JEE Main 2026 April 2 shift 2

Under Newton's law, the gravitational force between two point masses is $F=\dfrac{G\,m_1 m_2}{r^{2}}$ Writing the dimensions: $[F]=MLT^{-2},\qquad [m_1]=[m_2]=M,\qquad[r]=L$ Hence $MLT^{-2}=G\,M\;M\;L^{-2}$ which gives the dimensional formula of the universal gravitational constant as $[G]=M^{-1}L^{3}T^{-2}$ Planck's constant has the well-known dimensions of action (energy \times time): $[h]=ML^{2}T^{-1}$ We want to rewrite $[G]$ \in the form $h^{a}L^{b}M^{c}T^{d}$. Assume $[G]=[h]^{a}L^{b}M^{c}T^{d}$ Substituting $[h]=ML^{2}T^{-1}$: $M^{-1}L^{3}T^{-2}=(M^{1}L^{2}T^{-1})^{a}L^{b}M^{c}T^{d}$ $\Rightarrow M^{-1}L^{3}T^{-2}=M^{a+c}\;L^{2a+b}\;T^{-a+d}$ Equating the exponents of each fundamental dimension: For mass $M$ : $a+c=-1$ For length $L$ : $2a+b=3$ For time $T$ : $-a+d=-2$ A very simple choice that satisfies all three equations is $a=1,\; b=1,\; c=-2,\; d=-1$ Therefore $[G]=[h]^{1}L^{1}M^{-2}T^{-1}=h\,T^{-1}L\,M^{-2}$ That matches Option B. Answer: Option B which is: $[hT^{-1}LM^{-2}]$

Let the mass of the block be $m = 0.5 \text{ kg}$ and the radius of the cylindrical drum be $r = 4 \text{ m}$. The drum rotates with a minimum angular speed $\omega = 5 \text{ rad s}^{-1}$ so that the block just stays stuck to the wall without sliding down. 1. Identify the forces on the block \bullet Weight acts vertically downward: $mg$. \bullet Normal reaction from the wall acts horizontally towards the axis: $N$. \bullet Static friction acts vertically upward (opposing the possible downward motion): $f_s$, with the limiting value $f_{\max} = \mu N$. 2. Centripetal force requirement The normal reaction is the only horizontal force and therefore provides the necessary centripetal force: $N = m\omega^{2}r \quad -(1)$ 3. Condition for vertical equilibrium At the minimum speed, static friction is at its limiting value and exactly balances the weight: $f_{\max} = mg \quad\Longrightarrow\quad \mu N = mg \quad -(2)$ 4. Substitute $N$ from (1) into (2) $\mu \bigl(m\omega^{2}r\bigr) = mg$ $\mu = \frac{mg}{m\omega^{2}r} = \frac{g}{\omega^{2}r}$ 5. Insert the numerical values $\mu = \frac{10}{(5)^{2} \times 4} = \frac{10}{25 \times 4} = \frac{10}{100} = 0.1$ Hence, the coefficient of friction between the block and the drum is $0.1$. Option A which is: 0.1

Let the heavier block have mass $m_1 = 2\; \text{kg}$ and the lighter block have mass $m_2 = 1\; \text{kg}$. The system is an ideal Atwood machine (light, inextensible string and light, frictionless pulley). Step 1 - Acceleration of the system For an Atwood machine, the magnitude of the common acceleration is $a = \frac{(m_1 - m_2) g}{m_1 + m_2}$ $\Rightarrow a = \frac{(2 - 1)\,g}{2 + 1} = \frac{g}{3} = \frac{10}{3}\;\text{m s}^{-2}$ The 2 kg block accelerates downward, the 1 kg block upward, both with the same magnitude $a$. Step 2 - Individual displacements \in 2 s Starting from rest, the distance travelled \in time $t$ under constant acceleration $a$ is $s = \frac{1}{2}\,a\,t^{2}$ For $t = 2\;\text{s}$, $s = \frac{1}{2}\,\left(\frac{10}{3}\right)\,(2)^{2} = \frac{1}{2}\,\left(\frac{10}{3}\right)\,4 = \frac{20}{3}\;\text{m} \approx 6.67\;\text{m}$ Thus, after 2 s: \bullet the 2 kg block has moved $+s$ (downward) \bullet the 1 kg block has moved $-s$ (upward) Step 3 - Position of the centre of mass Choose the initial horizontal level as origin and take downward as positive. The centre-of-mass coordinate after 2 s is $y_{\text{cm}} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2}$ $\Rightarrow y_{\text{cm}} = \frac{\,2(+s) + 1(-s)}{2 + 1} = \frac{(2 - 1)s}{3} = \frac{s}{3}$ Substituting $s = \frac{20}{3}\;\text{m}$: $y_{\text{cm}} = \frac{1}{3}\left(\frac{20}{3}\right) = \frac{20}{9}\;\text{m} \approx 2.22\;\text{m}$ The centre of mass has shifted downward by $\mathbf{2.22\;m}$ \in 2 s. Option C which is: $2.22$

