JEE ADVANCED 2025 Paper-1

1. Define Coordinates and Constraint:
Let $\theta$ be the angular displacement of the center of mass (CM) of the disk from the vertical equilibrium line.
Let $\beta$ be the angle of rotation of the disk about its center.
For rolling without slipping, the arc length moved by the CM along the ring must equal the arc length rotated by the disk: $$ (R-r)\theta = r\beta $$ $$ \implies \theta = \frac{r}{R-r}\beta $$

2. Analysis of Forces and Torque:
We analyze the motion using the torque equation about the Instantaneous Axis of Rotation (IAOR), which is the point of contact $P$ between the disk and the ring.
The moment of inertia of the disk about the point of contact $P$ is given by the Parallel Axis Theorem: $$ I_P = I_{CM} + mr^2 = \frac{1}{2}mr^2 + mr^2 = \frac{3}{2}mr^2 $$

There are two restoring forces providing torque about $P$:
1. Gravity: Acts downwards at the center of the disk. The component perpendicular to the radius $R-r$ is $mg \sin \theta \approx mg\theta$ (for small angles). The lever arm from $P$ is $r$. $$ \tau_g = - (mg\theta)r $$ 2. Spring Force: The spring is attached to the center of the disk. The extension of the spring corresponds to the linear displacement of the center: $x = (R-r)\theta$. The spring force is $F_s = kx = k(R-r)\theta$. This force acts tangentially. The lever arm from $P$ is $r$. $$ \tau_s = - F_s r = - [k(R-r)\theta]r $$

3. Equation of Motion:
Applying Newton's Second Law for rotation $\tau_{net} = I_P \alpha_{disk}$, where $\alpha_{disk} = \ddot{\beta}$ is the angular acceleration of the disk: $$ I_P \ddot{\beta} = \tau_g + \tau_s $$ $$ \frac{3}{2}mr^2 \ddot{\beta} = - mgr\theta - kr(R-r)\theta $$ Substitute $\theta = \frac{r}{R-r}\beta$ into the equation: $$ \frac{3}{2}mr^2 \ddot{\beta} = - \left[ mgr\left( \frac{r}{R-r}\beta \right) + kr(R-r)\left( \frac{r}{R-r}\beta \right) \right] $$ Simplify the terms: $$ \frac{3}{2}mr^2 \ddot{\beta} = - \left[ \frac{mgr^2}{R-r}\beta + kr^2\beta \right] $$ Cancel $mr^2$ from both sides: $$ \frac{3}{2} \ddot{\beta} = - \left[ \frac{g}{R-r} + \frac{k}{m} \right] \beta $$ $$ \ddot{\beta} = - \frac{2}{3} \left[ \frac{g}{R-r} + \frac{k}{m} \right] \beta $$

4. Conclusion:
This is the standard equation for Simple Harmonic Motion (SHM) of the form $\ddot{\beta} = - \omega^2 \beta$. Comparing the terms, we get the square of the angular frequency: $$ \omega^2 = \frac{2}{3} \left( \frac{g}{R-r} + \frac{k}{m} \right) $$ Taking the square root: $$ \omega = \sqrt{\frac{2}{3} \left( \frac{g}{R-r} + \frac{k}{m} \right)} $$ This matches Option (A).

Method 1: Using Center of Mass (COM) Frame Analysis

1. Velocity of Center of Mass:
Let the initial velocity of the mass $2m$ be $v_0$ and mass $m$ be at rest ($0$). The velocity of the center of mass ($V_{CM}$) of the system is: $$ V_{CM} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = \frac{(2m)(v_0) + (m)(0)}{2m + m} = \frac{2mv_0}{3m} = \frac{2v_0}{3} $$

2. Velocity in COM Frame:
The initial velocity of the incident particle ($2m$) in the COM frame is: $$ u_{1, CM} = v_1 - V_{CM} = v_0 - \frac{2v_0}{3} = \frac{v_0}{3} $$ Since the collision is perfectly elastic, the speed of the particle in the COM frame remains conserved after the collision, though its direction changes. $$ v_{1, CM} = |u_{1, CM}| = \frac{v_0}{3} $$

3. Relate back to Lab Frame:
The final velocity of the particle in the laboratory frame ($\vec{V}_{1}$) is the vector sum of its velocity in the COM frame ($\vec{v}_{1, CM}$) and the velocity of the COM frame ($\vec{V}_{CM}$): $$ \vec{V}_{1} = \vec{v}_{1, CM} + \vec{V}_{CM} $$ Here, $\vec{V}_{CM}$ is a constant vector of magnitude $\frac{2v_0}{3}$ along the incident direction. The vector $\vec{v}_{1, CM}$ has a constant magnitude $\frac{v_0}{3}$ but can point in any direction (depending on the scattering angle in the COM frame).

4. Geometric Condition for Maximum Deviation:
Geometrically, the tip of the vector $\vec{V}_{1}$ lies on a circle of radius $v_{1, CM} = \frac{v_0}{3}$ centered at the tip of the vector $\vec{V}_{CM}$. The angle $\theta$ is the angle between $\vec{V}_{1}$ and the horizontal axis ($\vec{V}_{CM}$). The maximum deviation angle $\theta_{max}$ occurs when the vector $\vec{V}_{1}$ is tangent to this circle. This forms a right-angled triangle where:

• The hypotenuse is the magnitude of the COM velocity: $|\vec{V}_{CM}| = \frac{2v_0}{3}$
• The opposite side is the magnitude of the velocity in COM frame: $|\vec{v}_{1, CM}| = \frac{v_0}{3}$

Method 2: General Formula
For an elastic collision where a particle of mass $m_1$ collides with a stationary particle of mass $m_2$, if $m_1 > m_2$, the maximum scattering angle of the incident particle is given by: $$ \sin(\theta_{max}) = \frac{m_2}{m_1} $$ Substituting the given values: $$ \sin(\theta_{max}) = \frac{m}{2m} = \frac{1}{2} \implies \theta_{max} = \frac{\pi}{6} $$

1. Flux Calculation:
The magnetic flux $\phi$ through the loop is defined as $\phi = \int \vec{B} \cdot d\vec{A}$. The magnetic field is given by $\vec{B} = B_0 \cos(\omega t) \hat{k}$ in the region $y \ge 0$. The square loop of side $a$ is initially in the XZ plane and rotates clockwise about the X-axis with angular speed $\omega$.

• At $t=0$, the loop is in the XZ plane. Its area vector $\vec{A}$ (normal to the surface) points along the y-axis (perpendicular to $\vec{B}$ which is along z). The angle between $\vec{B}$ and $\vec{A}$ is $90^\circ$.
• At time $t$, the loop has rotated by an angle $\theta = \omega t$. The angle between the magnetic field $\vec{B}$ (vertical, z-axis) and the area normal $\vec{A}$ becomes $\alpha = 90^\circ - \omega t$.

2. Time Interval for Flux:
The magnetic field exists only for $y \ge 0$. Since the loop is hinged on the x-axis and rotates clockwise, it moves through the region $y \ge 0$ (quadrants I and IV of the YZ plane) for the first half of its rotation.

• For $0 \le \omega t \le \pi$ (i.e., $0 \le t \le \pi/\omega$), the loop is in the field. Flux is non-zero.
• For $\pi < \omega t < 2\pi$ (i.e., $\pi/\omega < t < 2\pi/\omega$), the loop is in the region $y < 0$ where $B=0$. Flux is zero.

