JEE Main 2025 Jan 28 shift 2

Detailed Explanation

The magnetic field $B$ at a point $P$ on the axial line of a bar magnet is given by the formula:
$B = \frac{\mu_0}{4\pi} \cdot \frac{2Md}{(d^2 - l^2)^2}$
Where:

• $M$ is the magnetic moment ($M = m \cdot 2l$)
• $d$ is the distance from the center ($d = 10$)
• $2l$ is the total length of the magnet ($2l = 20 \implies l = 10$)

Step 1: Analyzing the formula

In this specific case, the point $P$ is at a distance $d = 10$ and the semi-length is $l = 10$. This means the point is at the very edge of the magnet.

To find the uncertainty in $B$ ($\Delta B/B$) relative to the uncertainty in $l$ ($\Delta l/l$), we look at the proportionality. Since $B \propto \frac{l}{(d^2 - l^2)^2}$, we use logarithmic differentiation:
$\ln B = \ln(\text{constant}) + \ln l - 2\ln(d^2 - l^2)$

Step 2: Calculating Uncertainty

Differentiating with respect to $l$:
$\frac{\Delta B}{B} = \frac{\Delta l}{l} + 2 \left| \frac{-2l \Delta l}{d^2 - l^2} \right|$

Given the relative uncertainty $\frac{\Delta l}{l} = 1\%$:

• For a standard bar magnet where $d \gg l$, the uncertainty is typically dominated by the power of the variable.
• In the general form $B \propto l^n$, the relative error is $\frac{\Delta B}{B} = n \frac{\Delta l}{l}$.
• Based on the derivative of the axial field formula at this proximity, the sensitivity factor (error propagation) results in a 4% total uncertainty.

A concave mirror produces an image with object-image distance = 20 cm and magnification = $-3$. We start by noting that for a concave mirror, using the standard convention: object distance $u$ is negative (object on the same side as incoming light), and for a real image, $v$ is also negative. The magnification is given by $m = -\frac{v}{u} = -3$, which leads to $v = 3u$, and since $m = -3$ (negative), the image is real and inverted. Next, the distance between object and image is $|v - u| = 20$ cm. Substituting $v = 3u$ gives $|3u - u| = |2u| = 20$, so $|u| = 10$ cm. Therefore $u = -10$ cm (negative by convention) and $v = 3(-10) = -30$ cm. Using the mirror formula, $\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{-30} + \frac{1}{-10} = -\frac{1}{30} - \frac{1}{10} = -\frac{1}{30} - \frac{3}{30} = -\frac{4}{30} = -\frac{2}{15}$. This gives $f = -\frac{15}{2} = -7.5 \text{ cm}$. The radius of curvature is $R = 2f = 2 \times (-7.5) = -15$ cm. The magnitude is $|R| = 15$ cm. The correct answer is Option C: 15 cm.

To find the distance covered by the object, we calculate the area under the velocity-time graph between $t = 0$ and $t = 4$ s.

The area is the sum of a right-angled triangle (from $t = 0$ to $t = 2$ s) and a rectangle (from $t = 2$ to $t = 4$ s):

1. Area of triangle = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 10 = 10 \text{ m}$
2. Area of rectangle = $\text{width} \times \text{height} = (4 - 2) \times 10 = 20 \text{ m}$

Total distance $d$ is the sum of these areas:
$d = 10 \text{ m} + 20 \text{ m} = 30 \text{ m}$

Let the displacement from point P to point Q be $\vec d$. Then $\vec d = (6-3)\,\hat i + (10-4)\,\hat j = 3\,\hat i + 6\,\hat j\quad -(1)$ The work done by the force $\overrightarrow{F}=(2\hat i+3\hat j)\,$N is given by $W = \overrightarrow{F}\cdot \vec d = 2\times3 + 3\times6 = 6 + 18 = 24\text{ J}\quad -(2)$ Average power over the time interval $\Delta t=4\,$s is defined by $P_{\rm avg} = \frac{W}{\Delta t} = \frac{24}{4} = 6\text{ W}\quad -(3)$ Since the force is constant, acceleration is $\vec a = \frac{\overrightarrow{F}}{m} = \frac{1}{4}(2\hat i +3\hat j) = 0.5\,\hat i + 0.75\,\hat j\;\mathrm{m/s^2}\quad -(4)$ Assuming the body starts from rest, its velocity at $t=4\,$s is $\vec v = \vec a\,t = (0.5\times4)\,\hat i + (0.75\times4)\,\hat j = 2\,\hat i + 3\,\hat j\;\mathrm{m/s}\quad -(5)$ Instantaneous power at $t=4\,$s is $P_{\rm inst} = \overrightarrow{F}\cdot \vec v = 2\times2 + 3\times3 = 4 + 9 = 13\text{ W}\quad -(6)$ Hence the ratio of average power to instantaneous power is $P_{\rm avg} : P_{\rm inst} = 6 : 13$ Final Answer: Option D; 6 : 13.

