=0$. $m_1 = 30 \text{ g}, m_2 = 20 \text{ g}$. Length $L = 10 \text{ cm}$. $x_{CM} = \frac{m_1(0) + m_2(10)}{m_1+m_2} = \frac{200}{50} = 4 \text{ cm}$. So, the suspension point is 4 cm from the 30g mass and $10 - 4 = 6 \text{ cm}$ from the 20g mass.

2. Calculate Moment of Inertia (I):
The moment of inertia about the suspension axis (passing through CM): $I = m_1 r_1^2 + m_2 r_2^2$ $r_1 = 4 \text{ cm}, r_2 = 6 \text{ cm}$. $I = 30(4^2) + 20(6^2) = 30(16) + 20(36)$ $I = 480 + 720 = 1200 \text{ g cm}^2$. Convert to SI units: $I = 1200 \times 10^{-3} \text{ kg} \times (10^{-2} \text{ m})^2$ $I = 1.2 \times 10^{-4} \text{ kg m}^2$. (Or $12 \times 10^{-5} \text{ kg m}^2$ as in solution image).

3. Calculate Angular Frequency ($\omega$):
Formula for torsional pendulum: $\omega = \sqrt{\frac{K}{I}}$ Given torsional constant $K = 1.2 \times 10^{-8} \text{ Nm rad}^{-1}$. $$\omega = \sqrt{\frac{1.2 \times 10^{-8}}{1.2 \times 10^{-4}}} = \sqrt{10^{-4}} = 10^{-2} \text{ rad s}^{-1}$$ We need to express $\omega$ as $n \times 10^{-3}$. $$\omega = 10 \times 10^{-3} \text{ rad s}^{-1}$$ Thus, $n = 10$.

General Method:
Use conservation of mass number (A) and atomic number (Z). Let $x$ be number of $\alpha$ particles ($^4_2He$) and $y$ be number of $\beta^-$ ($-1$) or $\beta^+$ ($+1$) particles. Equation for A: $A_{parent} = A_{daughter} + 4x$. Equation for Z: $Z_{parent} = Z_{daughter} + 2x + (\text{charge of beta}) \cdot y$.

(P) $^{238}_{92}U \to ^{234}_{91}Pa$
Mass number change: $238 - 234 = 4$. This corresponds to 1 $\alpha$ particle ($x=1$). Atomic number change: $92 = 91 + 2(1) + \text{charge}$. $92 = 93 + \text{charge} \implies \text{charge} = -1$. So, 1 $\alpha$ and 1 $\beta^-$ particle. This matches Option (4). P $\to$ 4.

(Q) $^{214}_{82}Pb \to ^{210}_{82}Pb$
Mass change: $214 - 210 = 4 \implies 1 \, \alpha$. Z change: $82 = 82 + 2(1) + \text{charge}$. $82 = 84 + \text{charge} \implies \text{charge} = -2$. This means 2 $\beta^-$ particles. So, 1 $\alpha$ and 2 $\beta^-$. This matches Option (3). Q $\to$ 3.

(R) $^{210}_{81}Tl \to ^{206}_{82}Pb$
Mass change: $210 - 206 = 4 \implies 1 \, \alpha$. Z change: $81 = 82 + 2(1) + \text{charge}$. $81 = 84 + \text{charge} \implies \text{charge} = -3$. This means 3 $\beta^-$ particles. So, 1 $\alpha$ and 3 $\beta^-$. This matches Option (2). R $\to$ 2.

(S) $^{228}_{91}Pa \to ^{224}_{88}Ra$
Mass change: $228 - 224 = 4 \implies 1 \, \alpha$. Z change: $91 = 88 + 2(1) + \text{charge}$. $91 = 90 + \text{charge} \implies \text{charge} = +1$. This means 1 $\beta^+$ particle. So, 1 $\alpha$ and 1 $\beta^+$. This matches Option (1). S $\to$ 1.

Conclusion: Matching code: P-4, Q-3, R-2, S-1. This corresponds to Option (A).

1. Analyze List-II statements to find key Temperatures:

Item (1): Photoelectric emission from 4 eV metal. To emit electrons, photon energy $E \ge 4 \text{ eV}$. Wavelength $\lambda \le \frac{1240 \text{ eV nm}}{4 \text{ eV}} = 310 \text{ nm}$. Using Wien's Law $\lambda_m T = b$: $T \ge \frac{2.9 \times 10^{-3}}{310 \times 10^{-9}} = \frac{2.9 \times 10^6}{310} \approx 9354 \text{ K}$. This corresponds to $T = 10000 \text{ K}$. So (S) matches (1).

Item (2): Peak radiation visible to human eye. Visible range $\approx 400 \text{ nm} - 700 \text{ nm}$. $T_{min} = \frac{2.9 \times 10^{-3}}{700 \times 10^{-9}} \approx 4142 \text{ K}$. $T_{max} = \frac{2.9 \times 10^{-3}}{400 \times 10^{-9}} \approx 7250 \text{ K}$. The temperature $5000 \text{ K}$ falls in this range. So (R) matches (2).

Item (3): Widest central maximum in diffraction. Width $\propto \lambda_m$. Max width means max $\lambda_m$. Since $\lambda_m \propto 1/T$, max $\lambda_m$ means min $T$. Lowest T in List-I is 2000 K. So (P) matches (3).

Item (4): Power emitted per unit area is 1/16 of blackbody at 6000 K. Stefan-Boltzmann Law: $P \propto T^4$. $\frac{P}{P_0} = \left(\frac{T}{6000}\right)^4 = \frac{1}{16} = \left(\frac{1}{2}\right)^4$. $\frac{T}{6000} = \frac{1}{2} \implies T = 3000 \text{ K}$. So (Q) matches (4).