The net Lorentz force on a charged particle is given by $\mathbf{F}=q\left(\mathbf{E}+\mathbf{v}\times \mathbf{B}\right)$ Given data: Charge $q = 10^{-9}\,{\rm C}$ Electric field $\mathbf{E}=0.4\,\hat{\jmath}\,{\rm N\,C^{-1}}$ Magnetic field $\mathbf{B}=4\times10^{-3}\,\hat{k}\,{\rm T}$ Force $\mathbf{F}=(4\hat{\imath}+2\hat{\jmath})\times10^{-10}\,{\rm N}$ Let the velocity \in the x-y plane be $\mathbf{v}=v_x\hat{\imath}+v_y\hat{\jmath}$. Its cross-product with $\mathbf{B}$ (which is along $\hat{k}$) is $\mathbf{v}\times \mathbf{B}=(v_x\hat{\imath}+v_y\hat{\jmath})\times (B\hat{k})$ $\qquad =B\bigl(v_y\hat{\imath}-v_x\hat{\jmath}\bigr)$ Substituting $B=4\times10^{-3}\,{\rm T}$, $\mathbf{v}\times \mathbf{B}=4\times10^{-3}\bigl(v_y\hat{\imath}-v_x\hat{\jmath}\bigr).$ Add the electric field: $\mathbf{E}+\mathbf{v}\times \mathbf{B}=0.4\hat{\jmath}+4\times10^{-3}(v_y\hat{\imath}-v_x\hat{\jmath}).$ Now apply $\mathbf{F}=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$ component-wise. i-component : $q\,(4\times10^{-3}v_y)=4\times10^{-10}$ $\Rightarrow (10^{-9})(4\times10^{-3})v_y=4\times10^{-10}$ $\Rightarrow 4\times10^{-12}v_y=4\times10^{-10}$ $\Rightarrow v_y=100\,{\rm m/s}.$ j-component : $q\left(0.4-4\times10^{-3}v_x\right)=2\times10^{-10}$ $\Rightarrow (10^{-9})\left(0.4-4\times10^{-3}v_x\right)=2\times10^{-10}$ $\Rightarrow 0.4-4\times10^{-3}v_x=0.2$ $\Rightarrow 4\times10^{-3}v_x=0.2$ $\Rightarrow v_x=\frac{0.2}{4\times10^{-3}}=50\,{\rm m/s}.$ Therefore $\mathbf{v}=50\hat{\imath}+100\hat{\jmath}\;{\rm m/s}.$ Option A which is: $50\hat{\imath}+100\hat{\jmath}\;{\rm m/s}$