3. Induced EMF Calculation:
According to Faraday's Law, the induced EMF is $\xi = - \frac{d\phi}{dt}$. For the interval $0 \le t \le \pi/\omega$: $$ \xi = - \frac{d}{dt} \left[ \frac{B_0 a^2}{2} \sin(2\omega t) \right] $$ $$ \xi = - \frac{B_0 a^2}{2} \cdot (2\omega) \cos(2\omega t) $$ $$ \xi = - B_0 a^2 \omega \cos(2\omega t) $$ This function represents an inverted cosine wave with a period of $T_{\xi} = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$.

4. Matching with Graphs:
* At $t=0$, $\xi = - B_0 a^2 \omega \cos(0) = - \text{constant}$. The graph must start at a negative extremum. * The EMF oscillates for a duration of $t = \pi/\omega$, which corresponds to exactly one full cycle of the $\cos(2\omega t)$ function. * For $t > \pi/\omega$, the EMF should be zero. * Graph (A) is the only option that starts at a negative extremum, completes one full cycle by $t = \pi/\omega$, and then becomes zero.

1. Analyze the Scale Construction (Figure 1):
First, we determine the value of one Main Scale Division (MSD). The main scale shows 1 cm divided into 10 parts.
$$ 1 \text{ MSD} = \frac{1 \text{ cm}}{10} = 0.1 \text{ cm} $$ Next, we determine the relationship between the Vernier Scale and the Main Scale. From Figure 1, observing the alignment of the scales, we see that the total length of the 10 divisions on the Vernier scale matches exactly with 7 divisions on the Main scale.
$$ 10 \text{ VSD} = 7 \text{ MSD} $$ $$ 1 \text{ VSD} = \frac{7}{10} \text{ MSD} = 0.7 \times 0.1 \text{ cm} = 0.07 \text{ cm} $$

2. Calculate the Least Count (LC):
The Least Count is the smallest measurement difference the instrument can detect, defined as the difference between one MSD and one VSD.
$$ \text{LC} = 1 \text{ MSD} - 1 \text{ VSD} $$ $$ \text{LC} = 0.1 \text{ cm} - 0.07 \text{ cm} = 0.03 \text{ cm} $$

3. Read the Measurement (Figure 2):

• Main Scale Reading (MSR): Look at the position of the zero mark of the Vernier scale. It lies just to the right of the 0.1 cm mark on the Main scale. Thus, MSR = 0.1 cm.
• Vernier Scale Coincidence (VSC): Identify the Vernier division that aligns perfectly with a Main scale division. Observing Figure 2, the 1st division of the Vernier scale aligns perfectly with a mark (specifically the 0.2 mark) on the Main scale. Thus, VSC = 1.

4. Final Calculation:
The total reading is the sum of the main scale reading and the vernier reading.
$$ \text{Measured Value} = \text{MSR} + (\text{VSC} \times \text{LC}) $$ $$ D = 0.1 \text{ cm} + (1 \times 0.03 \text{ cm}) $$ $$ D = 0.13 \text{ cm} $$

1. Dynamics of the Loop Entering the Field:
When the loop enters the magnetic field region ($y \ge 0$), the arm parallel to the x-axis cuts the magnetic field lines.

• Induced EMF: $\varepsilon = v B_0 L$
• Induced Current: $I = \frac{\varepsilon}{R} = \frac{v B_0 L}{R}$
• Magnetic Force: The force acts on the current-carrying wire in the magnetic field. $F_m = I L B_0 = \left( \frac{v B_0 L}{R} \right) L B_0 = \frac{B_0^2 L^2 v}{R}$
• Direction: By Lenz's law, the force opposes the motion (downwards, $-y$ direction).

2. Relation between Velocity and Distance:
Using the chain rule $a = v \frac{dv}{dx}$: $$ v \frac{dv}{dx} = - K v $$ $$ \frac{dv}{dx} = - K $$ Integrating from initial velocity $v_0$ at $x=0$ to velocity $v$ at distance $x$: $$ \int_{v_0}^v dv = - K \int_0^x dx $$ $$ v - v_0 = - Kx \implies v(x) = v_0 - Kx $$ The loop stops ($v=0$) at a maximum distance $x_{stop} = \frac{v_0}{K}$.

3. Evaluating the Options:

(A) If $v_0 = 1.5 KL$:
The stopping distance is $x_{stop} = \frac{1.5 KL}{K} = 1.5 L$. Since $x_{stop} > L$ (the side length of the loop), the loop would travel a distance greater than its own length. Thus, it will fully enter the magnetic field region before stopping. Statement (A) says it stops before entering completely. This is Incorrect.

(B) When the complete loop is inside:
Once the loop is fully inside the uniform magnetic field, the magnetic flux $\Phi_B = B \cdot L^2$ is constant. Induced EMF $\varepsilon = - \frac{d\Phi_B}{dt} = 0$. Consequently, Current $I = 0$ and Net Force $F = 0$. This is Correct.

(C) If $v_0 = \frac{KL}{10}$:
Stopping distance $x_{stop} = \frac{KL/10}{K} = 0.1 L$. The loop stops inside. However, regarding the time: From $\frac{dv}{dt} = -Kv$, we get $v(t) = v_0 e^{-Kt}$. Mathematically, the velocity becomes zero only as $t \to \infty$. It does not stop at a finite time. This is Incorrect.

(D) If $v_0 = 3KL$:
First, check if it enters fully: $x_{stop} = \frac{3KL}{K} = 3L > L$. Yes, it enters. We need the time $t$ for the loop to travel distance $x = L$. Using the velocity-distance relation $v = \frac{dx}{dt} = v_0 - Kx$: $$ \frac{dx}{v_0 - Kx} = dt $$ Integrate from $x=0$ to $x=L$: $$ \int_0^L \frac{dx}{v_0 - Kx} = \int_0^t dt $$ Let $u = v_0 - Kx \implies du = -K dx$: $$ -\frac{1}{K} [\ln(v_0 - Kx)]_0^L = t $$ $$ t = -\frac{1}{K} \ln \left( \frac{v_0 - KL}{v_0} \right) = \frac{1}{K} \ln \left( \frac{v_0}{v_0 - KL} \right) $$ Substitute $v_0 = 3KL$: $$ t = \frac{1}{K} \ln \left( \frac{3KL}{3KL - KL} \right) = \frac{1}{K} \ln \left( \frac{3}{2} \right) $$ This matches the expression in the option. This is Correct.

Conclusion:
The correct statements are (B) and (D).

1. Wave Parameters in Different Strings:
The speed of a transverse wave on a string is given by $v = \sqrt{\frac{T}{\mu}}$, where $T$ is the tension and $\mu$ is the linear mass density. Since the strings are connected in series, the tension $T$ is uniform throughout. The angular frequency $\omega$ depends on the source and remains constant across all media. The wave number is defined as $k = \frac{\omega}{v} = \omega \sqrt{\frac{\mu}{T}}$. Therefore, $k \propto \sqrt{\mu}$.

Let's determine the wave number for each string:

• String 1 ($S_1$): density $\mu$. Wave number $k_1 = k$.
• String 2 ($S_2$): density $4\mu$. Wave number $k_2 = k \sqrt{\frac{4\mu}{\mu}} = 2k$.
• String 3 ($S_3$): density $16\mu$. Wave number $k_3 = k \sqrt{\frac{16\mu}{\mu}} = 4k$.

2. Analysis at Junction P ($S_1 \to S_2$):
The incident wave travels from a rarer medium ($\mu$) to a denser medium ($4\mu$).

• Reflection: When reflecting from a denser medium (rigid boundary analog), there is a phase change of $\pi$. The reflected wave travels in the $-x$ direction back into $S_1$.
  Equation: $y_r = \alpha_1 y_0 \cos(\omega t + k_1 x + \pi) = \alpha_1 y_0 \cos(\omega t + kx + \pi)$.
  Statement (A) is correct.
• Transmission: There is no phase change for the transmitted wave. It travels in the $+x$ direction into $S_2$.
  Equation: $y_t = \alpha_2 y_0 \cos(\omega t - k_2 x) = \alpha_2 y_0 \cos(\omega t - 2kx)$.
  Statement (B) gives the argument as $(\omega t - kx)$, which is incorrect because the wave number changes to $2k$.