Matching Code: P-3, Q-4, R-2, S-1. This corresponds to Option (C).

1. Circuit Parameters:
Source $V = 45 \sin(\omega t)$, so $V_0 = 45 \text{ V}$. Resonant frequency $\omega_r = 10^5 \text{ rad/s}$. Inductance $L = 50 \text{ mH} = 50 \times 10^{-3} \text{ H}$. Capacitance $C = \frac{1}{\omega_r^2 L} = \frac{1}{(10^5)^2 (50 \times 10^{-3})} = \frac{1}{10^{10} \times 0.05} = 2 \times 10^{-9} \text{ F}$.

2. Condition at $\omega = 8 \times 10^4 \text{ rad/s}$:
Given current amplitude is $I = 0.05 I_0$. $I = \frac{V_0}{Z}$ and $I_0 = \frac{V_0}{R}$. So $\frac{V_0}{Z} = 0.05 \frac{V_0}{R} \implies Z = \frac{R}{0.05} = 20R$. Impedance $Z = \sqrt{R^2 + (X_C - X_L)^2}$. $X_L = \omega L = 8 \times 10^4 \times 50 \times 10^{-3} = 4000 \, \Omega$. $X_C = \frac{1}{\omega C} = \frac{1}{8 \times 10^4 \times 2 \times 10^{-9}} = \frac{10^5}{16} = 6250 \, \Omega$. Reactance $X = |X_L - X_C| = |4000 - 6250| = 2250 \, \Omega$. Substituting into impedance equation: $(20R)^2 = R^2 + (2250)^2$ $400R^2 - R^2 = 2250^2 \implies 399R^2 = 2250^2 \implies R \approx \frac{2250}{20} = 112.5 \, \Omega$. (Using approx $400R^2 \approx 2250^2$).

3. Analyze List Items:

(P) $I_0$ in mA: $I_0 = \frac{V_0}{R} = \frac{45}{112.5} = 0.4 \text{ A} = 400 \text{ mA}$. Matches (3). P $\to$ 3.

(Q) Quality Factor: $Q = \frac{\omega_r L}{R} = \frac{10^5 \times 0.05}{112.5} = \frac{5000}{112.5} \approx 44.4$. Matches (1). Q $\to$ 1.

(R) Bandwidth: $\Delta \omega = \frac{R}{L} = \frac{112.5}{0.05} = 2250 \text{ rad/s}$. Matches (4). R $\to$ 4.

(S) Peak Power dissipated at resonance: $P_{peak} = I_0^2 R = (0.4)^2 \times 112.5 = 0.16 \times 112.5 = 18 \text{ W}$. Wait, solution says $P = \frac{V^2}{R}$. At resonance, peak power is $V_0 I_0 = 45 \times 0.4 = 18$ W. Matches (2). S $\to$ 2.

Conclusion: Matching Code: P-3, Q-1, R-4, S-2. This corresponds to Option (B).

1. Equation of Motion:
Forces acting on the rod: Gravity ($mg$) downwards, Magnetic Force ($F_m = IlB = \frac{B^2 l^2 v}{R}$) upwards. $ma = mg - \frac{B^2 l^2 v}{R}$ $\frac{dv}{dt} = g - \frac{B^2 l^2}{mR} v$. Let $k = \frac{B^2 l^2}{mR}$. $m = 0.02 \text{ kg}, l = 0.25 \text{ m}, R = 10 \, \Omega, B = 4 \text{ T}$. $k = \frac{4^2 \times (0.25)^2}{0.02 \times 10} = \frac{16 \times 0.0625}{0.2} = \frac{1}{0.2} = 5 \text{ s}^{-1}$. Equation: $\frac{dv}{dt} = 10 - 5v$. Solution for $v(t)$ starting from rest: $v(t) = \frac{g}{k} (1 - e^{-kt}) = \frac{10}{5} (1 - e^{-5t}) = 2(1 - e^{-5t})$.

2. Analyze List Items at $t = 0.2$ s:
$e^{-5(0.2)} = e^{-1} = 0.4$. $v(0.2) = 2(1 - 0.4) = 2(0.6) = 1.2 \text{ m/s}$.

(P) Induced emf ($ \varepsilon $): $\varepsilon = Blv = 4 \times 0.25 \times 1.2 = 1.2 \text{ V}$. Matches (3). P $\to$ 3.

(Q) Magnetic Force ($F_m$): $F_m = IlB = \frac{\varepsilon}{R} l B = \frac{1.2}{10} \times 0.25 \times 4 = 0.12 \times 1 = 0.12 \text{ N}$. Matches (4). Q $\to$ 4.

(R) Power Dissipated ($P$): $P = I^2 R = (\frac{1.2}{10})^2 \times 10 = \frac{1.44}{10} = 0.144 \text{ W}$. Wait, check options. Closest is 0.14. Matches (2). R $\to$ 2.

(S) Terminal Velocity ($v_T$): $v_T = \frac{g}{k} = \frac{10}{5} = 2.00 \text{ m/s}$. Matches (5). S $\to$ 5.

Conclusion: Matching Code: P-3, Q-4, R-2, S-5. This corresponds to Option (D).

