JEE Main 2025 Jan 23 shift 1

AtPP, force on charge Q is zero if net electric field is zero. Field due to dipole 1 (axial point): $E_1=\frac{1}{4\pi\epsilon_0}\frac{2pr}{(r^{2-}a^2)^2}$ where p=2aq Direction is leftward (toward −q). Field due to dipole 2 (equatorial point): $E_2=\frac{1}{4\pi\epsilon_0}\frac{p}{(r^{2+}a^2)^{3/2}}$ Direction is rightward (opposite to dipole moment). For zero net force, $E_1=E_2$ So, $\frac{2r}{(r^{2-}a^2)^2}=\frac{1}{(r^{2+}a^2)^{3/2}}$ Let $x=\frac{a}{r}$ Then $r^{2-}a^2=r^2(1-x^2)$ $r^{2+}a^2=r^2(1+x^2)$ Substitute: $\frac{2}{(1-x^2)^2}=\frac{1}{(1+x^2)^{3/2}}$ Solving, $x=\sqrt{5}-2$

Let the spherical surface have its vertex at point $V$ on the optic axis, with the centre of curvature $C$ lying inside the glass medium at a distance $R$ from $V$. Choose the Cartesian sign convention: light travels from left (air) to right (glass). Hence, distances measured to the right of $V$ are taken as positive. Given data \bullet Refractive index \in air, $n_1 = 1$ \bullet Refractive index \in glass, $n_2 = 1.5$ \bullet Radius of curvature, $R$, is positive (centre is on the image side). The object $O$ is \in air (left side), so its distance from the vertex is $u = -PO$ (negative). The real image $I$ is \in glass (right side), so its distance from the vertex is $v = PI$ (positive). It is given that $PO = PI$, therefore the magnitudes of the object and image distances are equal: $|u| = v \quad\Longrightarrow\quad v = -u$ $-(1)$ The refraction at a spherical surface is governed by the formula $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$ $-(2)$ Substitute $n_1 = 1$, $n_2 = 1.5$ and use $(1)$, i.e. $u = -v$: $\frac{1.5}{v} - \frac{1}{-v} = \frac{1.5 - 1}{R}$ Simplify the left-hand side: $\frac{1.5}{v} + \frac{1}{v} = \frac{2.5}{v}$ Thus $\frac{2.5}{v} = \frac{0.5}{R}$ Solving for $v$: $v = \frac{2.5}{0.5}\,R = 5R$ The distance asked \in the problem is $PO$, and $PO = v = 5R$. Therefore, the correct option is Option A: $5R$.

To find the dimension of $\frac{ABC}{D}$, observe that $x(t) = A\sin t + B\cos^2 t + Ct^2 + D$ represents a position, so each term on the right must have dimension of length $[L]$. Since $\sin t$ and $\cos^2 t$ are dimensionless and $[t^2] = [T^2]$, we have $[A\sin t] = [L]$ implying $[A] = [L]$, $[B\cos^2 t] = [L]$ implying $[B] = [L]$, and $[Ct^2] = [L]$ so $[C]\$[T^2]\$ = [L]$ and thus $[C] = [LT^{-2}]$. The constant $D$ must satisfy $[D] = [L].$ Therefore, $\left[\frac{ABC}{D}\right] = \frac{[L][L]\$[LT^{-2}]\$}{[L]} = \frac{L^3 T^{-2}}{L} = L^2 T^{-2}$ The correct answer is Option (1): $L^2T^{-2}$.

