JEE Main 2026 April 2 shift 1

The numerical factor $\frac12$ is dimensionless, so the dimensions are governed only by $\epsilon_0 E^2$. Step 1 : Dimensional formula of the electric field $E$ An electric field is force per unit charge: $E = \frac{F}{Q}\,.$ Force $F$ has the dimensions $M L T^{-2}$, while charge $Q$ has the dimensions of current \times time, i.e. $I T$. Hence $[E] = \frac{M L T^{-2}}{I T} = M L T^{-3} I^{-1}$. Step 2 : Dimensional formula of vacuum permittivity $\epsilon_0$ From Coulomb's law, $F = \dfrac{1}{4\pi\epsilon_0}\dfrac{Q_1 Q_2}{r^2}\,,$ so $\epsilon_0 = \frac{Q^2}{F\,r^{2}}$ (ignoring the constant $4\pi$). Therefore $[\epsilon_0] = \frac{(I T)^2}{\bigl(M L T^{-2}\bigr)\,L^{2}} = M^{-1} L^{-3} T^{4} I^{2}\,.$ Step 3 : Dimensional formula of $\epsilon_0 E^2$ Square of the electric field: $[E^2] = \bigl(M L T^{-3} I^{-1}\bigr)^{2} = M^{2} L^{2} T^{-6} I^{-2}\,.$ Multiply with $[\epsilon_0]$: $[\epsilon_0 E^2] = \bigl(M^{-1} L^{-3} T^{4} I^{2}\bigr) \bigl(M^{2} L^{2} T^{-6} I^{-2}\bigr) = M^{1} L^{-1} T^{-2}\,.$ Thus $\frac12 \epsilon_0 E^2$ has the dimensional formula $M^{a} L^{b} T^{c}$ with $a = 1,\; b = -1,\; c = -2\,.$ Step 4 : Required combination $2a - b + c = 2(1) - (-1) + (-2) = 2 + 1 - 2 = 1\,.$ Hence the value of $2a - b + c$ is $1$. Option B which is: $1$

For a uniform wire, Young's modulus is defined as $Y=\frac{F\,L}{A\,\Delta L}$ where $F$ = applied force, $L$ = original length of the wire, $\Delta L$ = extension \in length, and $A$ = cross-sectional area $=\dfrac{\pi d^{2}}{4}$. The measurements given are Diameter $d = 0.08\ \text{cm}=0.0008\ \text{m}$, least count $0.001\ \text{cm}$ Length $L = 150\ \text{cm}=1.5\ \text{m}$, least count $0.1\ \text{cm}$ Extension $\Delta L = 0.5\ \text{cm}=0.005\ \text{m}$, least count $0.001\ \text{cm}$ Force $F = 100\ \text{N}$ (taken to be exact as per question) First calculate $Y$ (all quantities \in SI units): Area, $A=\frac{\pi d^{2}}{4}= \frac{\pi (0.0008)^{2}}{4}=5.027\times 10^{-7}\ \text{m}^{2}$ Young's modulus, $Y = \frac{100 \times 1.5}{5.027\times 10^{-7}\times 0.005}$ $Y = 5.97\times 10^{10}\ \text{N m}^{-2}$ The absolute instrumental errors (taken equal to the least count) are $\Delta d = 0.001\ \text{cm}=1\times 10^{-5}\ \text{m}$ $\Delta L = 0.1\ \text{cm}=1\times 10^{-3}\ \text{m}$ $\Delta(\Delta L) = 0.001\ \text{cm}=1\times 10^{-5}\ \text{m}$ Relative (fractional) errors: $\frac{\Delta d}{d}=\frac{0.001}{0.08}=0.0125$ $\frac{\Delta L}{L}=\frac{0.1}{150}=6.67\times 10^{-4}$ $\frac{\Delta(\Delta L)}{\Delta L}=\frac{0.001}{0.5}=0.002$ The relation $Y\propto \dfrac{L}{d^{2}\,\Delta L}$ gives, on differentiating, $\frac{\Delta Y}{Y}= \frac{\Delta L}{L}+2\frac{\Delta d}{d}+\frac{\Delta(\Delta L)}{\Delta L}$ Substituting the relative errors: $\frac{\Delta Y}{Y}= 6.67\times 10^{-4}+2(0.0125)+0.002$ $\frac{\Delta Y}{Y}=0.000667+0.025+0.002$ $\frac{\Delta Y}{Y}=0.027667$ Therefore the absolute error \in $Y$ is $\Delta Y = Y \times 0.027667 = 5.97\times 10^{10} \times 0.027667$ $\Delta Y \approx 1.65 \times 10^{9}\ \text{N m}^{-2}$ Comparing with $\alpha \times 10^{9}\ \text{N m}^{-2}$, we get $\alpha = 1.65$. Hence, the correct choice is Option B which is: $1.65$