Analysis of the Statements:

• Statement (A): Malachite is \( \text{CuCO}_3 \cdot \text{Cu(OH)}_2 \). When roasted or calcined, it decomposes to form Copper(II) oxide (CuO), which is black, not Cuprite (\( \text{Cu}_2\text{O} \), red).
  Reaction: \( \text{CuCO}_3 \cdot \text{Cu(OH)}_2 \xrightarrow{\Delta} 2\text{CuO} + \text{CO}_2 + \text{H}_2\text{O} \)
  Statement (A) is Incorrect.
• Statement (B): Calamine is \( \text{ZnCO}_3 \). Calcination (heating in the absence of air) produces Zinc oxide (ZnO), which is known as Zincite.
  Reaction: \( \text{ZnCO}_3 \xrightarrow{\Delta} \text{ZnO} + \text{CO}_2 \)
  Statement (B) is Correct.
• Statement (C): Copper pyrites (\( \text{CuFeS}_2 \)) contain iron as an impurity. During extraction, it is roasted to convert sulfides to oxides. In the reverberatory furnace, silica (\( \text{SiO}_2 \)) is added as a flux to remove Iron(II) oxide (\( \text{FeO} \)) as slag (Iron silicate).
  Reaction: \( \text{FeO} + \text{SiO}_2 \rightarrow \text{FeSiO}_3 \text{ (Slag)} \)
  Statement (C) is Correct.
• Statement (D): This describes the Mac-Arthur Forrest Cyanide process for silver extraction. Silver is leached with \( \text{NaCN} \) or \( \text{KCN} \) in the presence of air (oxygen) to form a soluble complex, which is then reduced by Zinc.
  Leaching: \( 4\text{Ag} + 8\text{KCN} + 2\text{H}_2\text{O} + \text{O}_2 \rightarrow 4\text{K}[\text{Ag(CN)}_2] + 4\text{KOH} \)
  Recovery: \( 2\text{K}[\text{Ag(CN)}_2] + \text{Zn} \rightarrow \text{K}_2[\text{Zn(CN)}_4] + 2\text{Ag} \downarrow \)
  Statement (D) is Correct.

Thus, the correct statements are (B), (C), and (D).

Step-by-step Structural Analysis:

• Reaction for P:
  Reactant: \( \text{CH}_3\text{CH}_2\text{CH}(\text{CH}_3)\text{CH}_2\text{CN} \) (Chiral at C3).
  Step (i): \( \text{PhMgBr} / \text{H}_3\text{O}^+ \) converts \( -\text{CN} \) to ketone \( -\text{COPh} \).
  Step (ii): \( \text{PhMgBr} / \text{H}_2\text{O} \) converts ketone to tertiary alcohol \( -\text{C(OH)Ph}_2 \).
  Product P: \( \text{CH}_3\text{CH}_2\mathbf{C^*}\text{H}(\text{CH}_3)\text{CH}_2\text{C(OH)(Ph)}_2 \).
  The original chiral center (\( \mathbf{C^*} \)) is unaffected. The new alcohol carbon has two Phenyl groups, so it is achiral.
  Result: P has an asymmetric carbon.
• Reaction for Q:
  Reactant: Benzene + \( \text{CH}_3\text{COCl} \xrightarrow{\text{AlCl}_3} \text{Acetophenone} (\text{Ph-CO-CH}_3) \).
  Step (ii): \( \text{PhMgBr} / \text{H}_2\text{O} \) yields tertiary alcohol \( \text{Ph-C(OH)(CH}_3\text{)-Ph} \).
  Product Q: \( \text{Ph}_2\text{C(OH)CH}_3 \).
  The central carbon is bonded to two identical Phenyl groups.
  Result: Q has NO asymmetric carbon.
• Reaction for R:
  Reactant: \( \text{CH}_3\text{CH}_2\text{COCl} \).
  Step (i): \( (\text{PhCH}_2)_2\text{Cd} \) yields ketone \( \text{CH}_3\text{CH}_2\text{COCH}_2\text{Ph} \).
  Step (ii): \( \text{PhMgBr} / \text{H}_2\text{O} \) yields alcohol.
  Product R: \( \text{CH}_3\text{CH}_2-\mathbf{C^*}(\text{OH})(\text{Ph})-\text{CH}_2\text{Ph} \).
  The central carbon is bonded to four different groups: Ethyl, Benzyl, Phenyl, and Hydroxyl.
  Result: R has an asymmetric carbon.
• Reaction for S:
  Reactant: \( \text{PhCH}_2\text{CHO} \).
  Sequence: Grignard \( \rightarrow \) Alcohol \( \rightarrow \) Oxidation to Ketone (\( \text{PhCH}_2\text{COPh} \)) \( \rightarrow \) Cyanohydrin (\( \text{PhCH}_2\text{C(OH)(CN)Ph} \)) \( \rightarrow \) Hydrolysis/Dehydration.
  Dehydration gives an \(\alpha,\beta\)-unsaturated acid.
  Product S: \( \text{Ph-CH}=\text{C(Ph)-COOH} \).
  The molecule is planar around the double bond and contains no sp³ chiral carbons.
  Result: S has NO asymmetric carbon.

Evaluating Options:

• (A) P and Q have asymmetric carbons? False (Q does not).
• (B) Q and R have asymmetric carbons? False (Q does not).
• (C) P and R have asymmetric carbons? True.
• (D) P has asymmetric carbon, S does not? True.

Analysis of Reactions:

1. Identification of Cation (Mn²⁺):
   Reaction of metal halide with aqueous NaOH yields a white precipitate (P).
   \( \text{MnCl}_2 + 2\text{NaOH} \rightarrow \text{Mn(OH)}_2 \downarrow (\text{White ppt P}) + 2\text{NaCl} (\text{Filtrate Q}) \)
2. Confirmation Test for P (Formation of X):
   The precipitate P reacts with \( \text{PbO}_2 \) and conc. \( \text{H}_2\text{SO}_4 \) upon heating to form a colored species X. This is the standard confirmation test for Manganese(II), where it is oxidized to permanganate (purple/pink).
   \( 2\text{Mn(OH)}_2 + 5\text{PbO}_2 + 10\text{H}^+ \xrightarrow{\Delta} 2\text{MnO}_4^- (\text{Purple solution X}) + 5\text{Pb}^{2+} + 8\text{H}_2\text{O} \)
   Thus, X is \( \text{MnO}_4^- \).
3. Identification of Filtrate Q and Product Y:
   The filtrate Q contains \( \text{NaCl} \). It reacts with \( \text{MnO(OH)}_2 \) (hydrated \( \text{MnO}_2 \)) and conc. \( \text{H}_2\text{SO}_4 \) to release a gas Y.
   \( 2\text{NaCl} + \text{MnO}_2 + 3\text{H}_2\text{SO}_4 \rightarrow 2\text{NaHSO}_4 + \text{MnSO}_4 + 2\text{H}_2\text{O} + \text{Cl}_2 \uparrow (\text{Gas Y}) \)
   The gas Y turns starch-iodide paper blue, which confirms it is Chlorine (\( \text{Cl}_2 \)).
   \( \text{Cl}_2 + 2\text{I}^- \rightarrow \text{I}_2 + 2\text{Cl}^- \)
   \( \text{I}_2 + \text{Starch} \rightarrow \text{Blue Complex} \)

