JEE Main 2025 April 3 shift 2

A magnetic dipole with dipole moment $M$ \in a uniform magnetic field $B$ experiences a torque $\tau = MB\sin\theta$ and has potential energy $U = -MB\cos\theta$. Given $\tau = 80\sqrt{3}$ N m and $\theta = 60°$: $MB\sin 60° = 80\sqrt{3}$ $MB \cdot \dfrac{\sqrt{3}}{2} = 80\sqrt{3}$ $MB = 160$ The potential energy is: $U = -MB\cos 60° = -160 \cdot \dfrac{1}{2} = -80$ J Hence, the correct answer is Option D.

For a closed pipe (closed at one end), the fundamental frequency is $f = \dfrac{v}{4L}$. For the 100 cm pipe: $f_1 = \dfrac{v}{4 \times 1.00} = \dfrac{v}{4}$ For the 120 cm pipe: $f_2 = \dfrac{v}{4 \times 1.20} = \dfrac{v}{4.8}$ The beat frequency is $|f_1 - f_2| = 15$ Hz. $\dfrac{v}{4} - \dfrac{v}{4.8} = 15$ $v\left(\dfrac{1}{4} - \dfrac{1}{4.8}\right) = 15$ $v\left(\dfrac{4.8 - 4}{4 \times 4.8}\right) = 15$ $v \cdot \dfrac{0.8}{19.2} = 15$ $v = 15 \times \dfrac{19.2}{0.8} = 15 \times 24 = 360$ m/s Hence, the correct answer is Option D.

Since the vessels have equal cross-sectional areas and are connected at the bottom, water will flow until the heights are equal. The total volume is conserved. $h_f = \frac{h_1 + h_2}{2} = \frac{10 + 6}{2} = 8 \text{ m}$ The potential energy of a liquid column of mass $M$ and height $h$ is $U = Mg\frac{h}{2}$ (where $h/2$ is the center of mass). In terms of density ($\rho$) and area ($A$), this is $U = \frac{1}{2} \rho Ag h^2$. $U_i = \frac{1}{2} \rho Ag (h_1^2 + h_2^2)$ $U_i = \frac{1}{2} \times 10^3 \times 2 \times 10 \times (10^2 + 6^2) = 10^4 \times (100 + 36) = 136 \times 10^4 \text{ J}$ $U_f = \frac{1}{2} \rho Ag (h_f^2 + h_f^2) = \rho Ag h_f^2$ $U_f = 10^3 \times 2 \times 10 \times (8^2)$ $U_f = 2 \times 10^4 \times 64 = 128 \times 10^4 \text{ J}$ $W_g = U_i - U_f$ (work done by gravity is the negative change \in potential energy) $W_g = (136 - 128) \times 10^4 = 8 \times 10^4 \text{ J}$

In Young's double slit experiment, the intensity from each slit is proportional to the slit width. If one slit has width $w$ and the other has width $w/2$, then $I_1 = I$ and $I_2 = I/2$ (say). The maximum intensity is $I_{max} = \left(\sqrt{I_1} + \sqrt{I_2}\right)^2 = \left(\sqrt{I} + \sqrt{I/2}\right)^2 = I\left(1 + \dfrac{1}{\sqrt{2}}\right)^2$ The minimum intensity is $I_{min} = \left(\sqrt{I_1} - \sqrt{I_2}\right)^2 = \left(\sqrt{I} - \sqrt{I/2}\right)^2 = I\left(1 - \dfrac{1}{\sqrt{2}}\right)^2$ The ratio is: $\dfrac{I_{max}}{I_{min}} = \dfrac{\left(1 + \dfrac{1}{\sqrt{2}}\right)^2}{\left(1 - \dfrac{1}{\sqrt{2}}\right)^2} = \dfrac{\left(\sqrt{2} + 1\right)^2}{\left(\sqrt{2} - 1\right)^2}$ $= \dfrac{2 + 2\sqrt{2} + 1}{2 - 2\sqrt{2} + 1} = \dfrac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}}$ Hence, the correct answer is Option B.