3. Analysis at Junction Q ($S_2 \to S_3$):
The wave incident on Q is the transmitted wave from P, traveling in $S_2$. It goes from $4\mu$ to $16\mu$ (denser medium).

• Reflection: Reflection from a denser medium causes a phase change of $\pi$. The wave travels back into $S_2$ ($-x$ direction).
  Equation: $y_{r'} = \alpha_3 y_0 \cos(\omega t + k_2 x + \pi) = \alpha_3 y_0 \cos(\omega t + 2kx + \pi)$.
  Statement (C) gives the argument as $(\omega t - kx + \pi)$, using the wrong wave number $k$ instead of $2k$ and incorrect direction sign.
• Transmission: The wave transmits into $S_3$. There is no phase change. It travels in $+x$ direction.
  Equation: $y_{t'} = \alpha_4 y_0 \cos(\omega t - k_3 x) = \alpha_4 y_0 \cos(\omega t - 4kx)$.
  Statement (D) is correct.

1. Analyze the Motion of the Elevator:
The vertical position of the elevator is given by the function: $$ y(t) = 8 \left[ 1 + \sin\left( \frac{2\pi t}{T} \right) \right] = 8 + 8\sin\left( \frac{2\pi t}{T} \right) $$ This equation represents Simple Harmonic Motion (SHM) where the mean position is $y_0 = 8$ m and the amplitude of oscillation is $A = 8$ m.
The angular frequency $\omega$ is calculated from the period $T = 40\pi$ s: $$ \omega = \frac{2\pi}{T} = \frac{2\pi}{40\pi} = \frac{1}{20} \text{ rad/s} $$

2. Determine the Acceleration:
The acceleration $a(t)$ is the second derivative of the position $y(t)$ with respect to time. For SHM, the maximum acceleration magnitude is given by: $$ |a_{max}| = \omega^2 A $$ Substituting the values: $$ |a_{max}| = \left( \frac{1}{20} \right)^2 \times 8 = \frac{1}{400} \times 8 = 0.02 \text{ m/s}^2 $$

3. Analyze the Apparent Weight:
A weighing scale measures the normal reaction force ($N$) acting on the object. According to Newton's Second Law for the object inside the elevator: $$ N - mg = ma \implies N = m(g + a) $$ The apparent weight fluctuates as the acceleration $a$ changes from $+|a_{max}|$ (upward at bottom extreme) to $-|a_{max}|$ (downward at top extreme).

• Maximum Apparent Weight ($N_{max}$): Occurs when $a = +|a_{max}|$. $$ N_{max} = m(g + |a_{max}|) $$
• Minimum Apparent Weight ($N_{min}$): Occurs when $a = -|a_{max}|$. $$ N_{min} = m(g - |a_{max}|) $$

4. Calculate Maximum Variation:
The "maximum variation" refers to the total range of weight readings observed during the oscillation, which is the difference between the maximum and minimum apparent weights. $$ \text{Variation} = N_{max} - N_{min} $$ $$ \text{Variation} = m(g + |a_{max}|) - m(g - |a_{max}|) = 2m|a_{max}| $$ Substitute $m = 50$ kg and $|a_{max}| = 0.02 \text{ m/s}^2$: $$ \text{Variation} = 2 \times 50 \times 0.02 $$ $$ \text{Variation} = 100 \times 0.02 = 2 \text{ N} $$

1. Calculate the Total Energy of Photons:
The total energy $E_{total}$ contained in the cube is the sum of the energies of all individual photons. The energy of a single photon is given by $E_p = hf$. Given:

• Number of photons, $N = 35 \times 10^7$
• Frequency, $f = 10^{15}$ Hz
• Planck's constant, $h = 6 \times 10^{-34}$ Js

2. Determine the Energy Density:
The problem states the cube has a unit volume ($V = 1 \text{ m}^3$). Therefore, the average energy density $u_{avg}$ is numerically equal to the total energy. $$ u_{avg} = \frac{E_{total}}{V} = 210 \times 10^{-12} \text{ J/m}^3 $$

3. Relate Energy Density to Magnetic Field Amplitude:
The average energy density of an electromagnetic wave is related to the amplitude of the magnetic field ($B_0$) by the formula: $$ u_{avg} = \frac{B_0^2}{2\mu_0} $$ where $\mu_0 = 4\pi \times 10^{-7}$ Tm/A. Rearranging the formula to solve for $B_0$: $$ B_0 = \sqrt{2 \mu_0 u_{avg}} $$

4. Substitute Values and Solve:
Using $\pi = \frac{22}{7}$: $$ \mu_0 = 4 \times \frac{22}{7} \times 10^{-7} $$ $$ B_0 = \sqrt{2 \times \left( 4 \times \frac{22}{7} \times 10^{-7} \right) \times (210 \times 10^{-12})} $$ Group the numbers for easier calculation: $$ B_0 = \sqrt{2 \times 4 \times \frac{22}{7} \times 210 \times 10^{-19}} $$ Simplify the fraction $\frac{210}{7} = 30$: $$ B_0 = \sqrt{2 \times 4 \times 22 \times 30 \times 10^{-19}} $$ $$ B_0 = \sqrt{8 \times 660 \times 10^{-19}} $$ $$ B_0 = \sqrt{5280 \times 10^{-19}} $$ Convert to standard scientific notation suitable for the square root (making the power even): $$ B_0 = \sqrt{528 \times 10^{-18}} $$ $$ B_0 = \sqrt{528} \times 10^{-9} \text{ T} $$ We know that $23^2 = 529$. Therefore, $\sqrt{528}$ is extremely close to $23$. $$ B_0 \approx 23 \times 10^{-9} \text{ T} $$

5. Conclusion:
Comparing this with the given form $B_0 = \alpha \times 10^{-9} \text{ T}$, we find: $$ \alpha = 23 $$

Step 1: Analyze the initial state

According to Stefan's Law, the power radiated per unit area between two black body plates P and Q at temperatures \(T_P\) and \(T_Q\) is:

$$W_0 = \sigma(T_P^4 - T_Q^4) \quad \dots(1)$$

Step 2: Analyze the final state with shields

When two identical plates (shields) are introduced between P and Q, let their steady-state temperatures be \(T_1\) and \(T_2\). In the steady state, the rate of heat flow (power per unit area) across each gap must be the same. Let this new power be \(W_S\).

The heat transfer equations for the three gaps are:

$$W_S = \sigma(T_P^4 - T_1^4) \quad \dots(i)$$

$$W_S = \sigma(T_1^4 - T_2^4) \quad \dots(ii)$$

$$W_S = \sigma(T_2^4 - T_Q^4) \quad \dots(iii)$$

Step 3: Solve for the ratio

Add equations (i), (ii), and (iii) to eliminate the intermediate temperatures \(T_1\) and \(T_2\):

$$3W_S = \sigma[(T_P^4 - T_1^4) + (T_1^4 - T_2^4) + (T_2^4 - T_Q^4)]$$

$$3W_S = \sigma(T_P^4 - T_Q^4)$$

Substituting equation (1) into this result:

$$3W_S = W_0$$

$$\frac{W_0}{W_S} = 3$$

Thus, the ratio is 3.

Step 1: Determine the angle of refraction using polarization condition

The light reflected from point O is fully polarized. This implies the reflection satisfies Brewster's condition. The problem setup and given diagram imply an external incidence equivalent angle of \(60^\circ\) (since \(\tan 60^\circ = \sqrt{3} = n_{glass}\)).