Conclusion:
X = \( \text{MnO}_4^- \)
Y = \( \text{Cl}_2 \)
This corresponds to option (C).

Step-by-Step Derivation:

For a weak monobasic acid (HX), the dissociation constant \( K_a \) is given by: \[ K_a = \frac{c \alpha^2}{1 - \alpha} \]

The degree of dissociation \( \alpha \) is related to molar conductivity \( \Lambda_m \) and limiting molar conductivity \( \Lambda_m^0 \) by: \[ \alpha = \frac{\Lambda_m}{\Lambda_m^0} \]

Substituting \( \alpha \) into the \( K_a \) expression: \[ K_a = \frac{c (\Lambda_m / \Lambda_m^0)^2}{1 - (\Lambda_m / \Lambda_m^0)} = \frac{c \Lambda_m^2}{\Lambda_m^0 (\Lambda_m^0 - \Lambda_m)} \]

Rearranging to find a linear relationship between \( \frac{1}{\Lambda_m} \) (y-axis) and \( c\Lambda_m \) (x-axis): \[ K_a \Lambda_m^0 (\Lambda_m^0 - \Lambda_m) = c \Lambda_m^2 \] \[ K_a (\Lambda_m^0)^2 - K_a \Lambda_m^0 \Lambda_m = c \Lambda_m^2 \] Dividing the entire equation by \( \Lambda_m \cdot K_a (\Lambda_m^0)^2 \): \[ \frac{1}{\Lambda_m} - \frac{1}{\Lambda_m^0} = \frac{c \Lambda_m}{K_a (\Lambda_m^0)^2} \] \[ \frac{1}{\Lambda_m} = \frac{1}{K_a (\Lambda_m^0)^2} (c \Lambda_m) + \frac{1}{\Lambda_m^0} \]

This is in the form of a straight line equation \( y = S x + P \), where:

• \( y = 1 / \Lambda_m \)
• \( x = c \Lambda_m \)
• Slope (S) = \( \frac{1}{K_a (\Lambda_m^0)^2} \)
• y-intercept (P) = \( \frac{1}{\Lambda_m^0} \)

Calculating the Ratio P/S: \[ \frac{P}{S} = \frac{1 / \Lambda_m^0}{1 / (K_a (\Lambda_m^0)^2)} \] \[ \frac{P}{S} = \frac{1}{\Lambda_m^0} \times K_a (\Lambda_m^0)^2 \] \[ \frac{P}{S} = K_a \Lambda_m^0 \]

Thus, the correct option is (A).

Mathematical Analysis:

1. Solubility Relationship:
Let solubility be \( S \). The equilibria are:
\( \text{MX}(s) \rightleftharpoons \text{M}^+ + \text{X}^- \) ( \( K_{sp} \) )
\( \text{X}^- + \text{H}^+ \rightleftharpoons \text{HX} \) ( \( 1/K_a \) )
Mass balance: \( S = [\text{M}^+] = [\text{X}^-] + [\text{HX}] \).
Using \( [\text{HX}] = [\text{H}^+][\text{X}^-]/K_a \), we get:
\( S = [\text{X}^-] \left( 1 + \frac{[\text{H}^+]}{K_a} \right) \implies [\text{X}^-] = \frac{S}{1 + \frac{[\text{H}^+]}{K_a}} \)
The solubility product expression becomes:
\( K_{sp} = [\text{M}^+][\text{X}^-] = S \cdot \frac{S}{1 + \frac{[\text{H}^+]}{K_a}} = \frac{S^2}{1 + \frac{[\text{H}^+]}{K_a}} \)

2. Case 1: pH = 7 (\( [\text{H}^+] = 10^{-7} \))
Solubility \( S_1 = 10^{-4} \).
Assuming the weak acid has \( K_a \gg 10^{-7} \) (so \( 1 + \frac{10^{-7}}{K_a} \approx 1 \)):
\( K_{sp} \approx (10^{-4})^2 = 10^{-8} \).
(We will verify this assumption later).

3. Case 2: pH = 2 (\( [\text{H}^+] = 10^{-2} \))
Solubility \( S_2 = 10^{-3} \).
\( K_{sp} = \frac{(10^{-3})^2}{1 + \frac{10^{-2}}{K_a}} = \frac{10^{-6}}{1 + \frac{10^{-2}}{K_a}} \)

4. Solving for \( K_a \):
Equating \( K_{sp} \) from both cases:
\( 10^{-8} = \frac{10^{-6}}{1 + \frac{10^{-2}}{K_a}} \)
\( 1 + \frac{10^{-2}}{K_a} = \frac{10^{-6}}{10^{-8}} = 100 \)
\( \frac{10^{-2}}{K_a} = 99 \approx 100 \)
\( K_a = \frac{10^{-2}}{100} = 10^{-4} \)

5. Verification and Final Answer:
Check assumption for pH 7: \( \frac{10^{-7}}{K_a} = \frac{10^{-7}}{10^{-4}} = 10^{-3} \).
\( 1 + 0.001 \approx 1 \). The assumption holds true.
So, \( K_a = 10^{-4} \).
\( pK_a = -\log(10^{-4}) = 4 \).