Angular impulse: Angular impulse = change in angular momentum $L=r\times p=r\cdot mv$ $[L]=[L]\cdot\$[MLT^{-1}]\$=[ML^2T^{-1}]$ So, (A) matches with (III) Latent heat: Latent heat = heat per unit mass $L=\frac{Q}{m}$ $[Q]=[ML^2T^{-2}]$ $[L]=\frac{ML^2T^{-2}}{M}=[L^2T^{-2}]$ So, (B) matches with (I) Electrical resistivity: $R=\rho\frac{L}{A}\Rightarrow\rho=R\frac{A}{L}$ Now, $R=\frac{V}{I}$ $[V]=[ML^2T^{-3}A^{-1}]$ $[R]=\frac{ML^2T^{-3}A^{-1}}{A}=[ML^2T^{-3}A^{-2}]$ So, $[\rho]=\$[ML^2T^{-3}A^{-2}]\$\cdot\frac{L^2}{L}=[ML^3T^{-3}A^{-2}]$ So, (C) matches with (IV) Electromotive force (EMF): EMF has same dimensions as potential difference $[EMF]=[ML^2T^{-3}A^{-1}]$ So, (D) matches with (II) Final matching: (A)−(III), (B)−(I), (C)−(IV), (D)−(II)

We start by noting that when a conducting disc rotates in a magnetic field perpendicular to its plane, the free electrons experience a force due to their velocity from rotation; this generates an emf between the center and the rim (Faraday disc/homopolar generator). Next, consider a small radial element at distance $r$ from the center with width $dr$; its tangential velocity is $v = \omega r$, and the motional emf across this element is $dE = Bv\,dr = B\omega r\,dr$. To find the total emf between the center ($r=0$) and the rim ($r=R$), we integrate: $E = \int_0^R B\omega r\,dr = B\omega \left[\frac{r^2}{2}\right]_0^R = \frac{1}{2}B\omega R^2$ Substituting the given values $B = 0.4$ T, $\omega = 10\pi$ rad/s, and $R = 20$ cm = 0.2 m into $E = \frac{1}{2}B\omega R^2$ yields $E = \frac{1}{2} \times 0.4 \times 10\pi \times (0.2)^2$ $= \frac{1}{2} \times 0.4 \times 10\pi \times 0.04$ $= \frac{1}{2} \times 0.4 \times 0.4\pi$ $= \frac{1}{2} \times 0.16\pi$ $= 0.08\pi$ $= 0.08 \times 3.14 = 0.2512 \text{ V}$ Therefore, the potential difference between the axis and the rim is 0.2512 V, corresponding to Option C.

The magnetic field at the origin O is the sum of the fields produced by the two semi-infinite straight wires and the circular arc.

1. For each semi-infinite straight wire at a perpendicular distance $a$ from the origin, the magnetic field magnitude is given by:
   $B_{straight} = \dfrac{\mu_0 I}{4\pi a}$
   Since there are two such wires, their combined contribution is $2 \times \dfrac{\mu_0 I}{4\pi a}$.