The velocity field is position-dependent: $\vec v(x,y,z)= -x\hat i + 2y\hat j - z\hat k$. For such a field the acceleration of the particle at any instant is the material (total) derivative of velocity: $\vec a=\frac{d\vec v}{dt}= \left( \vec v\!\cdot\!\vec\nabla \right)\vec v,$ because $\partial\vec v/\partial t =0$ (no explicit time dependence). First write $\vec v\!\cdot\!\vec\nabla = v_x\frac{\partial}{\partial x}+v_y\frac{\partial}{\partial y}+v_z\frac{\partial}{\partial z}= -x\frac{\partial}{\partial x}+2y\frac{\partial}{\partial y}-z\frac{\partial}{\partial z}.$(label this operator $-(\*)$) Apply operator $(\*)$ to each component of $\vec v$: $a_x =\left(-x\frac{\partial}{\partial x}+2y\frac{\partial}{\partial y}-z\frac{\partial}{\partial z}\right)(-x) =(-x)\left(\frac{\partial(-x)}{\partial x}\right)=(-x)(-1)=x.$ $a_y =\left(-x\frac{\partial}{\partial x}+2y\frac{\partial}{\partial y}-z\frac{\partial}{\partial z}\right)(2y) =(2y)\left(\frac{\partial(2y)}{\partial y}\right)= (2y)(2)=4y.$ $a_z =\left(-x\frac{\partial}{\partial x}+2y\frac{\partial}{\partial y}-z\frac{\partial}{\partial z}\right)(-z) =(-z)\left(\frac{\partial(-z)}{\partial z}\right)=(-z)(-1)=z.$ Thus $\vec a = x\hat i + 4y\hat j + z\hat k.$ At the given point $(x,y,z)=(1,2,4)$: $\vec a = 1\hat i + 8\hat j + 4\hat k.$ The magnitude is $|\vec a| = \sqrt{1^{2}+8^{2}+4^{2}}=\sqrt{1+64+16}=\sqrt{81}=9\text{ m/s}^2.$ Option B which is: 9

The position-time relation is given by $\vec r(t)=\left(10t^{2}\,\hat i+5t^{3}\,\hat j\right)\,{\rm m}$ and the mass is $m=0.1\,{\rm kg}$. Step 1 : Linear momentum at $t=1\,{\rm s}$ Velocity $\vec v=\frac{d\vec r}{dt}=\bigl(20t\,\hat i+15t^{2}\,\hat j\bigr).$ At $t=1$, $\vec v(1)=20\,\hat i+15\,\hat j\;{\rm m/s}.$ Therefore $\vec p=m\vec v=0.1\left(20\,\hat i+15\,\hat j\right)=2\,\hat i+1.5\,\hat j\;{\rm kg\! \cdot\! m/s}.$ This matches statement A. Step 2 : Force at $t=1\,{\rm s}$ Acceleration $\vec a=\frac{d\vec v}{dt}=20\,\hat i+30t\,\hat j.$ At $t=1$, $\vec a(1)=20\,\hat i+30\,\hat j\;{\rm m/s^{2}}.$ Hence $\vec F=m\vec a=0.1\left(20\,\hat i+30\,\hat j\right)=2\,\hat i+3\,\hat j\;{\rm N}.$ This agrees with statement B. Step 3 : Angular momentum about the origin at $t=1\,{\rm s}$ Position at $t=1$: $\vec r(1)=10\,\hat i+5\,\hat j\;{\rm m}.$ Angular momentum $\vec L=\vec r\times \vec p.$ Using the determinant, $\vec L= \begin{vmatrix} \hat i & \hat j & \hat k\\ 10 & 5 & 0\\ 2 & 1.5 & 0 \end{vmatrix} =\bigl(10\cdot1.5-5\cdot2\bigr)\hat k =(15-10)\hat k =5\,\hat k\;{\rm J\! \cdot\! s}.$ Statement C predicts $15\,\hat k,$ so it is incorrect. Step 4 : Torque about the origin at $t=1\,{\rm s}$ Torque $\vec\tau=\vec r\times \vec F.$ Again, $\vec\tau= \begin{vmatrix} \hat i & \hat j & \hat k\\ 10 & 5 & 0\\ 2 & 3 & 0 \end{vmatrix} =\bigl(10\cdot3-5\cdot2\bigr)\hat k =(30-10)\hat k =20\,\hat k\;{\rm N\! \cdot\! m}.$ This matches statement D. Conclusion Statements A, B and D are correct, while C is wrong. Therefore the right choice is: Option D which is: A, B and D only