Assume positive-logic levels: LOW = $0\text{ V}$, HIGH = $+5\text{ V}$. In the given network the anodes of two identical silicon diodes are tied to the input lines $X$ and $Y$, while their cathodes meet at node $O$. Node $O$ is pulled up to $+5\text{ V}$ through a resistor $R_L$ (the load resistor) and is the output. Step 1 - Input combination $X = 0$, $Y = 0$ Both anodes are at $0\text{ V}$. Each diode is forward-biased (anode $0\text{ V}$, cathode initially $5\text{ V}$). They conduct and clamp node $O$ close to $0.7\text{ V}$ (the diode drop). The output is interpreted as LOW. Step 2 - Input combination $X = 0$, $Y = 1$ The diode connected to $X$ is forward-biased (anode $0\text{ V}$), so it again pulls node $O$ to about $0.7\text{ V}$. The other diode is reverse-biased, but the conducting one is enough to keep the output LOW. Step 3 - Input combination $X = 1$, $Y = 0$ This is symmetrical to Step 2. The diode linked with $Y$ conducts, forcing the output LOW (\approx $0.7\text{ V}$). Step 4 - Input combination $X = 1$, $Y = 1$ Now both diode anodes and cathodes sit at $+5\text{ V}$, so each diode is reverse-biased. No current flows through the diodes. The pull-up resistor $R_L$ therefore lifts node $O$ to $+5\text{ V}$, producing a HIGH output. Step 5 - Truth table

The body starts from rest at infinity, so its initial mechanical energy is zero (both kinetic and potential energies are zero at infinity). When it reaches the Earth's surface, let its speed be $v$. Its kinetic energy is $\dfrac12\,mv^{2}$ and its gravitational potential energy is $-\dfrac{GMm}{R}$, where $G$ is the universal gravitational constant, $M$ is Earth's mass, $R$ is Earth's radius. Using conservation of mechanical energy (initial energy = final energy): $0 = \dfrac12\,mv^{2} - \dfrac{GMm}{R}$ $\Rightarrow \dfrac12\,mv^{2} = \dfrac{GMm}{R}$ $\Rightarrow v^{2} = \dfrac{2GM}{R} \qquad -(1)$ Recall that $g = \dfrac{GM}{R^{2}}$. Multiplying both sides by $R$ gives $gR = \dfrac{GM}{R}$. Substitute this \in $(1)$: $v^{2} = 2gR \qquad -(2)$ Numerical substitution (convert radius to metres): $R = 6400\text{ km} = 6400 \times 10^{3}\text{ m} = 6.4 \times 10^{6}\text{ m}$ Using $g = 9.8\text{ m s}^{-2}$, from $(2)$ we get $v = \sqrt{2 \times 9.8 \times 6.4 \times 10^{6}}$ $v = \sqrt{1.2544 \times 10^{8}}$ $v = 1.12 \times 10^{4}\text{ m s}^{-1}$ $v = 11.2\text{ km s}^{-1}$ The kinetic energy on arrival is $k = \dfrac12\,mv^{2}$ For $m = 1\text{ kg}$, $k = \dfrac12 \times 1 \times (1.12 \times 10^{4})^{2}$ $k = \dfrac12 \times 1.2544 \times 10^{8}$ $k = 6.272 \times 10^{7}\text{ J}$ $k \approx 6.27 \times 10^{7}\text{ J}$ Thus, $v = 11.2\text{ km s}^{-1}$ and $k = 6.27 \times 10^{7}\text{ J}$. Option A which is: 11.2 km/s; 6.27 × 107 J

The pitch of the screw gauge is given as $0.1\text{ mm}$ and the circular scale has $100$ equal divisions. Least count (L.C.) of the screw gauge is therefore $\text{L.C.}= \frac{\text{pitch}}{\text{number of circular divisions}} = \frac{0.1\text{ mm}}{100}=0.001\text{ mm}$ Zero error : When the two studs just touch, the zero on the main scale should coincide with the zero on the circular scale. Instead, the 5th circular division coincides with the reference line. Thus the screw shows an excess of $5 \times \text{L.C.}=5 \times 0.001 = 0.005\text{ mm}$ Because the instrument reads more than the true length, the zero error is positive $+0.005\text{ mm}$. It must be subtracted from subsequent readings. Reading with the sphere \in place Main-scale reading (MSR) $= 5\text{ mm}$ Circular-scale reading (CSR) $= 50$ divisions Contribution of CSR $= 50 \times \text{L.C.}=50 \times 0.001 = 0.050\text{ mm}$ Observed diameter (before correction) $= \text{MSR} + \text{CSR} = 5.000\text{ mm} + 0.050\text{ mm}=5.050\text{ mm}$ Corrected diameter $\text{True diameter} = \text{observed reading} - (\text{positive zero error})$ $= 5.050\text{ mm} - 0.005\text{ mm}=5.045\text{ mm}$ Hence, the diameter of the sphere is $5.045\text{ mm}$. Option A which is: $5.045$ mm