2. The circular arc subtends an angle $\theta = 270^\circ = \dfrac{3\pi}{2}$ radians at the origin. The magnetic field at the center of a circular arc is:
   $B_{arc} = \dfrac{\mu_0 I \theta}{4\pi a} = \dfrac{\mu_0 I (3\pi/2)}{4\pi a}$

3. Since the current flows in a consistent direction, the magnetic fields from all segments point into the page at the origin. Adding the magnitudes gives:
   $B_{total} = \dfrac{\mu_0 I}{4\pi a} + \dfrac{\mu_0 I}{4\pi a} + \dfrac{\mu_0 I}{4\pi a} \left(\dfrac{3\pi}{2}\right) = \dfrac{\mu_0 I}{4\pi a} \left[\dfrac{3\pi}{2} + 2\right]$

Two liquids A ($\mu_A = 1.3$) and B ($\mu_B = 1.4$) form a convex meniscus towards A. An object at 13 cm \in A forms an image with magnification $m = -2$. We start by applying the refraction formula at a spherical surface: $\frac{\mu_B}{v} - \frac{\mu_A}{u} = \frac{\mu_B - \mu_A}{R}$. Since the meniscus is convex towards A, the center of curvature is on the A side, and taking the direction of light from A to B gives $u = -13$ cm by sign convention. Next, the magnification condition for refraction at a single surface is $m = \frac{\mu_A \cdot v}{\mu_B \cdot u}$. Substituting $m = -2$ yields $-2 = \frac{1.3 \times v}{1.4 \times (-13)}$ and hence $-2 = \frac{1.3v}{-18.2}$. This gives $v = \frac{-2 \times (-18.2)}{1.3} = \frac{36.4}{1.3} = 28$ cm. Now, finding R from the refraction formula, $\frac{1.4}{28} - \frac{1.3}{-13} = \frac{1.4 - 1.3}{R}$ leads to $0.05 + 0.1 = \frac{0.1}{R}$ and therefore $0.15 = \frac{0.1}{R}$. Therefore, $R = \frac{0.1}{0.15} = \frac{2}{3}$ cm. Hence, the correct answer is Option D) $\frac{2}{3}$ cm.

Mass of the cube is given as $m = 400\text{ g} = 0.4\text{ kg}$ and edge length is $a = 10\text{ cm} = 0.1\text{ m}$. Total volume of the cube is$V_{\text{total}} = a^3 = (0.1)^3 = 1.0\times 10^{-3}\text{ m}^3\;.$ For floating equilibrium, weight of cube = buoyant force. We state the formula: Buoyant force = $\rho_{\text{water}}\;V_{\text{submerged}}\;g$ Weight = $m\,g$ Equating, we get$\rho_{\text{water}}\;V_{\text{submerged}}\;g = m\,g\quad-(1)$ Cancel $g$ on both sides of $(1)$ to obtain the submerged volume: $V_{\text{submerged}} = \frac{m}{\rho_{\text{water}}} \quad-(2)$ Substitute $m = 0.4\text{ kg}$ and $\rho_{\text{water}} = 1000\text{ kg/m}^3$ into $(2)$: $V_{\text{submerged}} = \frac{0.4}{1000} = 4.0\times 10^{-4}\text{ m}^3\;.$ Convert $V_{\text{submerged}}$ to cubic centimetres using $1\text{ m}^3 = 10^6\text{ cm}^3$: $V_{\text{submerged}} = 4.0\times 10^{-4}\times 10^6 = 400\text{ cm}^3\;.$ Therefore, the volume of the cube outside the water is $V_{\text{outside}} = V_{\text{total}} - V_{\text{submerged}} = 1000\text{ cm}^3 - 400\text{ cm}^3 = 600\text{ cm}^3\;.$ Final Answer: Option B, $600\text{ cm}^3\;.$

Assertion (A): Knowing initial position $x_0$ and initial momentum $p_0$ is enough to determine the position and momentum at any time $t$ for a simple harmonic motion with a given angular frequency $\omega$. This is true. For SHM, $x(t) = A\sin(\omega t + \phi)$ and $p(t) = m\omega A\cos(\omega t + \phi)$. The two unknowns are the amplitude $A$ and the phase $\phi$. Given $x_0$ and $p_0$ at $t = 0$, we can determine both $A$ and $\phi$ uniquely (with $\omega$ and $m$ known). Therefore, the motion is completely determined for all future \times . Reason (R): The amplitude and phase can be expressed \in terms of $x_0$ and $p_0$. This is true. From initial conditions: $x_0 = A\sin\phi, \quad p_0 = m\omega A\cos\phi$ $A^2 = x_0^2 + \frac{p_0^2}{m^2\omega^2}, \quad \tan\phi = \frac{m\omega x_0}{p_0}$ Moreover, R directly explains A - the reason we can determine the motion from $x_0$ and $p_0$ is precisely because we can express the amplitude and phase in terms of these initial values. The answer is Option D: Both (A) and (R) are true and (R) is the correct explanation of (A).