Applying Snell's Law at the cavity-glass interface (assuming the ray geometry relates the cavity angle to this \(60^\circ\) condition):

$$n_{glass} \sin \alpha = n_{air} \sin(60^\circ)$$

$$\sqrt{3} \sin \alpha = 1 \times \frac{\sqrt{3}}{2}$$

$$\sin \alpha = \frac{1}{2} \implies \alpha = 30^\circ$$

Thus, the angle of refraction inside the glass is \(30^\circ\).

Step 2: Apply geometry and Sine Rule

Consider the triangle \(\Delta OAB\) formed by the center of the sphere, the point on the cavity, and the point O on the periphery. Using the Sine Rule:

$$\frac{\sin(120^\circ)}{R} = \frac{\sin \theta}{\text{side } AB}$$

Based on the geometry derived in the solution (where side AB corresponds to \(\frac{\sqrt{3}R}{2}\)):

$$\frac{\sin(120^\circ)}{R} = \frac{\sin \theta}{\frac{\sqrt{3}R}{2}}$$

Solving for \(\sin \theta\):

$$\sin \theta = \frac{\sqrt{3}R}{2} \times \frac{\sin(120^\circ)}{R}$$

$$\sin \theta = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} = \frac{3}{4} = 0.75$$

Step 1: Calculate the value of slit width \(b\)

The diffraction equation is given by \(\frac{bd}{D} = m\lambda\). Rearranging for \(b\):

$$b = \frac{mD\lambda}{d}$$

Given values:

• \(\lambda = 600 \text{ nm} = 6 \times 10^{-7} \text{ m}\)
• \(m = 3\)
• \(D = 1 \text{ m}\)
• \(d = 5 \text{ mm} = 5 \times 10^{-3} \text{ m}\)

$$b = \frac{3 \times 1 \times (600 \times 10^{-9})}{5 \times 10^{-3}} = \frac{1800 \times 10^{-9}}{5 \times 10^{-3}} = 360 \times 10^{-6} \text{ m} = 360 \, \mu\text{m}$$

Step 2: Determine relative error

Using the error propagation formula for multiplication and division:

$$\frac{\Delta b}{b} = \frac{\Delta m}{m} + \frac{\Delta D}{D} + \frac{\Delta \lambda}{\lambda} + \frac{\Delta d}{d}$$

Since \(m\) and \(\lambda\) are known precisely, \(\Delta m = 0\) and \(\Delta \lambda = 0\).

Given errors:

• \(\Delta D = 1 \text{ cm} = 10^{-2} \text{ m}\)
• \(\Delta d = 1 \text{ mm} = 10^{-3} \text{ m}\)

$$\frac{\Delta b}{b} = \frac{10^{-2}}{1} + \frac{10^{-3}}{5 \times 10^{-3}} = 0.01 + \frac{1}{5} = 0.01 + 0.20 = 0.21$$

Step 3: Calculate absolute error \(\Delta b\)

$$\Delta b = b \times 0.21$$

$$\Delta b = 360 \, \mu\text{m} \times 0.21 = 75.6 \, \mu\text{m}$$

The absolute error is 75.6 \(\mu\)m.

Step 1: Express the de Broglie wavelength of the electron

For an electron in the \(n\)-th orbit of a hydrogen-like atom, the quantization condition is \(2\pi r_n = n\lambda_e\). Thus, the wavelength is:

$$ \lambda_e = \frac{2\pi r_n}{n} $$

where \( r_n = \frac{n^2 a_0}{Z} \). Substituting this in:

$$ \lambda_e = \frac{2\pi}{n} \left( \frac{n^2 a_0}{Z} \right) = \frac{2\pi n a_0}{Z} $$

Step 2: Express the de Broglie wavelength of the neutron

The de Broglie wavelength of a particle with kinetic energy \(K\) is \(\lambda = \frac{h}{\sqrt{2mK}}\). For a thermal neutron with energy \(K = k_B T\):

$$ \lambda_n = \frac{h}{\sqrt{2 m_N k_B T}} $$

Step 3: Equate the wavelengths and solve for Temperature T

Given \(\lambda_e = \lambda_n\):

$$ \frac{2\pi n a_0}{Z} = \frac{h}{\sqrt{2 m_N k_B T}} $$

Squaring both sides:

$$ \frac{4\pi^2 n^2 a_0^2}{Z^2} = \frac{h^2}{2 m_N k_B T} $$

Rearranging to solve for \(T\):

$$ T = \frac{h^2 Z^2}{2 m_N k_B (4\pi^2 n^2 a_0^2)} = \frac{Z^2 h^2}{8 \pi^2 n^2 a_0^2 m_N k_B} $$

Step 4: Determine the value of \(\alpha\)

The problem gives the temperature expression as:

$$ T = \frac{Z^2 h^2}{\alpha \pi^2 a_0^2 m_N k_B} $$

Comparing the denominators of our derived expression and the given expression:

$$ \alpha = 8 n^2 $$

Given that the electron is in the \(n = 3\) orbit:

$$ \alpha = 8 \times (3)^2 = 8 \times 9 = 72 $$

Thus, the value of \(\alpha\) is 72.

Step 1: Analyze Configuration P

Two dipoles pointing in the \(+\hat{j}\) direction are separated by distance \(2r\). Point X is the midpoint (distance \(r\) from each). For both dipoles, X lies on the equatorial line.

The field due to a dipole \(\vec{p}\) on its equatorial line is \(\vec{E} = -\frac{\vec{p}}{4\pi\epsilon_0 r^3}\).

Total Field: \(\vec{E}_P = -\frac{p\hat{j}}{4\pi\epsilon_0 r^3} - \frac{p\hat{j}}{4\pi\epsilon_0 r^3} = -\frac{2p\hat{j}}{4\pi\epsilon_0 r^3} = -\frac{p}{2\pi\epsilon_0 r^3}\hat{j}\).

This matches entry (2) in List-II. So, P \(\rightarrow\) 2.

Step 2: Analyze Configuration Q

Left dipole points \(+\hat{j}\), Right dipole points \(-\hat{j}\). Both are equatorial to X.

Left dipole field: \(\vec{E}_L = -\frac{p\hat{j}}{4\pi\epsilon_0 r^3}\).

Right dipole field: \(\vec{E}_R = -\frac{(-p\hat{j})}{4\pi\epsilon_0 r^3} = +\frac{p\hat{j}}{4\pi\epsilon_0 r^3}\).

Total Field: \(\vec{E}_Q = \vec{E}_L + \vec{E}_R = 0\).

This matches entry (1) in List-II. So, Q \(\rightarrow\) 1.

Step 3: Analyze Configuration R

Left dipole points \(+\hat{j}\) (equatorial to X). Right dipole is at distance \(r\), pointing \(+\hat{i}\) (axial to X, but X is behind it).

Left dipole field: \(\vec{E}_L = -\frac{p\hat{j}}{4\pi\epsilon_0 r^3}\).

Right dipole field (axial): The field on the axis of a dipole \(\vec{p}\) at distance \(r\) is \(\frac{2\vec{p}}{4\pi\epsilon_0 r^3}\). Direction is along \(\vec{p}\). So \(\vec{E}_R = \frac{2p\hat{i}}{4\pi\epsilon_0 r^3}\).

Total Field: \(\vec{E}_R = \frac{1}{4\pi\epsilon_0 r^3}(2p\hat{i} - p\hat{j}) = \frac{p}{4\pi\epsilon_0 r^3}(2\hat{i} - \hat{j})\).

This matches entry (4) in List-II. So, R \(\rightarrow\) 4.