Initial state: $(P_0,V_0)$ For one complete cycle, $\Delta U=0$ So net heat exchanged equals net work done: $Q=W$ Now calculate work \in each process. Isothermal expansion from $V_0\ ​to\ 4V_0$ For isothermal process, $W_1=nRT\ln\frac{4V_0}{V_0}$ Since initially $nRT=P_0V_0$ So $W_1=P_0V_0\ln4$ Isobaric compression After isothermal expansion pressure becomes $P=\frac{P_0V_0}{4V_0}=\frac{P_0}{4}$ Compression occurs at constant pressure $\frac{P_0}{4}\ from\ 4V_0​\ to\ V_0​$. Work done: W_2=P\Delta V=\frac{P_0}{4}(V_0-4V_0)$$$$=-\frac{3}{4}P_0V_0 Isochoric heating $W_3=0$ Net work: $W=P_0V_0\ln4-\frac{3}{4}P_0V_0$

Two light beams have intensities in the ratio $I_1 : I_2 = 1 : 9$. Let $I_1 = I$ and $I_2 = 9I$. The maximum intensity is $I_{max} = \left(\sqrt{I_1} + \sqrt{I_2}\right)^2 = \left(\sqrt{I} + 3\sqrt{I}\right)^2 = 16I$ The minimum intensity is $I_{min} = \left(\sqrt{I_1} - \sqrt{I_2}\right)^2 = \left(\sqrt{I} - 3\sqrt{I}\right)^2 = 4I$ The ratio is $\dfrac{I_{max}}{I_{min}} = \dfrac{16I}{4I} = \dfrac{4}{1}$ Hence, the correct answer is Option D.

Assertion A: The Bohr model is applicable to hydrogen and hydrogen-like atoms only. This is TRUE. The Bohr model works for systems with a single electron orbiting a nucleus (hydrogen, He$^+$, Li$^{2+}$, etc.). Reason R: The formulation of the Bohr model does not include repulsive force between electrons. This is TRUE. Bohr's model considers only the attractive Coulomb force between the nucleus and a single electron. It does not account for electron-electron repulsion. The reason R correctly explains the assertion A. The Bohr model is limited to hydrogen-like atoms precisely because it does not incorporate electron-electron repulsive interactions. For multi-electron atoms, the inter-electron repulsion makes the Bohr model inapplicable. Hence, the correct answer is Option C.

The first capacitor has $C_1 = 100\,$pF and is initially charged to $V_{1i}=60\,$V. Initial charge on this capacitor: $Q_{\text{initial}} = C_1\,V_{1i} = 100\,\text{pF}\times 60\,\text{V} = 6000\,\text{pC}$ After disconnecting the battery, this charge cannot escape; it is conserved inside the closed system formed by the two capacitors. An uncharged capacitor of capacitance $C_2$ is then connected \in parallel. In a parallel connection the final voltage on both capacitors is the same. The problem states that this common voltage is $V_f = 20\,$V. Let $Q_{1f}$ and $Q_{2f}$ be the final charges on $C_1$ and $C_2$ respectively. Using $Q = C\,V$: $Q_{1f} = C_1\,V_f$ $Q_{2f} = C_2\,V_f$ Because total charge is conserved (battery is removed), $Q_{\text{initial}} = Q_{1f} + Q_{2f}$ $\Rightarrow 6000 = C_1\,V_f + C_2\,V_f$ Divide both sides by $V_f$ to isolate $C_2$: $\frac{6000}{V_f} = C_1 + C_2$ $\Rightarrow C_2 = \frac{6000}{20} - C_1$ Substitute the known values: $C_2 = 300 - 100 = 200\,\text{pF}$ Therefore, the capacitance of the second capacitor is $200\,$pF, which corresponds to Option B .