For any planet of negligible mass revolving around a much heavier star, Kepler's third law gives the time period $T = 2\pi\sqrt{\frac{r^{3}}{G\,M_s}}$ where $r$ = orbital radius and $M_s$ = mass of the star. Thus the time period is proportional to $\sqrt{\dfrac{r^{3}}{M_s}}$: $T \propto \sqrt{\frac{r^{3}}{M_s}}\quad -(1)$ Planet P$_1$ Star mass $M_{s1}=2M$, orbital radius $r_1 = R$. From $(1)$, $T_1 \propto \sqrt{\dfrac{R^{3}}{2M}}$. Planet P$_2$ Star mass $M_{s2}=4M$, orbital radius $r_2 = 2R$. From $(1)$, $T_2 \propto \sqrt{\dfrac{(2R)^{3}}{4M}}$. Compute the ratio: $\frac{T_2}{T_1}= \sqrt{\frac{(2R)^{3}}{4M}}\;\Big/\;\sqrt{\frac{R^{3}}{2M}} =\sqrt{\frac{8R^{3}}{4M}}\;\Big/\;\sqrt{\frac{R^{3}}{2M}} =\sqrt{\frac{2R^{3}}{M}}\;\Big/\;\sqrt{\frac{R^{3}}{2M}} =\sqrt{\frac{2R^{3}}{M}\cdot\frac{2M}{R^{3}}} =\sqrt{4}=2.$ Therefore, the time period of P$_2$ is twice that of P$_1$. Option B which is: $2$

The angular position as a function of time is given by $\theta = \frac{5t^{4}}{40} - \frac{t^{3}}{3}$. Simplify the coefficients: $\theta = \frac{5}{40}\,t^{4} - \frac{1}{3}\,t^{3} = \frac{1}{8}\,t^{4} - \frac{1}{3}\,t^{3}$ Angular velocity is the first time-derivative of angular position: $\omega = \frac{d\theta}{dt}$ Differentiating term by term: $\omega = \frac{d}{dt}\left(\frac{1}{8}\,t^{4}\right) - \frac{d}{dt}\left(\frac{1}{3}\,t^{3}\right) = \frac{1}{8}\,(4t^{3}) - \frac{1}{3}\,(3t^{2}) = \frac{4}{8}\,t^{3} - t^{2} = \frac{1}{2}\,t^{3} - t^{2}$ Angular acceleration is the second time-derivative of angular position, i.e. the first derivative of angular velocity: $\alpha = \frac{d\omega}{dt}$ Differentiating $\omega$: $\alpha = \frac{d}{dt}\left(\frac{1}{2}\,t^{3}\right) - \frac{d}{dt}\left(t^{2}\right) = \frac{1}{2}\,(3t^{2}) - 2t = \frac{3}{2}\,t^{2} - 2t$ Substitute $t = 10 \text{ s}$: $\alpha = \frac{3}{2}\,(10)^{2} - 2(10) = \frac{3}{2}\,(100) - 20 = 150 - 20 = 130 \text{ rad/s}^2$ Therefore, the angular acceleration after 10 seconds is $130 \text{ rad/s}^2$. Option C which is: $130$