Thus, the correct option is (B).

Step-by-Step Reaction Analysis:

1. Cleavage of Ether (P to Q):
   The reactant P is a phenyl alkyl ether (likely anisole or phenetole). Reaction with HI leads to the cleavage of the ether bond. Since the phenyl-oxygen bond has partial double bond character due to resonance, it does not break. Instead, the alkyl-oxygen bond breaks, forming Phenol (Q) and an alkyl iodide.
   \( \text{Ph-OR} + \text{HI} \rightarrow \text{Ph-OH} \text{ (Q)} + \text{R-I} \)
2. Kolbe's Reaction (Q to R):
   Phenol (Q) is treated with (i) NaOH, (ii) CO₂, and (iii) H₃O⁺. This is the classic Kolbe-Schmitt reaction, which introduces a carboxyl group ortho to the hydroxyl group.
   The product R is Salicylic acid (2-Hydroxybenzoic acid).
3. Acetylation (R to S):
   Salicylic acid (R) reacts with acetic anhydride, \( (\text{CH}_3\text{CO})_2\text{O} \), in acidic medium. This acetylates the hydroxyl group (-OH) to form an ester.
   The product S is Acetylsalicylic acid, commonly known as Aspirin.
   \( \text{Structure: } \text{2-CH}_3\text{COO-C}_6\text{H}_4\text{-COOH} \)

Properties of Product S (Aspirin):

• It is a non-narcotic analgesic (pain reliever).
• It acts as an anti-inflammatory and antipyretic drug.
• Mechanism of Action: It inhibits the synthesis of prostaglandins (chemicals that cause inflammation and pain) by inhibiting the cyclooxygenase (COX) enzymes.

Therefore, the correct statement is (B).

Step-by-Step Calculation:

1. Moles of Reactant:
Reactant: Dimethyldichlorosilane, \( (\text{CH}_3)_2\text{SiCl}_2 \)
Molar Mass \( = 28 (\text{Si}) + 2 \times 12 (\text{C}) + 6 \times 1 (\text{H}) + 2 \times 35.5 (\text{Cl}) \)
\( = 28 + 24 + 6 + 71 = 129 \text{ g/mol} \)
Moles of Reactant \( = \frac{\text{Given Mass}}{\text{Molar Mass}} = \frac{516}{129} = 4 \text{ mol} \)

2. Reaction Stoichiometry:
Hydrolysis of \( (\text{CH}_3)_2\text{SiCl}_2 \) followed by condensation produces cyclic silicones.
The formation of the tetrameric cyclic product X, \( [(\text{CH}_3)_2\text{SiO}]_4 \), follows the equation:
\( 4 (\text{CH}_3)_2\text{SiCl}_2 + 4 \text{H}_2\text{O} \rightarrow [(\text{CH}_3)_2\text{SiO}]_4 + 8 \text{HCl} \)
From the stoichiometry: 4 moles of reactant produce 1 mole of tetramer X.

3. Theoretical and Actual Yield:
Theoretical moles of X from 4 moles of reactant = 1 mol.
Given Percentage Yield = 75%.
Actual moles of X \( = 1 \text{ mol} \times \frac{75}{100} = 0.75 \text{ mol} \)

4. Mass of Product X:
Formula of X: \( [(\text{CH}_3)_2\text{SiO}]_4 \) (a cyclic ring of 4 units)
Molar Mass of one unit \( (\text{CH}_3)_2\text{SiO} = 2(12) + 6(1) + 28 + 16 = 24 + 6 + 28 + 16 = 74 \text{ g/mol} \)
Molar Mass of Tetramer X \( = 74 \times 4 = 296 \text{ g/mol} \)
Mass of X obtained \( = \text{Moles} \times \text{Molar Mass} \)
\( = 0.75 \times 296 = \frac{3}{4} \times 296 = 3 \times 74 = 222 \text{ g} \)

Answer: 222

Step-by-Step Calculation:

1. Calculate the Pressure (x):
Using the Compressibility Factor formula: \( Z = \frac{PV_m}{RT} \)
Given:
\( Z = 0.5 \)
\( V_m = 0.4 \text{ dm}^3\text{mol}^{-1} = 0.4 \text{ L mol}^{-1} \)
\( T = 800 \text{ K} \)
\( R = 8 \times 10^{-2} = 0.08 \text{ L atm K}^{-1}\text{mol}^{-1} \)
\( P = x \text{ atm} \)

Substituting values:
\( 0.5 = \frac{x \times 0.4}{0.08 \times 800} \)
\( 0.5 = \frac{0.4x}{64} \)
\( x = \frac{0.5 \times 64}{0.4} \)
\( x = \frac{32}{0.4} = 80 \text{ atm} \)

2. Calculate the Ideal Molar Volume (y):
For an ideal gas at the same temperature and pressure, \( PV_{ideal} = RT \).
Alternatively, we can use the definition of Z:
\( Z = \frac{V_{real}}{V_{ideal}} \)
Given \( V_{real} = 0.4 \) and \( Z = 0.5 \):
\( 0.5 = \frac{0.4}{y} \)
\( y = \frac{0.4}{0.5} = 0.8 \text{ dm}^3\text{mol}^{-1} \)

3. Calculate the Ratio x/y:
\( \frac{x}{y} = \frac{80}{0.8} \)
\( \frac{x}{y} = 100 \)

Answer: 100

Step-by-Step Calculation:

1. Analyze conditions at 500 K:

• From the graph, at \( \frac{1}{T} = 0.002 \text{ K}^{-1} \) (which means \( T = \frac{1}{0.002} = 500 \text{ K} \)), the value of \( \log k_f = 9 \).
• Given \( \log K = 6 \) at 500 K. Since \( K = \frac{k_f}{k_b} \), we have: \[ \log K = \log k_f - \log k_b \] \[ 6 = 9 - \log k_b \implies \log k_b = 3 \]

2. Determine Activation Energy for Backward Reaction:

• Use the Arrhenius equation in logarithmic form: \( \log k = \log A - \frac{E_a}{2.303RT} \).
• For the backward reaction at 500 K: \[ \log k_b = \log A_b - \frac{(E_a)_b}{2.303 R (500)} \]
• Given \( A_b = 10^{11} \), so \( \log A_b = 11 \). \[ 3 = 11 - \frac{(E_a)_b}{2.303 R (500)} \] \[ \frac{(E_a)_b}{2.303 R (500)} = 8 \] \[ \frac{(E_a)_b}{2.303 R} = 8 \times 500 = 4000 \text{ K} \]

3. Calculate \( \log k_b \) at 250 K:

• At \( T = 250 \text{ K} \): \[ \log k_b (250 \text{ K}) = \log A_b - \frac{(E_a)_b}{2.303 R (250)} \]
• Substitute the value of \( \frac{(E_a)_b}{2.303 R} = 4000 \): \[ \log k_b (250 \text{ K}) = 11 - \frac{4000}{250} \] \[ \log k_b = 11 - 16 = -5 \]

4. Final Answer:
We need the value of \( | \log k_b | \). \[ | -5 | = 5 \]

Step-by-Step Derivation:

1. Analyze Process A \(\rightarrow\) B (Reversible Adiabatic Expansion):
For a monoatomic ideal gas, \( C_{p,m} = \frac{5}{2}R \), so \( C_{v,m} = \frac{3}{2}R \) and \( \gamma = \frac{C_p}{C_v} = \frac{5}{3} \).
Adiabatic relation: \( T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \)
Given: \( T_1 = 600 \text{ K} \), \( V_1 = 10 \text{ m}^3 \), \( T_2 = 60 \text{ K} \). \[ 600 \times (10)^{\frac{5}{3} - 1} = 60 \times (V_2)^{\frac{5}{3} - 1} \] \[ 10 \times (10)^{2/3} = (V_2)^{2/3} \] \[ (10)^{1 + \frac{2}{3}} = V_2^{2/3} \implies (10)^{5/3} = V_2^{2/3} \] Raising both sides to the power of \( \frac{3}{2} \): \[ V_2 = (10^{5/3})^{3/2} = 10^{5/2} \]

2. Analyze Process B \(\rightarrow\) C (Reversible Isothermal Expansion):
Temperature is constant at \( T_2 = 60 \text{ K} \).
Heat absorbed in isothermal process: \( q_{BC} = nRT_2 \ln\left(\frac{V_3}{V_2}\right) \). Since \( n=1 \), \( q_{BC} = RT_2 \ln\left(\frac{V_3}{V_2}\right) \).
Heat absorbed in adiabatic process A \(\rightarrow\) B is zero (\( q_{AB} = 0 \)).
Given total heat absorbed \( q_{total} = RT_2 \ln 10 \). \[ q_{total} = q_{AB} + q_{BC} \] \[ RT_2 \ln 10 = 0 + RT_2 \ln\left(\frac{V_3}{V_2}\right) \] \[ \ln 10 = \ln\left(\frac{V_3}{V_2}\right) \implies \frac{V_3}{V_2} = 10 \implies V_3 = 10 V_2 \]

3. Calculate \( V_3 \) and Final Value:
Substitute \( V_2 = 10^{5/2} \): \[ V_3 = 10 \times 10^{5/2} = 10^{1 + 2.5} = 10^{3.5} = 10^{7/2} \] We need to find the value of \( 2 \log V_3 \): \[ 2 \log(10^{7/2}) = 2 \times \frac{7}{2} \log 10 \] \[ = 7 \times 1 = 7 \]

Answer: 7

Step-by-Step Calculation:

1. Calculate Equilibrium Constants:
Reaction: \( \text{A}(g) \rightleftharpoons \text{P}(g) \)
Initial moles of A = 6, Volume = 1 L. So initial concentration [A] = 6 M.

• At Temperature \( T_1 \):
  From graph, equilibrium concentration of P, \( [\text{P}]_{eq} = 4 \text{ M} \).
  \( [\text{A}]_{eq} = [\text{A}]_{initial} - [\text{P}]_{eq} = 6 - 4 = 2 \text{ M} \).
  \( K_1 = \frac{[\text{P}]}{[\text{A}]} = \frac{4}{2} = 2 \).
• At Temperature \( T_2 \):
  From graph, equilibrium concentration of P, \( [\text{P}]_{eq} = 2 \text{ M} \).
  \( [\text{A}]_{eq} = 6 - 2 = 4 \text{ M} \).
  \( K_2 = \frac{[\text{P}]}{[\text{A}]} = \frac{2}{4} = 0.5 = \frac{1}{2} \).

2. Calculate Standard Gibbs Free Energy Change (\( \Delta G^\circ \)):
Using formula \( \Delta G^\circ = -RT \ln K \).
Given \( T_1 = 2T_2 \).

• \( \Delta G_1^\circ = -R T_1 \ln K_1 = -R(2T_2) \ln 2 = -2 R T_2 \ln 2 \).
• \( \Delta G_2^\circ = -R T_2 \ln K_2 = -R T_2 \ln(1/2) = -R T_2 (-\ln 2) = R T_2 \ln 2 \).