A thin convex lens with refractive index $\mu_2$ is immersed \in a liquid of refractive index $\mu_1$ ($\mu_1 < \mu_2$), and its second surface is silvered, effectively forming a combination of a lens and a mirror. We seek the object distance for which the image coincides with the object position. In this silvered lens system light passes through the lens, reflects off the silvered surface, and passes through the lens again, so the system acts as an equivalent mirror whose power is given by $P_{eq} = 2P_{lens} + P_{mirror}$. For the thin lens \in the surrounding medium, the lens maker's formula yields $\frac{1}{f_{lens}} = \left(\frac{\mu_2}{\mu_1} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \left(\frac{\mu_2}{\mu_1} - 1\right)\left(\frac{1}{|R_1|} + \frac{1}{|R_2|}\right) = \frac{(\mu_2 - \mu_1)}{\mu_1}\cdot\frac{|R_1| + |R_2|}{|R_1|\cdot |R_2|}.$ The silvered surface behaves as a concave mirror of radius $|R_2|$, so $\frac{1}{f_{mirror}} = \frac{2}{|R_2|}.$ Therefore, the equivalent focal length satisfies $\frac{1}{f_{eq}} = \frac{2}{f_{lens}} + \frac{1}{f_{mirror}} = \frac{2\mu_2(|R_1| + |R_2|) - 2\mu_1|R_2|}{\mu_1|R_1|\cdot |R_2|}$ For the image to coincide with the object, the object must be placed at the center of curvature of the equivalent mirror, i.e., at $R_{eq} = 2f_{eq}$. Hence the required object distance is $R_{eq} = 2f_{eq} = \frac{2\mu_1|R_1|\cdot |R_2|}{2[\mu_2(|R_1| + |R_2|) - \mu_1|R_2|]} = \frac{\mu_1|R_1|\cdot |R_2|}{\mu_2(|R_1| + |R_2|) - \mu_1|R_2|},$ which matches Option A.

In the given circuit, the path from node $D$ to the negative terminal (ground) splits into two parallel branches: one through the $4\Omega$ resistor and the other through the forward-biased diode and the $4\Omega$ resistor $R_{AB}$.
The effective resistance of the parallel network is $R_p = 4\Omega \parallel 4\Omega = \frac{4 \times 4}{4 + 4} = 2\Omega$.
The total circuit resistance is $R_{total} = R_{CD} + R_p = 4\Omega + 2\Omega = 6\Omega$ (Observation A).
The total current measured by the ammeter is $I = \frac{E}{R_{total}} = \frac{6V}{6\Omega} = 1A$ (Observation B).
The potential drop across $CD$ is $V_{CD} = I \times R_{CD} = 1A \times 4\Omega = 4V$ (Observation D).
The potential at node $D$ is $V_D = E - V_{CD} = 6V - 4V = 2V$. Since the diode is forward-biased, the potential across $AB$ is equal to the potential at $D$ relative to ground, which is $2V$, making Observation C incorrect.

We need to evaluate two statements about fluid properties. Statement I: "Hot water flows faster than cold water." This is TRUE . The viscosity of a liquid decreases with increasing temperature because higher kinetic energy of the molecules helps them overcome intermolecular attractive forces more easily. Since viscosity resists flow, lower viscosity means hot water flows faster than cold water. Statement II: "Soap water has higher surface tension as compared to fresh water." This is FALSE . Soap is a surfactant that accumulates at the water surface and disrupts the cohesive hydrogen bonds between water molecules, thereby reducing the surface tension. Fresh water has a surface tension of about $0.073$ N/m at room temperature, and adding soap lowers this value significantly. This is precisely why soap helps in cleaning - the reduced surface tension allows water to spread and wet surfaces more effectively. The correct answer is Option (1) : Statement I is true but Statement II is false.

A circular disc of radius 20 cm has a circular hole of radius 5 cm cut such that the edge of the hole touches the edge of the disc. We need the distance of the center of mass of the remaining disc from the origin. Place the center of the original disc at the origin. Since the edge of the hole touches the edge of the disc, the center of the hole is at distance $20 - 5 = 15$ cm from the origin along the x-axis, so the hole center is at $(15, 0)$. Let $\sigma$ be the surface mass density. For a uniform disc, the mass of the original disc is $M = \sigma \pi (20)^2 = 400\pi\sigma$, the mass of the removed portion is $m = \sigma \pi (5)^2 = 25\pi\sigma$, and the mass of the remaining disc is $M - m = 375\pi\sigma$. Using the principle of superposition, we write $M \cdot 0 = (M-m) \cdot x_{cm} + m \cdot 15$, which gives $0 = 375\pi\sigma \cdot x_{cm} + 25\pi\sigma \cdot 15$ and hence $x_{cm} = -\frac{25 \times 15}{375} = -\frac{375}{375} = -1$ cm. The distance from the origin is $|x_{cm}| = 1$ cm, so the correct answer is Option 3: 1.0 cm.