The frequency of monochromatic light never changes when it passes from one medium to another. Given frequency $f = 5 \times 10^{14}\, \text{Hz}$ remains the same \in every medium. Step 1 : Wavelength \in air Speed of light \in air is practically the same as \in vacuum, $c = 3 \times 10^{8}\,\text{m s}^{-1}$. Using $\lambda = \dfrac{v}{f}$, the wavelength \in air is $\lambda_{\text{air}} = \frac{c}{f} = \frac{3 \times 10^{8}}{5 \times 10^{14}} = 6 \times 10^{-7}\,\text{m} = 600\,\text{nm}.$ Step 2 : Speed \in the second medium Refractive index $\mu$ of the medium is defined by $\mu = \frac{c}{v}$. For $\mu = 2$, $v = \frac{c}{\mu} = \frac{3 \times 10^{8}}{2} = 1.5 \times 10^{8}\,\text{m s}^{-1}.$ Step 3 : Wavelength \in the second medium Again using $\lambda = \dfrac{v}{f}$, $\lambda_{\text{medium}} = \frac{v}{f} = \frac{1.5 \times 10^{8}}{5 \times 10^{14}} = 3 \times 10^{-7}\,\text{m} = 300\,\text{nm}.$ Thus the wavelength of the light inside the medium of refractive index $2$ is $300\,\text{nm}$. Hence, the correct choice is Option A .

First step is to find out $v_f$ which is the final velocity with which both the blocks along with the spring move Using conservation of momentum:- so $m_1v_1=\left(m_1+m_2\right)v_f$ $v_f=\frac{m_1v_1}{m_1+m_2}$ On substituting the given values $v_f=\frac{10\times \ 3}{10+5}$ $v_f=\frac{30}{15}=2\ \frac{m}{s}$ Now lets conserve Energy Initital K.E = $\frac{1}{2}m_1v_1^2$ $=\frac{1}{2}\times \ 10\times \left(\ 3^2\right)$ $=5\times9$ $K.E_i=45$ Final K.E is $\frac{1}{2}\left(m_1+m_2\right)v_f^2$ $=\frac{1}{2}\left(10+5\right)\times \ 2^2$ $=15\times \ 2$ $K.E_f=30$ Now difference between these is the energy stored \in the spring K.E_i-K.E_f$$$$K.E_i-K.E_f\ =\ \ 45-\ 30\ =15 Now $\frac{1}{2}kx^2=15$ $\frac{1}{2}\times \ 3000\times \ x^2=15$ $\ x^2=\frac{30}{3000}$ $\ x^2=\frac{1}{100}=10^{-2}$ $\ x=10^{-1}=0.1m$

The standard formulae for a projectile launched with speed $u$ at an angle $\theta$ above the horizontal are: Horizontal range: $R = \frac{u^{2}\sin 2\theta}{g}$ Maximum height: $H = \frac{u^{2}\sin^{2}\theta}{2g}$ The condition given \in the problem is that the range is three \times the maximum height: $R = 3H$ Substitute the expressions for $R$ and $H$: $\frac{u^{2}\sin 2\theta}{g} = 3\left(\frac{u^{2}\sin^{2}\theta}{2g}\right)$ Cancel the common factors $u^{2}$ and $g$: $\sin 2\theta = \frac{3}{2}\sin^{2}\theta$ $-(1)$ Use the trigonometric identity $\sin 2\theta = 2\sin\theta\cos\theta$ and substitute it into $(1)$: $2\sin\theta\cos\theta = \frac{3}{2}\sin^{2}\theta$ Divide both sides by $\sin\theta$ (since $\theta \neq 0^{\circ},90^{\circ}$ for a projectile with non-zero range): $2\cos\theta = \frac{3}{2}\sin\theta$ Rearrange to get the cotangent of the launch angle: $\frac{\cos\theta}{\sin\theta} = \frac{3}{4} \quad\Longrightarrow\quad \cot\theta = \frac{3}{4}$ Hence $\tan\theta = \frac{4}{3}$ From the right-triangle definition, this gives $\sin\theta = \frac{4}{5},\qquad \cos\theta = \frac{3}{5}$ Now compute $\sin 2\theta$ using $\sin 2\theta = 2\sin\theta\cos\theta$: $\sin 2\theta = 2\left(\frac{4}{5}\right)\left(\frac{3}{5}\right) = \frac{24}{25}$ Finally, substitute $\sin 2\theta$ into the range formula: $R = \frac{u^{2}\sin 2\theta}{g} = \frac{u^{2}}{g}\left(\frac{24}{25}\right) = \frac{24\,u^{2}}{25g}$ The question writes the range as $\dfrac{\nu^{2}}{25g}$. Comparing gives $n = 24$. Therefore, the correct option is Option D (24).

