JEE Main 2025 April 3 shift 1

For any plane wave front, the light travels in the direction perpendicular (normal) to that plane. A general plane can be written as $ax + by + cz = \text{constant}$, whose normal vector is $\mathbf{n} = a\,\hat i + b\,\hat j + c\,\hat k$. The given wave fronts are $x + y + z = \text{constant}$. Comparing with $ax + by + cz = \text{constant}$, we get the normal (and hence the direction of wave propagation) $\mathbf{n} = 1\,\hat i + 1\,\hat j + 1\,\hat k = \langle 1,\,1,\,1\rangle$. Let $\theta$ be the angle between the direction of propagation and the x-axis. The x-axis is represented by the unit vector $\hat i = \langle 1,\,0,\,0\rangle$. First, write the unit vector along the direction of propagation: $\hat u = \frac{\mathbf{n}}{|\mathbf{n}|} = \frac{\langle 1,1,1\rangle}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{\langle 1,1,1\rangle}{\sqrt{3}} = \left\langle \frac{1}{\sqrt{3}},\,\frac{1}{\sqrt{3}},\,\frac{1}{\sqrt{3}}\right\rangle$. Using the dot-product formula $\hat u \cdot \hat i = |\hat u|\,|\hat i| \cos\theta$. Since $|\hat u| = |\hat i| = 1$, we get $\cos\theta = \hat u \cdot \hat i = \left\langle \frac{1}{\sqrt{3}},\,\frac{1}{\sqrt{3}},\,\frac{1}{\sqrt{3}}\right\rangle \cdot \langle 1,0,0\rangle = \frac{1}{\sqrt{3}}$. Therefore, $\theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right).$ The correct choice is Option A.

From the van der Waals equation, we know that $P$ has the same dimension as $a/V^2$. So, $[P] = [a]/[V^2]$, which means $[a] = [P][V^2]$.
Also, $V$ has the same dimension as $b$. So, $[b] = [V]$.
Now, we need to find the dimension of $ab^{-2}$.
[ab^{-2}] = [a]/\[b^2]$ = ([P]$[V^2]$)/$[V^2]$ = [P]$. Pressure has the dimension of Force per unit Area, which is$[F]/[A]$. Energy density is Energy per unit Volume, which is$[E]/[V]$. Since$[E] = [F][L]$and$[V] = [L^3]$,$[E]/[V] = [F][L]/$[L^3]$ = [F]/$[L^2]$ = [F]/[A]$. Thus, the dimension of$ab^{-2}$ is equivalent to that of pressure, which is also equivalent to energy density.

Since the wheel has negligible mass spokes, its mass is concentrated at the rim (treating it as a hoop/thin ring): $I = MR^2$ $I = 10 \text{ kg} \times (0.1 \text{ m})^2 = 0.1 \text{ kg m}^2$ The work done by the pull (F) over the unwound distance (s) equals the change \in rotational kinetic energy: $\text{Work} = \Delta KE_{rot}$ $F \cdot s = \frac{1}{2} I \omega^2$ $20 \times 1 = \frac{1}{2} \times 0.1 \times \omega^2$ $\omega = 20 \text{ rad/s}$

The object AB has coordinates A(0,0) and B(1,2) with respect to A as origin. The length of the object is $\sqrt{1^{2+}2^2} = \sqrt{5}$ cm. The angle $\alpha$ is $\arctan(2/1) = \arctan(2)$. The object is placed at $u = -30$ cm from the convex lens ($f=20$ cm).
Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$, for point A, $u_A = -30$ cm, $v_A = 60$ cm.
For point B, we consider its components.
For the x-component (along principal axis): $dx = 1$ cm, $M_L = (\frac{v}{u})^2 = (\frac{60}{-30})^2 = 4$. So, $dx' = M_L dx = 4 \times 1 = 4$ cm.
For the y-component (perpendicular to principal axis): $dy = 2$ cm, $M_T = \frac{v}{u} = \frac{60}{-30} = -2$. So, $dy' = M_T dy = -2 \times 2 = -4$ cm.
The image coordinates relative to A' are $(4, -4)$. The angle $\theta$ made by the image with the principal axis is $\arctan(\frac{-4}{4}) = \arctan(-1) = -45^\circ$.