The electric flux is given as $\phi = \alpha \sigma + \beta \lambda$, where $\sigma$ is the surface charge density and $\lambda$ is the linear charge density. We need to find what $\frac{\alpha}{\beta}$ represents. Electric flux has dimensions derived from Gauss's law. The dimensional formula for electric flux is $[\phi] = [\text{kg} \cdot \text{m}^3 \cdot \text{s}^{-2} \cdot \text{C}^{-1}]$. The surface charge density $\sigma = \frac{\text{charge}}{\text{area}}$, so its dimensional formula is $[\sigma] = [\text{C} \cdot \text{m}^{-2}]$. The linear charge density $\lambda = \frac{\text{charge}}{\text{length}}$, so its dimensional formula is $[\lambda] = [\text{C} \cdot \text{m}^{-1}]$. In the expression $\phi = \alpha \sigma + \beta \lambda$, both terms must have the same dimensions as flux. For the term $\alpha \sigma$: $[\alpha \sigma] = [\alpha] \cdot [\sigma] = [\alpha] \cdot \$[\text{C} \cdot \text{m}^{-2}]\$ = [\text{kg} \cdot \text{m}^3 \cdot \text{s}^{-2} \cdot \text{C}^{-1}]$ Solving for $[\alpha]$: $[\alpha] = \frac{[\text{kg} \cdot \text{m}^3 \cdot \text{s}^{-2} \cdot \text{C}^{-1}]\$}{[\text{C} \cdot \text{m}^{-2}]\$} = [\text{kg} \cdot \text{m}^5 \cdot \text{s}^{-2} \cdot \text{C}^{-2}]$ For the term $\beta \lambda$: $[\beta \lambda] = [\beta] \cdot [\lambda] = [\beta] \cdot \$[\text{C} \cdot \text{m}^{-1}]\$ = [\text{kg} \cdot \text{m}^3 \cdot \text{s}^{-2} \cdot \text{C}^{-1}]$ Solving for $[\beta]$: $[\beta] = \frac{[\text{kg} \cdot \text{m}^3 \cdot \text{s}^{-2} \cdot \text{C}^{-1}]\$}{[\text{C} \cdot \text{m}^{-1}]\$} = [\text{kg} \cdot \text{m}^4 \cdot \text{s}^{-2} \cdot \text{C}^{-2}]$ Now, the ratio $\frac{\alpha}{\beta}$ has dimensions: $\left[\frac{\alpha}{\beta}\right] = \frac{[\alpha]}{[\beta]} = \frac{[\text{kg} \cdot \text{m}^5 \cdot \text{s}^{-2} \cdot \text{C}^{-2}]\$}{[\text{kg} \cdot \text{m}^4 \cdot \text{s}^{-2} \cdot \text{C}^{-2}]\$} = [\text{m}]$ The dimension $[\text{m}]$ corresponds to length. Comparing with the options: A. Electric field has dimensions $[\text{kg} \cdot \text{m} \cdot \text{s}^{-2} \cdot \text{C}^{-1}]$ B. Area has dimensions $[\text{m}^2]$ C. Charge has dimensions $[\text{C}]$ D. Displacement has dimensions $[\text{m}]$ Displacement has the same dimension as length, so $\frac{\alpha}{\beta}$ represents displacement. Therefore, the correct answer is displacement (option D).

We need to find the de Broglie wavelength of the particle and identify which type of electromagnetic radiation it corresponds to. The de Broglie wavelength formula is $\lambda = \frac{h}{mv}$. Since the mass of the particle is $m = 10^{-30}$ kg, its speed is $v = 2.21 \times 10^{6}$ m/s, and Planck's constant is $h = 6.63 \times 10^{-34}$ J·s, substituting these values yields: $\lambda = \frac{6.63 \times 10^{-34}}{10^{-30} \times 2.21 \times 10^{6}}$ $\lambda = \frac{6.63 \times 10^{-34}}{2.21 \times 10^{-24}}$ $\lambda = 3 \times 10^{-10} \text{ m} = 3 \text{ Å}$ Now, to classify this wavelength, note that visible radiation spans $4000 - 7000$ Å; infrared radiation corresponds to greater than $7000$ Å; X-rays are \in the range $0.1 - 100$ Å; and \gamma rays are below $0.1$ Å. Since $\lambda = 3$ Å falls in the X-ray range, the particle will behave closely like X-rays. Therefore, the correct answer is Option 4: X-rays.

