JEE ADVANCED 2025 Paper-2

Given the condition $e^{x_0} + x_0 = 0$, we have $e^{x_0} = -x_0$.

We need to find $\alpha$ such that the limit exists and evaluate it. Let's test the option $\alpha = 3$.

The function is defined as:

$$g(x) = \frac{3xe^x + 3x - \alpha e^x - \alpha x}{3(e^x + 1)}$$

Substitute $\alpha = 3$:

$$g(x) = \frac{3xe^x + 3x - 3e^x - 3x}{3(e^x + 1)} = \frac{3xe^x - 3e^x}{3(e^x + 1)} = \frac{3e^x(x - 1)}{3(e^x + 1)} = \frac{e^x(x - 1)}{e^x + 1}$$

Now, substitute this simplified $g(x)$ into the limit expression:

$$L = \lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = \lim_{x \to x_0} \left| \frac{\frac{e^x(x - 1)}{e^x + 1} + e^{x_0}}{x - x_0} \right|$$

Combine terms in the numerator:

$$L = \lim_{x \to x_0} \left| \frac{e^x(x - 1) + e^{x_0}(e^x + 1)}{(e^x + 1)(x - x_0)} \right|$$

Let the numerator function be $N(x) = e^x(x - 1) + e^{x_0}(e^x + 1)$.

Substitute $e^{x_0} = -x_0$ into $N(x)$:

$$N(x) = e^x(x - 1) - x_0(e^x + 1) = xe^x - e^x - x_0e^x - x_0$$

Check if $N(x_0) = 0$ to apply L'Hopital's Rule:

$$N(x_0) = x_0e^{x_0} - e^{x_0} - x_0e^{x_0} - x_0 = -(e^{x_0} + x_0) = 0$$

The denominator $D(x) = (e^x + 1)(x - x_0)$ also approaches 0 as $x \to x_0$. Thus, we have a $0/0$ form. Apply L'Hopital's Rule:

Differentiate the numerator with respect to $x$:

$$N'(x) = \frac{d}{dx}(xe^x - e^x - x_0e^x - x_0) = (e^x + xe^x) - e^x - x_0e^x = xe^x - x_0e^x = e^x(x - x_0)$$

Differentiate the denominator:

$$D'(x) = \frac{d}{dx}((e^x + 1)(x - x_0)) = e^x(x - x_0) + (e^x + 1)(1)$$

Now evaluate the limit ratio at $x = x_0$:

$$L = \left| \frac{N'(x_0)}{D'(x_0)} \right| = \left| \frac{e^{x_0}(x_0 - x_0)}{e^{x_0}(x_0 - x_0) + (e^{x_0} + 1)} \right| = \left| \frac{0}{0 + e^{x_0} + 1} \right| = 0$$

Since the limit is 0 for $\alpha = 3$, statement (C) is TRUE.

We are asked to find the area of the region $R$ defined by the inequalities:

1. $x > 0$
2. $y > \frac{1}{x}$ (Region above the hyperbola $xy=1$)
3. $5x - 4y - 1 > 0 \implies 4y < 5x - 1 \implies y < \frac{5x-1}{4}$ (Region below Line 1)
4. $4x + 4y - 17 < 0 \implies 4y < 17 - 4x \implies y < \frac{17-4x}{4}$ (Region below Line 2)

Step 1: Find Intersection Points

• Intersection of Line 1 and Hyperbola ($y = 1/x$):
  $\frac{5x - 1}{4} = \frac{1}{x} \implies 5x^2 - x - 4 = 0 \implies (5x + 4)(x - 1) = 0$. Since $x > 0$, we get $x = 1$.
  Point A: $(1, 1)$.
• Intersection of Line 1 and Line 2:
  $\frac{5x - 1}{4} = \frac{17 - 4x}{4} \implies 5x - 1 = 17 - 4x \implies 9x = 18 \implies x = 2$.
  $y = \frac{5(2) - 1}{4} = \frac{9}{4}$.
  Point B: $(2, 9/4)$.
• Intersection of Line 2 and Hyperbola ($y = 1/x$):
  $\frac{17 - 4x}{4} = \frac{1}{x} \implies 17x - 4x^2 = 4 \implies 4x^2 - 17x + 4 = 0 \implies (4x - 1)(x - 4) = 0$.
  For the region defined by $x > 0$, the relevant intersection is $x = 4$ (looking at the geometry relative to point B).
  Point C: $(4, 1/4)$.

Step 2: Set up the Integral
The area is bounded below by the hyperbola and above by the two lines. The boundary changes from Line 1 to Line 2 at $x = 2$.

$$Area = \int_{1}^{2} \left( y_{Line1} - y_{hyp} \right) dx + \int_{2}^{4} \left( y_{Line2} - y_{hyp} \right) dx$$ $$Area = \int_{1}^{2} \left( \frac{5x - 1}{4} - \frac{1}{x} \right) dx + \int_{2}^{4} \left( \frac{17 - 4x}{4} - \frac{1}{x} \right) dx$$

Step 3: Evaluate the Integral
Part 1:

$$\int_{1}^{2} \left( \frac{5}{4}x - \frac{1}{4} - \frac{1}{x} \right) dx = \left[ \frac{5x^2}{8} - \frac{x}{4} - \ln x \right]_{1}^{2}$$ $$= \left( \frac{20}{8} - \frac{2}{4} - \ln 2 \right) - \left( \frac{5}{8} - \frac{1}{4} - 0 \right)$$ $$= (2.5 - 0.5 - \ln 2) - (0.625 - 0.25) = 2 - \ln 2 - 0.375 = 1.625 - \ln 2$$

Part 2:

$$\int_{2}^{4} \left( \frac{17}{4} - x - \frac{1}{x} \right) dx = \left[ \frac{17x}{4} - \frac{x^2}{2} - \ln x \right]_{2}^{4}$$ $$= \left( 17 - 8 - \ln 4 \right) - \left( \frac{34}{4} - 2 - \ln 2 \right)$$ $$= (9 - 2\ln 2) - (8.5 - 2 - \ln 2) = 9 - 6.5 - \ln 2 = 2.5 - \ln 2$$

Step 4: Total Area

$$Total Area = (1.625 - \ln 2) + (2.5 - \ln 2) = 4.125 - 2\ln 2$$ $$4.125 = \frac{33}{8} \quad \text{and} \quad 2\ln 2 = \ln(2^2) = \ln 4 = \log_e 4$$ $$Area = \frac{33}{8} - \log_e 4$$

We are given the equation:

$$\theta = \tan^{-1}(2\tan\theta) - \frac{1}{2}\sin^{-1}\left( \frac{6\tan\theta}{9 + \tan^2\theta} \right)$$

Let $t = \tan\theta$. The equation becomes:

$$\theta = \tan^{-1}(2t) - \frac{1}{2}\sin^{-1}\left( \frac{6t}{9 + t^2} \right)$$

Simplify the argument of the inverse sine function:

$$\frac{6t}{9 + t^2} = \frac{6t}{9(1 + t^2/9)} = \frac{\frac{2t}{3}}{1 + \left(\frac{t}{3}\right)^2} = \frac{2(t/3)}{1 + (t/3)^2}$$

Recall the identity $\sin^{-1}\left( \frac{2x}{1+x^2} \right) = 2\tan^{-1}x$ for $|x| \le 1$.
Let $x = t/3$. If $|t/3| \le 1$ (i.e., $|\tan\theta| \le 3$), then:

$$\frac{1}{2}\sin^{-1}\left( \frac{6t}{9 + t^2} \right) = \frac{1}{2} \cdot 2\tan^{-1}\left(\frac{t}{3}\right) = \tan^{-1}\left(\frac{\tan\theta}{3}\right)$$

Substitute this back into the original equation:

$$\theta = \tan^{-1}(2\tan\theta) - \tan^{-1}\left(\frac{\tan\theta}{3}\right)$$

Apply tangent to both sides:

$$\tan\theta = \tan\left( \tan^{-1}(2\tan\theta) - \tan^{-1}\left(\frac{\tan\theta}{3}\right) \right)$$

Using the formula $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$:

$$t = \frac{2t - \frac{t}{3}}{1 + 2t \cdot \frac{t}{3}}$$ $$t = \frac{\frac{5t}{3}}{1 + \frac{2t^2}{3}} = \frac{5t}{3 + 2t^2}$$ $$t(3 + 2t^2) = 5t$$ $$3t + 2t^3 - 5t = 0$$ $$2t^3 - 2t = 0 \implies 2t(t^2 - 1) = 0$$

The solutions for $t$ are $t = 0$, $t = 1$, and $t = -1$.
All these values satisfy the condition $|t| \le 3$ assumed for the inverse sine identity.

Now, find $\theta$ in the range $(-\pi/2, \pi/2)$ corresponding to these values:

• $t = 0 \implies \tan\theta = 0 \implies \theta = 0$
• $t = 1 \implies \tan\theta = 1 \implies \theta = \pi/4$
• $t = -1 \implies \tan\theta = -1 \implies \theta = -\pi/4$

Therefore, there are 3 real solutions.

(Note: For $|t| > 3$, substitution gives no solutions, as shown by contradiction in algebraic steps.)

Step 1: Determine the Locus S

We are given the pair of lines:

1. $4x - 3y = 12\alpha$
2. $4\alpha x + 3\alpha y = 12 \implies 4x + 3y = \frac{12}{\alpha}$

To find the locus of the intersection, we eliminate the parameter $\alpha$ by multiplying the two equations:

$$(4x - 3y)(4x + 3y) = (12\alpha) \cdot \left(\frac{12}{\alpha}\right)$$ $$16x^2 - 9y^2 = 144$$

Dividing by 144, we get the equation of a hyperbola:

$$\frac{x^2}{9} - \frac{y^2}{16} = 1$$

Here, $a^2 = 9$ and $b^2 = 16$.

Step 2: Find the Equation of the Tangent T

The tangent $T$ is parallel to the line $4x - \frac{3}{\sqrt{2}}y = 0$. The slope $m$ of this line is:

$$m = \frac{4}{3/\sqrt{2}} = \frac{4\sqrt{2}}{3}$$

The equation of a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with slope $m$ is given by:

$$y = mx \pm \sqrt{a^2m^2 - b^2}$$

Substitute the values:

$$a^2m^2 - b^2 = 9\left(\frac{4\sqrt{2}}{3}\right)^2 - 16 = 9\left(\frac{32}{9}\right) - 16 = 32 - 16 = 16$$

Thus, the tangent equations are:

$$y = \frac{4\sqrt{2}}{3}x \pm \sqrt{16} \implies y = \frac{4\sqrt{2}}{3}x \pm 4$$

The tangent passes through $(0, q)$ with $q > 0$. Therefore, the y-intercept must be positive. We choose the $+$ sign:

$$y = \frac{4\sqrt{2}}{3}x + 4$$

Step 3: Find Coordinates p and q

• y-intercept (set x=0): $y = 4$. Since the point is $(0, q)$, we have $q = 4$.
• x-intercept (set y=0): The point is $(p, 0)$.

$$0 = \frac{4\sqrt{2}}{3}p + 4 \implies \frac{4\sqrt{2}}{3}p = -4 \implies p = -\frac{3}{\sqrt{2}} = -\frac{3\sqrt{2}}{2}$$

Step 4: Calculate pq

$$pq = \left( -\frac{3\sqrt{2}}{2} \right)(4) = -6\sqrt{2}$$

Step 1: Set up the Matrix Equations
We are given matrices $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$, $P = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}$, and $Q = \begin{pmatrix} x & y \\ z & 4 \end{pmatrix}$.
We are given that $QR = RP$ for a $2 \times 2$ matrix $R = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ with non-zero entries.

Step 2: Expand the Matrix Multiplication
Calculate $QR$:

$$QR = \begin{pmatrix} x & y \\ z & 4 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} xa+yc & xb+yd \\ za+4c & zb+4d \end{pmatrix}$$

Calculate $RP$:

$$RP = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} 2a & 3b \\ 2c & 3d \end{pmatrix}$$

Step 3: Solve for Relationships between Variables
Equate corresponding elements:

1. $xa + yc = 2a \implies (x-2)a + yc = 0$
2. $xb + yd = 3b \implies (x-3)b + yd = 0$
3. $za + 4c = 2c \implies za = -2c \implies \frac{c}{a} = -\frac{z}{2}$ (Since $a \neq 0$)
4. $zb + 4d = 3d \implies zb = -d \implies \frac{d}{b} = -z$ (Since $b \neq 0$)

Substitute $c = -za/2$ into equation (1):

$$(x-2)a + y(-\frac{za}{2}) = 0$$

Dividing by $a$ (since $a \neq 0$):

$$x - 2 - \frac{yz}{2} = 0 \implies 2x - 4 - yz = 0 \quad \dots(\text{i})$$

Substitute $d = -zb$ into equation (2):

$$(x-3)b + y(-zb) = 0$$

Dividing by $b$ (since $b \neq 0$):

$$x - 3 - yz = 0 \implies yz = x - 3 \quad \dots(\text{ii})$$

Step 4: Solve for x, y, z
Substitute (ii) into (i):

$$2x - 4 - (x - 3) = 0 \implies x - 1 = 0 \implies x = 1$$

Then from (ii):

$$yz = 1 - 3 = -2$$

So, we have $x = 1$ and $yz = -2$. The matrix $Q$ is $\begin{pmatrix} 1 & y \\ z & 4 \end{pmatrix}$.

Step 5: Evaluate the Options

• Option (A): Determinant of $Q - 2I$

  $$Q - 2I = \begin{pmatrix} 1-2 & y \\ z & 4-2 \end{pmatrix} = \begin{pmatrix} -1 & y \\ z & 2 \end{pmatrix}$$ $$|Q - 2I| = (-1)(2) - yz = -2 - (-2) = 0$$

  This statement is TRUE.