Field due to an infinite sheet of charge is $E=\frac{\sigma}{2\varepsilon_0}$ For the left plate (+\sigma ): Field points away from positive sheet, so at midpoint it is toward right. $E_1=\frac{\sigma}{2\varepsilon_0}$ For the right plate (−2\sigma ): Field points toward negative sheet, so at midpoint it is also toward right. Magnitude $E_2=\frac{2\sigma}{2\varepsilon_0}=\frac{\sigma}{\varepsilon_0}$ Net field $E=E_1+E_2$ $=\frac{\sigma}{2\varepsilon_0}+\frac{\sigma}{\varepsilon_0}$ $=\frac{3\sigma}{2\varepsilon_0}$ Force on charge +q: F=qE so $F=\frac{3q\sigma}{2\varepsilon_0}$

The river flows due east with speed $v_r = 9 \text{ km h}^{-1}$. The boat can move \in still water with maximum speed $v_b = 27 \text{ km h}^{-1}$ and is steered at an angle $150^{\circ}$ to the river flow (angle measured counter-clockwise from east). Take east as the $+x$-axis and north as the $+y$-axis. Components of the boat's velocity relative to water: $v_{bx} = v_b \cos 150^{\circ} = 27 \cos 150^{\circ} = 27 \left(-\frac{\sqrt{3}}{2}\right) = -\frac{27\sqrt{3}}{2} \text{ km h}^{-1}$ $-(1)$ $v_{by} = v_b \sin 150^{\circ} = 27 \sin 150^{\circ} = 27 \left(\frac{1}{2}\right) = \frac{27}{2} \text{ km h}^{-1}$ $-(2)$ The river velocity has only an $x$-component: $v_{rx} = +9 \text{ km h}^{-1}, \qquad v_{ry}=0$ Ground (resultant) velocity of the boat: $v_{gx} = v_{bx} + v_{rx} = -\frac{27\sqrt{3}}{2} + 9 \text{ km h}^{-1}$ $-(3)$ $v_{gy} = v_{by} + v_{ry} = \frac{27}{2} \text{ km h}^{-1}$ $-(4)$ To cross the river we need the northward component $v_{gy}$ only. Hence the time taken $t$ and river width $D$ are related by $D = v_{gy}\, t$ $-(5)$ The boat crosses \in half a minute: $t = \frac{1}{2} \text{ min} = \frac{1}{2}\times \frac{1}{60} \text{ h} = \frac{1}{120} \text{ h}$ $-(6)$ Substituting $v_{gy}$ from $(4)$ into $(5)$: $D = \frac{27}{2} \times \frac{1}{120} \text{ km}$ Simplify: $D = \frac{27}{2 \times 120} \text{ km} = \frac{27}{240} \text{ km} = 0.1125 \text{ km}$ Convert kilometres to metres: $D = 0.1125 \times 1000 = 112.5 \text{ m}$ Therefore, the width of the river is $112.5 \text{ m}$, which corresponds to Option B .

Let the first charge $+q$ be at the origin $O(0,0,0)$ and the second charge $+9q$ be at $P(d,0,0)$ on the x-axis. Because both charges are positive, the electric field produced by each charge at any point on the line joining them points away from the respective charge. Hence, at a point \in between the two charges the field contributions will be \in opposite directions and can cancel out. Assume the required point $R$ is at $x$ (with $0 \lt x \lt d$). Position of $R$: $(x,0,0)$. Magnitude of field at $R$ due to $+q$ at the origin: $E_1 = \frac{kq}{x^{2}}$ directed along $+\hat{i}$. Magnitude of field at $R$ due to $+9q$ at $P(d,0,0)$: $E_2 = \frac{k(9q)}{(d-x)^{2}}$ directed along $-\hat{i}$ (to the left). For the net field to vanish, the magnitudes must be equal: $E_1 = E_2 \Longrightarrow \frac{kq}{x^{2}} = \frac{k(9q)}{(d-x)^{2}}$. Cancelling the common factors $kq$ gives $\frac{1}{x^{2}} = \frac{9}{(d-x)^{2}}$. Taking square roots: $\frac{1}{x} = \frac{3}{d-x} \Longrightarrow d - x = 3x$. Solving for $x$: $d = 4x \Longrightarrow x = \frac{d}{4}$. Therefore the electric field is zero at the point $(d/4, 0, 0)$. Hence, the correct option is Option B .