For a small sphere moving slowly through a viscous liquid, Stokes' law gives the viscous drag force as $F_{v} = 6 \pi \eta r v$, where $\eta$ is the viscosity, $r$ the radius of the sphere and $v$ its speed. At terminal velocity, the sphere moves with constant speed, so the net force on it is zero. The downward forces (weight) are balanced by the upward forces (buoyant force + viscous drag): $\text{Weight}\;-\;\text{Buoyant force}\;=\;\text{Viscous drag}$ $\bigl(\rho_{s} V g - \rho_{o} V g\bigr)\;=\;6 \pi \eta r v_{t}$ Here $\rho_{s} = 7825\;\text{kg m}^{-3}$ (density of steel) $\rho_{o} = 925\;\text{kg m}^{-3}$ (density of oil) $V = \tfrac{4}{3}\pi r^{3}$ is the volume of the sphere $v_{t} = 2.45 \times 10^{-2}\;\text{m s}^{-1}$ is the terminal velocity. Cancelling the common factor $V g$ and simplifying: $(\rho_{s}-\rho_{o})\;\tfrac{4}{3}\pi r^{3} g = 6 \pi \eta r v_{t}$ Divide both sides by $\pi r$ to isolate $\eta$: $\eta = \frac{2 r^{2} g\,(\rho_{s}-\rho_{o})}{9 v_{t}}\quad -(1)$ Now substitute the numerical data. Diameter of ball $= 3.6\;\text{mm} \Rightarrow r = 1.8\;\text{mm} = 1.8 \times 10^{-3}\;\text{m}$ Density difference: $(\rho_{s}-\rho_{o}) = 7825 - 925 = 6900\;\text{kg m}^{-3}$ Using $g = 9.8\;\text{m s}^{-2}$ and $v_{t} = 2.45 \times 10^{-2}\;\text{m s}^{-1}$ \in equation $(1)$: $\eta = \frac{2\,(1.8 \times 10^{-3})^{2}\,(9.8)\,(6900)}{9\,(2.45 \times 10^{-2})}$ Step-wise calculation: $r^{2} = (1.8 \times 10^{-3})^{2} = 3.24 \times 10^{-6}\;\text{m}^{2}$ $2 r^{2} = 2 \times 3.24 \times 10^{-6} = 6.48 \times 10^{-6}$ $6.48 \times 10^{-6} \times 6900 = 4.471 \times 10^{-2}$ $4.471 \times 10^{-2} \times 9.8 = 0.438\;(\text{approximately})$ Denominator $= 9 \times 2.45 \times 10^{-2} = 0.2205$ Finally, $\eta = \frac{0.438}{0.2205} \approx 1.99\;\text{Pa·s}$ The viscosity of the oil is therefore $\mathbf{1.99\;Pa\cdot s}$. Hence, the correct option is Option D .

The circuit consists of an OR gate followed by an AND gate. The inputs to the OR gate are A and B, so its output is $A+B$. This output and input A are then fed into the AND gate. The final output C is given by the Boolean expression $C = (A+B) \cdot A$. Using the distributive law, this expression expands to $C = (A \cdot A) + (A \cdot B)$. As per Boolean algebra, $A \cdot A = A$, so the expression becomes $C = A + (A \cdot B)$. By the absorption law ($X+XY=X$), this simplifies to $C=A$. This means the output C is always equal to the input A, which is correctly depicted in the truth table of option B.

The displacement of the particle is given as $x = c_0\,(t^{2}-2) + c\,(t-2)^{2} \qquad -(1)$ Step 1 - Find velocity Velocity is the time-derivative of displacement: $v = \frac{dx}{dt}$ Differentiating $-(1)$ term by term: \bullet For $c_0\,(t^{2}-2)$: $\frac{d}{dt}\big(c_0\,(t^{2}-2)\big)=c_0\,(2t)=2c_0t$ \bullet For $c\,(t-2)^{2}$: $\frac{d}{dt}\big(c\,(t-2)^{2}\big)=c\,(2\,(t-2))=2c\,(t-2)$ Therefore $v = 2c_0t + 2c\,(t-2) \qquad -(2)$ Step 2 - Find acceleration Acceleration is the time-derivative of velocity: $a = \frac{dv}{dt}$ Differentiating $-(2)$: \bullet $\frac{d}{dt}\big(2c_0 t\big)=2c_0$ \bullet $\frac{d}{dt}\big(2c\,(t-2)\big)=2c$ Hence $a = 2c_0 + 2c = 2\,(c_0 + c) \qquad -(3)$ Acceleration is a constant equal to $2(c_0 + c)$. Step 3 - Check other statements Initial velocity (at $t=0$) from $-(2)$: $v(0)=2c_0\,(0) + 2c\,(0-2) = -4c$ Thus the initial velocity is $-4c$, not $4c$. Conclusion From $-(3)$, the only correct statement is: "the acceleration of the particle is $2(c + c_0)$," which corresponds to Option D.