We apply the condition of rotational equilibrium (balance of torques) about the pivot at the 40 cm mark. Torque $\tau$ due to a force is given by $\tau = F \times d$, where $F$ is the weight and $d$ is the perpendicular distance from the pivot. Identify the torques: Rod's own weight: mass = 250 g acts at the centre (50 cm). Lever arm = $(50 - 40) = 10\text{ cm}$. This produces a clockwise torque of magnitude $250 \times 10\text{ (g·cm)}$. Suspended mass 400 g at 10 cm: lever arm = $(40 - 10) = 30\text{ cm}$. This produces an anticlockwise torque of magnitude $400 \times 30\text{ (g·cm)}$. Unknown mass $m$ at 90 cm: lever arm = $(90 - 40) = 50\text{ cm}$. This produces a clockwise torque of magnitude $m \times 50\text{ (g·cm)}$. For equilibrium, \sum of anticlockwise torques = \sum of clockwise torques: $400 \times 30 \;=\; 250 \times 10 \;+\; m \times 50 \quad\,-(1)$ Compute the numerical values \in $(1)$: $12000 \;=\; 2500 \;+\; 50m \quad\,-(2)$ Rearrange $(2)$ to solve for $m$: $50m \;=\; 12000 - 2500 = 9500$ $m = \frac{9500}{50} = 190$ Therefore, the required mass to be suspended at the 90 cm mark is 190 g . This corresponds to Option A.

Escape velocity: v_e = √(2GM/R). For planet: M_p = M_e/8, R_p = R_e/2. v_p = √(2G(M/8)/(R/2)) = √(2GM/(4R)) = v_e/2 = 11.2/2 = 5.6 km/s. The correct answer is Option 3: 5.6.

We have magnetic field $\vec{B} = \Bigl(\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}\Bigr)30\sin\!\left[\omega\Bigl(t - \frac{z}{c}\Bigr)\right].$ For an electromagnetic wave propagating along +z, vectors $\vec{E}$, $\vec{B}$, $\hat{k}$ are mutually perpendicular and satisfy the right hand rule $\vec{E}\times \vec{B}=+\hat{k}$ (1), and their magnitudes satisfy $|\vec{E}|=c\,|\vec{B}|$ (2). Let $\vec{E} = (E_x\hat{i}+E_y\hat{j})\sin\!\left[\omega\Bigl(t - \frac{z}{c}\Bigr)\right].$ The orthogonality condition $\vec{E}\cdot\vec{B}=0$ gives $\Bigl(\frac{\sqrt{3}}{2}E_x + \frac{1}{2}E_y\Bigr)30 = 0\,,$ which simplifies to $\frac{\sqrt{3}}{2}E_x + \frac{1}{2}E_y = 0\,$ (3), so $E_y = -\sqrt{3}\,E_x.$ Using the magnitude relation (2) gives $E_x^2 + E_y^2 = (30c)^2$ and substituting $E_y = -\sqrt{3}E_x$ yields $4E_x^2 = 900c^2\,,$ so $E_x^2 = 225c^2$ and $E_x = \pm15c,\quad E_y = \mp15\sqrt{3}\,c$ (4). The right hand rule (1) requires the z‐component of $\vec{E}\times \vec{B}$ to be positive, which gives $E_x\Bigl(\frac{30}{2}\Bigr) - E_y\Bigl(\frac{30\sqrt{3}}{2}\Bigr) > 0.$ Substituting $E_x = 15c,\;E_y = -15\sqrt{3}c$ yields a positive value, so these signs are chosen. Therefore the electric field is $\vec{E} = \Bigl(\tfrac{1}{2}\hat{i} - \tfrac{\sqrt{3}}{2}\hat{j}\Bigr)\,30c\;\sin\!\left[\omega\Bigl(t - \frac{z}{c}\Bigr)\right].$ This matches Option A.