The battery is specified by two numbers: voltage $V = 4.2\text{ V}$ and capacity $C = 5800\text{ mAh}$. Step 1 Convert the capacity into ampere-hours (Ah) Since $1000\text{ mAh} = 1\text{ Ah}$, we have $C = \frac{5800}{1000}\text{ Ah} = 5.8\text{ Ah}$ Step 2 Convert ampere-hours to coulombs (C) By definition, $1\text{ A} = 1\text{ C s}^{-1}$ and there are $3600\text{ s}$ \in one hour. Hence $1\text{ Ah} = 1\text{ A}\times3600\text{ s} = 3600\text{ C}$ Therefore, $Q = 5.8\text{ Ah}\times3600\text{ C Ah}^{-1} = 20880\text{ C}$ Step 3 Calculate the stored energy Electrical energy stored, $E = VQ$, so $E = 4.2\text{ V}\times20880\text{ C} = 87696\text{ J}$ Step 4 Express the result \in kilojoules $1\text{ kJ} = 10^{3}\text{ J}$, hence $E = \frac{87696}{10^{3}}\text{ kJ} \approx 87.7\text{ kJ}$ Thus, the energy stored \in the fully charged mobile phone battery is about $87.7\text{ kJ}$. Option C is correct.

The net displacement of the particle is the algebraic sum of the two given SHMs that have the same angular frequency $\omega = 5 \text{ s}^{-1}$: $x = x_1 + x_2 = \sqrt{7}\,\sin 5t + 2\sqrt{7}\,\sin\left(5t + \frac{\pi}{3}\right) \text{ cm}$ When two SHMs with the same $\omega$ are added, the result is again an SHM with the same $\omega$ but with a new amplitude $R$ obtained from the phasor (vector) addition formula: $R = \sqrt{A_1^{2} + A_2^{2} + 2A_1A_2\cos\phi}$ where $A_1 = \sqrt{7}\text{ cm},\quad A_2 = 2\sqrt{7}\text{ cm},\quad\phi = \frac{\pi}{3}$ Compute each term: $A_1^{2} = 7,\quad A_2^{2} = 28,$ $2A_1A_2\cos\phi = 2(\sqrt{7})(2\sqrt{7})\cos\frac{\pi}{3} = 4\cdot7\cdot\frac12 = 14$ Therefore, $R = \sqrt{7 + 28 + 14} = \sqrt{49} = 7\text{ cm}$ The combined motion can thus be written as $x = 7\sin(5t + \delta)\text{ cm}$ for some phase constant $\delta$. The acceleration \in SHM is given by $a = -\omega^{2}x$. Hence, the magnitude of the maximum acceleration is $a_{\max} = \omega^{2}R$ Convert the amplitude to metres: $R = 7\text{ cm} = 7\times10^{-2}\text{ m}$. Now substitute $\omega = 5\text{ s}^{-1}$: $a_{\max} = (5)^{2}\,(7\times10^{-2}) = 25 \times 0.07 = 1.75\text{ m s}^{-2}$ The question states $a_{\max} = x \times 10^{-2}\text{ m s}^{-2}$. Write $1.75$ \in that form: $1.75 = 175 \times 10^{-2}$ Hence, $x = 175$. Option A is correct.

The magnetisation $M$ produced \in a material kept \in an external magnetic field $H$ is written as $M = \chi\,H$, where $\chi$ is called the magnetic susceptibility. The magnetic induction $B$ inside the same material is given by $B = \mu\,H$, where $\mu$ is the magnetic permeability of the material. For vacuum (or free space) the induction is $B_0 = \mu_0\,H$, so the relative permeability of the material is defined as $\mu_r = \frac{\mu}{\mu_0}$. By definition we can also write $B = \mu_0\,(H + M)$. Substituting $M = \chi\,H$ gives $B = \mu_0\,(H + \chi H) = \mu_0\,(1 + \chi)\,H$. Comparing the two expressions for $B$: $\mu\,H = \mu_0\,(1 + \chi)\,H$. Cancelling the common factor $H$, we get $\mu = \mu_0\,(1 + \chi)$. Divide both sides by $\mu_0$: $\frac{\mu}{\mu_0} = 1 + \chi$. The left side is just $\mu_r$, so $\mu_r = 1 + \chi \quad$(relation between $\mu_r$ and $\chi$). Rearranging for susceptibility: $\chi = \mu_r - 1 = \frac{\mu}{\mu_0} - 1$. Thus the correct option is $\chi = \frac{\mu}{\mu_0} - 1$ → Option A.