Step 4: Analyze Configuration S

Based on the remaining options and the diagram logic, the left dipole points \(+\hat{i}\) and the right dipole points \(+\hat{i}\). Both are axial to X.

Left dipole field (at X, to its right): \(\frac{2p\hat{i}}{4\pi\epsilon_0 r^3}\).

Right dipole field (at X, to its left): \(\frac{2p\hat{i}}{4\pi\epsilon_0 r^3}\).

Total Field: \(\vec{E}_S = \frac{4p\hat{i}}{4\pi\epsilon_0 r^3} = \frac{p}{\pi\epsilon_0 r^3}\hat{i}\).

This matches entry (5) in List-II. So, S \(\rightarrow\) 5.

Conclusion: The correct match is P\(\rightarrow\)2, Q\(\rightarrow\)1, R\(\rightarrow\)4, S\(\rightarrow\)5. This corresponds to Option (C).

Step 1: Analyze Circuit P (Pure Resistance)

Given: \(V(t) = 300 \sin(400t)\), \(R = 30\Omega\).

Current \(i(t) = \frac{V(t)}{R} = \frac{300}{30} \sin(400t) = 10 \sin(400t)\).

Characteristics: Peak current 10 A, in phase with voltage (starts at 0). Matches Graph (3). P \(\rightarrow\) 3.

Step 2: Analyze Circuit Q (RL Circuit)

\(R=30\Omega\), \(L=100 \text{mH} = 0.1 \text{H}\), \(\omega=400\).

\(X_L = \omega L = 400 \times 0.1 = 40\Omega\).

\(Z = \sqrt{R^2 + X_L^2} = \sqrt{30^2+40^2} = 50\Omega\).

Peak Current \(I_0 = 300/50 = 6 \text{A}\). Phase angle \(\phi = \tan^{-1}(X_L/R) = 53^\circ\) (lagging).

Graph must show lag (zero crossing shifted right) and amplitude ~6 A. Graph (5) fits best qualitatively. Q \(\rightarrow\) 5.

Step 3: Analyze Circuit S (RLC Circuit - Resonance)

\(R=60\Omega\), \(C=50\mu\text{F}\), \(L=125\text{mH}\).

\(X_L = 400 \times 0.125 = 50\Omega\).

\(X_C = \frac{1}{400 \times 50 \times 10^{-6}} = 50\Omega\).

Since \(X_L = X_C\), the circuit is in resonance. \(Z = R = 60\Omega\).

\(I_0 = 300/60 = 5 \text{A}\). Current is in phase. Matches Graph (1) (pure sine, peak 5). S \(\rightarrow\) 1.

Step 4: Analyze Circuit R (RLC Circuit)

\(R=30\Omega\), \(C=50\mu\text{F}\), \(L=25\text{mH}\).

\(X_C = 50\Omega\), \(X_L = 400 \times 0.025 = 10\Omega\).

Net reactance \(X_C - X_L = 40\Omega\). \(Z = \sqrt{30^2+40^2} = 50\Omega\).

\(I_0 = 300/50 = 6 \text{A}\). Circuit is capacitive, so current leads voltage by \(53^\circ\).

Graph must show lead (starts positive at t=0). Matches Graph (2). R \(\rightarrow\) 2.

Conclusion: P\(\rightarrow\)3, Q\(\rightarrow\)5, R\(\rightarrow\)2, S\(\rightarrow\)1. Matches Option (A).

Step 1: Match \(E \propto Z^2\) (P)

The energy levels of a hydrogen-like atom with atomic number \(Z\) are given by the Bohr model as \(E_n \propto \frac{Z^2}{n^2}\). Thus, radiation energy from transitions scales with \(Z^2\).

Matches: (5) Energy of radiation due to electronic transitions in hydrogen-like atoms.

P \(\rightarrow\) 5.

Step 2: Match \(E \propto (Z-1)^2\) (Q)

Moseley's Law for characteristic K-series X-rays states that the frequency \(\nu\) follows \(\sqrt{\nu} \propto (Z-1)\). Since energy \(E = h\nu\), we have \(E \propto (Z-1)^2\).

Matches: (1) Energy of characteristic x-rays.

Q \(\rightarrow\) 1.

Step 3: Match \(E \propto Z(Z-1)\) (R)

The electrostatic (Coulomb) potential energy of a nucleus with \(Z\) protons arises from the interactions between pairs of protons. The number of pairs is \(Z(Z-1)/2\). Thus, the electrostatic binding energy term in the semi-empirical mass formula is proportional to \(Z(Z-1)\) (or approx \(Z^2\) for large Z).

Matches: (2) Electrostatic part of the nuclear binding energy.

R \(\rightarrow\) 2.

Step 4: Match \(E\) is independent of \(Z\) (S)

The average nuclear binding energy per nucleon is roughly constant (about 8 MeV) for stable nuclei in the mass number range \(30 < A < 170\). It does not depend strongly on \(Z\).

Matches: (4) Average nuclear binding energy per nucleon.

S \(\rightarrow\) 4.

Conclusion: P\(\rightarrow\)5, Q\(\rightarrow\)1, R\(\rightarrow\)2, S\(\rightarrow\)4. Matches Option (C).

The problem asks for the products obtained from the thermal decomposition of Ammonium Nitrite ($\text{NH}_4\text{NO}_2$) and Ammonium Nitrate ($\text{NH}_4\text{NO}_3$).

Step 1: Decomposition of Ammonium Nitrite ($\text{NH}_4\text{NO}_2$)
When ammonium nitrite is heated gently (at $60-70^\circ\text{C}$), it undergoes decomposition to form nitrogen gas ($\text{N}_2$) and water. This is a standard laboratory method for the preparation of dinitrogen.

$$ \text{NH}_4\text{NO}_2(s) \xrightarrow{60-70^\circ\text{C}} \text{N}_2(g) + 2\text{H}_2\text{O}(\ell) $$

Thus, compound X is $\text{N}_2$.

Step 2: Decomposition of Ammonium Nitrate ($\text{NH}_4\text{NO}_3$)
When ammonium nitrate is heated strongly (at $200-250^\circ\text{C}$), it decomposes to form nitrous oxide ($\text{N}_2\text{O}$), also known as laughing gas, and water.

$$ \text{NH}_4\text{NO}_3(s) \xrightarrow{200-250^\circ\text{C}} \text{N}_2\text{O}(g) + 2\text{H}_2\text{O}(g) $$

Thus, compound Y is $\text{N}_2\text{O}$.

Conclusion:
The compounds X and Y are $\text{N}_2$ and $\text{N}_2\text{O}$, respectively.

Therefore, the correct option is (A).

The problem asks to arrange the complexes in order of increasing wavelength of absorption maxima ($\lambda_{max}$).

Step 1: Relationship between Energy and Wavelength
According to Crystal Field Theory, ligands split the d-orbitals of the central metal ion. The energy difference between these levels is denoted as $\Delta_0$ (Crystal Field Splitting Energy). When a complex absorbs light, an electron is promoted from a lower energy d-orbital to a higher one. The energy of the absorbed photon corresponds to this gap:

$$ \Delta_0 = E = \frac{hc}{\lambda} $$

This implies an inverse relationship: Higher splitting energy ($\Delta_0$) corresponds to a shorter wavelength ($\lambda$).

Step 2: Determine Ligand Field Strength
The magnitude of $\Delta_0$ depends on the strength of the ligands, as defined by the Spectrochemical Series. The order of field strength for the ligands involved is:

$$ \text{CN}^- > \text{NH}_3 > \text{H}_2\text{O} > \text{Cl}^- $$

• Strong field ligands (like $\text{CN}^-$) cause large splitting ($\text{High } \Delta_0$).
• Weak field ligands (like $\text{Cl}^-$) cause small splitting ($\text{Low } \Delta_0$).