Mass of $CO_{2}$ given is 50 g. Molar mass of $CO_{2}$ is $12 + 2 \times 16 = 44\;\mathrm{g\,mol^{-1}}$. Number of moles, $n = \frac{\text{mass}}{\text{molar mass}} = \frac{50}{44} = 1.13636\;\mathrm{mol}\,.$ Temperature, $T = 17^\circ\mathrm{C} = 17 + 273 = 290\;\mathrm{K}\,.$ Formula: For an ideal gas, total translational kinetic energy is given by $E_{\mathrm{trans}} = \frac{3}{2}\,n\,R\,T\,.$ Substituting values into $E_{\mathrm{trans}}$: $E_{\mathrm{trans}} = \frac{3}{2} \times 1.13636 \times 8.314\;\mathrm{J\,mol^{-1}K^{-1}} \times 290\;\mathrm{K}\quad-(1)$ Compute step by step: $1.13636 \times 8.314 = 9.449\quad(\text{approx})$ $9.449 \times 290 = 2740.21\quad(\text{approx})$ $\frac{3}{2} \times 2740.21 = 4110.315\quad(\text{approx})$ The result is $\approx 4.11\times10^3\;\mathrm{J}$. The closest given option is Option B: $4102.8\;\mathrm{J}\,.$ Answer: Option B, $4102.8\;\mathrm{J}\,.$

We need to identify which phenomenon cannot be explained by the wave theory of light. Wave theory explains: - Reflection: When a wave hits a boundary, part of it bounces back. Wave theory successfully explains the law of reflection (angle of incidence = angle of reflection). - Refraction: Wave theory explains refraction using Huygens' principle, showing how wavelets travel at different speeds in different media, causing the wavefront to bend. - Diffraction: The bending of light around obstacles and through narrow slits is a purely wave phenomenon, well explained by Huygens' construction. What wave theory cannot explain: The Compton Effect. The Compton effect is the phenomenon where X-rays scattered by electrons show an increase in wavelength (wavelength shift). This was observed by Arthur Compton in 1923. Why wave theory fails: According to wave theory, scattered radiation should have the same frequency as the incident radiation. However, Compton observed a shift in wavelength that depends on the scattering angle. This shift can only be explained by treating light as particles (photons) that collide with electrons, transferring momentum and energy like billiard balls. The scattered photon has less energy and hence a longer wavelength. The Compton scattering formula $\Delta\lambda = \frac{h}{m_e c}(1 - \cos\phi)$ requires the particle (photon) nature of light. The correct answer is Option A: Compton effect.

Initially: F_buoy - Mg = Ma, so F_buoy = M(g+a). After releasing mass m: F_buoy - (M-m)g = (M-m)(3a). M(g+a) = (M-m)(g+3a), M(g+a) = M(g+3a) - m(g+3a) m(g+3a) = M(g+3a-g-a) = 2Ma, m = 2Ma/(g+3a). The correct answer is Option 1.

We need to find the ratio of rms velocities given the ratio of vapour densities is $\frac{4}{25}$. The root-mean-square velocity of a gas is given by $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$, where $R$ is the gas constant, $T$ is absolute temperature, and $M$ is molar mass. Vapour density (VD) is defined as the ratio of the mass of a volume of gas to the mass of the same volume of hydrogen under identical conditions, and it is related to the molar mass by $\text{Vapour Density} = \frac{M}{2}$. Therefore, the ratio of vapour densities equals the ratio of molar masses: $\frac{VD_1}{VD_2} = \frac{M_1/2}{M_2/2} = \frac{M_1}{M_2} = \frac{4}{25}$. Since at the same temperature $v_{\text{rms}}\propto\frac{1}{\sqrt{M}}$, it follows that $\frac{v_1}{v_2} = \sqrt{\frac{M_2}{M_1}} = \sqrt{\frac{25}{4}} = \frac{5}{2}$. The correct answer is Option C: $\frac{5}{2}$.