The zener diode keeps the voltage across itself and the parallel load fixed at $V_Z = 5\text{ V}$. The source voltage is $V_S = 25\text{ V}$ and the series resistor is $R_S = 400\,\Omega$. Hence the voltage across $R_S$ equals $V_S - V_Z = 25 - 5 = 20\text{ V}$. Current through the series resistor (which is the \sum of zener current $I_Z$ and load current $I_L$) is therefore $I_S = \frac{20}{400} = 0.05\text{ A} = 50\text{ mA}$ $-(1)$ Given that the zener current is four \times the load current, $I_Z = 4 I_L$ $-(2)$ From $I_S = I_Z + I_L$ and substituting from $(2)$: $I_S = 4 I_L + I_L = 5 I_L$ $-(3)$ Using $(1)$ and $(3)$: $5 I_L = 50\text{ mA} \;\Longrightarrow\; I_L = 10\text{ mA}$ The load resistance is found from Ohm's law using the regulated voltage: $R_L = \frac{V_Z}{I_L} = \frac{5\text{ V}}{10\text{ mA}} = 500\,\Omega$ Thus, $I_L = 10\text{ mA}$ and $R_L = 500\,\Omega$. This matches Option D.

For any thermodynamic process the first law of thermodynamics states $dQ = dU + dW$ where $dQ$ = heat supplied to the system, $dU$ = change \in internal energy and $dW$ = work done by the system. Adiabatic process: By definition no heat is exchanged with the surroundings, hence $dQ = 0$ $-(1)$ Putting $dQ = 0$ \in the first-law equation $-(1)$ gives $0 = dU + dW \;\;\Longrightarrow\;\; dU = -\,dW$ $-(2)$ Equation $-(2)$ shows that the work done by the gas is numerically equal to the decrease (negative) \in its internal energy, or equivalently, the increase \in internal energy is numerically equal to the negative of the work done. Now recall the definition of molar heat capacity for any path: $C = \frac{1}{n}\,\frac{dQ}{dT}$ where $n$ is the number of moles. For an adiabatic process $dQ = 0$, therefore $C = \frac{1}{n}\,\frac{0}{dT} = 0$ $-(3)$ Hence the molar heat capacity along an adiabatic path is zero. Let us test each option: Option A The molar heat capacity is infinite - contradicts $-(3)$ (false). Option B Work done by the gas equals the increase \in internal energy - according to $-(2)$ the signs are opposite (false). Option C The molar heat capacity is zero - matches $-(3)$ (true). Option D The internal energy of the gas decreases as the temperature increases - no such rule exists; internal energy is directly related to temperature for an ideal gas (false). Therefore the correct statement is Option C: the molar heat capacity in an adiabatic process is zero.

For the lamina to remain stationary, the net torque about the pivot O must be zero: $\sum \tau_O = 0$ $\tau=\text{Force}\times \text{Perpendicular distance}$ $At\ Point\ A:The\ vertical\ 10\text{ N}\ (down)\ creates\ a\ clockwise\ torque:\ \tau_1=-10\cdot L$ $At\ Point\ B:The\ vertical\ 10\text{ N }(up)\ creates\ a\ counter\ clockwise\ torque:\ \tau_2=+10\cdot L$ $The\ horizontal\ 10\text{ N}\ (right)\ creates\ a\ clockwise\ torque:\ \tau_3=-10\cdot L$ $At\ Point\ C:The\ horizontal\ force\ F\ (left)\ creates\ a\ counter\ clockwise\ torque:\ \tau_4=+F\cdot L$ $The\ vertical\ force\ passes\ through\ O,\ so\ its\ torque\ is\ 0.$ $\sum \tau_O = (-10 \cdot L) + (10 \cdot L) + (-10 \cdot L) + (F \cdot L) = 0$ $-10L + FL = 0$ $F = 10 \text{ N}$