The rating 100 W - 220 V means the bulb consumes power $P = 100\ \text{W}$ when an rms voltage $V_{\text{rms}} = 220\ \text{V}$ is applied. Step 1: Find the resistance of the filament. For an ac (or dc) load, $P = \frac{V_{\text{rms}}^{\,2}}{R}$ $-(1)$ Rearranging $-(1)$ gives $R = \frac{V_{\text{rms}}^{\,2}}{P}$. Substituting the given values: $R = \frac{(220)^{2}}{100} = \frac{48400}{100} = 484\ \Omega$. Step 2: Calculate the rms current through the bulb. Ohm's law for rms values is $I_{\text{rms}} = \frac{V_{\text{rms}}}{R}$ $-(2)$. Using $R = 484\ \Omega$: $I_{\text{rms}} = \frac{220}{484}\ \text{A} \approx 0.455\ \text{A}$. Step 3: Convert rms current to peak (maximum) current. For a sinusoidal current, $I_{0} = \sqrt{2}\,I_{\text{rms}}$ $-(3)$. Applying $-(3)$: $I_{0} = \sqrt{2}\times 0.455\ \text{A} \approx 1.414 \times 0.455\ \text{A} \approx 0.64\ \text{A}$. Hence, the peak value of current through the bulb is $0.64\ \text{A}$. Option A is correct.

The dimensional analysis will be carried out for each physical quantity in LIST-I and then compared with the four alternatives given in LIST-II. Case A : Boltzmann constant $k_B$ The average translational energy of one molecule \in an ideal gas is $E = \tfrac{3}{2}\,k_B\,T$. Hence $k_B = \dfrac{E}{T}$. Energy has dimension $ML^{2}T^{-2}$ and temperature contributes the factor $K$ \in the denominator. Therefore $[k_B] = \dfrac{ML^{2}T^{-2}}{K} = ML^{2}T^{-2}K^{-1}$, which corresponds to LIST-II entry III. Case B : Coefficient of viscosity $\eta$ Newton's law of viscous flow states $\tau = \eta \,\dfrac{dv}{dy}$, where $\tau$ is shear stress. Shear stress $\tau = \dfrac{\text{Force}}{\text{Area}}$ has dimension $\dfrac{MLT^{-2}}{L^{2}} = ML^{-1}T^{-2}$. Velocity gradient $\dfrac{dv}{dy}$ has dimension $T^{-1}$. Thus $[\eta] = \dfrac{ML^{-1}T^{-2}}{T^{-1}} = ML^{-1}T^{-1}$, matching LIST-II entry IV. Case C : Planck's constant $h$ The photon energy is $E = h\nu$, and frequency $\nu$ has dimension $T^{-1}$. Hence $h = \dfrac{E}{\nu}$. Taking $[E] = ML^{2}T^{-2}$ and dividing by $T^{-1}$ gives $[h] = ML^{2}T^{-1}$, i.e. LIST-II entry I. Case D : Thermal conductivity $K$ Fourier's law of heat conduction is $\dfrac{Q}{t} = -K\,A\,\dfrac{\Delta T}{L}$. Heat conducted per unit time $\dfrac{Q}{t}$ is power with dimension $ML^{2}T^{-3}$. Area $A$ contributes $L^{2}$ and temperature gradient $\dfrac{\Delta T}{L}$ contributes $K\,L^{-1}$ \in the denominator. Therefore $[K] = \dfrac{ML^{2}T^{-3}}{L^{2}\,K\,L^{-1}} = \dfrac{ML^{2}T^{-3}}{L^{1}K} = MLT^{-3}K^{-1}$, corresponding to LIST-II entry II. Collecting the results: A → III, B → IV, C → I, D → II This is exactly the pairing given in Option A. Hence, the correct answer is Option A .