Let us analyze each statement about the Moving Coil Galvanometer (MCG): Statement A: The torsional constant has dimensions $[ML^{2}T^{-2}]$. The restoring torque is $\tau = k\theta$, where $k$ is the torsional constant and $\theta$ is dimensionless (\in radians). So $[k] = [\tau] = [ML^{2}T^{-2}]$. Statement A is correct. Statement B: Increasing the current sensitivity may not necessarily increase the voltage sensitivity. Current sensitivity: $I_s = \frac{NAB}{k}$ Voltage sensitivity: $V_s = \frac{I_s}{R} = \frac{NAB}{kR}$ If we increase $N$ to increase current sensitivity, the resistance $R$ also increases (since $R \propto N$). So voltage sensitivity may not increase. Statement B is correct. Statement C: If $N$ is doubled to $2N$, voltage sensitivity doubles. When $N$ is doubled, $R$ also roughly doubles. So $V_s = \frac{NAB}{kR}$. Since both $N$ and $R$ double, $V_s$ remains approximately the same. Statement C is incorrect. Statement D: MCG can be converted into an ammeter by introducing a shunt resistance of large value \in parallel. To convert a galvanometer into an ammeter, a small (low) resistance shunt is connected \in parallel, not a large resistance. Statement D is incorrect. Statement E: Current sensitivity depends inversely on number of turns. Current sensitivity $I_s = \frac{NAB}{k}$, which is directly proportional to $N$, not inversely. Statement E is incorrect. Therefore, only statements A and B are correct. The correct answer is Option 4: A, B Only.

We are given the electric field $\vec{E} = 57 \cos\left[7.5 \times 10^{6}t - 5 \times 10^{-3}(3x + 4y)\right](4\hat{i} - 3\hat{j}) \text{ N/C}$. The direction of propagation is along the wave vector $\vec{k}$, which from the phase is $\hat{k} = \frac{3\hat{i} + 4\hat{j}}{5}$. Since the direction of $\vec{E}$ is along $(4\hat{i} - 3\hat{j})$, we verify that it is perpendicular to $\hat{k}$ by computing $(3\hat{i} + 4\hat{j}) \cdot (4\hat{i} - 3\hat{j}) = 12 - 12 = 0$, confirming orthogonality. For an electromagnetic wave, the magnetic field $\vec{B}$ is given by $\vec{B} = \frac{\hat{k} \times \vec{E}}{c}$. Now we compute the cross product: $\hat{k} \times (4\hat{i} - 3\hat{j}) = \frac{1}{5}(3\hat{i} + 4\hat{j}) \times (4\hat{i} - 3\hat{j}) = \frac{1}{5}\left[3(-3)(\hat{i} \times \hat{j}) + 4(4)(\hat{j} \times \hat{i})\right] = \frac{1}{5}\left[-9\hat{k} - 16\hat{k}\right] = \frac{-25\hat{k}}{5} = -5\hat{k}.$ Substituting back into the expression for $\vec{B}$ gives $\vec{B} = \frac{57}{c} \cos\left[7.5 \times 10^{6}t - 5 \times 10^{-3}(3x + 4y)\right](-5\hat{k})$. Therefore, using $c = 3 \times 10^{8}\text{ m/s}$, we have $\vec{B} = -\frac{57}{3 \times 10^{8}} \cos\left[7.5 \times 10^{6}t - 5 \times 10^{-3}(3x + 4y)\right](5\hat{k})$ T $.$ The correct answer is Option 3.

We need to find the mass of the bullet that melts completely after penetrating the wooden block. Since the total heat required has two components, the first part is the heat to raise the temperature from 300 K to the melting point at 600 K, which is given by $Q_1 = mc\Delta T$, and the second part is the heat to melt the bullet at 600 K, given by $Q_2 = mL$. Given that $c = 125 \text{ J kg}^{-1}\text{K}^{-1}$, $L = 2.5 \times 10^{4} \text{ J kg}^{-1}$, and $\Delta T = 600 - 300 = 300 \text{ K}$, the total heat supplied is $Q = 625 \text{ J}$. Substituting these values into the expression for total heat, we have $Q = mc\Delta T + mL$, which becomes $625 = m\bigl(125 \times 300 + 2.5 \times 10^{4}\bigr)$. Therefore, $625 = m(37500 + 25000)$, leading to $625 = m \times 62500$. This gives $m = \frac{625}{62500} = 0.01 \text{ kg} = 10 \text{ grams}$. Therefore, the correct answer is Option 1: 10 grams.