• Option (B): Determinant of $Q - 6I$

  $$Q - 6I = \begin{pmatrix} 1-6 & y \\ z & 4-6 \end{pmatrix} = \begin{pmatrix} -5 & y \\ z & -2 \end{pmatrix}$$ $$|Q - 6I| = (-5)(-2) - yz = 10 - (-2) = 12$$

  This statement is TRUE.

• Option (C): Determinant of $Q - 3I$

  $$|Q - 3I| = \begin{vmatrix} -2 & y \\ z & 1 \end{vmatrix} = -2 - yz = 0$$

  The option claims it is 15. This is FALSE.

• Option (D): $yz = 2$
  We found $yz = -2$. This statement is FALSE.

Step 1: Determine the Locus S
Let the midpoint of a chord of the parabola $y^2 = x$ be $(h, k)$. The equation of the chord with midpoint $(h, k)$ is given by $T = S_1$:

$$yk - \frac{1}{2}(x + h) = k^2 - h$$ $$2yk - x - h = 2k^2 - 2h \implies x = 2yk - (2k^2 - h)$$

To find the area enclosed by the parabola and this chord, we can use the standard formula for the area of a segment of a parabola cut by a chord. The area is given by:

$$Area = \frac{1}{6a} |y_2 - y_1|^3$$

For $y^2 = x$, $4a = 1 \implies a = 1/4$. The intersection points' y-coordinates $y_1, y_2$ are roots of substituting line into curve: $y^2 = 2yk - 2k^2 + h \implies y^2 - 2ky + (2k^2 - h) = 0$. The difference of roots is:

$$|y_1 - y_2| = \frac{\sqrt{D}}{1} = \sqrt{(-2k)^2 - 4(1)(2k^2 - h)} = \sqrt{4k^2 - 8k^2 + 4h} = \sqrt{4(h - k^2)} = 2\sqrt{h - k^2}$$

Substituting into the area formula:

$$\frac{4}{3} = \frac{1}{6(1/4)} (2\sqrt{h - k^2})^3 = \frac{4}{6} \cdot 8 (h - k^2)^{3/2} = \frac{16}{3} (h - k^2)^{3/2}$$

Wait, standard formula check: Area $= \frac{(y_2-y_1)^3}{6}$? No, it's $\frac{1}{6} |coefficient \ of \ y^2 \ in \ x-eq| \dots$ Using integration: Area $= \int_{y_1}^{y_2} (x_{line} - x_{parabola}) dy = \frac{1}{6}(1)(y_2-y_1)^3$. So,

$$\frac{4}{3} = \frac{1}{6} (2\sqrt{h - k^2})^3 = \frac{8}{6} (h - k^2)^{3/2} = \frac{4}{3} (h - k^2)^{3/2}$$ $$(h - k^2)^{3/2} = 1 \implies h - k^2 = 1 \implies h = k^2 + 1$$

Replacing $(h, k)$ with $(x, y)$, the locus $S$ is $x = y^2 + 1$ or $y^2 = x - 1$.

Step 2: Check Points on S

• Option A: $(4, \sqrt{3})$. Substitute: $(\sqrt{3})^2 = 3$. $x - 1 = 4 - 1 = 3$. LHS = RHS. TRUE.
• Option B: $(5, \sqrt{2})$. Substitute: $(\sqrt{2})^2 = 2$. $x - 1 = 5 - 1 = 4$. $2 \neq 4$. FALSE.

Step 3: Calculate Area of Region R
Region R is in the first quadrant, bounded by $y^2 = x$ (upper curve, $y = \sqrt{x}$), $y^2 = x - 1$ (lower curve, $y = \sqrt{x-1}$), $x = 1$, and $x = 4$.

$$Area(R) = \int_{1}^{4} (\sqrt{x} - \sqrt{x-1}) dx$$ $$= \int_{1}^{4} x^{1/2} dx - \int_{1}^{4} (x-1)^{1/2} dx$$ $$= \left[ \frac{2}{3}x^{3/2} \right]_{1}^{4} - \left[ \frac{2}{3}(x-1)^{3/2} \right]_{1}^{4}$$ $$= \frac{2}{3} \left( 4^{3/2} - 1^{3/2} \right) - \frac{2}{3} \left( (3)^{3/2} - 0^{3/2} \right)$$ $$= \frac{2}{3} (8 - 1) - \frac{2}{3} (3\sqrt{3})$$ $$= \frac{14}{3} - 2\sqrt{3}$$

This matches Option C.

Step 1: Determine coordinates of points R and S on the circle.
The circle is $x^2 + y^2 = 9$ (radius 3). M is $(3, 0)$.
$\angle ROM = \pi/6$. The coordinates of R are $(3\cos(\pi/6), 3\sin(\pi/6)) = \left(\frac{3\sqrt{3}}{2}, \frac{3}{2}\right)$.
Since the line $x = x_1$ passes through R, $x_1 = \frac{3\sqrt{3}}{2}$.
$\angle SOM = \pi/3$. The coordinates of S are $(3\cos(\pi/3), 3\sin(\pi/3)) = \left(\frac{3}{2}, \frac{3\sqrt{3}}{2}\right)$.
Since the line $x = x_2$ passes through S, $x_2 = \frac{3}{2}$.

Step 2: Determine coordinates of points P and Q on the ellipse.
The ellipse is $\frac{x^2}{9} + \frac{y^2}{4} = 1$.
For P $(x_1, y_1)$ where $x_1 = \frac{3\sqrt{3}}{2}$ and $y_1 > 0$:

$$\frac{(3\sqrt{3}/2)^2}{9} + \frac{y_1^2}{4} = 1 \Rightarrow \frac{27/4}{9} + \frac{y_1^2}{4} = 1 \Rightarrow \frac{3}{4} + \frac{y_1^2}{4} = 1 \Rightarrow y_1^2 = 1 \Rightarrow y_1 = 1$$

So, $P = \left(\frac{3\sqrt{3}}{2}, 1\right)$.

For Q $(x_2, y_2)$ where $x_2 = \frac{3}{2}$ and $y_2 > 0$:

$$\frac{(3/2)^2}{9} + \frac{y_2^2}{4} = 1 \Rightarrow \frac{9/4}{9} + \frac{y_2^2}{4} = 1 \Rightarrow \frac{1}{4} + \frac{y_2^2}{4} = 1 \Rightarrow y_2^2 = 3 \Rightarrow y_2 = \sqrt{3}$$

So, $Q = \left(\frac{3}{2}, \sqrt{3}\right)$.

Step 3: Equation of line PQ.
Slope $m = \frac{\sqrt{3} - 1}{\frac{3}{2} - \frac{3\sqrt{3}}{2}} = \frac{\sqrt{3} - 1}{\frac{3}{2}(1 - \sqrt{3})} = \frac{-1(\sqrt{3} - 1)}{\frac{3}{2}(\sqrt{3} - 1)} = -\frac{2}{3}$.
Equation passing through P:

$$y - 1 = -\frac{2}{3}\left(x - \frac{3\sqrt{3}}{2}\right)$$ $$3y - 3 = -2x + 3\sqrt{3} \Rightarrow 2x + 3y = 3(1 + \sqrt{3})$$

This matches statement (A).

Step 4: Analyze Ratios.
$N_1$ is $(x_1, 0) = (\frac{3\sqrt{3}}{2}, 0)$ and $N_2$ is $(x_2, 0) = (\frac{3}{2}, 0)$.
Lengths:
$|N_1P| = y_1 = 1$.
$|N_1R| = y_R = \frac{3}{2}$.
$|N_2Q| = y_2 = \sqrt{3}$.
$|N_2S| = y_S = \frac{3\sqrt{3}}{2}$.

Check (C): $3|N_2Q| = 3\sqrt{3}$ and $2|N_2S| = 2(\frac{3\sqrt{3}}{2}) = 3\sqrt{3}$.
$3\sqrt{3} = 3\sqrt{3}$, so statement (C) is TRUE.

Check (D): $9|N_1P| = 9(1) = 9$ and $4|N_1R| = 4(\frac{3}{2}) = 6$.
$9 \neq 6$, so statement (D) is FALSE.

Step 1: Differentiate the Function
Given $f(x) = \frac{6x + \sin x}{2x + \sin x}$. Using the quotient rule:

$$f'(x) = \frac{(6 + \cos x)(2x + \sin x) - (6x + \sin x)(2 + \cos x)}{(2x + \sin x)^2}$$

Simplifying the numerator:

$$Numerator = 12x + 6\sin x + 2x\cos x + \sin x\cos x - (12x + 6x\cos x + 2\sin x + \sin x\cos x)$$ $$= 4\sin x - 4x\cos x = 4\cos x (\tan x - x)$$ $$f'(x) = \frac{4\cos x(\tan x - x)}{(2x + \sin x)^2}$$

Step 2: Analyze Local Maxima/Minima at x = 0
For small $x > 0$: $\tan x > x \implies \tan x - x > 0$. Also $\cos x > 0$. So $f'(0^+) > 0$.
For small $x < 0$: $\tan x < x \implies \tan x - x < 0$. Also $\cos x> 0$. So $f'(0^-) < 0$.
Since $f'(x)$ changes from negative to positive, $x = 0$ is a point of local minima. Statement (B) is True.

Step 3: Analyze Critical Points in $[\pi, 6\pi]$
Critical points occur where $f'(x) = 0$, i.e., $\tan x = x$ (roots $\lambda_k$) or $\cos x = 0$ (no, because $\tan x$ would be undefined, but looking at the original simplified numerator $4(\sin x - x \cos x)$, roots are $\tan x = x$).
The roots of $\tan x = x$ occur in intervals $(k\pi, k\pi + \pi/2)$.
Let's check the sign change of $f'(x)$ around roots $\lambda_k$:

• For odd $k$ (e.g., in $(\pi, 1.5\pi)$): $\cos x < 0$. $(\tan x - x)$ changes from $-$ to $+$. Product changes $+\to-$. Maxima.
• For even $k$ (e.g., in $(2\pi, 2.5\pi)$): $\cos x > 0$. $(\tan x - x)$ changes from $-$ to $+$. Product changes $-\to+$. Minima.

Interval $[\pi, 6\pi]$ covers $k=1, 2, 3, 4, 5$.

• $k=1, 3, 5$ correspond to Maxima (3 points).
• $k=2, 4$ correspond to Minima (2 points).

Statement (C) claims 3 local maxima in $[\pi, 6\pi]$. This is TRUE.

Step 4: Analyze Interval $[2\pi, 4\pi]$
This interval covers $k=2$ and $k=3$.

• $k=2$ is a local Minima.
• $k=3$ is a local Maxima.

The number of local minima is 1. Statement (D) claims 1 local minima. This is TRUE.

Given the differential equation:

$$x^2 \frac{dy}{dx} + xy = x^2 + y^2$$

Divide by $x^2$:

$$\frac{dy}{dx} = \frac{x^2 + y^2 - xy}{x^2} = 1 + \left(\frac{y}{x}\right)^2 - \frac{y}{x}$$

This is a homogeneous equation. Let $y = vx$, so $\frac{dy}{dx} = v + x \frac{dv}{dx}$. Substitute into the equation:

$$v + x \frac{dv}{dx} = 1 + v^2 - v$$ $$x \frac{dv}{dx} = v^2 - 2v + 1 = (v - 1)^2$$

Separate variables:

$$\frac{dv}{(v - 1)^2} = \frac{dx}{x}$$

Integrate both sides:

$$-\frac{1}{v - 1} = \ln x + C$$

Substitute $v = y/x$ back:

$$-\frac{1}{y/x - 1} = \ln x + C \implies \frac{x}{x - y} = \ln x + C$$

Use the condition $y(1) = 0$:

$$\frac{1}{1 - 0} = \ln 1 + C \implies 1 = 0 + C \implies C = 1$$

So, the solution is:

$$\frac{x}{x - y} = \ln x + 1 \implies x - y = \frac{x}{\ln x + 1} \implies y = x - \frac{x}{\ln x + 1} = \frac{x \ln x}{\ln x + 1}$$

Now, compute $y(e)$ and $y(e^2)$:

$$y(e) = \frac{e \ln e}{1 + \ln e} = \frac{e(1)}{2} = \frac{e}{2}$$ $$y(e^2) = \frac{e^2 \ln(e^2)}{1 + \ln(e^2)} = \frac{e^2(2)}{1 + 2} = \frac{2e^2}{3}$$

Finally, calculate the required value:

$$2 \frac{(y(e))^2}{y(e^2)} = 2 \frac{(e/2)^2}{2e^2/3} = 2 \frac{e^2/4}{2e^2/3} = \frac{e^2/2}{2e^2/3} = \frac{e^2}{2} \cdot \frac{3}{2e^2} = \frac{3}{4} = 0.75$$

Step 1: Identify the general term

The given expansion is $(1 + \frac{2}{5}x)^{23} = \sum_{i=0}^{23} a_i x^i$.

Using the binomial theorem, the general term involving $x^r$ is given by:

$$T_{r+1} = \binom{23}{r} (1)^{23-r} \left(\frac{2}{5}x\right)^r = \binom{23}{r} \left(\frac{2}{5}\right)^r x^r$$

Thus, the coefficient $a_r$ is:

$$a_r = \binom{23}{r} \left(\frac{2}{5}\right)^r$$

Step 2: Determine the condition for the largest coefficient

We want to find $r$ such that $a_r$ is maximized. This is equivalent to finding the numerically greatest term in the expansion of $(1 + y)^{23}$ where $y = 2/5$.

Let's consider the ratio of consecutive coefficients:

$$\frac{a_r}{a_{r-1}} = \frac{\binom{23}{r} (2/5)^r}{\binom{23}{r-1} (2/5)^{r-1}} = \frac{23 - (r-1)}{r} \cdot \frac{2}{5} = \frac{24 - r}{r} \cdot \frac{2}{5}$$

For $a_r$ to be increasing ($a_r \ge a_{r-1}$), we need:

$$\frac{24 - r}{r} \cdot \frac{2}{5} \ge 1$$ $$2(24 - r) \ge 5r$$ $$48 - 2r \ge 5r$$ $$48 \ge 7r$$ $$r \le \frac{48}{7} \approx 6.85$$

Step 3: Identify the value of r

Since $r$ must be an integer, the coefficients increase for $r = 1, 2, \dots, 6$ and decrease for $r > 6$.