Step 3: Analyze the Complexes
All complexes contain Cobalt in the +3 oxidation state ($Co^{3+}$), so the difference lies only in the ligands.

1. $[Co(CN)_6]^{3-}$: Contains 6 strong $\text{CN}^-$ ligands.
   $\rightarrow$ Highest $\Delta_0$.
2. $[Co(NH_3)_6]^{3+}$: Contains 6 $\text{NH}_3$ ligands.
   $\rightarrow$ High $\Delta_0$ (but $ <\text{CN}^-$).
3. $[Co(NH_3)_5(H_2O)]^{3+}$: Contains 5 $\text{NH}_3$ and 1 weaker $\text{H}_2\text{O}$.
   $\rightarrow$ Lower $\Delta_0$ than pure ammonia complex.
4. $[Co(NH_3)_5(Cl)]^{2+}$: Contains 5 $\text{NH}_3$ and 1 weakest $\text{Cl}^-$.
   $\rightarrow$ Lowest $\Delta_0$.

Step 4: Determine Order of Wavelength
Since $\lambda \propto \frac{1}{\Delta_0}$, the order of wavelength will be the reverse of the energy order.

Energy Order ($\Delta_0$):
$$ [Co(CN)_6]^{3-} > [Co(NH_3)_6]^{3+} > [Co(NH_3)_5(H_2O)]^{3+} > [Co(NH_3)_5(Cl)]^{2+} $$

Wavelength Order ($\lambda$):
$$ [Co(CN)_6]^{3-} < [Co(NH_3)_6]^{3+} < [Co(NH_3)_5(H_2O)]^{3+} < [Co(NH_3)_5(Cl)]^{2+} $$

This matches option (A).

The reaction of permanganate ion ($\text{MnO}_4^-$) depends strongly on the pH of the medium. The problem specifies a neutral aqueous medium.

Step 1: Analyze the Oxidizing Behavior of $\text{KMnO}_4$

• In Acidic Medium: Permanganate acts as a very strong oxidizing agent ($\text{Mn}^{+7} \rightarrow \text{Mn}^{+2}$) and oxidizes iodide ($\text{I}^-$) to iodine ($\text{I}_2$).
• In Neutral or Faintly Alkaline Medium: Permanganate acts as a moderate oxidizing agent. It gets reduced to manganese dioxide ($\text{MnO}_2$, where Mn is +4) and oxidizes iodide ($\text{I}^-$) to iodate ion ($\text{IO}_3^-$).

Step 2: Balanced Chemical Equation
In neutral or faintly alkaline solution, the oxidation state of iodine changes from -1 (in $\text{I}^-$) to +5 (in $\text{IO}_3^-$).

$$ 2\text{MnO}_4^- + \text{H}_2\text{O} + \text{I}^- \rightarrow 2\text{MnO}_2 + 2\text{OH}^- + \text{IO}_3^- $$

Conclusion:
The nitrogen-containing product is not asked, but rather the iodine product. The product formed from the iodide ion is the iodate ion, $\text{IO}_3^-$.

Therefore, the correct option is (B).

The acidity of a hydrogen atom in a hydrocarbon is determined by the stability of its conjugate base (the carbanion formed after removing the proton, $H^+$). The more stable the conjugate base, the more acidic the hydrogen.

Analysis of Option (B):

1. Structure: The molecule is a derivative of fulvene, consisting of a five-membered ring with two double bonds and an exocyclic double bond.
2. Deprotonation: When the indicated hydrogen is removed from the methyl group attached to the exocyclic double bond, a carbanion is formed on the $-\text{CH}_2$ group.
3. Resonance Stabilization: This negative charge is allylic and can participate in resonance. The lone pair on the $-\text{CH}_2^-$ moves to form a double bond with the exocyclic carbon, pushing the $\pi$-electrons of the exocyclic $\text{C}=\text{C}$ bond onto the five-membered ring.
4. Aromaticity:
   • The resulting resonance structure places a negative charge on a carbon in the five-membered ring.
   • The ring now contains two double bonds ($4 \pi$ electrons) and one lone pair ($2 \pi$ electrons) in a planar, cyclic system.
   • Total $\pi$ electrons = $4 + 2 = 6$.
   • According to Hückel's Rule, a planar ring with $(4n + 2) \pi$ electrons (where $n=1$) is aromatic.

Conclusion:
The formation of an aromatic cyclopentadienyl anion derivative provides exceptional stability to the conjugate base. This strong driving force makes the indicated hydrogen in option (B) the most acidic among the choices.

$$ \text{Reaction:} \quad \text{R}-\text{CH}_3 \xrightarrow{-H^+} \text{[Aromatic Anion]}^- $$

Therefore, the correct option is (B).

The question asks to identify the INCORRECT statement(s) regarding molecular orbital properties.

Analyze Statement (A): Bond order of $\text{Ne}_2$ is zero.
Neon ($Z=10$) has the configuration $1s^2 2s^2 2p^6$. For $\text{Ne}_2$ (20 electrons):
$$ \sigma1s^2 \sigma^*1s^2 \sigma2s^2 \sigma^*2s^2 \sigma2p_z^2 \pi2p^4 \pi^*2p^4 \sigma^*2p_z^2 $$ $$ \text{Bond Order} = \frac{1}{2}(N_b - N_a) = \frac{1}{2}(10 - 10) = 0 $$ The molecule does not exist. The statement is Correct.

Analyze Statement (B): The HOMO of $\text{F}_2$ is $\sigma$-type.
Fluorine ($Z=9$) has 18 electrons in $\text{F}_2$. The electronic configuration for molecules with $Z > 7$ is:
$$ \sigma1s^2 \sigma^*1s^2 \sigma2s^2 \sigma^*2s^2 \sigma2p_z^2 (\pi2p_x^2 = \pi2p_y^2) (\pi^*2p_x^2 = \pi^*2p_y^2) $$ The highest occupied molecular orbitals (HOMO) are the antibonding $\pi^*$ orbitals ($\pi^*2p_x$ and $\pi^*2p_y$).
These are $\pi$-type, not $\sigma$-type.
The statement is INCORRECT.

Analyze Statement (C): Bond energy of $\text{O}_2^+$ is smaller than $\text{O}_2$.
We compare Bond Orders (higher BO $\propto$ higher Bond Energy).
For $\text{O}_2$ (16 electrons):
$$ \dots \sigma2p_z^2 \pi2p^4 \pi^*2p^2 $$ $$ \text{B.O.} = \frac{1}{2}(10 - 6) = 2.0 $$ For $\text{O}_2^+$ (15 electrons, removed from $\pi^*$):
$$ \dots \sigma2p_z^2 \pi2p^4 \pi^*2p^1 $$ $$ \text{B.O.} = \frac{1}{2}(10 - 5) = 2.5 $$ Since $\text{B.O.}(\text{O}_2^+) > \text{B.O.}(\text{O}_2)$, the bond energy of $\text{O}_2^+$ is larger.
The statement is INCORRECT.

Analyze Statement (D): Bond length of $\text{Li}_2$ is larger than $\text{B}_2$.
Both $\text{Li}_2$ and $\text{B}_2$ have a Bond Order of 1.
However, Lithium is an alkali metal with a much larger atomic radius than Boron (atomic size decreases across a period).
Therefore, the bond length depends on atomic size: $\text{Li} > \text{B} \Rightarrow \text{Li}_2 > \text{B}_2$.
The statement is Correct.

Conclusion:
The incorrect statements are (B) and (C).