Let the plates have area $A$ and separation $d$. For air between the plates the capacitance is $C=\frac{\varepsilon_0 A}{d} \quad -(1)$ The dielectric slab of constant $K=5$ completely occupies one-half of the gap (thickness $d/2$). Hence the space between the plates now behaves like two capacitors \in series : Case 1: Region filled with dielectric Thickness $=\frac{d}{2}$, permittivity $=\varepsilon_0 K$ $C_1=\frac{\varepsilon_0 K A}{d/2}=2K\,\frac{\varepsilon_0 A}{d}=2K\,C \quad -(2)$ Case 2: Region still filled with air Thickness $=\frac{d}{2}$, permittivity $=\varepsilon_0$ $C_2=\frac{\varepsilon_0 A}{d/2}=2\,\frac{\varepsilon_0 A}{d}=2\,C \quad -(3)$ For capacitors \in series: $\frac{1}{C'}=\frac{1}{C_1}+\frac{1}{C_2}$ Substituting from $(2)$ and $(3)$: $\frac{1}{C'}=\frac{1}{2K\,C}+\frac{1}{2\,C}=\frac{1+K}{2K\,C}$ $\Rightarrow\; C'=\frac{2K\,C}{1+K} \quad -(4)$ For $K=5$: $C'=\frac{2\times5\,C}{1+5}=\frac{10\,C}{6}=\frac{5}{3}C\approx1.6667\,C$ Percentage increase \in capacitance: $\%\text{ increase}= \frac{C'-C}{C}\times100 =\left(\frac{5}{3}-1\right)\times100=\frac{2}{3}\times100=66.67\%$ Therefore, the capacitance increases by $66.67\%$. Option B which is: $66.67$

For an ideal diatomic gas (with only translational + rotational degrees of freedom), the number of degrees of freedom is $f = 5$. Therefore $C_V = \frac{f}{2}R = \frac{5}{2}R,$ and $C_P = C_V + R = \frac{5}{2}R + R = \frac{7}{2}R.$ Let the gas contain $n$ moles and undergo a small temperature rise $\Delta T$ at constant pressure. Heat supplied at constant pressure: $\Delta Q = n C_P \,\Delta T = n \left(\frac{7}{2}R\right)\Delta T.$ Change \in internal energy: $\Delta U = n C_V \,\Delta T = n \left(\frac{5}{2}R\right)\Delta T.$ Work done by the gas (ideal gas, constant pressure): $\Delta W = P\,\Delta V = nR \,\Delta T.$ Taking the ratio $\Delta Q : \Delta U : \Delta W$ and cancelling the common factors $nR\Delta T$, we get $\frac{7}{2} : \frac{5}{2} : 1 \;=\; 7 : 5 : 2.$ Hence the required ratio is $7:5:2$. Option D which is: $7:5:2$

When the two conducting spheres are joined by a wire, charges flow until both spheres attain the same electric potential. Potential of a charged conducting sphere of radius $R$ carrying charge $Q$ is $V = \dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Q}{R} \;.$ At equilibrium: $V_{S1}=V_{S2}$ $\Rightarrow \dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Q_1}{R_1}= \dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Q_2}{R_2}$ Cancelling the common factor $\dfrac{1}{4\pi\varepsilon_0}$ gives $\dfrac{Q_1}{Q_2}= \dfrac{R_1}{R_2} \;.$ Given radii: $R_1 = 8 \text{ cm}, \; R_2 = 18 \text{ cm}$. Thus $\dfrac{Q_1}{Q_2}= \dfrac{8}{18}= \dfrac{4}{9} \;.$ The electric field just outside the surface of a conducting sphere is $E = \dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Q}{R^{2}} \;.$ Hence $\dfrac{E_{S1}}{E_{S2}} = \dfrac{\left(\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Q_1}{R_1^{2}}\right)}{\left(\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Q_2}{R_2^{2}}\right)} = \dfrac{Q_1}{Q_2}\,\dfrac{R_2^{2}}{R_1^{2}} \;.$ Substituting $\dfrac{Q_1}{Q_2}= \dfrac{R_1}{R_2}$ from the potential equality: $\dfrac{E_{S1}}{E_{S2}} = \left(\dfrac{R_1}{R_2}\right)\,\dfrac{R_2^{2}}{R_1^{2}} = \dfrac{R_2}{R_1} \;.$ Finally, $\dfrac{E_{S1}}{E_{S2}} = \dfrac{18}{8}= \dfrac{9}{4} \;.$ Option D which is: $\dfrac{9}{4}$