3. Calculate the Difference and Solve for x: \[ \Delta G_2^\circ - \Delta G_1^\circ = (R T_2 \ln 2) - (-2 R T_2 \ln 2) \] \[ = R T_2 \ln 2 + 2 R T_2 \ln 2 \] \[ = 3 R T_2 \ln 2 \] Using the property \( a \ln b = \ln (b^a) \): \[ = R T_2 \ln (2^3) = R T_2 \ln 8 \] We are given \( (\Delta G_2^\circ - \Delta G_1^\circ) = R T_2 \ln x \). Comparing the terms: \[ x = 8 \]

Answer: 8

Step-by-Step Analysis:

1. Reaction 1: Reduction
   The starting material is ethane-1,1,2,2-tetracarbonitrile: \( (\text{NC})_2\text{CH}-\text{CH}(\text{CN})_2 \).
   Reaction with excess \( \text{LiAlH}_4 \) followed by hydrolysis reduces all four nitrile groups (\( -\text{CN} \)) to primary amine groups (\( -\text{CH}_2\text{NH}_2 \)).
   Intermediate Product: \( (\text{H}_2\text{N-CH}_2)_2\text{CH}-\text{CH}(\text{CH}_2\text{-NH}_2)_2 \).
2. Reaction 2: Condensation with Acetophenone
   The amine reacts with excess Acetophenone (\( \text{Ph-C(=O)CH}_3 \)) to form imines (Schiff bases).
   Reaction: \( \text{R-NH}_2 + \text{O=C(CH}_3)\text{Ph} \rightarrow \text{R-N=C(CH}_3)\text{Ph} + \text{H}_2\text{O} \)
   Since there are 4 amine groups, 4 molecules of acetophenone will react.
3. Structure of Product P:
   The product contains 4 imine units attached to the central ethane backbone via methylene bridges.
   General structure of the substituent group: \( -\text{CH}_2-\text{N}=\text{C}(\text{CH}_3)(\text{Ph}) \)
4. Counting \( sp^2 \) Hybridized Carbon Atoms:
   We count the \( sp^2 \) carbons in one substituent unit and then multiply by 4.
   • Imine carbon (\( \text{C=N} \)): 1 carbon (hybridization is \( sp^2 \)).
   • Phenyl ring (\( \text{Ph} \)): 6 carbons (all are part of the aromatic ring, so all are \( sp^2 \)).
   
   Total \( sp^2 \) carbons per unit = \( 1 + 6 = 7 \).
   Total units = 4.
   Total \( sp^2 \) carbons in P = \( 4 \times 7 = 28 \).

Note: The carbons in the backbone (\( \text{CH-CH} \)), the methylene bridges (\( \text{CH}_2 \)), and the methyl groups (\( \text{CH}_3 \)) on the imine are all \( sp^3 \) hybridized.

Answer: 28

Reaction Analysis:

• (P) \( \text{P}_2\text{O}_3 + 3\text{H}_2\text{O} \rightarrow \)
  \( \text{P}_2\text{O}_3 \) is the anhydride of phosphorous acid. Hydrolysis yields phosphorous acid.
  Reaction: \( \text{P}_2\text{O}_3 + 3\text{H}_2\text{O} \rightarrow 2\text{H}_3\text{PO}_3 \)
  Matches with (2).
• (Q) \( \text{P}_4 + 3\text{NaOH} + 3\text{H}_2\text{O} \rightarrow \)
  This is the disproportionation reaction of white phosphorus in a basic medium.
  Reaction: \( \text{P}_4 + 3\text{NaOH} + 3\text{H}_2\text{O} \rightarrow \text{PH}_3 + 3\text{NaH}_2\text{PO}_2 \)
  One of the products is Phosphine (\( \text{PH}_3 \)).
  Matches with (3).
• (R) \( \text{PCl}_5 + \text{CH}_3\text{COOH} \rightarrow \)
  \( \text{PCl}_5 \) is a chlorinating agent reacting with carboxylic acid.
  Reaction: \( \text{CH}_3\text{COOH} + \text{PCl}_5 \rightarrow \text{CH}_3\text{COCl} + \text{POCl}_3 + \text{HCl} \)
  One of the products is Phosphoryl chloride (\( \text{POCl}_3 \)).
  Matches with (4).
• (S) \( \text{H}_3\text{PO}_2 + 2\text{H}_2\text{O} + 4\text{AgNO}_3 \rightarrow \)
  Hypophosphorous acid (\( \text{H}_3\text{PO}_2 \)) is a strong reducing agent. It reduces silver nitrate to metallic silver and oxidizes to phosphoric acid.
  Reaction: \( \text{H}_3\text{PO}_2 + 4\text{AgNO}_3 + 2\text{H}_2\text{O} \rightarrow 4\text{Ag} \downarrow + \text{H}_3\text{PO}_4 + 4\text{HNO}_3 \)
  Matches with (5).

Matching Code:
P \(\rightarrow\) 2; Q \(\rightarrow\) 3; R \(\rightarrow\) 4; S \(\rightarrow\) 5.

Thus, option (D) is correct.