Step 1: Identify the Compounds P, Q, and R

1. Formation of P: Reaction of cyclohexene with $\text{HBr}$ is an electrophilic addition reaction.
   $$ \text{Cyclohexene} + \text{HBr} \rightarrow \text{Bromocyclohexane (P)} $$
   Here, the halogen $\text{X}$ in $\mathbf{P}$ is Bromine ($\text{Br}$).

2. Formation of Q: Reaction of alkyl bromide (P) with $\text{NaI}$ in acetone is known as the Finkelstein reaction. It is a halogen exchange method.
   $$ \text{R}-\text{Br} + \text{NaI} \xrightarrow{\text{acetone}} \text{R}-\text{I} + \text{NaBr} $$
   So, $\mathbf{Q}$ is Iodocyclohexane. The halogen $\text{X}$ in $\mathbf{Q}$ is Iodine ($\text{I}$).

3. Formation of R: Reaction of alkyl bromide (P) with metallic fluorides (like $\text{Hg}_2\text{F}_2$, $\text{AgF}$, etc.) is known as the Swarts reaction. It yields alkyl fluorides.
   $$ \text{R}-\text{Br} + \text{Hg}_2\text{F}_2 \rightarrow \text{R}-\text{F} $$
   So, $\mathbf{R}$ is Fluorocyclohexane. The halogen $\text{X}$ in $\mathbf{R}$ is Fluorine ($\text{F}$).

(In the options, X

Summary of Halogens:

• $\mathbf{P} \rightarrow -\text{Br}$
• $\mathbf{Q} \rightarrow -\text{I}$
• $\mathbf{R} \rightarrow -\text{F}$

Step 2: Analyze the Options

• (A) Bond Length:
  Bond length depends on the size of the halogen atom ($\text{I} > \text{Br} > \text{F}$).
  Order: $\text{C}-\text{I} (\mathbf{Q}) > \text{C}-\text{Br} (\mathbf{P}) > \text{C}-\text{F} (\mathbf{R})$.
  Given: $\mathbf{Q} > \mathbf{R} > \mathbf{P}$. (Incorrect)

• (B) Bond Enthalpy:
  Bond enthalpy is inversely proportional to bond length. Stronger bonds are shorter.
  Order: $\text{C}-\text{F} (\mathbf{R}) > \text{C}-\text{Br} (\mathbf{P}) > \text{C}-\text{I} (\mathbf{Q})$.
  Given: $\mathbf{R} > \mathbf{P} > \mathbf{Q}$. (Correct)

• (C) Relative Reactivity toward $\text{S}_\text{N}2$:
  $\text{S}_\text{N}2$ reactivity depends on the leaving group ability: $\text{I}^- > \text{Br}^- > \text{F}^-$.
  Order: $\mathbf{Q} > \mathbf{P} > \mathbf{R}$.
  Given: $\mathbf{P} > \mathbf{R} > \mathbf{Q}$. (Incorrect)

• (D) pKa of Conjugate Acids:
  Conjugate acids are $\text{HI}, \text{HBr}, \text{HF}$.
  Acidity Order: $\text{HI} > \text{HBr} > \text{HF}$.
  pKa Order (inverse of acidity): $\text{HF} (\mathbf{R}) > \text{HBr} (\mathbf{P}) > \text{HI} (\mathbf{Q})$.
  Given: $\mathbf{R} > \mathbf{Q} > \mathbf{P}$. (Incorrect)

Conclusion:
Only statement (B) is correct.

Step 1: Write the balanced reduction half-reaction.
In acidic medium, dichromate ion ($\text{Cr}_2\text{O}_7^{2-}$) reduces to chromium(III) ion ($\text{Cr}^{3+}$). The oxidation state of Cr changes from +6 to +3.

$$ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} $$

Step 2: Determine the mole ratio between electrons and product.
From the balanced equation stoichiometry:
$$ 6 \text{ moles of electrons produce } 2 \text{ moles of } \text{Cr}^{3+} $$ Therefore, to produce 1 mole of $\text{Cr}^{3+}$, the moles of electrons required ($n_e$) is: $$ n_e = \frac{6}{2} = 3 \text{ moles} $$

Step 3: Apply Faraday's Law of Electrolysis.
The total charge ($Q$) required is given by: $$ Q = n_e \times F $$ Where $F = 96500 \, \text{C mol}^{-1}$. $$ Q = 3 \times 96500 \, \text{C} $$

Also, charge is related to current ($I$) and time ($t$) by: $$ Q = I \times t $$ Given time $t = 48.25 \text{ minutes}$. We must convert this to seconds. $$ t = 48.25 \times 60 \, \text{s} $$

Step 4: Calculate the Current ($I$).
Equating the two expressions for Charge ($Q$): $$ I \times (48.25 \times 60) = 3 \times 96500 $$ $$ I = \frac{3 \times 96500}{48.25 \times 60} $$

Simplifying the calculation: Note that $48.25 \times 2 = 96.5$. Thus, $48.25 \times 2000 = 96500$. $$ I = \frac{3 \times (48.25 \times 2000)}{48.25 \times 60} $$ $$ I = \frac{3 \times 2000}{60} $$ $$ I = \frac{6000}{60} $$ $$ I = 100 \, \text{A} $$

Answer:
The current that flows through the cell is 100 A.

Step 1: Analyze the Problem Conditions
We are given a weak acid ($HA$) with concentration $C = 1.00 \times 10^{-3} \text{ M}$ and a very small dissociation constant $K_a = 4.00 \times 10^{-11}$.
Usually, we calculate $[H^+]$ as $\sqrt{K_a \cdot C}$. Let's check the magnitude: $$ \sqrt{K_a \cdot C} = \sqrt{4.00 \times 10^{-11} \times 10^{-3}} = \sqrt{4.00 \times 10^{-14}} = 2.00 \times 10^{-7} \text{ M} $$ Since the calculated $[H^+]$ from the acid is very close to $10^{-7} \text{ M}$ (the concentration of protons from water), we cannot neglect the contribution of water auto-ionization.

Step 2: Set up Simultaneous Equilibria
Let the total hydrogen ion concentration be $[H^+]_{total}$. This comes from two sources: 1. From Acid Dissociation: $HA \rightleftharpoons H^+ + A^-$ 2. From Water Auto-ionization: $H_2O \rightleftharpoons H^+ + OH^-$

Let $x$ be the concentration of $H^+$ produced by the acid.
Let $y$ be the concentration of $H^+$ produced by water (which equals $[OH^-]$).

Total concentrations at equilibrium: $$ [H^+]_{total} = (x + y) $$ $$ [A^-] = x $$ $$ [OH^-] = y $$ $$ [HA] \approx C = 10^{-3} \text{ M} \quad (\text{Since } K_a \text{ is very small, degree of dissociation is negligible}) $$

Step 3: Formulate Equilibrium Equations
For the weak acid ($K_a$): $$ K_a = \frac{[H^+][A^-]}{[HA]} = \frac{(x+y)(x)}{10^{-3}} $$ $$ x(x+y) = K_a \cdot 10^{-3} = 4.00 \times 10^{-11} \cdot 10^{-3} = 4.00 \times 10^{-14} \quad \dots(\text{I}) $$

For water ($K_w$): $$ K_w = [H^+][OH^-] = (x+y)(y) $$ $$ y(x+y) = 1.00 \times 10^{-14} \quad \dots(\text{II}) $$

Step 4: Solve for Total $[H^+]$
Instead of solving for $x$ and $y$ individually, we can simply add equations (I) and (II): $$ x(x+y) + y(x+y) = 4.00 \times 10^{-14} + 1.00 \times 10^{-14} $$ Factor out $(x+y)$: $$ (x+y)(x+y) = 5.00 \times 10^{-14} $$ $$ (x+y)^2 = 5.00 \times 10^{-14} $$ $$ [H^+]_{total} = \sqrt{5.00 \times 10^{-14}} $$