Specifically:

• For $r=6$: $\frac{a_6}{a_5} \ge 1 \implies a_6 > a_5$.
• For $r=7$: $\frac{a_7}{a_6} = \frac{24-7}{7} \cdot \frac{2}{5} = \frac{17}{7} \cdot \frac{2}{5} = \frac{34}{35} < 1 \implies a_7 < a_6$.

Therefore, the largest coefficient is $a_6$.

Value of $r = 6$.

Step 1: Define Events and Probabilities

Let $E_1, E_2, E_3$ be the events that a bulb is produced by unit $M_1, M_2, M_3$ respectively.
Let $A$ be the event that a bulb is defective.

Given proportions: $2:2:1$.

• $P(E_1) = \frac{2}{5}$, $P(E_2) = \frac{2}{5}$, $P(E_3) = \frac{1}{5}$.

Given defective information:

• Total probability of defective bulb: $P(A) = 20\% = 0.2 = \frac{1}{5}$.
• $M_1$ defective rate: $P(A|E_1) = 15\% = 0.15$.
• Posterior probability for $M_2$: $P(E_2|A) = \frac{2}{5}$.

Step 2: Use Bayes' Theorem to find $P(A|E_2)$

$$P(E_2|A) = \frac{P(A|E_2)P(E_2)}{P(A)}$$ Substitute known values: $$\frac{2}{5} = \frac{P(A|E_2) \cdot \frac{2}{5}}{\frac{1}{5}}$$ $$\frac{2}{5} = 2 \cdot P(A|E_2)$$ $$P(A|E_2) = \frac{1}{5} = 0.2$$

Step 3: Use Total Probability Theorem to find $P(A|E_3)$

$$P(A) = P(A|E_1)P(E_1) + P(A|E_2)P(E_2) + P(A|E_3)P(E_3)$$ Substitute known values: $$0.2 = (0.15)\left(\frac{2}{5}\right) + (0.2)\left(\frac{2}{5}\right) + P(A|E_3)\left(\frac{1}{5}\right)$$ Multiply the entire equation by 5 to simplify: $$1.0 = 0.30 + 0.40 + P(A|E_3)$$ $$1.0 = 0.70 + P(A|E_3)$$ $$P(A|E_3) = 1.0 - 0.70 = 0.30$$

The probability that a bulb chosen from $M_3$ is defective is $0.30$.

Step 1: Check Linear Independence of Basis Vectors
Calculate the scalar triple product $[\vec{x} \vec{y} \vec{z}]$:

$$[\vec{x} \vec{y} \vec{z}] = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{vmatrix}$$ $$= 1(6 - 1) - 2(4 - 3) + 3(2 - 9)$$ $$= 1(5) - 2(1) + 3(-7)$$ $$= 5 - 2 - 21 = -18 \neq 0$$

Since the base vectors are linearly independent, the vectors $\vec{X}, \vec{Y}, \vec{Z}$ are coplanar if and only if the determinant of the transformation matrix is zero.

Step 2: Formulate the Coplanarity Condition
The vectors are defined as:

$$\vec{X} = \alpha \vec{x} + \beta \vec{y} - \vec{z}$$ $$\vec{Y} = -\vec{x} + \alpha \vec{y} + \beta \vec{z}$$ $$\vec{Z} = \beta \vec{x} - \vec{y} + \alpha \vec{z}$$

Wait, checking indices carefully from the question image:

$$\vec{Y} = \alpha\vec{y} + \beta\vec{z} - \vec{x}$$ $$\vec{Z} = \alpha\vec{z} + \beta\vec{x} - \vec{y}$$

Ordering coefficients for $(\vec{x}, \vec{y}, \vec{z})$:

• Row 1 ($\vec{X}$): $\alpha, \beta, -1$
• Row 2 ($\vec{Y}$): $-1, \alpha, \beta$
• Row 3 ($\vec{Z}$): $\beta, -1, \alpha$

The determinant must be zero:

$$D = \begin{vmatrix} \alpha & \beta & -1 \\ -1 & \alpha & \beta \\ \beta & -1 & \alpha \end{vmatrix} = 0$$

Step 3: Expand the Determinant

$$D = \alpha(\alpha^2 - (-\beta)) - \beta(-\alpha - \beta^2) + (-1)(1 - \alpha\beta)$$ $$D = \alpha^3 + \alpha\beta + \alpha\beta + \beta^3 - 1 + \alpha\beta$$ $$D = \alpha^3 + \beta^3 - 1 + 3\alpha\beta = 0$$

Step 4: Solve for $\alpha$ and $\beta$
We use the algebraic identity: $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab - bc - ca)$.
Here, let $a = \alpha$, $b = \beta$, and $c = -1$.
The equation becomes:

$$\alpha^3 + \beta^3 + (-1)^3 - 3(\alpha)(\beta)(-1) = 0$$ $$(\alpha + \beta - 1)(\alpha^2 + \beta^2 + 1 - \alpha\beta + \beta + \alpha) = 0$$

Since $\alpha, \beta$ are distinct positive real numbers, the second factor is:

$$\frac{1}{2}[(\alpha-\beta)^2 + (\alpha+1)^2 + (\beta+1)^2]$$

This term is strictly positive (greater than 0) because $\alpha, \beta > 0$.
Therefore, the only solution is:

$$\alpha + \beta - 1 = 0 \implies \alpha + \beta = 1$$

Step 5: Calculate Final Value
We need to find $\alpha + \beta - 3$.

$$\alpha + \beta - 3 = 1 - 3 = -2$$

Step 1: Simplify the Summation
We need to evaluate the sum $S = \sum_{n=1}^{2025} (-\omega)^n$.
This is a geometric progression with first term $a = -\omega$, common ratio $r = -\omega$, and number of terms $N = 2025$.

$$S = \frac{a(r^N - 1)}{r - 1} = \frac{-\omega((-\omega)^{2025} - 1)}{-\omega - 1}$$

Since 2025 is an odd number, $(-\omega)^{2025} = -(\omega^{2025})$.
Since $\omega$ is a cube root of unity, $\omega^3 = 1$. Also, $2025$ is divisible by 3 ($2+0+2+5=9$).
So, $\omega^{2025} = (\omega^3)^{675} = 1^{675} = 1$.
Thus, $(-\omega)^{2025} = -1$.

Substitute this back into the sum:

$$S = \frac{-\omega(-1 - 1)}{-(1 + \omega)} = \frac{2\omega}{-(-\omega^2)} = \frac{2\omega}{\omega^2} = \frac{2}{\omega} = 2\omega^2$$

(Using the identity $1 + \omega + \omega^2 = 0 \implies 1 + \omega = -\omega^2$ and $\frac{1}{\omega} = \omega^2$)

Step 2: Determine the Argument
Given $\omega = e^{i2\pi/3} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$.
Then $\omega^2 = e^{i4\pi/3} = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$.
So, $S = 2\omega^2 = 2\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = -1 - i\sqrt{3}$.

The complex number $z = -1 - i\sqrt{3}$ lies in the third quadrant (both real and imaginary parts are negative).
The principal argument $\alpha$ is given by:

$$\alpha = \arg(z) = -\pi + \tan^{-1}\left(\left|\frac{-\sqrt{3}}{-1}\right|\right) = -\pi + \frac{\pi}{3} = -\frac{2\pi}{3}$$

Step 3: Calculate the Final Value
We need to find the value of $\frac{3\alpha}{\pi}$.

$$\frac{3\alpha}{\pi} = \frac{3}{\pi} \left(-\frac{2\pi}{3}\right) = -2$$

Step 1: Understand the Goal
We need to find the derivative of the composite function $(f \circ g^{-1})(x)$ at $x = 2$. Let $h(x) = g^{-1}(x)$. We want to evaluate $\frac{d}{dx} f(h(x)) \big|_{x=2}$. Using the chain rule:

$$\frac{d}{dx} f(h(x)) = f'(h(x)) \cdot h'(x)$$

At $x=2$, this is $f'(h(2)) \cdot h'(2)$.

Step 2: Find $h(2)$
$h(2) = g^{-1}(2)$. Let $y = h(2)$, so $g(y) = 2$.

$$\frac{4}{1 + e^{-2y}} = 2 \implies 1 + e^{-2y} = 2 \implies e^{-2y} = 1 \implies -2y = 0 \implies y = 0$$

So, $h(2) = 0$.

Step 3: Find $h'(2)$
Since $h(x) = g^{-1}(x)$, the derivative of the inverse function is given by:

$$h'(x) = \frac{1}{g'(h(x))}$$ $$h'(2) = \frac{1}{g'(h(2))} = \frac{1}{g'(0)}$$

Compute $g'(x)$:

$$g(x) = 4(1 + e^{-2x})^{-1}$$ $$g'(x) = 4(-1)(1 + e^{-2x})^{-2} \cdot (e^{-2x}) \cdot (-2) = \frac{8e^{-2x}}{(1 + e^{-2x})^2}$$

Evaluate at $x=0$:

$$g'(0) = \frac{8e^0}{(1 + e^0)^2} = \frac{8(1)}{(1+1)^2} = \frac{8}{4} = 2$$

Thus, $h'(2) = \frac{1}{2}$.

Step 4: Find $f'(0)$
$f(x) = \log_e(x^2 + 2x + 4)$.

$$f'(x) = \frac{1}{x^2 + 2x + 4} \cdot (2x + 2)$$

Evaluate at $x = h(2) = 0$:

$$f'(0) = \frac{2(0) + 2}{0^2 + 2(0) + 4} = \frac{2}{4} = \frac{1}{2}$$

Step 5: Calculate the Final Result

$$\text{Derivative} = f'(0) \cdot h'(2) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} = 0.25$$

Step 1: Simplify the General Term
The given series is $\alpha = \sum \frac{1}{\sin(k^\circ)\sin(k+1)^\circ}$.
Using the difference identity for sine: $\sin(B - A) = \sin B \cos A - \cos B \sin A$.
We can write:

$$\frac{1}{\sin A \sin B} = \frac{\sin(B-A)}{\sin(B-A) \sin A \sin B} = \frac{1}{\sin(B-A)} \left( \frac{\sin B \cos A - \cos B \sin A}{\sin A \sin B} \right)$$ $$= \frac{1}{\sin(B-A)} (\cot A - \cot B)$$

Here, the difference between angles in the denominator is $(k+1) - k = 1^\circ$.
So each term becomes $\frac{1}{\sin 1^\circ} [\cot k^\circ - \cot(k+1)^\circ]$.

Step 2: Analyze the Series Terms
The series consists of terms like: $\frac{1}{\sin 60^\circ \sin 61^\circ} + \frac{1}{\sin 62^\circ \sin 63^\circ} + \dots + \frac{1}{\sin 118^\circ \sin 119^\circ}$.

This can be written as:

$$\alpha = \frac{1}{\sin 1^\circ} \left[ (\cot 60^\circ - \cot 61^\circ) + (\cot 62^\circ - \cot 63^\circ) + \dots + (\cot 118^\circ - \cot 119^\circ) \right]$$

Convert to tangent using $\cot \theta = \tan(90^\circ - \theta)$:

• $\cot 60^\circ - \cot 61^\circ = \tan 30^\circ - \tan 29^\circ$
• $\cot 62^\circ - \cot 63^\circ = \tan 28^\circ - \tan 27^\circ$
• ...
• $\cot 118^\circ - \cot 119^\circ = \tan(90^\circ-118^\circ) - \tan(90^\circ-119^\circ) = \tan(-28^\circ) - \tan(-29^\circ) = -\tan 28^\circ + \tan 29^\circ$.

Step 3: Sum the Series
The sum inside the bracket is:

$$S = (\tan 30^\circ - \tan 29^\circ) + (\tan 28^\circ - \tan 27^\circ) + \dots + (\tan 2^\circ - \tan 1^\circ) + (\tan 0^\circ - \tan(-1)^\circ) + \dots + (-\tan 28^\circ + \tan 29^\circ)$$

Notice the symmetry. For every positive tangent term like $-\tan 29^\circ$ in the first group, there is a corresponding $+\tan 29^\circ$ in the last group.
Specifically, the terms form a telescoping-like pattern where all intermediate terms cancel out perfectly.

• $\tan 29^\circ$ cancels with $-\tan 29^\circ$.
• $\tan 28^\circ$ cancels with $-\tan 28^\circ$.
• ...