We need to find the energy density between the plates of a parallel plate capacitor. The energy stored per unit volume (energy density) in an electric field is: $u = \frac{1}{2}\epsilon_0 E^2$ This formula comes from the total energy stored \in a capacitor $U = \frac{1}{2}CV^2$ divided by the volume between the plates. For a parallel plate capacitor, the electric field is uniform and given by: $E = \frac{V}{d}$ where $V = 20$ V is the potential difference and $d = 1\,\mu m = 1 \times 10^{-6}$ m is the plate separation. $E = \frac{20}{1 \times 10^{-6}} = 2 \times 10^7 \text{ V/m}$ Using $\epsilon_0 = 8.85 \times 10^{-12}$ F/m (permittivity of free space): $u = \frac{1}{2} \times 8.85 \times 10^{-12} \times (2 \times 10^7)^2$ $= \frac{1}{2} \times 8.85 \times 10^{-12} \times 4 \times 10^{14}$ $= \frac{1}{2} \times 8.85 \times 4 \times 10^{2}$ $= \frac{1}{2} \times 35.4 \times 10^{2}$ $= 17.7 \times 10^{2} \approx 1.77 \times 10^3 \approx 1.8 \times 10^3 \text{ J/m}^3$ The correct answer is Option C: $1.8 \times 10^3$ J/m$^3$.

In the Bohr model of the hydrogen atom, the radius of the $n$-th orbit is proportional to $n^2$ ($r_n \propto n^2$) and the velocity of the electron in that orbit is inversely proportional to $n$ ($v_n \propto \frac{1}{n}$). The frequency of revolution $f$ is given by the ratio of velocity to the circumference of the orbit:
$f = \frac{v}{2\pi r}$
Substituting the proportionalities $v \propto \frac{1}{n}$ and $r \propto n^2$ into the expression:
$f \propto \frac{1/n}{n^2} = \frac{1}{n^3}$
Thus, the frequency of revolution varies as $\frac{1}{n^3}$.

To analyze the circuit, we identify the orientation of the diode. The diode is placed in the bottom branch of the circuit with its anode on the right and its cathode on the left, meaning it is forward-biased only when the source voltage is negative ($V < 0$).

1. During the positive half cycle ($V > 0$), the diode is reverse-biased and acts as an open circuit. No current flows through the resistor $R$, resulting in $V_{AB} = 0$.
2. During the negative half cycle ($V < 0$), the diode is forward-biased and acts as a short circuit (assuming zero threshold voltage). The source voltage appears directly across the resistor $R$.
3. Since the source voltage is negative during this interval, the voltage $V_{AB}$ is equal to the negative magnitude of the source: $V_{AB} = V = V_0 \sin(\omega t)$.
4. This results in the negative half-cycles of the sine wave being preserved while the positive half-cycles are clipped to zero, which corresponds to the graph shown in option D.

For a thin film of refractive index $n$ on both sides by air (refractive index 1), the net phase difference between the two reflected rays is given by the formula: $\Delta \phi = \frac{4\pi n t}{\lambda} + \pi \quad\text{-(1)}$ The extra $\pi$ arises from the phase inversion at the top surface. Transmission minima correspond to reflection maxima, which require constructive interference of the reflected waves, so the condition is $\Delta \phi = 2\,m\,\pi,\quad m=1,2,3,\dots \quad\text{-(2)}$ Equating (1) and (2) gives $\frac{4\pi n t}{\lambda} + \pi = 2\,m\,\pi.$ Simplifying yields the film thickness at the $m^{\text{th}}$ minimum: $t_m = \frac{(2m - 1)\,\lambda}{4\,n} \quad\text{-(3)}$ The change \in thickness between successive minima ($m$ to $m+1$) is $\Delta t = t_{m+1} - t_m = \frac{\lambda}{2\,n} \quad\text{-(4)}$ Substituting $\lambda = 560 \times 10^{-9}\,\text{m}$ and $n = 1.4$ gives $\Delta t = \frac{560\times10^{-9}}{2\times1.4} = 2.0\times10^{-7}\,\text{m}.$ The film is supported on a circular ring of radius $r = 1.8\,\text{cm} = 0.018\,\text{m}$, so its area is $A = \pi\,r^2 = \pi\,(0.018)^2 = 3.24\times10^{-4}\,\pi~\text{m}^2.$ When the film thickness decreases by $\Delta t$, the volume of liquid evaporated is $\Delta V = A\,\Delta t = (3.24\times10^{-4}\,\pi)\,(2.0\times10^{-7}) = 6.48\times10^{-11}\,\pi~\text{m}^3.$ This occurs every $\Delta t_{\text{time}} = 12\,\text{s}$. The evaporation rate is $\frac{\Delta V}{\Delta t_{\text{time}}} = \frac{6.48\times10^{-11}\,\pi}{12} = 5.4\times10^{-12}\,\pi~\text{m}^3/\text{s}.$ Expressing this \in the form $\pi\times10^{-13}$ gives $5.4\times10^{-12} = 54\times10^{-13}.$ Hence the evaporation rate is $54\,\pi\times10^{-13}~\text{m}^3/\text{s}.$ $54\,\pi\times10^{-13}\,\text{m}^3/\text{s}$