Crystal Field Theory Analysis:

• (P) \( t_{2g}^6 e_g^0 \):
  This configuration represents an octahedral complex with 6 d-electrons fully paired in the lower energy \( t_{2g} \) orbitals. This corresponds to a Low Spin \( d^6 \) complex.
  Looking at the options:
  \( [\text{Co(NH}_3)_6]^{3+} \): \( \text{Co}^{3+} \) is \( 3d^6 \). \( \text{NH}_3 \) is a strong field ligand, causing pairing.
  Match: (3).
• (Q) \( t_{2g}^3 e_g^2 \):
  This configuration has 5 electrons singly occupied (3 in \( t_{2g} \), 2 in \( e_g \)). This is a High Spin \( d^5 \) octahedral complex.
  Looking at the options:
  \( [\text{Mn(H}_2\text{O})_6]^{2+} \): \( \text{Mn}^{2+} \) is \( 3d^5 \). \( \text{H}_2\text{O} \) is a weak field ligand, so no pairing occurs.
  Match: (2).
• (R) \( e^2 t_2^3 \):
  This notation (\( e \) lower than \( t_2 \)) indicates a Tetrahedral geometry. The configuration shows 5 electrons singly occupied. This corresponds to a High Spin \( d^5 \) tetrahedral complex.
  Looking at the options:
  \( [\text{FeCl}_4]^- \): \( \text{Fe}^{3+} \) is \( 3d^5 \). \( \text{Cl}^- \) is a weak ligand and coordination number 4 implies tetrahedral geometry.
  Match: (4).
• (S) \( t_{2g}^4 e_g^2 \):
  This configuration has 6 electrons in an octahedral field: 3 singly in \( t_{2g} \), 2 singly in \( e_g \), and the 6th electron paired in \( t_{2g} \). This is a High Spin \( d^6 \) complex.
  Looking at the options:
  \( [\text{Fe(H}_2\text{O})_6]^{2+} \): \( \text{Fe}^{2+} \) is \( 3d^6 \). \( \text{H}_2\text{O} \) is a weak field ligand.
  Match: (1).

Matching Code:
P \(\rightarrow\) 3; Q \(\rightarrow\) 2; R \(\rightarrow\) 4; S \(\rightarrow\) 1.

Thus, option (D) is correct.

Detailed Reaction Analysis:

• (P) (-)-1-Bromo-2-ethylpentane (SN2):
  Structure: \(\text{CH}_3\text{CH}_2\text{CH}_2-\text{CH}(\text{CH}_2\text{CH}_3)-\text{CH}_2\text{Br}\).
  The chiral center is at C2. The reaction occurs at C1 (primary carbon).
  Since the reaction center is not the chiral center, the configuration at the chiral center remains unchanged.
  Result: Retention of configuration (2).
• (Q) (-)-2-Bromopentane (SN2):
  Structure: \(\text{CH}_3-\text{CH(Br)}-\text{CH}_2\text{CH}_2\text{CH}_3\).
  The reaction occurs directly at the chiral center (C2).
  SN2 mechanism proceeds with backside attack, leading to inversion of the stereocenter (Walden inversion).
  Result: Inversion of configuration (1).
• (R) (-)-3-Bromo-3-methylhexane (SN1):
  Structure: A tertiary alkyl halide with a chiral center.
  SN1 mechanism involves the formation of a planar carbocation intermediate.
  The nucleophile (\(\text{OH}^-\)) can attack from either side with equal probability, producing a racemic mixture (50:50 mixture of enantiomers).
  Result: Mixture of enantiomers (3).
• (S) Reaction of a diastereomeric bromide (SN1):
  The reactant has two chiral centers. Let's say configurations are (A, B). The leaving group is on one center (B).
  SN1 reaction forms a carbocation at center B. The other chiral center (A) is unaffected.
  Attack on the planar carbocation yields two configurations at B (say B' and B'').
  Products: (A, B') and (A, B''). Since one center is identical and the other is inverted/retained, these are diastereomers, not enantiomers.
  Result: Mixture of diastereomers (5).

Matching Code:
P \(\rightarrow\) 2; Q \(\rightarrow\) 1; R \(\rightarrow\) 3; S \(\rightarrow\) 5.

Thus, option (B) is correct.

Step-by-Step Matching:

The question asks to match named reactions (List-I) with the reactant generated by reactions in List-II.

• (P) Etard Reaction:
  Reactant required: Toluene.
  Toluene is oxidized by chromyl chloride (\(\text{CrO}_2\text{Cl}_2\)) to benzaldehyde.
  Look at List-II to find where Toluene is produced.
  Reaction (3): Benzene + \(\text{CH}_3\text{Cl}\) (anhyd. \(\text{AlCl}_3\)) \(\rightarrow\) Toluene (Friedel-Crafts Alkylation).
  So, P matches with 3.
• (Q) Gattermann Reaction:
  Reactant required: Benzene Diazonium Chloride (\(\text{Ph-N}_2^+\text{Cl}^-\)).
  Reaction: \(\text{Ph-N}_2^+\text{Cl}^- + \text{Cu}/\text{HCl} \rightarrow \text{Ph-Cl}\).
  Look at List-II to find where Diazonium salt is produced.
  Reaction (4): Aniline + \(\text{NaNO}_2/\text{HCl}\) (at 0-5°C) \(\rightarrow\) Benzene Diazonium Chloride.
  So, Q matches with 4.
• (R) Gattermann-Koch Reaction:
  Reactant required: Benzene.
  Reaction: Benzene + CO + HCl (anhyd. \(\text{AlCl}_3\)) \(\rightarrow\) Benzaldehyde.
  Look at List-II to find where Benzene is produced.
  Reaction (5): Phenol + Zn dust (\(\Delta\)) \(\rightarrow\) Benzene.
  So, R matches with 5.
• (S) Rosenmund Reduction:
  Reactant required: Benzoyl Chloride (\(\text{Ph-COCl}\)).
  Reaction: \(\text{Ph-COCl} + \text{H}_2\) (\(\text{Pd}/\text{BaSO}_4\)) \(\rightarrow\) Benzaldehyde.
  Look at List-II to find where Benzoyl Chloride is produced.
  Reaction (2): Toluene \(\xrightarrow{\text{KMnO}_4}\) Benzoic Acid \(\xrightarrow{\text{SOCl}_2}\) Benzoyl Chloride.
  So, S matches with 2.

Matching Code:
P \(\rightarrow\) 3; Q \(\rightarrow\) 4; R \(\rightarrow\) 5; S \(\rightarrow\) 2.

Thus, option (D) is correct.