The only term that does not have a pair to cancel is the very first term, $\tan 30^\circ$, because the sequence wraps around zero symmetrically.
Thus, the sum inside the bracket is simply $\tan 30^\circ$.

$$\alpha = \frac{1}{\sin 1^\circ} (\tan 30^\circ) = \frac{1}{\sqrt{3} \sin 1^\circ}$$

Step 4: Calculate the Final Value
We need to find $\left( \frac{\text{cosec } 1^\circ}{\alpha} \right)^2$.

$$\frac{\text{cosec } 1^\circ}{\alpha} = \frac{1/\sin 1^\circ}{1/(\sqrt{3}\sin 1^\circ)} = \sqrt{3}$$ $$\left( \sqrt{3} \right)^2 = 3$$

Step 1: Apply Substitution to the Integral
Let the given integral be: $$ \alpha = \int_{1/2}^{2} \frac{\tan^{-1}x}{2x^2 - 3x + 2} dx $$ Substitute $x = \frac{1}{t}$, which implies $dx = -\frac{1}{t^2} dt$.
The limits change as follows: when $x = 1/2, t = 2$ and when $x = 2, t = 1/2$. $$ \alpha = \int_{2}^{1/2} \frac{\tan^{-1}(1/t)}{2(1/t)^2 - 3(1/t) + 2} \left(-\frac{1}{t^2}\right) dt $$ Simplify the denominator and the differential term: $$ \alpha = \int_{2}^{1/2} \frac{\cot^{-1}t}{\frac{2 - 3t + 2t^2}{t^2}} \left(-\frac{1}{t^2}\right) dt = \int_{1/2}^{2} \frac{\cot^{-1}t}{2t^2 - 3t + 2} dt $$ Since the variable of integration is a dummy variable, we can write: $$ \alpha = \int_{1/2}^{2} \frac{\cot^{-1}x}{2x^2 - 3x + 2} dx $$

Step 2: Add the Two Expressions for $\alpha$
Adding the original integral and the transformed integral: $$ 2\alpha = \int_{1/2}^{2} \frac{\tan^{-1}x + \cot^{-1}x}{2x^2 - 3x + 2} dx $$ Using the identity $\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}$ for $x > 0$: $$ 2\alpha = \frac{\pi}{2} \int_{1/2}^{2} \frac{1}{2x^2 - 3x + 2} dx $$ Factor out 2 from the denominator: $$ 2\alpha = \frac{\pi}{4} \int_{1/2}^{2} \frac{1}{x^2 - \frac{3}{2}x + 1} dx $$

Step 3: Evaluate the Integral
Complete the square in the denominator: $$ x^2 - \frac{3}{2}x + 1 = \left(x - \frac{3}{4}\right)^2 + 1 - \frac{9}{16} = \left(x - \frac{3}{4}\right)^2 + \frac{7}{16} = \left(x - \frac{3}{4}\right)^2 + \left(\frac{\sqrt{7}}{4}\right)^2 $$ Substitute this back into the integral expression: $$ 2\alpha = \frac{\pi}{4} \int_{1/2}^{2} \frac{dx}{\left(x - \frac{3}{4}\right)^2 + \left(\frac{\sqrt{7}}{4}\right)^2} $$ Using the standard integral $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right)$: $$ 2\alpha = \frac{\pi}{4} \cdot \frac{1}{\sqrt{7}/4} \left[ \tan^{-1}\left(\frac{x - 3/4}{\sqrt{7}/4}\right) \right]_{1/2}^{2} $$ $$ 2\alpha = \frac{\pi}{\sqrt{7}} \left[ \tan^{-1}\left(\frac{4x - 3}{\sqrt{7}}\right) \right]_{1/2}^{2} $$ Apply limits: $$ 2\alpha = \frac{\pi}{\sqrt{7}} \left( \tan^{-1}\left(\frac{8 - 3}{\sqrt{7}}\right) - \tan^{-1}\left(\frac{2 - 3}{\sqrt{7}}\right) \right) $$ $$ 2\alpha = \frac{\pi}{\sqrt{7}} \left( \tan^{-1}\left(\frac{5}{\sqrt{7}}\right) - \tan^{-1}\left(\frac{-1}{\sqrt{7}}\right) \right) $$ Since $\tan^{-1}(-x) = -\tan^{-1}x$: $$ 2\alpha = \frac{\pi}{\sqrt{7}} \left( \tan^{-1}\left(\frac{5}{\sqrt{7}}\right) + \tan^{-1}\left(\frac{1}{\sqrt{7}}\right) \right) $$

Step 4: Simplify the Trigonometric Expression
Rearrange the equation to isolate the term required for the final answer: $$ \frac{2\alpha\sqrt{7}}{\pi} = \tan^{-1}\left(\frac{5}{\sqrt{7}}\right) + \tan^{-1}\left(\frac{1}{\sqrt{7}}\right) $$ Apply the sum formula $\tan^{-1}A + \tan^{-1}B = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$: $$ A = \frac{5}{\sqrt{7}}, \quad B = \frac{1}{\sqrt{7}} $$ $$ A + B = \frac{6}{\sqrt{7}} $$ $$ 1 - AB = 1 - \frac{5}{7} = \frac{2}{7} $$ $$ \frac{A+B}{1-AB} = \frac{6/\sqrt{7}}{2/7} = \frac{6}{\sqrt{7}} \cdot \frac{7}{2} = 3\sqrt{7} $$ So, $$ \frac{2\alpha\sqrt{7}}{\pi} = \tan^{-1}(3\sqrt{7}) $$ Taking the tangent of both sides: $$ \tan\left(\frac{2\alpha\sqrt{7}}{\pi}\right) = 3\sqrt{7} $$

Step 5: Final Calculation
We need to find the value of $\sqrt{7} \tan\left(\frac{2\alpha\sqrt{7}}{\pi}\right)$. $$ \sqrt{7} (3\sqrt{7}) = 3 \times 7 = 21 $$

We need to determine the dimensional formula for the quantity \( Z = \frac{S^2 \sigma}{k} \).

Step 1: Determine the dimensions of individual quantities.

1. \( S \) (Seebeck coefficient):
   Defined as e.m.f per unit temperature difference.
   \( S = \frac{V}{\Delta T} \)
   Dimension of Potential \( V = \frac{\text{Work}}{\text{Charge}} = \frac{[ML^2T^{-2}]}{[IT]} = [ML^2T^{-3}I^{-1}] \)
   Dimension of Temperature \( \Delta T = [K] \)
   Therefore, \( [S] = [ML^2T^{-3}I^{-1}K^{-1}] \)

2. \( \sigma \) (Electrical conductivity):
   Inverse of resistivity \( \rho \). From \( R = \rho \frac{L}{A} \), we have \( \sigma = \frac{1}{\rho} = \frac{L}{R A} \).
   Using Ohm's law \( V = IR \), so \( R = V/I \).
   \( \sigma = \frac{L \cdot I}{V \cdot A} \)
   \( [\sigma] = \frac{[L][I]}{[ML^2T^{-3}I^{-1}][L^2]} = [M^{-1}L^{-3}T^3I^2] \)

3. \( \kappa \) (Thermal conductivity):
   From the heat conduction equation \( \frac{dQ}{dt} = k A \frac{\Delta T}{L} \).
   \( k = \frac{\text{Power} \cdot L}{A \cdot \Delta T} \)
   \( [\kappa] = \frac{[ML^2T^{-3}][L]}{[L^2][K]} = [MLT^{-3}K^{-1}] \)

Step 2: Substitute dimensions into the formula for \( Z \).

\[ Z = \frac{S^2 \sigma}{k} \]

\[ [Z] = \frac{[ML^2T^{-3}I^{-1}K^{-1}]^2 \cdot [M^{-1}L^{-3}T^3I^2]}{[MLT^{-3}K^{-1}]} \]

Let's simplify the numerator first: \[ \text{Numerator} = [M^2L^4T^{-6}I^{-2}K^{-2}] \cdot [M^{-1}L^{-3}T^3I^2] = [M^{2-1} L^{4-3} T^{-6+3} I^{-2+2} K^{-2}] = [MLT^{-3}K^{-2}] \]

Now divide by the denominator: \[ [Z] = \frac{[MLT^{-3}K^{-2}]}{[MLT^{-3}K^{-1}]} = [M^{1-1} L^{1-1} T^{-3-(-3)} K^{-2-(-1)}] \]

\[ [Z] = [M^0 L^0 T^0 K^{-1}] \]

Alternatively, using algebraic simplification before dimensional substitution: \( Z = \frac{(V/T)^2 \cdot (L/RA)}{(P \cdot L / (A \cdot T))} \) where \( P \) is power. Using \( P = V^2/R \), the expression simplifies to \( \frac{1}{T} \). The dimension is simply \( K^{-1} \).

We need to calculate the electric flux through a rectangular plane embedded in the dielectric material between two cylinders.

1. Electric Field in the Annular Region:
Using Gauss's Law for a cylinder of infinite length (approximation since \( l \) is large compared to radii, or simply applying the formula for coaxial cable): The electric field at a distance \( r \) from the axis is given by: \[ E = \frac{\lambda}{2 \pi \epsilon_0 \kappa r} \] where \( \lambda = Q/l \) is the linear charge density, and \( \kappa = 5 \) is the dielectric constant. So, \( E = \frac{Q}{2 \pi \epsilon_0 (5) l r} = \frac{Q}{10 \pi \epsilon_0 l r} \).

2. Setup for Flux Calculation:
The flux \( \phi = \int \mathbf{E} \cdot d\mathbf{A} \). The plane is parallel to the axis at a distance \( R \) from the center. Let's set up a coordinate system where the plane is vertical. Consider a thin strip on this plane at a vertical height \( y \) from the horizontal axis. The strip has length \( l \) and width \( dy \). Area element \( dA = l \cdot dy \). The distance of this strip from the center (radius) is \( r = \sqrt{R^2 + y^2} \). The electric field vector is radial. The angle \( \theta \) between the electric field vector and the surface normal (which is horizontal) satisfies \( \cos \theta = \frac{R}{r} = \frac{R}{\sqrt{R^2 + y^2}} \). Thus, the flux through the strip is: \[ d\phi = E \cos \theta dA = \left( \frac{\lambda}{2 \pi \epsilon_0 \kappa r} \right) \left( \frac{R}{r} \right) (l dy) = \frac{\lambda l R}{2 \pi \epsilon_0 \kappa} \frac{dy}{r^2} \] Substituting \( r^2 = R^2 + y^2 \): \[ d\phi = \frac{\lambda l R}{2 \pi \epsilon_0 \kappa} \frac{dy}{R^2 + y^2} \]

3. Limits of Integration:
The flux is restricted to the region between the inner and outer cylinders. At inner cylinder: \( r = \sqrt{2}R \implies \sqrt{R^2 + y^2} = \sqrt{2}R \implies y^2 = R^2 \implies y = R \). At outer cylinder: \( r = 2R \implies \sqrt{R^2 + y^2} = 2R \implies y^2 = 3R^2 \implies y = \sqrt{3}R \). The total flux is through both the upper and lower parts of the plane (symmetry), so we integrate from \( R \) to \( \sqrt{3}R \) and multiply by 2. \[ \phi = 2 \int_{R}^{\sqrt{3}R} \frac{\lambda l R}{2 \pi \epsilon_0 \kappa} \frac{dy}{R^2 + y^2} \]

4. Evaluate Integral:
\[ \phi = \frac{\lambda l R}{\pi \epsilon_0 \kappa} \int_{R}^{\sqrt{3}R} \frac{dy}{R^2 + y^2} \] Recall \( \int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1}(x/a) \). \[ \phi = \frac{\lambda l R}{\pi \epsilon_0 \kappa} \left[ \frac{1}{R} \tan^{-1}\left(\frac{y}{R}\right) \right]_{R}^{\sqrt{3}R} \] \[ \phi = \frac{\lambda l}{\pi \epsilon_0 \kappa} \left( \tan^{-1}(\sqrt{3}) - \tan^{-1}(1) \right) \] Substitute \( \lambda l = Q \) and \( \kappa = 5 \): \[ \phi = \frac{Q}{5 \pi \epsilon_0} \left( \frac{\pi}{3} - \frac{\pi}{4} \right) \] \[ \phi = \frac{Q}{5 \pi \epsilon_0} \left( \frac{\pi}{12} \right) = \frac{Q}{60 \epsilon_0} \]

We need to compare the oscillation frequencies \( f_1 \) and \( f_2 \) for the two arrangements.

General Formula: For angular SHM, the restoring torque is \( \tau = -C \theta \). The angular frequency is \( \omega = \sqrt{\frac{C}{I}} \), where \( I \) is the moment of inertia about the hinge. Here, \( I = \frac{m l^2}{3} \).

Case I: One spring at the midpoint (\( l/2 \)) and one at the top end (\( l \)). Both springs act to restore the rod. Displacement at midpoint: \( x_1 = \frac{l}{2} \theta \). Force \( F_1 = k x_1 \). Displacement at top: \( x_2 = l \theta \). Force \( F_2 = k x_2 \). Torque \( \tau_1 = - [ F_1 \cdot \frac{l}{2} + F_2 \cdot l ] \) \( \tau_1 = - [ k(\frac{l}{2}\theta)(\frac{l}{2}) + k(l\theta)(l) ] \) \( \tau_1 = - k l^2 \theta [ \frac{1}{4} + 1 ] = - \frac{5}{4} k l^2 \theta \) Torsional constant \( C_1 = \frac{5}{4} k l^2 \). \( \omega_1^2 = \frac{C_1}{I} = \frac{\frac{5}{4} k l^2}{\frac{m l^2}{3}} = \frac{15k}{4m} \)

Case II: Two springs connected at the midpoint (\( l/2 \)), effectively in parallel for rotational displacement. Displacement: \( x = \frac{l}{2} \theta \). Both springs provide restoring force in the same direction. Total Force \( F = 2 \cdot k \cdot x = 2k(\frac{l}{2}\theta) = k l \theta \). Torque \( \tau_2 = - F \cdot \frac{l}{2} = - (k l \theta)(\frac{l}{2}) = - \frac{1}{2} k l^2 \theta \) Torsional constant \( C_2 = \frac{1}{2} k l^2 \). \( \omega_2^2 = \frac{C_2}{I} = \frac{\frac{1}{2} k l^2}{\frac{m l^2}{3}} = \frac{3k}{2m} \)

Calculate Ratio: \( \frac{f_1}{f_2} = \sqrt{\frac{\omega_1^2}{\omega_2^2}} = \sqrt{\frac{15k/4m}{3k/2m}} \) \( \frac{f_1}{f_2} = \sqrt{\frac{15}{4} \cdot \frac{2}{3}} = \sqrt{\frac{5}{2}} \)

This matches option (C).

System Setup:

• Point charge \( q = 10^{-8} \) C.
• Sphere radius \( R = 10 \) cm = 0.1 m.
• Distance from center \( d_1 = 20 \) cm = 0.2 m.
• \( k = 9 \times 10^9 \) Nm\(^2\)/C\(^2\).