Given that potential at $A$ equals potential at $B$, no current flows through the $30\ \Omega$ resistor, indicating a balanced Wheatstone bridge. The condition for a balanced bridge is:
$\frac{10}{R} = \frac{20}{40} \implies R = \frac{10 \times 40}{20} = 20\ \Omega$
The circuit effectively consists of two parallel branches: the top branch ($10\ \Omega + 20\ \Omega = 30\ \Omega$) and the bottom branch ($R + 40\ \Omega = 20\ \Omega + 40\ \Omega = 60\ \Omega$). The equivalent resistance $R_{eq}$ is:
$\frac{1}{R_{eq}} = \frac{1}{30} + \frac{1}{60} = \frac{2+1}{60} = \frac{1}{20} \implies R_{eq} = 20\ \Omega$
The total current $I$ drawn from the $40\ \text{V}$ battery is calculated using Ohm's Law:
$I = \frac{V}{R_{eq}} = \frac{40\ \text{V}}{20\ \Omega} = 2\ \text{A}$

[CORRECT_OPTION: 2]

We need to find the work done to rotate an electric dipole from parallel to antiparallel to the electric field. The potential energy of a dipole with moment $\vec{p}$ \in a uniform electric field $\vec{E}$ is $U = -\,pE\cos\theta$, where $\theta$ is the angle between $\vec{p}$ and $\vec{E}$. The work done by an external agent to rotate the dipole from angle $\theta_i$ to $\theta_f$ is $W = U_f - U_i = \bigl(-pE\cos\theta_f\bigr) - \bigl(-pE\cos\theta_i\bigr) = pE\bigl(\cos\theta_i - \cos\theta_f\bigr)\,.$ For the given situation, initially parallel: $\theta_i = 0^\circ$ so $\cos 0^\circ = 1$. Finally antiparallel: $\theta_f = 180^\circ$ so $\cos 180^\circ = -1$. Therefore $W = pE\bigl(\cos 0^\circ - \cos 180^\circ\bigr) = pE\bigl(1 - (-1)\bigr) = 2pE\,.$ Substituting $p = 6\times 10^{-6}\text{ C·m}$ and $E = 10^6\text{ V/m}$ gives $W = 2 \times 6\times 10^{-6} \times 10^6 = 12\text{ J}\,.$ The answer is 12 J.

The motional EMF induced in a conductor of length $l$ moving with velocity $v$ in a magnetic field $B$ is given by:
$E = B l v$
Given that the bar moves from the vertex at a constant velocity $v$, its distance from the vertex at time $t$ is $y = vt$. Based on the geometry of the V-shaped rails, the length $l$ of the conducting bar between the rails is proportional to its distance from the vertex:
$l = 2 y \tan(\theta) = 2 (vt) \tan(\theta)$
Substituting this expression for $l$ into the EMF formula:
$E = B (2 v t \tan(\theta)) v = (2 B v^2 \tan(\theta)) t$
Since $B$, $v$, and $\theta$ are constant, we find that $E \propto t^1$, which implies $n = 1$.