A soap bubble has TWO liquid surfaces (inner & outer). Hence, to increase its size, the work done equals the increase in surface energy of both surfaces. Surface tension is the energy per unit surface area. Therefore $\text{Work} = 2 \, T \, \Delta A$ where $T = 0.03 \, \text{N m}^{-1}$ and $\Delta A$ is the change \in surface area of ONE spherical surface. Initial radius: $r_1 = \frac{2 \text{ cm}}{2} = 1 \text{ cm} = 0.01 \text{ m}$ Final radius: $r_2 = \frac{6 \text{ cm}}{2} = 3 \text{ cm} = 0.03 \text{ m}$ Surface area of a sphere is $A = 4\pi r^2$, so $\Delta A = 4\pi\left(r_2^{2} - r_1^{2}\right)$ Substituting into the work-expression gives W = 2T \, \Delta A = 2T \Bigl\$[4\pi\left(r_2^{2} - r_1^{2}\right)\Bigr]\$ = 8\pi T\left(r_2^{2} - r_1^{2}\right) Calculate $r_2^{2} - r_1^{2}$: $r_2^{2} - r_1^{2} = (0.03)^2 - (0.01)^2 = 0.0009 - 0.0001 = 0.0008 = 8 \times 10^{-4}$ Now compute the work: $W = 8\pi (0.03)\,(8 \times 10^{-4})$ $\;\;= (8 \times 0.03 \times 8)\,\pi \times 10^{-4}$ $\;\;= 1.92\,\pi \times 10^{-4} \text{ J}$ The question states the work as $\alpha\pi \times 10^{-4} \text{ J}$, so $\alpha = 1.92$ Option C which is: $1.92$

The ideal-gas equation is $PV = nRT$, where $n$ is the total number of moles present. Convert the given data to SI units: Volume : $V = 8310\ \text{cm}^3 = 8310 \times 10^{-6}\ \text{m}^3 = 8.31 \times 10^{-3}\ \text{m}^3$ Pressure : $P = 100\ \text{kPa} = 100 \times 10^{3}\ \text{Pa} = 1.00 \times 10^{5}\ \text{Pa}$ Temperature : $T = 300\ \text{K}$ Gas constant : $R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}$ Calculate the total moles \in the mixture: $n_{\text{total}} = \frac{PV}{RT} = \frac{(1.00 \times 10^{5}) (8.31 \times 10^{-3})}{8.31 \times 300}$ Numerator : $1.00 \times 10^{5} \times 8.31 \times 10^{-3} = 8.31 \times 10^{2} = 831$ Denominator : $8.31 \times 300 = 2493$ $n_{\text{total}} = \frac{831}{2493} \approx 0.333\ \text{mol}$ Let $n_1$ be the moles of $CO_2$ (molar mass $M_{CO_2}=44\ \text{g mol}^{-1}$) and $n_2$ be the moles of $O_2$ (molar mass $M_{O_2}=32\ \text{g mol}^{-1}$). Number-of-moles relation: $n_1 + n_2 = 0.333 \quad -(1)$ Mass relation (total mass given is $13.2\ \text{g}$): $44 n_1 + 32 n_2 = 13.2 \quad -(2)$ From $(1)$, $n_2 = 0.333 - n_1$. Substitute into $(2)$: $44 n_1 + 32 (0.333 - n_1) = 13.2$ $44 n_1 + 10.666 - 32 n_1 = 13.2$ $12 n_1 + 10.666 = 13.2$ $12 n_1 = 13.2 - 10.666 = 2.534$ $n_1 = \frac{2.534}{12} \approx 0.211\ \text{mol}$ Using $(1)$ again: $n_2 = 0.333 - 0.211 \approx 0.122\ \text{mol}$ Hence, moles of $CO_2 \approx 0.21\ \text{mol}$ and moles of $O_2 \approx 0.12\ \text{mol}$. Option C which is: $0.21$ and $0.12$