Step 1: Before Grounding (Analysis of Option A) The sphere is neutral. The potential at the center (and everywhere on/inside the conductor) is due to the external charge \( q \) plus the induced charges. The net potential due to induced charges at the center is zero (sum of \( q_{ind}/R \) where \( \sum q_{ind} = 0 \)). No, wait. Potential of a neutral conducting sphere in the field of a point charge: The potential is constant throughout the sphere. Potential at center = Potential due to \( q \) at distance \( d_1 \) + Potential due to induced charges. Wait, for a neutral sphere, the average potential is not necessarily just \( kq/d_1 \). Actually, the potential of an isolated neutral conductor is determined by the potential at its center, which is \( V_c = \frac{kq}{d_1} + V_{induced} \). Since the total induced charge is zero and distributed on the surface, the potential at the center due to induced charges is \( \int \frac{k dq_{ind}}{R} = \frac{k}{R} \int dq_{ind} = 0 \). So, \( V_{sphere} = \frac{kq}{d_1} \). Calculation: \( V = \frac{9 \times 10^9 \times 10^{-8}}{0.2} = \frac{90}{0.2} = 450 \) V. Statement (A) is Correct.

Step 2: Grounding Process (Analysis of Option B) The sphere is grounded, so its potential becomes 0. Let the new charge on the sphere be \( Q \). Potential \( V = \frac{kQ}{R} + \frac{kq}{d_1} = 0 \). \( \frac{Q}{R} = - \frac{q}{d_1} \) \( Q = - q \frac{R}{d_1} = - 10^{-8} \times \frac{0.1}{0.2} = - \frac{1}{2} \times 10^{-8} = -5 \times 10^{-9} \) C. Initial charge was 0. Final charge is \( -5 \times 10^{-9} \) C. Charge flowed from ground to sphere is \( -5 \times 10^{-9} \) C. Therefore, charge flowed from sphere to ground is \( +5 \times 10^{-9} \) C. Statement (B) is Correct.

Step 3: Grounding Removed (Analysis of Option C) The grounding is removed. The charge \( Q = -5 \times 10^{-9} \) C is trapped on the sphere. Statement (C) is Correct.

Step 4: Point Charge Moved (Analysis of Option D) The point charge \( q \) is moved 10 cm further away. New distance \( d_2 = 20 + 10 = 30 \) cm = 0.3 m. The charge on the sphere is still \( Q = -5 \times 10^{-9} \) C. New Potential of the sphere \( V_{final} = \frac{kQ}{R} + \frac{kq}{d_2} \). \( V_{final} = \frac{9 \times 10^9 \times (-5 \times 10^{-9})}{0.1} + \frac{9 \times 10^9 \times 10^{-8}}{0.3} \) Term 1: \( \frac{-45}{0.1} = -450 \) V. Term 2: \( \frac{90}{0.3} = 300 \) V. \( V_{final} = -450 + 300 = -150 \) V. The statement says 300 V. Statement (D) is Incorrect.

Correct Options: A, B, C

1. Analyze the Optical Path and Shift:

The light source \( S \) is embedded at the center of a glass slab of thickness \( t \) and refractive index \( n_0 \). Rays traveling from \( S \) towards either mirror pass through a thickness of \( t/2 \) of the glass.

Refraction at the plane surface of the slab causes an apparent shift in the position of the object. For an object at a real depth \( d_{real} \) inside a medium of refractive index \( \mu \), the apparent depth is \( d_{app} = \frac{d_{real}}{\mu} \).

The shift in the position of the object in the direction of light propagation is given by:

$$ \text{Shift} = d_{real} - d_{app} = \frac{t}{2} - \frac{t/2}{n_0} = \frac{t}{2} \left( 1 - \frac{1}{n_0} \right) $$

This means the object \( S \) appears closer to the mirror surface by this shift amount.

2. Define Distances:

Let \( d \) be the distance between the two mirrors. Since the slab is equidistant from the mirrors, the geometric distance from the center (source \( S \)) to either mirror is \( d/2 \).

The apparent distance of the source \( S \) from the mirror, \( u \), is:

$$ u = \text{Geometric Distance} - \text{Shift} $$ $$ u = \frac{d}{2} - \frac{t}{2} \left( 1 - \frac{1}{n_0} \right) $$

3. Conditions for Image to Form on \( S \) itself:

For the image to form back at the source \( S \), the rays must retrace their path to the original position. There are two primary cases for a concave mirror where this symmetry can occur in this setup:

Case A: Rays strike the mirror normally.

This happens when the object (apparent position of \( S \)) is located at the center of curvature of the mirror. Thus, \( u = 2f \).

$$ \frac{d}{2} - \frac{t}{2} \left( 1 - \frac{1}{n_0} \right) = 2f $$ $$ \frac{d}{2} = 2f + \frac{t}{2} \left( 1 - \frac{1}{n_0} \right) $$ $$ d = 4f + t \left( 1 - \frac{1}{n_0} \right) $$

This matches Option (A).

Case B: Rays strike the mirror and become parallel to the principal axis.

This happens when the object (apparent position of \( S \)) is located at the focal point of the mirror. Thus, \( u = f \).

If the rays become parallel after reflection from the first mirror, they will travel across to the second identical mirror. Parallel rays striking a concave mirror focus at its focal point. Due to the symmetry of the setup, the focal point of the second mirror coincides with the apparent position of \( S \) as seen from that side. Thus, the image forms at \( S \).

$$ \frac{d}{2} - \frac{t}{2} \left( 1 - \frac{1}{n_0} \right) = f $$ $$ \frac{d}{2} = f + \frac{t}{2} \left( 1 - \frac{1}{n_0} \right) $$ $$ d = 2f + t \left( 1 - \frac{1}{n_0} \right) $$

This matches Option (B).

Therefore, both distances are possible correct configurations.

Analysis of Electric Fields due to Infinite Sheets:

The electric field \( E \) due to a single infinite sheet with charge density \( \sigma \) is given by \( E = \frac{\sigma}{2\epsilon_0} \), directed away from the sheet if \( \sigma > 0 \) and towards it if \( \sigma < 0 \). The net field in any region is the vector sum of fields from all sheets.

Option (A): Region 4 of Configuration I

Configuration I has charge densities: \( +\sigma_0, -\sigma_0, +\sigma_0, -\sigma_0, +\sigma_0, -\sigma_0 \).

Region 4 is located between the 4th sheet (\(-\sigma_0\)) and the 5th sheet (\(+\sigma_0\)).

• Sheets to the left (1, 2, 3, 4): Sum of charge densities = \( \sigma_0 - \sigma_0 + \sigma_0 - \sigma_0 = 0 \).
• Sheets to the right (5, 6): Sum of charge densities = \( \sigma_0 - \sigma_0 = 0 \).

Using the superposition principle, the net field is proportional to the difference between the net charge density to the left and to the right. Since both are zero, the electric field \( E_4 = 0 \).
Alternatively, summing individual fields (Right is positive): \( E = \frac{\sigma_0}{2\epsilon_0}(+1 -1 +1 -1 -1 +1) = 0 \).
Statement (A) is correct.

Option (B): Region 3 of Configuration II

Configuration II has charge densities: \( +\frac{\sigma_0}{2}, -\sigma_0, +\sigma_0, -\sigma_0, +\sigma_0, -\frac{\sigma_0}{2} \).

Region 3 is between the 3rd (\(+\sigma_0\)) and 4th (\(-\sigma_0\)) sheets.

• Sheets to the left (1, 2, 3): Net \( \sigma_L = \frac{\sigma_0}{2} - \sigma_0 + \sigma_0 = \frac{\sigma_0}{2} \).
• Sheets to the right (4, 5, 6): Net \( \sigma_R = -\sigma_0 + \sigma_0 - \frac{\sigma_0}{2} = -\frac{\sigma_0}{2} \).

The field points from positive net charge to negative net charge (Left to Right). Magnitude \( E = \frac{1}{2\epsilon_0} (\sigma_{net, left} - \sigma_{net, right}) = \frac{1}{2\epsilon_0} (\frac{\sigma_0}{2} - (-\frac{\sigma_0}{2})) = \frac{\sigma_0}{2\epsilon_0} \).

Statement (B) claims \( E = \frac{\sigma_0}{\epsilon_0} \), which is incorrect.

Option (C): Potential Difference in Configuration I

We need \( V_1 - V_6 = \int E \, dx = \sum E_i d \). The sequence of fields in regions 1 to 5 for Config I:

• Region 1 (between 1&2): Left net \(\sigma_0\), Right net \(-\sigma_0\). \( E = \sigma_0/\epsilon_0 \).
• Region 2 (between 2&3): Left net \(0\), Right net \(0\). \( E = 0 \).
• Region 3 (between 3&4): Left net \(\sigma_0\), Right net \(-\sigma_0\). \( E = \sigma_0/\epsilon_0 \).
• Region 4 (between 4&5): \( E = 0 \).
• Region 5 (between 5&6): \( E = \sigma_0/\epsilon_0 \).

Total Potential Difference \( \Delta V = (\frac{\sigma_0}{\epsilon_0} + 0 + \frac{\sigma_0}{\epsilon_0} + 0 + \frac{\sigma_0}{\epsilon_0})d = \frac{3\sigma_0 d}{\epsilon_0} \).

Substituting values: \( \Delta V = 3 \times \frac{(9\times 10^{-6})(10^{-6})}{9\times 10^{-12}} = 3 \times 1 = 3 \, \text{V} \).

Statement (C) claims 5 V, which is incorrect.

Option (D): Potential Difference in Configuration II

Calculating fields in Config II:

• Region 1: \( \sigma_L = \sigma_0/2 \). \( E_1 = \frac{\sigma_0}{2\epsilon_0} \).
• Region 2: \( \sigma_L = -\sigma_0/2 \). \( E_2 = -\frac{\sigma_0}{2\epsilon_0} \).
• Region 3: \( \sigma_L = \sigma_0/2 \). \( E_3 = \frac{\sigma_0}{2\epsilon_0} \).
• Region 4: \( \sigma_L = -\sigma_0/2 \). \( E_4 = -\frac{\sigma_0}{2\epsilon_0} \).
• Region 5: \( \sigma_L = \sigma_0/2 \). \( E_5 = \frac{\sigma_0}{2\epsilon_0} \).

Sum of fields: \( \sum E = \frac{\sigma_0}{2\epsilon_0} (1 - 1 + 1 - 1 + 1) = \frac{\sigma_0}{2\epsilon_0} \). \( \Delta V = E_{net} \cdot d = \frac{\sigma_0 d}{2\epsilon_0} = 0.5 \, \text{V} \).

Statement (D) claims 0 V, which is incorrect.

1. Carnot Engine Analysis:

Given parameters:

• Temperature of hot reservoir, \( T_{1} = 1000 \, \text{K} \)
• Efficiency, \( \eta = 0.4 \)
• Heat extracted from hot reservoir, \( Q_{1} = 150 \, \text{J} \)

Calculate Work Extracted (\( W \)):

$$ \eta = \frac{W}{Q_{1}} $$ $$ W = \eta \times Q_{1} = 0.4 \times 150 = 60 \, \text{J} $$

This matches Statement (A).

Calculate Cold Reservoir Temperature (\( T_{2} \)):

$$ \eta = 1 - \frac{T_{2}}{T_{1}} $$ $$ 0.4 = 1 - \frac{T_{2}}{1000} \implies \frac{T_{2}}{1000} = 0.6 $$ $$ T_{2} = 600 \, \text{K} $$

This matches Statement (B).

2. Heat Pump Analysis:

The work \( W = 60 \, \text{J} \) drives a heat pump.
Given parameters for the heat pump:

• Coefficient of Performance (COP), \( \beta = 10 \)
• Temperature of hot reservoir (where heat is pumped to), \( T_{H} = 300 \, \text{K} \)

Calculate Cold Reservoir Temperature (\( T_{L} \)):

For a Carnot heat pump, the COP is defined as:

$$ \beta = \frac{T_{H}}{T_{H} - T_{L}} $$ $$ 10 = \frac{300}{300 - T_{L}} $$ $$ 3000 - 10 T_{L} = 300 $$ $$ 10 T_{L} = 2700 \implies T_{L} = 270 \, \text{K} $$

This matches Statement (C).

Calculate Heat Supplied to Hot Reservoir (\( Q_{H} \)):

By definition of COP for a heat pump (\( \beta = Q_{H} / W \)):

$$ Q_{H} = \beta \times W = 10 \times 60 = 600 \, \text{J} $$

Statement (D) claims the heat supplied is 540 J. However, \( Q_{H} = 600 \, \text{J} \). Note that the heat extracted from the cold reservoir would be \( Q_{L} = Q_{H} - W = 600 - 60 = 540 \, \text{J} \), but the question asks for heat supplied to the hot reservoir.

Thus, Statement (D) is incorrect.

Conclusion: Statements A, B, and C are correct.

Step 1: Understand the Charge and Mass Distribution

• The problem specifies a conducting solid sphere. For a conductor, any excess charge \(Q\) resides entirely on the outer surface. Thus, the charge distribution behaves like a spherical shell.
• The mass \(M\) is that of a solid sphere, distributed uniformly throughout the volume.

Step 2: Calculate the Magnetic Moment (\(\mu\))

Consider the charge \(Q\) distributed uniformly on the surface of a sphere of radius \(R\). If the sphere rotates with angular velocity \(\omega\):

The magnetic moment of a rotating spherical shell is given by:

$$ \mu = \frac{1}{3} Q R^2 \omega $$

(Note: This can be derived by integrating the magnetic moments of infinitesimal rings on the surface: \(d\mu = (\text{Area}) \cdot dI\). For a shell, \(\mu = \frac{Q}{2M_{\text{shell}}} L_{\text{shell}}\). Since \(L_{\text{shell}} = \frac{2}{3} M_{\text{shell}} R^2 \omega\), we get \(\mu = \frac{Q}{2M_{\text{shell}}} (\frac{2}{3} M_{\text{shell}} R^2 \omega) = \frac{1}{3}QR^2\omega\)).