For a fixed amount of ideal gas in a closed rigid vessel, the volume remains constant. According to Gay-Lussac's law (also called the pressure law), at constant volume the pressure $P$ of an ideal gas is directly proportional to its absolute temperature $T$. Mathematically, $\frac{P}{T} = \text{constant} \quad\Longrightarrow\quad \frac{\Delta P}{P} = \frac{\Delta T}{T}$ The problem states: the pressure rises by $0.4\%$ upon heating the gas by $1^{\circ}\text{C}$. Writing these changes \in symbols, $\frac{\Delta P}{P} = 0.4\% = \frac{0.4}{100} = 0.004$ Since a change of $1^{\circ}\text{C}$ equals a change of $1\text{ K}$ on the absolute scale, $\Delta T = 1\text{ K}$ Using the proportionality relation: $\frac{\Delta T}{T} = \frac{\Delta P}{P}$ $\Rightarrow \quad \frac{1\text{ K}}{T} = 0.004$ Solving for the initial absolute temperature $T$: $T = \frac{1}{0.004} = 250 \text{ K}$ Hence the initial temperature of the gas is $250\text{ K}$. Converting to the Celsius scale, $250\text{ K} - 273 \approx -23^{\circ}\text{C}$, but none of the given options list this negative Celsius value; instead, option C directly offers $250\text{ K}$. Therefore, the correct choice is Option C (250 K) .

The motor is supplied with a voltage $V = 100\,\text{V}$ and draws a current $I = 1\,\text{A}$, so the electrical power input is $P_{\text{\in}} = V I = 100 \times 1 = 100\,\text{W}$ The efficiency of the motor is $\eta = 91.6\% = 0.916$. Efficiency is defined as $\eta = \frac{P_{\text{out}}}{P_{\text{\in}}}$ Hence the useful (output) power is $P_{\text{out}} = \eta P_{\text{\in}} = 0.916 \times 100 = 91.6\,\text{W}$ The power lost as heat (and other losses) is therefore $P_{\text{loss}} = P_{\text{\in}} - P_{\text{out}} = 100 - 91.6 = 8.4\,\text{W}$ To express this loss \in calories per second, use the relation $1\,\text{cal} = 4.186\,\text{J}$, so $1\,\text{cal/s} = 4.186\,\text{W}$. Converting: $P_{\text{loss}} = \frac{8.4\,\text{W}}{4.186\,\text{W per cal/s}} \approx 2.0\,\text{cal/s}$ Therefore, the power loss is approximately $2\,\text{cal/s}$. This corresponds to Option C .

The motion is along the +x-direction. Outside the rough strip the block is free; inside the strip (from $x_i = 0.1\;{\rm m}$ to $x_f = 1.9\;{\rm m}$) it experiences the position-dependent retarding force $F_r = -\,k\,x,$ with $k = 10\;{\rm N\,m^{-1}}$. (The minus sign shows that the force is opposite to the direction of motion, so it removes kinetic energy from the block.) Case 1 : Work done by the retarding force The infinitesimal work done by the force over a small displacement $dx$ is $dW = F_r\,dx = -k\,x\,dx.$ To obtain the total work between the entry and exit points, integrate from $x = 0.1$ m to $x = 1.9$ m:

$

$ Insert the numerical values:

$

$ The negative sign confirms that the force takes 18 J of mechanical energy out of the block. Case 2 : Applying the work-energy theorem The work-energy theorem states $W = K_f - K_i,$ where $K_i$ and $K_f$ are the initial and final kinetic energies, respectively. Initial kinetic energy: $K_i = \frac{1}{2} m v_i^{2} = \tfrac12 (1\;{\rm kg})(10\;{\rm m/s})^{2} = 50\;{\rm J}.$ Therefore,

$

\begin{aligned} -18;{\rm J} &= K_f - 50;{\rm J}\ K_f &= 50;{\rm J} - 18;{\rm J} = 32;{\rm J}. \end{aligned}