At steady (terminal) speed the upward buoyant force on the bubble is exactly balanced by the downward viscous drag exerted by the liquid. Buoyant force (because the weight of air in the bubble is negligible) is equal to the weight of the displaced liquid: $F_{\text{buoy}} = \rho V g = \rho\left(\frac{4}{3}\pi r^{3}\right)g$ For creeping (Stokes-flow) conditions, the viscous drag on a sphere of radius $r$ moving with speed $v$ \in a liquid of viscosity $\eta$ is $F_{\text{drag}} = 6\pi \eta r v$ At terminal velocity, $F_{\text{drag}} = F_{\text{buoy}}$, hence $6\pi \eta r v = \rho\left(\frac{4}{3}\pi r^{3}\right)g$ Simplifying (cancel $\pi$ and one factor of $r$): $6 \eta v = \frac{4}{3}\rho r^{2} g$ $\eta = \frac{2}{9}\,\frac{\rho g r^{2}}{v}$ $-(1)$ Insert the data: Radius $r = \frac{2\text{ mm}}{2} = 1\text{ mm} = 1\times10^{-3}\text{ m}$ Speed $v = 0.5\text{ cm s}^{-1} = 5\times10^{-3}\text{ m s}^{-1}$ Density $\rho = 2000\text{ kg m}^{-3}$ Acceleration $g = 10\text{ m s}^{-2}$ Compute the numerator of $(1)$: $\rho g r^{2} = 2000 \times 10 \times (1\times10^{-3})^{2} = 20000 \times 10^{-6} = 2\times10^{-2}$ Divide by $v$: $\frac{\rho g r^{2}}{v} = \frac{2\times10^{-2}}{5\times10^{-3}} = 4$ Finally, from $(1)$: $\eta = \frac{2}{9} \times 4 = \frac{8}{9}\text{ Pa·s} \approx 0.888\text{ Pa·s}$ Conversion to Poise (1 Pa·s = 10 Poise): $\eta \approx 0.888 \times 10 = 8.88\text{ Poise}$ Rounded to two significant figures, $\eta \approx 8.8\text{ Poise}$. Option B which is: $8.8$

The ball is released from rest, so its entire loss of gravitational potential energy is converted into the work done against the resisting force of the sand during the 10 cm penetration. Total vertical drop before coming to rest = free-fall height + penetration depth $h = 10 \text{ m} + 0.1 \text{ m} = 10.1 \text{ m}$ Loss of gravitational potential energy $\Delta U = m g h$ Here, $m = 2 \text{ kg}$ and (as usual for JEE) we take $g = 10 \text{ m s}^{-2}$. $\Delta U = 2 \times 10 \times 10.1 = 202 \text{ J}$ This entire 202 J is dissipated only while the ball moves the last $0.1 \text{ m}$ inside the sand. If $F_{\text{avg}}$ is the average upward (resisting) force exerted by the sand, the work it does is $W_{\text{sand}} = F_{\text{avg}} \times s$ where $s = 0.1 \text{ m}$ is the penetration depth. Because the ball finally stops, the work done by the sand equals the loss \in potential energy (energy conservation): $F_{\text{avg}} \times 0.1 = 202$ $\Rightarrow \; F_{\text{avg}} = \frac{202}{0.1} = 2020 \text{ N}$ Thus the average force exerted by the sand on the ball is Option B which is: $2020 \text{ N}$