Step 3: Calculate the Angular Momentum (\(L\))

The angular momentum is determined by the mass distribution. For a solid sphere of mass \(M\) and radius \(R\):

$$ L = I \omega = \frac{2}{5} M R^2 \omega $$

Step 4: Determine the Ratio and Solve for \(\alpha\)

Now, we take the ratio of the magnitude of the magnetic dipole moment to the angular momentum:

$$ \frac{\mu}{L} = \frac{\frac{1}{3} Q R^2 \omega}{\frac{2}{5} M R^2 \omega} $$ $$ \frac{\mu}{L} = \frac{1/3}{2/5} \cdot \frac{Q}{M} = \frac{5}{6} \frac{Q}{M} $$

The problem states that this ratio is given as \(\alpha \frac{Q}{2M}\). Equating the two expressions:

$$ \frac{5}{6} \frac{Q}{M} = \alpha \frac{Q}{2M} $$

Canceling common terms (\(\frac{Q}{M}\)):

$$ \frac{5}{6} = \frac{\alpha}{2} $$ $$ \alpha = \frac{10}{6} = \frac{5}{3} $$ $$ \alpha \approx 1.67 $$

Step 1: Analyze the First Process (Photoelectric Effect on Hydrogen)

A photon of frequency \(\nu_1\) ionizes a Hydrogen atom from its ground state. The energy conservation equation is:

$$ E_{\text{photon}} = \text{Ionization Energy} + \text{Kinetic Energy of electron} $$ $$ h\nu_1 = 13.6 \text{ eV} + 10 \text{ eV} $$ $$ h\nu_1 = 23.6 \text{ eV} $$

Step 2: Analyze the Second Process (Positronium Formation)

The ejected electron (KE = 10 eV) combines with a positron at rest to form a Positronium atom (bound state of \(e^-\) and \(e^+\)).

• Initial Energy (\(E_i\)): The total energy before combination is the kinetic energy of the electron (since the positron is at rest and potential energy at infinite separation is zero). $$ E_i = 10 \text{ eV} $$
• Final Energy (\(E_f\)): The system forms a Positronium atom moving with kinetic energy, and emits a photon \(h\nu_2\).
  • Positronium Kinetic Energy (\(KE_{Ps}\)) = 5 eV (Given).
  • Positronium Binding Energy: The reduced mass of the positronium system is \(\mu = \frac{m_e m_e}{m_e + m_e} = \frac{m_e}{2}\). The ground state energy is \(E_{Ps} = -13.6 \text{ eV} \times \frac{\mu}{m_e} = -6.8 \text{ eV}\).
  • Photon Energy = \(h\nu_2\).

Using Conservation of Energy:

$$ E_i = KE_{Ps} + E_{Ps} + h\nu_2 $$ $$ 10 \text{ eV} = 5 \text{ eV} + (-6.8 \text{ eV}) + h\nu_2 $$ $$ 10 = -1.8 + h\nu_2 $$ $$ h\nu_2 = 11.8 \text{ eV} $$

Step 3: Calculate the Difference

We need to find the difference between the two photon energies:

$$ h\nu_1 - h\nu_2 = 23.6 \text{ eV} - 11.8 \text{ eV} $$ $$ h\nu_1 - h\nu_2 = 11.8 \text{ eV} $$

Given Data:

• Monatomic gas (\(\gamma = 5/3\)).
• Volumes: \(V_W = 64\), \(V_X = 125\), \(V_Y = 250\) (units in \(cm^3\)).
• Condition at W: \(nRT_W = 1 \text{ J}\).

Step 1: Analyze Process W-X (Adiabatic Expansion)

For an adiabatic process involving an ideal gas:

$$ TV^{\gamma-1} = \text{constant} $$ $$ T_W V_W^{\gamma-1} = T_X V_X^{\gamma-1} $$

Substituting \(\gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3}\):

$$ T_W (64)^{2/3} = T_X (125)^{2/3} $$ $$ T_W (4^3)^{2/3} = T_X (5^3)^{2/3} $$ $$ T_W (4^2) = T_X (5^2) $$ $$ 16 T_W = 25 T_X \Rightarrow T_X = \frac{16}{25} T_W $$

Step 2: Analyze Process X-Y (Isobaric Expansion)

For an isobaric process, Charles's Law applies:

$$ \frac{V}{T} = \text{constant} \Rightarrow \frac{V_X}{T_X} = \frac{V_Y}{T_Y} $$ $$ T_Y = T_X \left(\frac{V_Y}{V_X}\right) $$ $$ T_Y = T_X \left(\frac{250}{125}\right) = 2T_X $$

Substitute \(T_X\):

$$ T_Y = 2 \left(\frac{16}{25} T_W\right) = \frac{32}{25} T_W $$

Step 3: Calculate Heat Absorbed in Process X-Y

For an isobaric process, heat absorbed is given by:

$$ Q_{XY} = n C_P \Delta T $$

For a monatomic gas, \(C_P = \frac{5}{2} R\).

$$ \Delta T = T_Y - T_X = \frac{32}{25}T_W - \frac{16}{25}T_W = \frac{16}{25}T_W $$ $$ Q_{XY} = n \left(\frac{5}{2}R\right) \left(\frac{16}{25}T_W\right) $$ $$ Q_{XY} = \frac{5}{2} \cdot \frac{16}{25} \cdot (nRT_W) $$

Simplifying the fraction:

$$ Q_{XY} = \frac{1}{1} \cdot \frac{8}{5} \cdot (nRT_W) = \frac{8}{5} nRT_W $$

Given that \(nRT_W = 1 \text{ J}\):

$$ Q_{XY} = \frac{8}{5} (1) = 1.6 \text{ J} $$

Step 1: Determine the Time Period of the Second Satellite

According to Kepler's Third Law, $T^2 \propto r^3$, or $T \propto r^{3/2}$.

For the geostationary satellite (Satellite 1):

• Period $T_1 = 24$ hours.
• Radius $r_1$.

For the second satellite (Satellite 2):

• Radius $r_2 = \frac{r_1}{1.21}$.

The ratio of their periods is:

$$ \frac{T_1}{T_2} = \left(\frac{r_1}{r_2}\right)^{3/2} = (1.21)^{3/2} $$ $$ \frac{T_1}{T_2} = (1.1^2)^{3/2} = (1.1)^3 = 1.331 $$ $$ T_2 = \frac{T_1}{1.331} $$

Step 2: Calculate the Relative Time Period

Let $t$ be the time period of the second satellite as observed from the first. This is the time between two consecutive encounters.

• Satellite 1 (Geostationary) moves with angular velocity $\omega_1 = \frac{2\pi}{T_1}$ in the direction of Earth's rotation (say, West to East).
• Satellite 2 moves in the opposite direction (East to West) with angular velocity $\omega_2 = \frac{2\pi}{T_2}$.

The relative angular velocity is the sum of their angular speeds because they are moving in opposite directions:

$$ \omega_{\text{rel}} = \omega_1 + \omega_2 $$ $$ \frac{2\pi}{t} = \frac{2\pi}{T_1} + \frac{2\pi}{T_2} $$ $$ \frac{1}{t} = \frac{1}{T_1} + \frac{1}{T_2} $$

Substituting $T_2 = \frac{T_1}{1.331}$:

$$ \frac{1}{t} = \frac{1}{T_1} + \frac{1.331}{T_1} = \frac{2.331}{T_1} $$ $$ t = \frac{T_1}{2.331} $$

Step 3: Solve for p

Given $T_1 = 24$ hours, we have:

$$ t = \frac{24}{2.331} \text{ hours} $$

The problem states the time period is $\frac{24}{p}$ hours. Comparing the two expressions:

$$ p = 2.331 $$

Rounding to two decimal places (as per the answer key range):

$$ p \approx 2.33 $$

Step 1: Analyze the Equilibrium State

• The container of length $L$ is divided equally at equilibrium, so the length of each compartment is $L/2$.
• Volume of each compartment $V = A(L/2)$.
• Left compartment: $n_L = 1.5$ moles. Pressure $P_L$.
• Right compartment: $n_R = 1$ mole. Pressure $P_R$.
• Since the piston is thermally conducting, both sides are at the same temperature $T$.

Step 2: Force Balance Equation

The piston is in equilibrium under the influence of gas pressures and the spring force.

• Spring extension: The natural length is $0.4L$ ($2L/5$). The current length is $0.5L$ ($L/2$).
• Extension $x = 0.5L - 0.4L = 0.1L = \frac{L}{10}$.
• The spring is stretched, so it pulls the piston to the left.
• Force balance (Rightward Forces = Leftward Forces): $$ P_L A = P_R A + kx $$

Step 3: Use Ideal Gas Law

Express pressures using $PV = nRT$:

$$ P_L = \frac{n_L RT}{V} = \frac{1.5 RT}{AL/2} = \frac{3RT}{AL} $$ $$ P_R = \frac{n_R RT}{V} = \frac{1 RT}{AL/2} = \frac{2RT}{AL} $$

Substitute these into the force balance equation:

$$ \left(\frac{3RT}{AL}\right)A = \left(\frac{2RT}{AL}\right)A + k\left(\frac{L}{10}\right) $$ $$ \frac{3RT}{L} = \frac{2RT}{L} + \frac{kL}{10} $$ $$ \frac{RT}{L} = \frac{kL}{10} \Rightarrow RT = \frac{kL^2}{10} $$

Step 4: Calculate Pressure in Right Compartment ($P_R$)

We need to find $P_R$ in terms of $kL/A$.

$$ P_R = \frac{2RT}{AL} $$

Substitute $RT = \frac{kL^2}{10}$:

$$ P_R = \frac{2}{AL} \left( \frac{kL^2}{10} \right) $$ $$ P_R = \frac{2kL}{10A} = \frac{1}{5} \frac{kL}{A} = 0.2 \frac{kL}{A} $$

Step 5: Identify $\alpha$

The problem states $P = \frac{kL}{A} \alpha$. Comparing this with our result:

$$ \alpha = 0.2 $$

Step 1: Understand the Geometry of the Wedges

The setup consists of two overlapping wedges forming a rectangular block of constant total thickness $12 \mu\text{m}$.

• Wedge A has refractive index $n_A = 1.7$.
• Wedge B has refractive index $n_B = 1.5$.
• The wedges are symmetric but inverted. Let's assume the combination forms a block of length $L$.
• From the diagram: $L = d + 2l = 2 \text{ mm} + 2(1 \text{ mm}) = 4 \text{ mm}$.

Step 2: Determine Thicknesses at the Slits

Due to the linear slope of the wedges and the symmetry of the arrangement:

• Let $t_1$ be the thickness of the thinner part of a wedge at the position of the first slit (distance $l$ from the edge).
• Let $t_2$ be the thickness of the thicker part of the same wedge at the position of the second slit (distance $3l$ from the edge).
• Using similar triangles: $\frac{t_1}{l} = \frac{t_2}{L-l} = \frac{t_2}{3l}$. Thus, $t_2 = 3t_1$.
• The sum of the thicknesses of the two complementary wedges at any point is constant ($12 \mu\text{m}$). Since the arrangement is symmetric, the thickness of the thick part of one wedge plus the thin part of the other is the total thickness. $$ t_1 + t_2 = 12 \mu\text{m} $$
• Substituting $t_2 = 3t_1$: $$ t_1 + 3t_1 = 12 \Rightarrow 4t_1 = 12 \Rightarrow t_1 = 3 \mu\text{m} $$ $$ t_2 = 9 \mu\text{m} $$

So at Slit 1 (top): Thickness of Wedge B ($n_B$) is $t_1 = 3 \mu\text{m}$. Thickness of Wedge A ($n_A$) is $t_2 = 9 \mu\text{m}$. At Slit 2 (bottom): Thickness of Wedge B ($n_B$) is $t_2 = 9 \mu\text{m}$. Thickness of Wedge A ($n_A$) is $t_1 = 3 \mu\text{m}$.

Step 3: Calculate Optical Path Difference ($\Delta x$)

The optical path length (OPL) added by a medium of thickness $t$ is $(n-1)t$. Path difference between the two rays emerging from the slits:

$$ \Delta x = \text{OPL}_2 - \text{OPL}_1 $$ $$ \text{OPL}_1 = (n_A - 1)t_2 + (n_B - 1)t_1 = (0.7)(9) + (0.5)(3) = 6.3 + 1.5 = 7.8 \mu\text{m} $$ $$ \text{OPL}_2 = (n_A - 1)t_1 + (n_B - 1)t_2 = (0.7)(3) + (0.5)(9) = 2.1 + 4.5 = 6.6 \mu\text{m} $$ $$ \Delta x = 6.6 - 7.8 = -1.2 \mu\text{m} $$

The magnitude of the path difference is $|\Delta x| = 1.2 \mu\text{m}$.