$ Case 3 : Determining the final speed Let the required speed be $v_f$. $K_f = \frac{1}{2} m v_f^{2} \; \Longrightarrow\; \frac{1}{2}(1)\,v_f^{2} = 32.$ Solve for $v_f$: $v_f^{2} = 64 \quad\Longrightarrow\quad v_f = 8\;{\rm m/s}.$ Hence, when the block leaves the rough region its speed is 8 m s−1. Final answer : 8 m/s (Option D)

When a charged particle of charge $q$ and linear momentum $p$ enters a uniform magnetic field $\mathbf{B}$ perpendicular to the field, the magnetic force $|q|vB$ provides the required centripetal force $\dfrac{mv^{2}}{r}$. Substituting $p = mv$ and $v = \dfrac{p}{m}$, the radius (called radius of curvature) is obtained as $r = \dfrac{p}{|q|B}$ $-(1)$ Case 1: Oxide ion $O^{2-}$ versus proton $H^{+}$ For $O^{2-}: |q| = 2e \quad(\text{two electronic charges})$ For $H^{+}: |q| = e$ Both ions are given to have the same momentum $p$ and the same field $B$, so using $(1)$: $r_{O^{2-}} = \dfrac{p}{2eB}, \qquad r_{H^{+}} = \dfrac{p}{eB} = 2\,r_{O^{2-}}$ Thus $r_{O^{2-}} \lt r_{H^{+}}$, i.e. the oxide ion describes a circle of smaller radius (larger bending) than the proton. Consequently its curvature radius is smaller. Hence Assertion A is correct. Case 2: Proton versus electron with the same linear momentum For a proton $|q_p| = e$ and for an electron $|q_e| = e$. With equal momenta $p$, equation $(1)$ gives $r_p = \dfrac{p}{eB}, \qquad r_e = \dfrac{p}{eB}$ The radii are identical; the proton does not form a path of smaller radius than the electron. Therefore Reason R is false. Conclusion: Assertion A is true, Reason R is false. The correct choice is Option A.

Take the pole of the spherical surface as the origin and measure all distances from this point along the principal axis (Cartesian sign convention). • Incident medium (air): $n_1 = 1$. \bullet Refracting medium (glass): $n_2 = 1.5$. \bullet Radius of curvature: the centre of curvature lies inside the glass, i.e. on the side to which light travels, so $R = +50\,\text{cm}$. \bullet Image is formed inside the glass at a distance $v = +200\,\text{cm}$ (positive because it is on the side of the refracted light). For refraction at a spherical surface, the Cartesian-form equation is $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \quad -(1)$ Substitute the given data into $(1)$: $\frac{1.5}{200} - \frac{1}{u} = \frac{1.5 - 1}{50}$ Simplify each term: $\frac{1.5}{200} = 0.0075$, $\frac{1.5 - 1}{50} = \frac{0.5}{50} = 0.01$. Hence, $0.0075 - \frac{1}{u} = 0.01$ Rearrange to solve for $\frac{1}{u}$: $-\frac{1}{u} = 0.01 - 0.0075 = 0.0025$ Therefore, $\frac{1}{u} = -0.0025 \;\Rightarrow\; u = -\frac{1}{0.0025} = -400\,\text{cm}$ The negative sign confirms that the object (light source) is on the side opposite to the direction of positive measurement, i.e. \in air \in front of the surface. Magnitude of the object distance: $|u| = 400\,\text{cm} = 4\,\text{m}$. Hence, the light source is $4\,\text{m}$ in front of the glass surface.

For a soap bubble in air, the excess pressure inside it is given by $\Delta P = \frac{4T}{r}$ where $T$ is the surface tension of the soap film and $r$ is the radius of the bubble. Let the radii of bubbles A and B be $r_A$ and $r_B$ respectively. Then $\Delta P_A = \frac{4T}{r_A}$ $\Delta P_B = \frac{4T}{r_B}$ The statement "excess pressure inside bubble A is half the excess pressure inside bubble B" translates to $\Delta P_A = \frac{1}{2}\,\Delta P_B$ Substituting from above, $\frac{4T}{r_A} = \frac{1}{2}\left(\frac{4T}{r_B}\right)$ Simplifying (the factor $4T$ cancels out): $\frac{1}{r_A} = \frac{1}{2}\,\frac{1}{r_B}$ which gives $r_B = \frac{1}{2}\,r_A$ or equivalently $r_A = 2\,r_B$. The volume of a spherical bubble is $V = \frac{4}{3}\,\pi r^{3}$. Therefore, the ratio of the volumes of bubbles A and B is $\frac{V_A}{V_B} = \left(\frac{r_A}{r_B}\right)^3 = \left(2\right)^3 = 8$. Given that $V_A = n\,V_B$, we have $n = 8$. Hence the required value of $n$ is $8$.