The wave is moving in the $+x$-direction, so for a plane electromagnetic (EM) wave \in free space the three vectors $\vec{E}, \vec{B}, \vec{k}$ are mutually perpendicular and obey the right-hand rule $\vec{E}\times \vec{B} \;=\; \dfrac{1}{\mu_0}\,(\text{Poynting vector}) \propto \vec{k},$ therefore $\vec{E}\times \vec{B}$ must point \in the $+x$-direction. Given $\vec{B}=2\times10^{-7}\,\hat{j}\; \text{T}$ (along $+y$), let the unknown electric field be $\vec{E}=E_z\,\hat{k}$ (along $\pm z$). Compute the cross product: $\vec{E}\times \vec{B}=E_z\,\hat{k}\;\times \;2\times10^{-7}\,\hat{j}$ Use the cyclic relation $\hat{i}\times \hat{j}=\hat{k}, \;\hat{j}\times \hat{k}=\hat{i}, \;\hat{k}\times \hat{i}=\hat{j}$, so $\hat{k}\times \hat{j}=-\hat{i}.$ Hence $\vec{E}\times \vec{B}=E_z\;(2\times10^{-7})\,(-\hat{i}).$ To make this result point along $+\hat{i}$ (the propagation direction), we need $E_z$ to be negative. Therefore the electric field must point along $-\hat{k}$. In free space the magnitudes satisfy $|\vec{E}| = c\,|\vec{B}|$, where $c = 3\times10^{8}\,\text{m/s}$. Thus $|\vec{E}| = c\,|\vec{B}| = (3\times10^{8})(2\times10^{-7}) = 6.0\times10^{1} \;\text{V/m} = 60 \;\text{V/m}.$ Adding the direction found earlier, $\boxed{\;\vec{E} = -60\,\hat{k}\;\text{V/m}\;}.$ Hence the correct option is: Option B which is: $-60\hat{k}$ V/m

The bulb is not connected directly across the battery; it is placed in parallel with the $400\;\Omega$ resistor, and this parallel pair is \in series with the $200\;\Omega$ resistor and the 100 V source. Step 1 Find the resistance of the bulb. The rating 200 V, 100 W means the bulb draws 100 W when 200 V is applied. Using $P = \frac{V^{2}}{R}$, the resistance of the bulb is $R_{b} = \frac{(200)^{2}}{100} = 400\;\Omega$. Step 2 Find the equivalent resistance of the parallel branch. The bulb ($R_{b}=400\;\Omega$) is \in parallel with the given $400\;\Omega$ resistor $R_{2}$. For two equal resistances \in parallel, $R_{\text{parallel}} = \frac{R_{b}\,R_{2}}{R_{b}+R_{2}} = \frac{400 \times 400}{400+400} = \frac{160000}{800} = 200\;\Omega$. Step 3 Find the total resistance seen by the battery. This parallel combination ($200\;\Omega$) is \in series with the $200\;\Omega$ resistor $R_{1}$: $R_{\text{total}} = R_{1} + R_{\text{parallel}} = 200 + 200 = 400\;\Omega$. Step 4 Calculate the circuit current. Using Ohm's law with the 100 V source: $I = \frac{V}{R_{\text{total}}} = \frac{100}{400} = 0.25\;\text{A}$. Step 5 Find the potential drop across the bulb. The bulb shares the same potential as its parallel partner, so $V_{\text{bulb}} = I \times R_{\text{parallel}} = 0.25 \times 200 = 50\;\text{V}$. Hence, the potential drop across the bulb is $50\;\text{V}$. Option B which is: $50\;\text{V}$

Density of oil $\rho = 1.5 \,\text{g cm}^{-3} = 1.5 \times 10^{3}\,\text{kg m}^{-3}$ Radius of droplet $r = 1\,\text{mm} = 1 \times 10^{-3}\,\text{m}$ Volume of a sphere $V = \frac{4}{3}\pi r^{3} = \frac{4}{3}\pi(1 \times 10^{-3})^{3} = \frac{4}{3}\pi \times 10^{-9}\,\text{m}^{3}$ Mass of one droplet $m = \rho V = 1.5 \times 10^{3} \times \frac{4}{3}\pi \times 10^{-9} = 2\pi \times 10^{-6}\,\text{kg}$ Weight acting downward $W = mg = 2\pi \times 10^{-6} \times 10 = 2\pi \times 10^{-5}\,\text{N} \approx 6.28 \times 10^{-5}\,\text{N}$ Charge on each droplet $q = 5\,\text{nC} = 5 \times 10^{-9}\,\text{C}$ For the droplet to float, the upward electric force must equal its weight: $qE = mg \quad\Longrightarrow\quad E = \frac{mg}{q} = \frac{6.28 \times 10^{-5}}{5 \times 10^{-9}} = 1.26 \times 10^{4}\,\text{V m}^{-1}$ Separation between the plates $d = \frac{12}{\pi}\,\text{cm} = \frac{12}{\pi} \times 10^{-2}\,\text{m} = \frac{0.12}{\pi}\,\text{m} \approx 3.82 \times 10^{-2}\,\text{m}$ Required potential difference $\Delta V = E d = (1.26 \times 10^{4})(3.82 \times 10^{-2}) \approx 4.8 \times 10^{2}\,\text{V} \approx 480\,\text{V}$ Direction of field: the electric force must act upward (toward plate A, which is on top). For a positive charge, $\mathbf{E}$ points from higher to lower potential. Hence the bottom plate B must be at the higher potential. Choosing convenient values that differ by $480\,\text{V}$ with $V_B \gt V_A$, $V_A = 100\,\text{V}, \quad V_B = 580\,\text{V}$ These match Option A. Option A which is: 100 V and 580 V