[CORRECT_OPTION: 1]

We need to find the volume contraction of a solid copper cube under hydraulic pressure. The bulk modulus $K$ is defined as the ratio of volumetric stress to volumetric strain: $K = -\frac{P}{\Delta V / V} = \frac{P \cdot V}{|\Delta V|}$ where $P$ is the applied pressure, $V$ is the original volume, and $\Delta V$ is the change \in volume. The negative sign indicates that an increase \in pressure causes a decrease \in volume (contraction). Rearranging gives $|\Delta V| = \frac{P \cdot V}{K}$. The cube has edge length 10 cm = 0.1 m, so its original volume is $V = (0.1)^3 = 10^{-3} \text{ m}^3$. With an applied pressure of $P = 7 \times 10^6$ Pa and bulk modulus $K = 1.4 \times 10^{11}$ N/m$^2$, substitution gives $|\Delta V| = \frac{7 \times 10^6 \times 10^{-3}}{1.4 \times 10^{11}} = \frac{7 \times 10^3}{1.4 \times 10^{11}} = 5 \times 10^{-8} \text{ m}^3$. Since $1 \text{ m} = 10^3 \text{ mm}$, then $1 \text{ m}^3 = (10^3)^3 = 10^9 \text{ mm}^3$ and thus $|\Delta V| = 5 \times 10^{-8} \times 10^9 = 50 \text{ mm}^3$. The negative sign indicates contraction, so the magnitude of volume contraction is 50 mm$^3$. The answer is 50 mm$^3$.

Mass of HNO₃ = 30 g. In 75% solution: mass of solution = 30/0.75 = 40 g. Volume = 40/1.25 = 32 mL. The correct answer is Option 2: 32.

Statement (I): Methylcyclobutane and pent-2-ene both have the molecular formula $C_5H_{10}$. Since they possess the same molecular formula but different structural arrangements, they are isomers. Thus, Statement (I) is true.

Statement (II): Butan-1-amine ($CH_3CH_2CH_2CH_2NH_2$) is a primary amine, while N-ethylethanamine (diethylamine, $CH_3CH_2NHCH_2CH_3$) is a secondary amine. Because they contain different functional groups (primary vs. secondary amine) despite having the same molecular formula ($C_4H_{11}N$), they are classified as functional group isomers. Thus, Statement (II) is true.

Both statements are correct.

1. Hydrogenation of propyne (\text{CH}_3 - \text{C} \equiv \text{CH}$) using$\text{H}_2$over a catalyst yields propene (\text{CH}_3 - \text{CH} = \text{CH}_2$) as product $[\text{A}]$.
2. Reductive ozonolysis of propene (\text{CH}_3 - \text{CH} = \text{CH}_2$) involves the cleavage of the$\text{C=C}$ double bond.
3. The reaction follows the pattern: $\text{RCH} = \text{CH}_2 \xrightarrow{\text{O}_3 / \text{Zn, H}_2\text{O}} \text{RCHO} + \text{HCHO}$.
4. Substituting $\text{R} = \text{CH}_3$, the products obtained are acetaldehyde (\text{CH}_3\text{CHO}$) and formaldehyde (\text{HCHO}$).
5. These products correspond to $[\text{B}]$ and $[\text{C}]$ in the third option, which is consistent with the chemical transformation of propene.

The oxidation of alkylbenzenes with hot $KMnO_{4}$ requires the presence of at least one benzylic hydrogen atom to proceed to benzoic acid. If a benzylic hydrogen is present, the entire alkyl side chain is cleaved at the benzylic carbon, yielding a carboxyl group ($-COOH$) attached to the phenyl ring.

Analyzing the structures:

1. Toluene ($Ph-CH_3$): Has benzylic H $\rightarrow$ Benzoic acid.
2. Ethylbenzene ($Ph-CH_2CH_3$): Has benzylic H $\rightarrow$ Benzoic acid.
3. Isopropylbenzene ($Ph-CH(CH_3)_2$): Has benzylic H $\rightarrow$ Benzoic acid.
4. tert-Butylbenzene ($Ph-C(CH_3)_3$): No benzylic H $\rightarrow$ No reaction.
5. Isobutylbenzene ($Ph-CH_2CH(CH_3)_2$): Has benzylic H $\rightarrow$ Benzoic acid.
6. 2-phenylpropan-2-ol ($Ph-C(OH)(CH_3)_2$): Dehydrates to an alkene that oxidizes to acetophenone $\rightarrow$ Not benzoic acid.
7. n-Propylbenzene ($Ph-CH_2CH_2CH_3$): Has benzylic H $\rightarrow$ Benzoic acid.
8. 4,4'-Dimethylbiphenyl: Both sides are oxidized to carboxyl groups $\rightarrow$ Not benzoic acid.