Case 1: Cells in series For series combination, the net emf is the algebraic sum $E_s = E_1 + E_2 = 1 + 2 = 3\;{\rm V}$ The net internal resistance is the arithmetic \sum $r_s = r_1 + r_2 = 2 + 1 = 3\;\Omega$ Total resistance \in the external circuit is $R_{\text{tot}} = r_s + R = 3 + 6 = 9\;\Omega$ Hence the current \in the series circuit is $I_1 = \frac{E_s}{R_{\text{tot}}} = \frac{3}{9} = \frac{1}{3}\;{\rm A}$ Case 2: Cells \in parallel For two unequal cells \in parallel, use the general formulae: Equivalent internal resistance: $\frac{1}{r_p} = \frac{1}{r_1} + \frac{1}{r_2}$ Equivalent emf: $E_p = \frac{\dfrac{E_1}{r_1} + \dfrac{E_2}{r_2}}{\dfrac{1}{r_1} + \dfrac{1}{r_2}}$ Compute $r_p$: $\frac{1}{r_p} = \frac{1}{2} + \frac{1}{1} = 0.5 + 1 = 1.5$ $r_p = \frac{1}{1.5} = \frac{2}{3}\;\Omega$ Compute $E_p$: $E_p = \frac{\dfrac{1}{2} + \dfrac{2}{1}}{1.5} = \frac{0.5 + 2}{1.5} = \frac{2.5}{1.5} = \frac{5}{3}\;{\rm V}$ Total resistance now is $R_{\text{tot}} = R + r_p = 6 + \frac{2}{3} = \frac{20}{3}\;\Omega$ Therefore the current \in the parallel configuration is $I_2 = \frac{E_p}{R_{\text{tot}}} = \frac{\dfrac{5}{3}}{\dfrac{20}{3}} = \frac{5}{20} = \frac{1}{4}\;{\rm A}$ Required ratio $\frac{I_1}{I_2} = \frac{\dfrac{1}{3}}{\dfrac{1}{4}} = \frac{1}{3} \times 4 = \frac{4}{3}$ The problem states $\dfrac{I_1}{I_2} = \dfrac{x}{3}$, so $x = 4$. Hence, the value of $x$ is 4 .

The energy of an electron in a hydrogen atom is given by $E_n = -\frac{13.6\text{ eV}}{n^{2}}$ "Fourth excited state'' means the electron starts from the level $n_i = 5$ (because $n = 1$ is ground, $n = 2$ first excited, ...). If the electron finally reaches an unknown level $n_f$ and emits a photon of energy $E_{\gamma}=2.86\text{ eV}$, then by energy conservation $E_{\gamma}=E_{n_i}-E_{n_f}$ Substituting the formula for $E_n$ gives $2.86 = \left(-\frac{13.6}{5^{2}}\right) - \left(-\frac{13.6}{n_f^{2}}\right)$ Simplify the brackets: $2.86 = 13.6\left(\frac{1}{n_f^{2}} - \frac{1}{25}\right)$ Divide both sides by $13.6$ to isolate the bracketed term: $\frac{2.86}{13.6} = \frac{1}{n_f^{2}} - \frac{1}{25}$ Calculate the numerical value: $\frac{2.86}{13.6} \approx 0.21$ Add $\frac{1}{25}=0.04$ to both sides: $0.21 + 0.04 = \frac{1}{n_f^{2}}$ $\frac{1}{n_f^{2}} \approx 0.25$ Taking reciprocal and square-root: $n_f^{2} \approx 4 \quad\Longrightarrow\quad n_f \approx 2$ Since $n_f$ must be an integer quantum number, we take $n_f = 2$. Therefore, the electron drops to the second energy level. The required integer value of $n$ is 2 .