For a circular coil of radius $R$ carrying current $I$, the standard results for magnetic field are: \bullet At the centre of the coil (on its plane) $B_1 = \frac{\mu_0 I}{2R} \quad -(1)$ \bullet On the axis of the coil at a distance $x$ from the centre $B_2 = \frac{\mu_0 I R^{2}}{2\left(R^{2}+x^{2}\right)^{3/2}} \quad -(2)$ To obtain the required ratio, divide $(2)$ by $(1)$: $\frac{B_2}{B_1} = \frac{\mu_0 I R^{2}}{2\left(R^{2}+x^{2}\right)^{3/2}} \times \frac{2R}{\mu_0 I} = \frac{R^{3}}{\left(R^{2}+x^{2}\right)^{3/2}} \quad -(3)$ Rewrite $(3)$ more compactly: $\frac{B_2}{B_1} = \left(\frac{R^{2}}{R^{2}+x^{2}}\right)^{3/2} \quad -(4)$ The problem specifies $x : R = 3 : 4$, so choose $x = 3k$ and $R = 4k$ for some constant $k$. Substitute these into $(4)$: $\frac{B_2}{B_1} = \left( \frac{(4k)^{2}}{(4k)^{2} + (3k)^{2}} \right)^{3/2} = \left( \frac{16k^{2}}{16k^{2} + 9k^{2}} \right)^{3/2} = \left( \frac{16}{25} \right)^{3/2} \quad -(5)$ Evaluate the power: $\left( \frac{16}{25} \right)^{3/2} = \frac{16^{3/2}}{25^{3/2}} = \frac{( \sqrt{16} )^{3}}{( \sqrt{25} )^{3}} = \frac{4^{3}}{5^{3}} = \frac{64}{125}$ Hence, $\displaystyle \frac{B_2}{B_1} = \frac{64}{125}$, or \in ratio form $64 : 125$. Therefore, the correct option is Option C (64 : 125) .

The energy of an electron in a Bohr orbit of any hydrogen-like species (single-electron ion) is given by $E_n = -\dfrac{13.6\,\text{eV}\,Z^{2}}{n^{2}}$ where $Z$ is the atomic number and $n = 1,2,3,\ldots$ is the principal quantum number. For the hydrogen atom $\left(Z = 1\right)$ \in its ground state $\left(n = 1\right)$: $E_{\text{H, ground}} = -\dfrac{13.6 \cdot 1^{2}}{1^{2}} = -13.6\ \text{eV}$ Case A: He$^{+}$ ion has $Z = 2$. First excited state means $n = 2$. $E_{\text{He}^{+},\,n=2} = -\dfrac{13.6 \cdot 2^{2}}{2^{2}} = -13.6\ \text{eV}$ This equals the hydrogen ground-state energy, so statement (A) is correct. Case B: Li$^{++}$ ion has $Z = 3$. Second excited state means $n = 3$. $E_{\text{Li}^{++},\,n=3} = -\dfrac{13.6 \cdot 3^{2}}{3^{2}} = -13.6\ \text{eV}$ Again equal to the hydrogen ground-state energy, so statement (B) is correct. Case C: He$^{+}$ ground state: $n = 1$. $E_{\text{He}^{+},\,n=1} = -\dfrac{13.6 \cdot 2^{2}}{1^{2}} = -54.4\ \text{eV}$ This is not equal to $-13.6\ \text{eV}$, so statement (C) is incorrect. Case D: He$^{+}$ first excited state energy has already been found as $-13.6\ \text{eV}$. Li$^{++}$ ground state: $n = 1$. $E_{\text{Li}^{++},\,n=1} = -\dfrac{13.6 \cdot 3^{2}}{1^{2}} = -122.4\ \text{eV}$ These are not equal, so statement (D) is incorrect. Only statements (A) and (B) are correct. Hence the correct option is Option B: (A), (B) only.