The electric flux $\Phi$ through any closed surface is given by Gauss's law: $\Phi = \frac{Q_{\text{enc}}}{\varepsilon_{0}}$ where $Q_{\text{enc}}$ is the net charge enclosed by the surface and $\varepsilon_{0}$ is permittivity of free space. Because $\varepsilon_{0}$ is the same for both spheres, the ratio of fluxes equals the ratio of the enclosed charges. Sphere of radius 3 cm Distance of each charge from the origin:$8\,\mu{\rm C}$ at $x = 2\text{ cm}$ (inside, since $2 \lt 3$), $-2\,\mu{\rm C}$ at $x = 4\text{ cm}$ (outside, since $4 \gt 3$). Net charge enclosed: $Q_{3} = 8\,\mu{\rm C}$. Sphere of radius 5 cm Both charges lie within 5 cm of the origin (2 cm and 4 cm), so Net charge enclosed: $Q_{5} = 8\,\mu{\rm C} + (-2\,\mu{\rm C}) = 6\,\mu{\rm C}$. Therefore, ratio of electric fluxes $\frac{\Phi_{3}}{\Phi_{5}} = \frac{Q_{3}}{Q_{5}} = \frac{8\,\mu{\rm C}}{6\,\mu{\rm C}} = \frac{4}{3}.$ Hence the required ratio is $4 : 3$. Option C which is: $4 : 3$.

For total internal reflection to occur, the refractive index of the medium (painted face, $n_2$) must be greater than the refractive index of the prism ($n_1 = 1.6$) multiplied by the sine of the critical angle. The critical angle $\theta_c$ can be calculated using $\sin(\theta_c) = \frac{n_2}{n_1}$, leading to $n_2 > n_1 \sin(\theta_c)$. Since the prism is equilateral, the critical angle is $\frac{\pi}{3}$ radians, therefore we derive $n_2 > n_1 \cdot \frac{\sqrt{3}}{2}$. Solving this gives $n_2$ at a minimum value of $\frac{3.2}{\sqrt{3}}$ which approximates to option D.