Step 4: Calculate the Shift of the Central Maximum

For the central maximum, the total path difference (geometric + optical) must be zero. $$ d \sin \theta + \Delta x = 0 $$ Using the approximation $\sin \theta \approx \tan \theta = y/D$: $$ \frac{dy}{D} = |\Delta x| $$ $$ y = \frac{D |\Delta x|}{d} $$ Substitute the values ($D = 2 \text{ m}$, $d = 2 \text{ mm} = 2 \times 10^{-3} \text{ m}$): $$ y = \frac{2}{2 \times 10^{-3}} (1.2 \times 10^{-6}) $$ $$ y = 10^3 \times 1.2 \times 10^{-6} = 1.2 \times 10^{-3} \text{ m} $$ $$ y = 1.2 \text{ mm} $$

Step 1: Set up the equation of motion for the horizontal component

The projectile experiences a drag force $\vec{F} = -c\vec{v}$. For the horizontal motion (x-direction), the force is $F_x = -cv_x$. According to Newton's Second Law:

$$ m a_x = -c v_x $$ $$ m \frac{dv_x}{dt} = -c v_x $$ $$ \frac{dv_x}{v_x} = -\frac{c}{m} dt $$

Step 2: Solve for horizontal velocity $v_x(t)$

Integrate both sides from time $0$ to $t$:

$$ \int_{v_{x0}}^{v_x} \frac{dv}{v} = -\frac{c}{m} \int_0^t dt $$ $$ \ln\left(\frac{v_x}{v_{x0}}\right) = -\frac{c}{m} t $$ $$ v_x(t) = v_{x0} e^{-\frac{c}{m} t} $$

The initial horizontal velocity is: $$ v_{x0} = v_0 \cos(60^\circ) = 270 \times \frac{1}{2} = 135 \text{ m/s} $$

Step 3: Solve for horizontal position $x(t)$

Since $v_x = \frac{dx}{dt}$, we integrate the velocity function:

$$ x(t) = \int_0^t v_x(\tau) d\tau = \int_0^t v_{x0} e^{-\frac{c}{m} \tau} d\tau $$ $$ x(t) = v_{x0} \left[ \frac{e^{-\frac{c}{m} \tau}}{-c/m} \right]_0^t $$ $$ x(t) = \frac{m v_{x0}}{c} \left( 1 - e^{-\frac{c}{m} t} \right) $$

Step 4: Substitute values and calculate distance at $t = 2$ s

Given parameters: $m = 200 \text{ g} = 0.2 \text{ kg}$ $c = 0.1 \text{ kg/s}$ $t = 2 \text{ s}$ $e = 2.7$

First, calculate the exponent coefficient: $$ \frac{c}{m} = \frac{0.1}{0.2} = 0.5 \text{ s}^{-1} $$ $$ \frac{c}{m} t = 0.5 \times 2 = 1 $$

Next, calculate the pre-factor: $$ \frac{m v_{x0}}{c} = \frac{0.2 \times 135}{0.1} = 2 \times 135 = 270 \text{ m} $$

Now, substitute into the position equation: $$ x(2) = 270 \left( 1 - e^{-1} \right) $$ $$ x(2) = 270 \left( 1 - \frac{1}{2.7} \right) $$ $$ x(2) = 270 \left( \frac{2.7 - 1}{2.7} \right) = 270 \left( \frac{1.7}{2.7} \right) = 270 \times \frac{17}{27} $$ $$ x(2) = 10 \times 17 = 170 \text{ m} $$

Step 1: Calculate the maximum speed of the bobs

Both the transmitter (source) and receiver (observer) are released from an angle $\theta_0$. They will attain their maximum velocity at the lowest point (equilibrium position) of the swing. Using the principle of conservation of energy:

$$ PE_{\text{loss}} = KE_{\text{gain}} $$ $$ mgL(1 - \cos\theta_0) = \frac{1}{2} m v_{max}^2 $$ $$ v_{max} = \sqrt{2gL(1 - \cos\theta_0)} $$

Given: $L = 8 \text{ m}$, $g = 10 \text{ m/s}^2$, $\cos\theta_0 = 0.9$ $$ v_{max} = \sqrt{2 \times 10 \times 8 \times (1 - 0.9)} $$ $$ v_{max} = \sqrt{160 \times 0.1} = \sqrt{16} = 4 \text{ m/s} $$ Both the source velocity ($v_s$) and observer velocity ($v_o$) at the lowest point are $4 \text{ m/s}$.

Step 2: Determine maximum and minimum received frequencies

The maximum frequency variation occurs between the case where they approach each other at maximum speed and the case where they recede from each other at maximum speed.

• Maximum Frequency ($f_{max}$): Both approach each other. Observer moves towards source (+ve $v_o$). Source moves towards observer (-ve $v_s$ in denominator). $$ f_{max} = f_0 \left( \frac{v_{sound} + v_o}{v_{sound} - v_s} \right) $$ $$ f_{max} = 660 \left( \frac{330 + 4}{330 - 4} \right) = 660 \left( \frac{334}{326} \right) \approx 676.20 \text{ Hz} $$
• Minimum Frequency ($f_{min}$): Both move away from each other. Observer moves away from source (-ve $v_o$). Source moves away from observer (+ve $v_s$ in denominator). $$ f_{min} = f_0 \left( \frac{v_{sound} - v_o}{v_{sound} + v_s} \right) $$ $$ f_{min} = 660 \left( \frac{330 - 4}{330 + 4} \right) = 660 \left( \frac{326}{334} \right) \approx 644.19 \text{ Hz} $$

Step 3: Calculate the variation

$$ \Delta f = f_{max} - f_{min} $$ $$ \Delta f = 676.20 - 644.19 = 32.01 \text{ Hz} $$

Correct Answer: (A)

The sodium nitroprusside test is a specific qualitative test for the detection of sulphide ions (\(S^{2-}\)).

1. The reagent used is sodium nitroprusside: \(Na_2[Fe(CN)_5NO]\). The complex ion is \([Fe(CN)_5NO]^{2-}\).
2. When sulphide ions are added to an aqueous solution of sodium nitroprusside, a purple/violet colored complex is formed.
3. The reaction involves the nucleophilic attack of the sulphide ion (\(S^{2-}\)) on the nitrogen atom of the nitrosyl (\(NO^+\)) ligand.
4. The reaction is: \[ [Fe(CN)_5NO]^{2-} + S^{2-} \rightarrow [Fe(CN)_5(NOS)]^{4-} \]
5. The resulting ligand formed is the thionitroprusside ion, \(NOS^-\).

Thus, the \(NO\) ligand is converted to \(NOS^-\).

Correct Answer: (A)

During the hydrolysis of interhalogen compounds (\(XX'_n\)), the larger/less electronegative halogen (X) forms the oxyacid, while the smaller/more electronegative halogen (X') forms the hydrohalic acid (HX'). The oxidation state of the central halogen remains unchanged.

• ICl (Iodine Monochloride): Iodine is in the +1 oxidation state. \[ ICl + H_2O \rightarrow \underset{\text{Hypoiodous acid}}{HOI} + HCl \] \(HOI\) dissociates to give \(H^+\) and \(IO^-\).
• ClF\(_3\) (Chlorine Trifluoride): Chlorine is in the +3 oxidation state. \[ ClF_3 + 2H_2O \rightarrow \underset{\text{Chlorous acid}}{HClO_2} + 3HF \] \(HClO_2\) dissociates to give \(H^+\) and \(ClO_2^-\).
• BrF\(_5\) (Bromine Pentafluoride): Bromine is in the +5 oxidation state. \[ BrF_5 + 3H_2O \rightarrow \underset{\text{Bromic acid}}{HBrO_3} + 5HF \] \(HBrO_3\) dissociates to give \(H^+\) and \(BrO_3^-\).

Therefore, the anions formed are \(IO^-\), \(ClO_2^-\), and \(BrO_3^-\).

Correct Answer: (D) S

We analyze each reaction scheme to find the number of unsaturated carbon atoms in the major product. Note: In this context (based on the provided solution key), "unsaturated carbon" refers to carbons in alkene double bonds (C=C) and aromatic rings, excluding carbonyl carbons (C=O).

Comparing the counts:
P=6, Q=8, R=8, S=10.
The highest number is in S.

Correct Answer: (C)

We analyze the reaction sequence for each option to determine the final product:

1. (A) Reaction of 2-chloroethanol:
   • \(HO-CH_2-CH_2-Cl \xrightarrow{NaCN} HO-CH_2-CH_2-CN\) (Nucleophilic substitution).
   • Hydrolysis (\(OH^-, H_3O^+\)) converts \(-CN\) to \(-COOH\).
   • Product: 3-hydroxypropanoic acid (\(HO-CH_2-CH_2-COOH\)). This is a hydroxy-acid, not a dicarboxylic acid.
2. (B) Oxidation of Glucose:
   • \(Br_2/H_2O\) is a mild oxidizing agent. It oxidizes the aldehyde group (\(-CHO\)) to carboxylic acid (\(-COOH\)) but leaves primary and secondary alcohol groups untouched.
   • Product: Gluconic acid (\(HOOC-(CHOH)_4-CH_2OH\)). This is a monocarboxylic acid.
3. (C) Reaction of Bromocyclohexane:
   • Step 1: \(KOH, EtOH\) causes dehydrohalogenation (E2 elimination) to form Cyclohexene.
   • Step 2: \(KMnO_4, H_2SO_4, \Delta\) causes vigorous oxidative cleavage of the double bond.
   • The cyclic alkene breaks at the double bond, and both carbons are oxidized to carboxylic acid groups.
   • Product: Hexanedioic acid (Adipic acid) (\(HOOC-CH_2-CH_2-CH_2-CH_2-COOH\)). This is a dicarboxylic acid.
4. (D) Oxidation of Hydroxy-ketone:
   • \(H_2CrO_4\) (Chromic acid) is a strong oxidizing agent. It oxidizes the primary alcohol to a carboxylic acid.
   • It does not typically cleave the carbon chain adjacent to the ketone under these conditions.
   • Product: A keto-acid (e.g., 2,4-dimethyl-4-oxopentanoic acid).

Thus, only reaction sequence (C) produces a dicarboxylic acid as the major product.

Correct Answer: (C) and (D)

We evaluate each statement regarding intermolecular forces:

• (A) Potential Energy distance dependence:
  • Ion-Ion interaction energy: \(E \propto \frac{1}{r}\).
  • Ion-Dipole interaction energy: \(E \propto \frac{1}{r^2}\).
  • As distance \(r \rightarrow \infty\), the term \(\frac{1}{r^2}\) approaches zero faster (more rapidly) than \(\frac{1}{r}\).
  • The statement claims the ion-ion energy (\(1/r\)) approaches zero more rapidly than ion-dipole (\(1/r^2\)). This is False.
• (B) Rotating Polar Molecules (Keesom Force):
  • The average potential energy for two freely rotating permanent dipoles is inversely proportional to \(r^6\) (i.e., \(E \propto \frac{1}{r^6}\)).
  • The statement claims \(1/r^3\) dependence (which is true for static/fixed dipoles, not rotating ones). This is False.
• (C) Dipole-Induced Dipole (Debye Force):
  • This interaction arises from the polarization of a non-polar molecule by a permanent dipole.
  • The interaction energy formula is \(E = -\frac{\mu^2 \alpha}{r^6}\), where \(\alpha\) is polarizability.
  • Unlike orientation forces (Keesom), induction does not depend on the thermal averaging of orientations in the same way, so it is generally considered independent of temperature. This is True.
• (D) Dispersion Forces (London Forces):
  • These forces arise from instantaneous dipole-induced dipole interactions.
  • They exist between all atoms and molecules, including nonpolar ones. This is True.

Correct Answer: (B) and (D)

We examine the structures of the given phosphorus oxyacids to count the number of P-H bonds. Note that hydrogen atoms attached directly to phosphorus (P-H) are non-ionizable (reducing), while those attached to oxygen (P-O-H) are acidic.

• (A) \(H_3PO_4\) (Orthophosphoric acid):
  Structure: One \(P=O\) bond, three \(P-OH\) bonds.
  Number of P-H bonds: 0.
• (B) \(H_3PO_3\) (Orthophosphorous acid):
  Structure: One \(P=O\) bond, two \(P-OH\) bonds, one \(P-H\) bond.
  Number of P-H bonds: 1.
• (C) \(H_4P_2O_7\) (Pyrophosphoric acid):
  Structure: Two \(P=O\) units connected by an oxygen bridge (\(P-O-P\)), with four \(P-OH\) groups total.
  Number of P-H bonds: 0.
• (D) \(H_3PO_2\) (Hypophosphorous acid):
  Structure: One \(P=O\) bond, one \(P-OH\) bond, two \(P-H\) bonds.
  Number of P-H bonds: 2.

Therefore, compounds (B) and (D) contain P-H bonds.

Correct Answer: (A), (C)

The reaction sequence describes the preparation of phthalimide and its subsequent use in the Gabriel Phthalimide Synthesis.

Evaluating Statements:

• (A): Both X and Y contain carbonyl oxygens. (True)
• (B): Y is an N-substituted imide, not a primary amine. The Carbylamine test (heating with \(CHCl_3/KOH\)) is specific to primary amines. Y will not give this test. (False)
• (C): Z is a primary amine (\(1^\circ\)). Primary amines react with Hinsberg's reagent (Benzenesulfonyl chloride) to form sulfonamides soluble in alkali. (True)
• (D): In the Gabriel synthesis, aryl halides (\(Ar-X\)) generally do not undergo the necessary nucleophilic substitution step to form Y. Therefore, R is typically an alkyl group, making Z an aliphatic primary amine, not aromatic. (False)

Correct Answer: (B), (C)

We track the structure and stereochemistry through the reaction scheme:

Correct Answer: 11.00

To find the density (\(d\)) of the metal, we use the formula for a cubic lattice:

\[ d = \frac{Z \cdot M}{a^3 \cdot N_A} \]

Given values:

• Lattice type: Cubic Close Packed (ccp) / Face Centered Cubic (fcc), so \( Z = 4 \).
• Edge length (\(a\)): \( 400 \text{ pm} = 400 \times 10^{-10} \text{ cm} = 4 \times 10^{-8} \text{ cm} \).
• Atomic Mass (\(M\)): \( 105.6 \text{ g/mol} \).
• Avogadro's Constant (\(N_A\)): \( 6 \times 10^{23} \text{ mol}^{-1} \).

Calculation:

1. Calculate volume of unit cell (\(a^3\)): \[ a^3 = (4 \times 10^{-8})^3 = 64 \times 10^{-24} \text{ cm}^3 \]
2. Substitute values into the density formula: \[ d = \frac{4 \times 105.6}{(64 \times 10^{-24}) \times (6 \times 10^{23})} \]
3. Simplify the denominator: \[ 64 \times 6 \times 10^{-24} \times 10^{23} = 384 \times 10^{-1} = 38.4 \]
4. Calculate final density: \[ d = \frac{422.4}{38.4} = 11 \text{ g cm}^{-3} \]

Step 1: Write the balanced chemical equation and calculate initial millimoles.