The given rod has mass $M$ and length $L$. For a slender rod, the moment of inertia about an axis passing through its centre and perpendicular to its length is $I = \frac{1}{12}\,M\,L^{2}$ The question denotes this value by $\alpha$, so write $\alpha = \frac{1}{12}\,M\,L^{2} \; \; -(1)$ Next, cut the rod into two equal parts. Each part therefore has length $=\frac{L}{2}$ mass $=\frac{M}{2}$ These two half-rods are joined at right angles through their centres to form a symmetric cross (a '+' shape). The required axis passes through the common centre and is perpendicular to the plane of the cross. For one half-rod of length $\tfrac{L}{2}$, the moment of inertia about this axis is again given by the same formula $I=\tfrac{1}{12}ML^{2}$ with $M \rightarrow \frac{M}{2}$ and $L \rightarrow \frac{L}{2}$: $I_{\text{one half}} = \frac{1}{12}\left(\frac{M}{2}\right)\left(\frac{L}{2}\right)^{2} = \frac{1}{12}\,\frac{M}{2}\,\frac{L^{2}}{4} = \frac{M\,L^{2}}{96}$ The cross consists of two such identical half-rods, so total moment of inertia is $I_{\text{cross}} = 2 \times \frac{M\,L^{2}}{96} = \frac{M\,L^{2}}{48}$ From $(1)$ we have $M\,L^{2} = 12\,\alpha$. Substituting this into the above result: $I_{\text{cross}} = \frac{12\,\alpha}{48} = \frac{\alpha}{4}$ Hence, the moment of inertia of the cross about the stated axis is $\alpha/4$. Answer: Option B $(\alpha/4)$

Using the refraction formula for a spherical surface: $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$.
Given $n_1 = 1$, $n_2 = 1.5$, $u = -0.2$ m (object to the left), $R = -0.4$ m (convex surface from the perspective of light entering from medium 1).
$\frac{1.5}{v} - \frac{1}{-0.2} = \frac{1.5 - 1}{-0.4}$
$\frac{1.5}{v} + 5 = \frac{0.5}{-0.4} = -1.25$
$\frac{1.5}{v} = -1.25 - 5 = -6.25$
$v = \frac{1.5}{-6.25} = -0.24$ m.
The negative sign indicates the image is formed to the left of the spherical surface.

The target answer is incorrect.

We first write one standard defining relation for each physical quantity and then deduce its dimensional formula. Case A: Coefficient of viscosity ($\eta$) Shear stress $\tau = \eta \dfrac{dv}{dy}\quad -(1)$ \bullet Dimension of shear stress $\tau$ is the same as pressure: $[M\,L^{-1}\,T^{-2}]$. \bullet Velocity gradient $\dfrac{dv}{dy}$ has dimension $T^{-1}$. Therefore, $\eta = \dfrac{\tau}{dv/dy} \; \Rightarrow \; \$[M\,L^{-1}\,T^{-2}]\$ \; / \; \$[T^{-1}]\$ = [M\,L^{-1}\,T^{-1}]$. Hence $\eta$ corresponds to List-II (IV). Case B: Intensity of wave ($I$) Intensity $I = \dfrac{\text{Power}}{\text{Area}} \quad -(2)$ \bullet Power $P = \dfrac{\text{Energy}}{\text{time}}$. Energy has dimension $[M\,L^{2}\,T^{-2}]$, so $P = [M\,L^{2}\,T^{-3}]$. \bullet Area has dimension $[L^{2}]$. Therefore, $I = \dfrac{[M\,L^{2}\,T^{-3}]\$}{[L^{2}]\$} = [M\,L^{0}\,T^{-3}]$. Hence $I$ corresponds to List-II (I). Case C: Pressure gradient ($\nabla P$) Pressure gradient means pressure per unit length: $\nabla P = \dfrac{P}{L} \quad -(3)$ \bullet Pressure $P$ has dimension $[M\,L^{-1}\,T^{-2}]$. \bullet Length $L$ has dimension $[L]$. Therefore, $\nabla P = \dfrac{[M\,L^{-1}\,T^{-2}]\$}{[L]} = [M\,L^{-2}\,T^{-2}]$. Hence $\nabla P$ corresponds to List-II (II). Case D: Compressibility ($\beta$) Compressibility is the reciprocal of bulk modulus (pressure): $\beta = \dfrac{1}{P} \quad -(4)$ Since the dimension of pressure is $[M\,L^{-1}\,T^{-2}]$, we get $\beta = [M^{-1}\,L\,T^{2}]$. Hence $\beta$ corresponds to List-II (III). Final matching (A) $\eta$ \to (IV) $[ML$^{-1}$T$^{-1}$]$ (B) $I$ \to (I) $[ML$^{0}$T$^{-3}$]$ (C) $\nabla P$ \to (II) $[ML$^{-2}$T$^{-2}$]$ (D) $\beta$ \to (III) $[M$^{-1}$LT$^{2}$]$ Thus the correct set is: (A)-(IV), (B)-(I), (C)-(II), (D)-(III), which is given in Option B. Answer : Option B