The source voltage is $v = V_0 \sin 300 t$, hence the angular frequency of the supply is $\omega = 300 \,\text{rad s}^{-1}$. The capacitance is given as $C = 100 \,\mu\text{F} = 1 \times 10^{-4}\,\text{F}$. The capacitive reactance is therefore $X_C = \frac{1}{\omega C} = \frac{1}{300 \times 1 \times 10^{-4}} = 33.33 \,\Omega$. Let the resistance \in the series circuit be $R$ (it remains the same \in both cases). For an LCR series circuit the phase angle $\phi$ between source voltage and current is given by $\tan\phi = \frac{X_L - X_C}{R} = \frac{\omega L - X_C}{R}\,\,\,\, -(1)$ Case 1: Switch $S_1$ closed (inductance $L_1$ active) and $S_2$ open gives $\phi_1 = 30^\circ$. Using $(1)$: $\tan 30^\circ = \frac{\omega L_1 - X_C}{R}$ $\frac{1}{\sqrt 3} = \frac{\omega L_1 - X_C}{R}$ $R = \sqrt 3\;(\omega L_1 - X_C)\,\,\,\, -(2)$ Case 2: Switch $S_2$ closed (inductance $L_2$ active) and $S_1$ open gives $\phi_2 = 60^\circ$. Again using $(1)$: $\tan 60^\circ = \frac{\omega L_2 - X_C}{R}$ $\sqrt 3 = \frac{\omega L_2 - X_C}{R}$ $R = \frac{\omega L_2 - X_C}{\sqrt 3}\,\,\,\, -(3)$ Equate the two expressions for $R$ from $(2)$ and $(3)$: $\sqrt 3\,(\omega L_1 - X_C) \;=\; \frac{\omega L_2 - X_C}{\sqrt 3}$ Multiplying by $\sqrt 3$, $3(\omega L_1 - X_C) = \omega L_2 - X_C$ $3\omega L_1 - \omega L_2 = 2X_C$ $\omega\,(3L_1 - L_2) = 2X_C\,\,\,\, -(4)$ Substitute $X_C = \dfrac{1}{\omega C}$ into $(4)$: $\omega\,(3L_1 - L_2) = \frac{2}{\omega C}$ $3L_1 - L_2 = \frac{2}{\omega^2 C}$ Insert the numerical values $\omega = 300\,\text{rad s}^{-1}$ and $C = 1 \times 10^{-4}\,\text{F}$: $\omega^2 = (300)^2 = 90000$ $\omega^2 C = 90000 \times 1 \times 10^{-4} = 9$ Hence $3L_1 - L_2 = \frac{2}{9}\,\text{H}$ Therefore, the required value is $\dfrac{2}{9}\,\text{H}$. Option B which is: $\dfrac{2}{9}$

Mutual inductance $M$ is defined as the flux through one circuit when the other carries unit current. We let a current $I$ flow \in the square loop (side $L$). The circular loop (radius $R,\,R\ll L$) lies \in the same plane with the same centre, so it intercepts the flux produced at the square's centre. 1. Magnetic field at the centre of a square loop (single turn) Consider one side of the square. The centre of the square is at a perpendicular distance $r=\dfrac{L}{2}$ from this straight segment of length $L$. For a finite straight wire carrying current $I$, the magnetic field at a point a perpendicular distance $r$ away is $B_{\text{segment}}=\frac{\mu_0 I}{4\pi r}\left(\sin\theta_1+\sin\theta_2\right)$ where $\theta_1$ and $\theta_2$ are the angles subtended by the two ends of the wire at the point. Here each half-length is $\dfrac{L}{2}$, so $\theta_1=\theta_2=\arctan\!\left(\frac{L/2}{\,L/2}\right)=45^{\circ}$, and $\sin45^{\circ}=\dfrac{1}{\sqrt2}.$ Therefore for one side $B_{\text{side}}=\frac{\mu_0 I}{4\pi \bigl(L/2\bigr)}\left(2\cdot\frac1{\sqrt2}\right) =\frac{\mu_0 I}{2\pi L}\,\sqrt2.$ The direction is perpendicular to the plane of the loop (by the right-hand rule) and is the same for all four sides. 2. Field at the centre due to the complete square Adding the contribution of the four sides, $B_{\text{centre}}=4B_{\text{side}} =4\left(\frac{\mu_0 I}{2\pi L}\,\sqrt2\right) =\frac{2\sqrt2\,\mu_0 I}{\pi L}.$ 3. Flux linked with the small circular loop Because $R\ll L$, the field is practically uniform over the area of the small circle, so $\Phi = B_{\text{centre}}\;(\text{area of circle}) =\frac{2\sqrt2\,\mu_0 I}{\pi L}\;\bigl(\pi R^2\bigr) =\frac{2\sqrt2\,\mu_0 I R^2}{L}.$ 4. Mutual inductance By definition, $M=\dfrac{\Phi}{I}$, hence $M=\frac{2\sqrt2\,\mu_0 R^2}{L}.$ 5. Choosing the correct option This value corresponds to Option D. Option D which is: $\dfrac{2\sqrt{2}\,\mu_0 R^2}{L}$