Consider a light ray incident at an angle $i$ on a glass slab of thickness $h$, refracting at an angle $r$. The length of the refracted ray segment ($d$) inside the slab is given by:
$d = \frac{h}{\cos r}$
The lateral shift $L$ is the perpendicular distance between the incident ray and the emergent ray. By observing the geometry of the rays, the angle between the incident ray direction and the refracted ray is $(i-r)$. In the right-angled triangle formed by the lateral shift, we have:
$\sin(i-r) = \frac{L}{d}$
Substituting $d$ into the equation:
$L = d \sin(i-r) = \frac{h \sin(i-r)}{\cos r}$

Since the decay constant of $n_2$ is three \times that of $n_1$, $\lambda_2 = 3\lambda_1$, we use the radioactive decay law: $N = N_0 e^{-\lambda t}$ After one half-life of $n_1$ (that is, $t = T_1 = \frac{\ln 2}{\lambda_1}$), the remaining amount of $n_1$ is $N_1 = N_0 e^{-\lambda_1 \cdot \frac{\ln 2}{\lambda_1}} = N_0 e^{-\ln 2} = \frac{N_0}{2}$ At the same time, the amount of $n_2$ is $N_2 = N_0 e^{-\lambda_2 \cdot \frac{\ln 2}{\lambda_1}} = N_0 e^{-3\lambda_1 \cdot \frac{\ln 2}{\lambda_1}} = N_0 e^{-3\ln 2} = N_0 \cdot 2^{-3} = \frac{N_0}{8}$ Hence, the ratio of the two amounts is $\frac{N_2}{N_1} = \frac{N_0/8}{N_0/2} = \frac{1}{4}$ The correct answer is Option 4: 1/4.

A light hollow cube of side $a = 10$ cm = 0.1 m and mass $m = 10$ g = 0.01 kg floats \in water. We need to find the time period of vertical SHM oscillations. At equilibrium the weight of the cube equals the buoyant force; if $h_0$ is the depth of immersion, then $mg = \rho_{water} \cdot a^2 \cdot h_0 \cdot g$ from which $h_0 = \frac{m}{\rho_{water} \cdot a^2} = \frac{0.01}{1000 \times 0.01} = \frac{0.01}{10} = 0.001 \text{ m}.$ When the cube is pushed down by a small distance $x$ from equilibrium, the additional buoyant force provides a restoring force given by $F_{restoring} = -\rho_{water} \cdot a^2 \cdot x \cdot g,$ where the negative sign indicates the force opposes the displacement. Comparing this with $F = -kx$ shows that the effective spring constant is $k = \rho_{water} \cdot a^2 \cdot g = 1000 \times (0.1)^2 \times 10 = 1000 \times 0.01 \times 10 = 100 \text{ N/m}.$ The time period of oscillation is therefore $T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{0.01}{100}} = 2\pi\sqrt{10^{-4}} = 2\pi \times 10^{-2} \text{ s}.$ This corresponds to $y = 2$, so the correct answer is Option 2: $y = 2$.

Let us analyze each statement about self-inductance: Statement A: The self-inductance of a coil depends on its geometry (shape, size, number of turns). This is correct. For example, for a solenoid, $L = \mu_0 n^2 A l$. Statement B: Self-inductance does not depend on the permeability of the medium. This is incorrect. Self-inductance is directly proportional to the permeability. For a solenoid with a core, $L = \mu n^2 A l$, where $\mu = \mu_0 \mu_r$. Statement C: Self-induced e.m.f. opposes any change \in the current \in a circuit. This is correct (Lenz's law). The self-induced emf is $\varepsilon = -L\frac{dI}{dt}$. Statement D: Self-inductance is the electromagnetic analogue of mass \in mechanics. This is correct. In the energy analogy, $\frac{1}{2}LI^2$ (electromagnetic) corresponds to $\frac{1}{2}mv^2$ (mechanical), so $L$ is analogous to mass $m$. Statement E: Work needs to be done against self-induced e.m.f. \in establishing the current. This is correct. The energy stored \in the inductor $\frac{1}{2}LI^2$ is the work done against the self-induced emf. Statements A, C, D, E are correct. The correct answer is Option 3: A, C, D, E only.