For an infinite non-conducting sheet, electric field is $E=\frac{\sigma}{2\varepsilon_0}$ Force on charged bob due to field: $F_e=qE$ $=q\frac{\sigma}{2\varepsilon_0}$ At equilibrium forces balance. Weight downward: mg Electric force horizontal. Tension along string. Resolving tension: $T\cos45^{\circ}=mg$ $T\sin45^{\circ}=F_e$ Dividing, $\tan45^{\circ}=\frac{F_e}{mg}$ Since $\tan45^{\circ}=1$ $F_e=mg$ So $q\frac{\sigma}{2\varepsilon_0}=mg$ Thus $\sigma=\frac{2\varepsilon_0mg}{q}$ Now Mass: $100mg=100\times10^{-6}kg=10^{-4}kg$ Charge: $q=10\mu C=10^{-5}C$ Substitute: \sigma $=\frac{2(8.85\times10^{-12})(10^{-4})(10)}{10^{-5}}$ $=17.7\times10^{-11}$ $=1.77\times10^{-10}\ \text{C/m}^2$

When monochromatic light of wavelength $\lambda$ is incident on a metal surface of work function $\phi$, the photo-electric equation is $h\nu = \phi + K_{\max}$ Using $\nu = \dfrac{c}{\lambda}$ we have Maximum kinetic energy of an emitted electron: $K_{\max}= \dfrac{hc}{\lambda} - \phi$ $-(1)$ Relating kinetic energy to speed: $K_{\max}= \dfrac{1}{2} m v^{2}$ Hence the maximum speed of the electron is $v = \sqrt{\dfrac{2K_{\max}}{m}} = \sqrt{\dfrac{2}{m}\left(\dfrac{hc}{\lambda} - \phi\right)}$ $-(2)$ The magnetic field $\mathbf{B}$ is perpendicular to the initial velocity, so the electron moves \in a uniform circular path. For a charge $e$ moving with speed $v$ perpendicular to $\mathbf{B}$, the radius of the circular path is given by $r = \dfrac{mv}{eB}$ $-(3)$ Substituting $v$ from $(2)$ into $(3)$: $r = \dfrac{m}{eB}\,\sqrt{\dfrac{2}{m}\left(\dfrac{hc}{\lambda} - \phi\right)} = \dfrac{\sqrt{2m\left(\dfrac{hc}{\lambda} - \phi\right)}}{eB}$ $-(4)$ Geometry of the motion: \bullet The electron is emitted normally from point A on the plate (the plate is the initial plane, say $z = 0$). \bullet Because $\mathbf{B}$ is \in the plane of the plate and $\mathbf{v}$ is perpendicular to the plate, the circular path lies \in a plane perpendicular to $\mathbf{B}$ and intersects the plate at two points. \bullet Starting from A, the electron completes half a circle (angle $\pi$) before arriving back at the plate, now at point B. \bullet The chord joining the two end points of a semicircle is the diameter, whose length is $2r$. Therefore the distance AB is $AB = 2r = 2 \times \dfrac{\sqrt{2m\left(\dfrac{hc}{\lambda} - \phi\right)}}{eB} = \dfrac{\sqrt{8m\left(\dfrac{hc}{\lambda} - \phi\right)}}{eB}$ $-(5)$ The expression \in $(5)$ matches Option C. Final answer: Option C