The reaction between barium nitrate and sodium iodate is:

$$ \text{Ba(NO}_3)_2 + 2\text{NaIO}_3 \rightarrow \text{Ba(IO}_3)_2 \downarrow + 2\text{NaNO}_3 $$

Initial millimoles (mmol) of reactants:

• For \( \text{Ba(NO}_3)_2 \): \( 200 \, \text{mL} \times 0.010 \, \text{M} = 2 \, \text{mmol} \)
• For \( \text{NaIO}_3 \): \( 100 \, \text{mL} \times 0.10 \, \text{M} = 10 \, \text{mmol} \)

Step 2: Determine the limiting reagent and concentration of excess ions.

From stoichiometry, 1 mole of \( \text{Ba}^{2+} \) reacts with 2 moles of \( \text{IO}_3^- \).

Therefore, 2 mmol of \( \text{Ba}^{2+} \) will react with \( 2 \times 2 = 4 \, \text{mmol} \) of \( \text{IO}_3^- \).

• \( \text{Ba}^{2+} \) is the limiting reagent and is completely precipitated (ignoring solubility for a moment).
• Remaining \( \text{IO}_3^- = 10 - 4 = 6 \, \text{mmol} \).

Total volume of the solution \( = 200 + 100 = 300 \, \text{mL} \).

Concentration of excess iodate ions:

$$ [\text{IO}_3^-] = \frac{6 \, \text{mmol}}{300 \, \text{mL}} = \frac{1}{50} \, \text{M} = 0.02 \, \text{M} $$

Step 3: Calculate the solubility using \( K_{sp} \).

Let the solubility of \( \text{Ba(IO}_3)_2 \) be \( S \). This corresponds to the concentration of \( \text{Ba}^{2+} \) in solution.

The equilibrium is: \( \text{Ba(IO}_3)_2(s) \rightleftharpoons \text{Ba}^{2+}(aq) + 2\text{IO}_3^-(aq) \)

Using the solubility product constant equation:

$$ K_{sp} = [\text{Ba}^{2+}][\text{IO}_3^-]^2 $$

Substitute the values (approximating total \( [\text{IO}_3^-] \) as the excess concentration due to the common ion effect):

$$ 1.58 \times 10^{-9} = S \times (0.02)^2 $$

$$ S = \frac{1.58 \times 10^{-9}}{4 \times 10^{-4}} $$

$$ S = \frac{1.58}{4} \times 10^{-5} = 0.395 \times 10^{-5} \, \text{M} $$

$$ S = 3.95 \times 10^{-6} \, \text{M} $$

Conclusion:

Comparing this with \( X \times 10^{-6} \text{ mol dm}^{-3} \), we get \( X = 3.95 \).

Step 1: Formulate equations using the Freundlich Adsorption Isotherm.

The Freundlich isotherm equation is given by:

$$ \frac{x}{m} = K \cdot C^{1/n} $$

Where \( x/m \) is the amount adsorbed per unit mass and \( C \) is the equilibrium concentration.

Given Data:

• Case 1: \( C_1 = 10 \), \( (x/m)_1 = 4 \)
• Case 2: \( C_2 = 16 \), \( (x/m)_2 = 10 \)

Substituting these into the equation:

(1) \( 4 = K(10)^{1/n} \)

(2) \( 10 = K(16)^{1/n} \)

Step 2: Solve for \( 1/n \).

Divide equation (2) by equation (1):

$$ \frac{10}{4} = \left( \frac{16}{10} \right)^{1/n} $$

$$ 2.5 = (1.6)^{1/n} $$

Taking log on both sides:

$$ \log(2.5) = \frac{1}{n} \log(1.6) $$

$$ \log\left(\frac{10}{4}\right) = \frac{1}{n} \log\left(\frac{16}{10}\right) $$

$$ (\log 10 - \log 4) = \frac{1}{n} (\log 16 - \log 10) $$

Using \( \log 2 \approx 0.3 \):

$$ (1 - 2(0.3)) = \frac{1}{n} (4(0.3) - 1) $$

$$ (1 - 0.6) = \frac{1}{n} (1.2 - 1) $$

$$ 0.4 = \frac{1}{n} (0.2) $$

$$ \frac{1}{n} = \frac{0.4}{0.2} = 2 $$

Step 3: Calculate the value for the third case.

We need to find \( x/m \) for \( C_3 = 20 \, \text{mg g}^{-1} \).

$$ \frac{x}{m} = K(20)^{1/n} $$

Dividing this by equation (1) \( [4 = K(10)^{1/n}] \):

$$ \frac{x/m}{4} = \left( \frac{20}{10} \right)^{1/n} $$

$$ \frac{x/m}{4} = (2)^2 $$

$$ \frac{x/m}{4} = 4 \implies \frac{x}{m} = 16 $$

Answer: The concentration of adsorbed phenol is 16.00.

Step 1: Define initial conditions and rate laws.

Reaction: \( A + R \rightarrow \text{Product} \)

True Rate Law: \( r = k[A][R] \)

Let initial concentration \( [A]_0 = a \). Then \( [R]_0 = 10a \).

Step 2: Calculate concentrations at 40% completion.

When the reaction is 40% complete, 0.4 of \( [A]_0 \) has reacted.

• \( [A] = a - 0.4a = 0.6a \)
• \( [R] = 10a - 0.4a = 9.6a \) (Since stoichiometry is 1:1)

Step 3: Calculate the True Rate (\( r_{true} \)).

$$ r_{true} = k(0.6a)(9.6a) = 5.76 k a^2 $$

Step 4: Calculate the Pseudo First Order Rate (\( r_{pseudo} \)).

In the pseudo first order approximation, the concentration of the excess reagent \( R \) is assumed to be constant at its initial value.

Assumption: \( [R] \approx [R]_0 = 10a \)

$$ r_{pseudo} = k[A][R]_0 = k(0.6a)(10a) = 6 k a^2 $$

Step 5: Calculate Relative Error.

Relative Error in rate is given by the formula:

$$ \text{Relative Error (\%)} = \frac{|r_{pseudo} - r_{true}|}{r_{true}} \times 100 $$

$$ \text{Error (\%)} = \frac{6 k a^2 - 5.76 k a^2}{5.76 k a^2} \times 100 $$

$$ \text{Error (\%)} = \frac{0.24}{5.76} \times 100 $$

$$ \text{Error (\%)} = \frac{2400}{576} \approx 4.166... $$

Answer: The relative error is approximately 4.17.

Step 1: Convert all units to the SI system.

Given parameters:

• Height of solution column ($h$) = $2.00 \text{ cm} = 0.02 \text{ m}$
• Density ($\rho$) = $1.00 \text{ g cm}^{-3} = 1000 \text{ kg m}^{-3}$
• Gravity ($g$) = $10 \text{ m s}^{-2}$
• Concentration ($C_{mass}$) = $2.00 \text{ g dm}^{-3} = 2.00 \text{ kg m}^{-3}$ (since $1 \text{ g/L} = 1 \text{ kg/m}^3$)

Step 2: Calculate the Osmotic Pressure ($\Pi$).

The osmotic pressure is equivalent to the hydrostatic pressure of the solution column:

$$ \Pi = h \rho g $$

$$ \Pi = 0.02 \text{ m} \times 1000 \text{ kg m}^{-3} \times 10 \text{ m s}^{-2} $$

$$ \Pi = 200 \text{ Pa (Pascals)} $$

Step 3: Calculate the Molar Mass ($M$).

Using the osmotic pressure equation $\Pi = C RT$, where $C$ is the molar concentration ($mol/m^3$):

$$ C = \frac{\text{Concentration}_{mass}}{M} $$

Substituting into the main equation:

$$ \Pi = \frac{C_{mass} R T}{M} $$

We rearrange to solve for $M$ (in kg/mol, since we are using SI units):

$$ M = \frac{C_{mass} R T}{\Pi} $$

$$ M = \frac{2.00 \times 8.3 \times 300}{200} $$

$$ M = \frac{4980}{200} = 24.9 \text{ kg mol}^{-1} $$

Step 4: Convert to g/mol and find $X$.

$$ M = 24.9 \text{ kg mol}^{-1} = 24900 \text{ g mol}^{-1} $$

Expressing in scientific notation as required ($X \times 10^4$):

$$ 24900 = 2.49 \times 10^4 $$

$$ X = 2.49 $$

Step 1: Write the balanced combustion reaction.

$$ \text{C}_4\text{H}_{10}(g) + \frac{13}{2}\text{O}_2(g) \rightarrow 4\text{CO}_2(g) + 5\text{H}_2\text{O}(l) $$

Step 2: Calculate $\Delta_r G^\circ$ for the reaction.

Using standard Gibbs energies of formation ($\Delta_f G^\circ$):

$$ \Delta_r G^\circ = \sum \Delta_f G^\circ (\text{Products}) - \sum \Delta_f G^\circ (\text{Reactants}) $$

$$ \Delta_r G^\circ = [4(-394) + 5(-237)] - [1(-18) + \frac{13}{2}(0)] $$

$$ \Delta_r G^\circ = [-1576 - 1185] + 18 $$

$$ \Delta_r G^\circ = -2761 + 18 = -2743 \text{ kJ mol}^{-1} $$

$$ \Delta_r G^\circ = -2743 \times 10^3 \text{ J mol}^{-1} $$

Step 3: Determine the number of electrons exchanged ($n$).

Consider the oxidation state changes:

• Carbon: In $\text{C}_4\text{H}_{10}$, average O.S. is $-2.5$. In $\text{CO}_2$, O.S. is $+4$. Change per C = $6.5$. Total for 4 carbons = $4 \times 6.5 = 26$.
• Oxygen: In $\text{O}_2$, O.S. is $0$. In Products, O.S. is $-2$. Total change = $13 \text{ atoms} \times 2 = 26$.

Thus, $n = 26$.

Step 4: Calculate the Cell Potential ($E^\circ_{\text{cell}}$).

Using the relationship between Gibbs energy and cell potential:

$$ \Delta_r G^\circ = -nFE^\circ_{\text{cell}} $$

$$ -2743 \times 10^3 = -26 \times F \times E^\circ_{\text{cell}} $$

$$ E^\circ_{\text{cell}} = \frac{2743}{26 F} \times 10^3 $$

Given $E_{\text{cell}} = \frac{X}{F} \times 10^3$, we can equate:

$$ X = \frac{2743}{26} = 105.5 $$

Step 1: Determine the oxidation state of Mn in both complexes.

Manganese ($Z=25$) has the electronic configuration $[\text{Ar}] 3d^5 4s^2$.

• For $[\text{Mn(Br)}_6]^{3-}$: $x + 6(-1) = -3 \Rightarrow x = +3$.
• For $[\text{Mn(CN)}_6]^{3-}$: $x + 6(-1) = -3 \Rightarrow x = +3$.

The $\text{Mn}^{3+}$ ion has a $3d^4$ configuration.

Step 2: Apply Crystal Field Theory (CFT).

Case 1: $[\text{Mn(Br)}_6]^{3-}$

• $\text{Br}^-$ is a Weak Field Ligand.
• Result: High Spin complex. Electrons do not pair up in $t_{2g}$ before filling $e_g$.
• Configuration: $t_{2g}^3 e_g^1$.
• Number of unpaired electrons ($n_1$) = 4.
• Magnetic Moment $\mu_1 = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90$ B.M.

Case 2: $[\text{Mn(CN)}_6]^{3-}$

• $\text{CN}^-$ is a Strong Field Ligand.
• Result: Low Spin complex. Electrons pair up in $t_{2g}$ due to large splitting energy ($\Delta_o$).
• Configuration: $t_{2g}^4 e_g^0$.
• Number of unpaired electrons ($n_2$) = 2.
• Magnetic Moment $\mu_2 = \sqrt{2(2+2)} = \sqrt{8} \approx 2.82$ B.M.

Step 3: Calculate the sum.

$$ \text{Sum} = \sqrt{24} + \sqrt{8} \approx 4.90 + 2.82 = 7.72 $$

Step 1: Understand the Hydrolysis Reaction.

An octasaccharide consists of 8 monosaccharide units. Being linear, these units are linked by 7 glycosidic bonds.

Complete hydrolysis requires the addition of one water molecule for each bond broken.

$$ \text{Octasaccharide} + 7\text{H}_2\text{O} \rightarrow \text{Mixture of Monosaccharides} $$

Step 2: Apply Conservation of Mass.

Calculate the total mass of the products formed from 1 mole of octasaccharide:

• Mass of Octasaccharide = $1024$ g/mol
• Mass of Water added ($7 \text{ moles}$) = $7 \times 18 = 126$ g/mol
• Total Mass of Products = $1024 + 126 = 1150$ g

Step 3: Determine the number of 2-deoxyribose units.

The problem states that 2-deoxyribose constitutes $58.26\%$ (w/w) of the products.

$$ \text{Mass of 2-deoxyribose} = \frac{58.26}{100} \times 1150 \text{ g} $$

$$ \text{Mass of 2-deoxyribose} \approx 670 \text{ g} $$

Given the molar mass of 2-deoxyribose is $134$ g/mol, the number of units ($z$) is:

$$ z = \frac{670}{134} = 5 \text{ units} $$

Step 4: Solve for Ribose units ($x$) and Glucose units ($y$).

We know the total number of units is 8. Since we have 5 units of 2-deoxyribose:

$$ x + y = 8 - 5 = 3 \text{ units} \quad \text{......(1)} $$

Calculate the remaining mass for Ribose and Glucose:

$$ \text{Remaining Mass} = \text{Total Mass} - \text{Mass of 2-deoxyribose} $$

$$ \text{Remaining Mass} = 1150 - 670 = 480 \text{ g} $$

Using the molar masses (Ribose = 150, Glucose = 180):

$$ 150x + 180y = 480 \quad \text{......(2)} $$

Simplify equation (2) by dividing by 30:

$$ 5x + 6y = 16 $$

From equation (1), substitute $y = 3 - x$:

$$ 5x + 6(3 - x) = 16 $$

$$ 5x + 18 - 6x = 16 $$

$$ -x = -2 \implies x = 2 $$

Answer: The number of ribose units is 2.