Let us divide this problem into 2 parts:- First 2 seconds where the velocity changes , which means acceleration is present Second part from 2 seconds to 30.5 (28.5 seconds in total) seconds where Velocity is constant We know that distance s:- $s=ut+\ \frac{\ 1}{2}at^2$ For first 2 seconds :- u=$200\ \ \frac{\ m}{s}$ a=$\ \frac{\ v_f=v_i}{t}$ a=$\ \frac{\ 400-200}{2}$ a=100 $\ \frac{\ m}{s^2}$ Now $s=200\times \ 2\ +\ \ \frac{\ 1}{2}\times \ 100\times \ \left(2\right)^2$ $s=600\ m$ For next 28.5 s a=0(constant velocity), Here u=400 $\ \frac{\ m}{s}$ so s=$ut$ s=$400\times \ 28.5$ s=11400 m now adding both $s_{total}$ = 600+ 11400 $s_{total}$= 12000 m = 12 Km

A. no current through resistor wrong. at the instant of closing, the capacitor behaves like a short circuit, so the circuit is complete. current flows through the resistor and is not zero. B. maximum current in wires correct. initially the capacitor has zero voltage across it, so the entire battery voltage drives current through the circuit. hence current is maximum at that instant. C. potential difference between plates A and B is minimum correct. since the capacitor is uncharged at the beginning, the voltage across its plates is zero, which is the minimum possible value. D. charge on capacitor plates is minimum correct. at the instant of closing, no charge has yet accumulated on the capacitor, so the charge is zero, i.e., minimum.

For a solid sphere rolling without slipping down an inclined plane of length $L$, energy conservation relates the loss of gravitational potential energy to the \sum of translational and rotational kinetic energies. The moment of inertia of a solid sphere is $I = \frac{2}{5}mr^2$. Specifically, if the sphere descends a vertical height $h$, then $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2\left(1 + \frac{I}{mr^2}\right) = \frac{1}{2}mv^2\left(1 + \frac{2}{5}\right) = \frac{7}{10}mv^2,$ leading to $v^2 = \frac{10gh}{7} = \frac{10g \cdot L\sin\theta}{7}.$ For an incline at $30°$, this gives $v_1^2 = \frac{10gL\sin 30°}{7} = \frac{10gL}{7} \cdot \frac{1}{2},$ and for $45°$, $v_2^2 = \frac{10gL\sin 45°}{7} = \frac{10gL}{7} \cdot \frac{1}{\sqrt{2}}.$ Hence the ratio of the squared speeds is $\frac{v_1^2}{v_2^2} = \frac{\sin 30°}{\sin 45°} = \frac{1/2}{1/\sqrt{2}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}},$ so that $v_1^2 : v_2^2 = 1 : \sqrt{2}$. Therefore, the correct answer is Option 1: $1:\sqrt{2}$.

We need to find the ratio of electrostatic force to gravitational force between the two ions. The electrostatic force (Coulomb's law) is given by $F_e = \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2}$. The gravitational force is $F_g = \frac{Gm_1 m_2}{r^2}$. Dividing these gives $\frac{F_e}{F_g} = \frac{q_1 q_2}{4\pi\epsilon_0 G m_1 m_2}$. The charges are $q_1 = 6.67 \times 10^{-19}$ C and $q_2 = 9.6 \times 10^{-10}$ C, and the masses are $m_1 = 19.2 \times 10^{-27}$ kg and $m_2 = 9 \times 10^{-27}$ kg. We also use $\frac{1}{4\pi\epsilon_0} = 9 \times 10^{9}$ Nm$^2$C$^{-2}$ and $G = 6.67 \times 10^{-11}$ Nm$^2$kg$^{-2}$. Substituting these values into the ratio yields: $\frac{F_e}{F_g} = \frac{9 \times 10^{9} \times 6.67 \times 10^{-19} \times 9.6 \times 10^{-10}}{6.67 \times 10^{-11} \times 19.2 \times 10^{-27} \times 9 \times 10^{-27}}$. The numerator simplifies as $9 \times 6.67 \times 9.6 \times 10^{9-19-10} = 9 \times 6.67 \times 9.6 \times 10^{-20} = 576.288 \times 10^{-20} = 5.76288 \times 10^{-18}$. The denominator becomes $6.67 \times 19.2 \times 9 \times 10^{-11-27-27} = 6.67 \times 19.2 \times 9 \times 10^{-65} = 1152.576 \times 10^{-65} = 1.152576 \times 10^{-62}$. Therefore, $\frac{F_e}{F_g} = \frac{5.76288 \times 10^{-18}}{1.152576 \times 10^{-62}} = 5 \times 10^{44} = 0.5 \times 10^{45}$. Hence, $P = 0.5$ and $10P = 5$, giving the final answer as 5.