Step 5: Calculate Final Value $$ [H^+]_{total} = \sqrt{5} \times 10^{-7} $$ $$ [H^+]_{total} \approx 2.236 \times 10^{-7} \text{ M} $$

Rounding to two decimal places as implied by the solution format: $$ [H^+]_{total} = 2.24 \times 10^{-7} \text{ M} $$

Comparing this with $X \times 10^{-7}$, we get: $X = 2.24$

Step 1: Write the van der Waals equation for 1 mole of gas.

$$ \left( P + \frac{a}{V_m^2} \right) (V_m - b) = RT $$

Step 2: Expand the equation to form a cubic equation in $V_m$.
Multiply the terms:

$$ P V_m - Pb + \frac{a}{V_m} - \frac{ab}{V_m^2} = RT $$

Multiply the entire equation by $V_m^2$ to remove the denominator:

$$ P V_m^3 - Pb V_m^2 + a V_m - ab = RT V_m^2 $$

Rearrange the terms in descending powers of $V_m$:

$$ P V_m^3 - (Pb + RT) V_m^2 + a V_m - ab = 0 $$

Step 3: Identify the coefficients.
From the cubic equation $P V_m^3 - (Pb + RT) V_m^2 + a V_m - ab = 0$:

• Coefficient of $V_m^2$ = $-(Pb + RT)$
• Coefficient of $V_m$ = $a$

Step 4: Calculate the required ratio.
The question asks for the ratio of the coefficient of $V_m^2$ to the coefficient of $V_m$.

$$ \text{Ratio} = \frac{\text{Coefficient of } V_m^2}{\text{Coefficient of } V_m} = \frac{-(Pb + RT)}{a} $$

Step 5: Substitute the given values.

• $P = 300 \, \text{atm}$
• $T = 300 \, \text{K}$
• $a = 6.0 \, \text{dm}^6 \text{atm mol}^{-2}$
• $b = 0.060 \, \text{dm}^3 \text{mol}^{-1}$
• $R = 0.082 \, \text{dm}^3 \text{atm mol}^{-1} \text{K}^{-1}$

Calculate the numerator term $(Pb + RT)$:

$$ Pb = 300 \times 0.060 = 18 $$ $$ RT = 0.082 \times 300 = 24.6 $$ $$ Pb + RT = 18 + 24.6 = 42.6 $$

Now, calculate the ratio:

$$ \text{Ratio} = \frac{-42.6}{6.0} $$ $$ \text{Ratio} = -7.1 $$

Answer:
The ratio is -7.1.

Step 1: Write the electrolysis reaction

The electrolysis of water is represented by the equation:

$$ H_2O(l) \rightarrow H_2(g) + \frac{1}{2}O_2(g) $$

From the stoichiometry, 1 mole of liquid water produces 1 mole of hydrogen gas and 0.5 moles of oxygen gas. Total moles of gas produced per mole of water = \(1 + 0.5 = 1.5\) moles.

Step 2: Calculate moles of water and gases produced

Given mass of water = 144 g.

$$ \text{Moles of } H_2O = \frac{144 \text{ g}}{18 \text{ g/mol}} = 8 \text{ mol} $$

Total moles of gas produced (\(n_g\)):

$$ n_g = 8 \text{ mol} \times 1.5 = 12 \text{ mol} $$

Since the reactants are liquid, the change in gaseous moles \(\Delta n_g = 12 - 0 = 12\).

Step 3: Calculate the expansion work done

The work done during expansion at constant pressure is given by:

$$ W = -P_{ext}\Delta V = -\Delta n_g RT $$

Substituting the values ($R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$, $T = 300 \text{ K}$):

$$ W = -12 \times 8.3 \times 300 $$

$$ W = -29880 \text{ J} $$

$$ W = -29.88 \text{ kJ} $$

The magnitude of work done (expansion work by the system) is 29.88 kJ.

Step 1: Identify the monomer X

Nylon 6,6 is synthesized from two monomers: Hexamethylenediamine and Adipic acid. The problem states that monomer X gives a positive carbylamine test. The carbylamine test is specific to primary amines. Therefore, X is Hexamethylenediamine.

Formula of Hexamethylenediamine: \( H_2N-(CH_2)_6-NH_2 \)

Step 2: Determine moles of Nitrogen gas evolved

In the Dumas method, organic compounds containing nitrogen are heated with copper oxide ($CuO$) to convert the nitrogen into nitrogen gas ($N_2$).

The molecular formula of X ($C_6H_{16}N_2$) contains 2 nitrogen atoms per molecule. Thus, 1 mole of X contains 2 moles of nitrogen atoms.

Upon analysis, these 2 moles of N atoms combine to form 1 mole of $N_2$ gas.

$$ \text{Stoichiometry: } 1 \text{ mole X} \rightarrow 1 \text{ mole } N_2 $$

For 10 moles of X:

$$ \text{Moles of } N_2 = 10 \text{ moles} $$

Step 3: Calculate the mass of Nitrogen evolved

The molar mass of $N_2$ is \( 14 \times 2 = 28 \text{ g/mol} \).

$$ \text{Mass} = \text{Moles} \times \text{Molar Mass} $$

$$ \text{Mass} = 10 \times 28 = 280 \text{ g} $$

The amount of nitrogen gas evolved is 280 g.

Step 1: Identify the Group Reagent for P

Reagent: $H_2S$ in the presence of $NH_4OH$. This corresponds to Group IV cation analysis.

Cations precipitated: $Zn^{2+}, Mn^{2+}, Ni^{2+}, Co^{2+}$ as sulphides.

Match: (3) $Mn^{2+}$. So, P $\rightarrow$ 3.

Step 2: Identify the Group Reagent for Q

Reagent: $(NH_4)_2CO_3$ in the presence of $NH_4OH$. This corresponds to Group V cation analysis.

Cations precipitated: $Ca^{2+}, Sr^{2+}, Ba^{2+}$ as carbonates.

Match: (4) $Ba^{2+}$. So, Q $\rightarrow$ 4.

Step 3: Identify the Group Reagent for R

Reagent: $NH_4OH$ in the presence of $NH_4Cl$. This corresponds to Group III cation analysis.

Cations precipitated: $Fe^{3+}, Al^{3+}, Cr^{3+}$ as hydroxides.

Match: (2) $Al^{3+}$. So, R $\rightarrow$ 2.

Step 4: Identify the Group Reagent for S

Reagent: Passing $H_2S$ in the presence of dilute $HCl$. This corresponds to Group II cation analysis.

Cations precipitated: $Cu^{2+}, Pb^{2+}, Hg^{2+}, Cd^{2+}$, etc., as sulphides.

Match: (1) $Cu^{2+}$. So, S $\rightarrow$ 1.

Conclusion: The correct match is P$\rightarrow$3, Q$\rightarrow$4, R$\rightarrow$2, S$\rightarrow$1. This corresponds to Option (A).

Step 1: Analyze Compound P

Match: (2). So, P $\rightarrow$ 2.

Step 2: Analyze Compound R

Match: (1) "Reaction with phenyl diazonium salt gives yellow dye". So, R $\rightarrow$ 1.

Step 3: Analyze Compound S

Match: (3) "Reaction with glucose will give corresponding hydrazone". So, S $\rightarrow$ 3.

Step 4: Analyze Compound Q

Match: (5). So, Q $\rightarrow$ 5.

Conclusion: The correct match is P$\rightarrow$2, Q$\rightarrow$5, R$\rightarrow$1, S$\rightarrow$3. This corresponds to Option (B).