Applying Kirchhoff's Voltage Law to the given RL circuit, the potential difference across the battery must equal the sum of the potential drops across the inductor and the resistor:
$E = L\frac{di}{dt} + IR$
Since the sliding contact is pulled outwards to increase the resistance, the current $I$ in the circuit is decreasing. Thus, the rate of change of current is $\frac{di}{dt} = -8 \, \text{A/s}$. Substituting the given values $E = 12 \, \text{V}$, $L = 3 \, \text{H}$, and $R = 12 \, \Omega$ into the equation:
$12 = 3(-8) + I(12)$
$12 = -24 + 12I$
$36 = 12I$
$I = 3 \, \text{A}$

[CORRECT_OPTION: N/A]

Let the two position vectors be $\vec{A} = 2\hat{i} + 3n\hat{j} + 2\hat{k}$ and $\vec{B} = 2\hat{i} - 2\hat{j} + 4p\hat{k}$. Since both particles are equidistant from the origin, their magnitudes are equal, that is $|\vec{A}| = |\vec{B}|$. Thus we have $4 + 9n^2 + 4 = 4 + 4 + 16p^2$ which simplifies to $9n^2 = 16p^2 \quad \cdots (1).$ Moreover, the perpendicularity condition $\vec{A}\cdot\vec{B}=0$ gives $(2)(2) + (3n)(-2) + (2)(4p) = 0$ or $4 - 6n + 8p = 0,$ hence $8p = 6n - 4$ and $p = \frac{6n - 4}{8} = \frac{3n - 2}{4} \quad \cdots (2).$ Substituting (2) into (1) yields $9n^2 = 16 \left(\frac{3n - 2}{4}\right)^2 = 16 \cdot \frac{(3n-2)^2}{16} = (3n-2)^2,$ so $9n^2 = 9n^2 - 12n + 4,$ which leads to $0 = -12n + 4$ and $n = \frac{1}{3}.$ Therefore, $n^{-1} = 3$, and the answer is 3.

For an adiabatic process (sudden compression), the relation between temperature and volume is $TV^{\gamma - 1} = \text{constant}$. In this case, $\gamma = \frac{C_p}{C_v} = \frac{3}{2}$, the initial temperature is $T_1 = 0°C = 273$ K, and the gas is compressed to one-fourth of its original volume, so $V_2 = \frac{V_1}{4}$. Applying $T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$ gives $T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma - 1} = 273 \times (4)^{3/2 - 1} = 273 \times 4^{1/2} = 273 \times 2 = 546 \text{ K}$. Therefore, the change \in temperature is $\Delta T = T_2 - T_1 = 546 - 273 = 273 \text{ K}$. The temperature rise is 273 K.

The work done by $\vec{F} = x^2 y\,\hat{i} + y^2\,\hat{j}$ along a displacement from $(0, 0)$ to $(4, 2)$ is $W = \int_C x^2 y\,dx + \int_C y^2\,dy$. Since the force acts \in the plane $x + y = 10$, we substitute $y = 10 - x$ \in the $x$-component of the force. The first integral becomes $\int_0^4 x^2(10 - x)\,dx = \int_0^4 (10x^2 - x^3)\,dx = \left[\dfrac{10x^3}{3} - \dfrac{x^4}{4}\right]_0^4 = \dfrac{640}{3} - 64 = \dfrac{448}{3}$. The second integral evaluates directly as $\int_0^2 y^2\,dy = \left[\dfrac{y^3}{3}\right]_0^2 = \dfrac{8}{3}$. Adding both contributions: $W = \dfrac{448}{3} + \dfrac{8}{3} = \dfrac{456}{3} = \boxed{152}$ Joule.

Let us analyze both statements: Statement I: Fructose does not contain an aldehydic group but still reduces Tollen's reagent. Fructose is a ketose sugar (contains a keto group, not an aldehyde group). However, in basic medium (Tollen's reagent provides a basic environment), fructose undergoes tautomerism and can reduce Tollen's reagent. This statement is true. Statement II: In the presence of base, fructose undergoes rearrangement to give glucose. In basic medium, fructose undergoes Lobry de Bruyn-van Ekenstein transformation, converting to glucose (and mannose) through an enediol intermediate. This statement is true. Both statements are true, and Statement II correctly explains Statement I (the base in Tollen's reagent converts fructose to glucose, which then reduces the reagent). The correct answer is Option 1: Both Statement I and Statement II are true.