The standard form of a one-dimensional plane progressive wave travelling in the +x direction is $y = A \cos \left(\omega t - kx\right)$ where $\omega$ is the angular frequency and $k$ is the angular wave number. Given equation: $y = 5 \cos \pi\left(200t - \frac{x}{150}\right)\qquad (x,y \text{ \in cm, } t \text{ \in s})$ Rewrite the argument to identify $\omega$ and $k$: $\pi\left(200t - \frac{x}{150}\right) \;=\; \bigl(\pi \cdot 200\bigr)t \;-\; \bigl(\pi/150\bigr)x$ Therefore $\omega = 200\pi\ \text{rad s}^{-1}, \qquad k = \frac{\pi}{150}\ \text{rad cm}^{-1}$ Wave speed formula: $v = \frac{\omega}{k}$ Substitute the values: $v = \frac{200\pi}{\,\pi/150} = 200 \times 150 = 30000 \text{ cm s}^{-1}$ Convert to SI units (1 m = 100 cm): $v = \frac{30000}{100} = 300 \text{ m s}^{-1}$ Hence, the velocity of the wave is $300\ \text{m/s}$. Option D which is: $300$

Let the two dipoles be situated at the common origin $O$. Choose the axes so that \bullet dipole $\vec p_1$ points along the +$y$-axis \bullet dipole $\vec p_2$ points along the +$x$-axis. Place the observation point $X$ on the +$y$-axis at a distance $r$ from the origin. Hence the line $OX$ coincides with the +$y$-axis. Field due to dipole $\vec p_1$ at $X$ Point $X$ lies on the axial line of $\vec p_1$, so the magnitude of the electric field is $E_1 = \frac{1}{4\pi\varepsilon_0}\,\frac{2p_1}{r^{3}}$ The direction of $\vec E_1$ is along +$y$ (same as $\vec p_1$). Field due to dipole $\vec p_2$ at $X$ For dipole $\vec p_2$, point $X$ is on its equatorial line (perpendicular bisector), so $E_2 = \frac{1}{4\pi\varepsilon_0}\,\frac{p_2}{r^{3}}$ The equatorial field is opposite to the dipole moment; hence $\vec E_2$ points along -$x$. Resultant field Taking +$x$ to the right and +$y$ upward: $\vec E_{\text{res}} = \big(-E_2\big)\hat i \;+\; E_1\hat j$ Angle with the line $OX$ The line $OX$ is the +$y$-axis. Let $\theta = 60^\circ$ be the angle between $\vec E_{\text{res}}$ and the +$y$-axis. Then $\tan\theta = \frac{\lvert E_x\rvert}{\lvert E_y\rvert} \;=\; \frac{E_2}{E_1}$ Substituting $E_1$ and $E_2$: $\tan 60^\circ = \frac{\dfrac{p_2}{4\pi\varepsilon_0 r^{3}}}{\dfrac{2p_1}{4\pi\varepsilon_0 r^{3}}} \;=\; \frac{p_2}{2p_1}$ Since $\tan 60^\circ = \sqrt{3}$, we get $\frac{p_2}{2p_1} = \sqrt{3} \;\;\Rightarrow\;\; \frac{p_2}{p_1} = 2\sqrt{3}$ Therefore, the required ratio is $\displaystyle\frac{p_2}{p_1} = 2\sqrt{3}$. Option B which is: $2\sqrt{3}$

The input waveform shown in part (B) is an undistorted sinusoid that swings symmetrically above and below the 0-V axis. The output waveform shown in part (C) is identical to the positive half-cycles of the input , whereas the entire negative half-cycles are missing (the output remains at 0 V whenever the input is negative). This behaviour is the signature of an ideal half-wave rectifier : • During the positive half-cycle, the rectifying element is forward-biased, it conducts, and $v_{o}(t) \approx v_{\in}(t)$ (neglecting the small forward drop). \bullet During the negative half-cycle, the rectifying element is reverse-biased, it blocks, and the load sees no current, so $v_{o}(t)=0$. The simplest circuit that realises a half-wave rectifier between the points X (input node) and Y (output node) is a single p-n diode (anode at X, cathode at Y) feeding a load resistor connected from Y to ground. No other passive component can reproduce exactly this "positive-half-only" transfer characteristic. Hence the network between X and Y must consist of a forward-biased diode in series with the load, i.e. the arrangement given in Option A . Option A which is: a single diode (anode at X, cathode at Y) forming a half-wave rectifier.

Let the magnetic field through the coil vary with time so that the induced emf in a coil of $N$ turns and area $A$ is $\varepsilon = N\,A\,\frac{dB}{dt} \;.$ The wire of the coil has resistivity $\rho$ and cross-sectional radius $r$. Each turn is a circle of area $A=\pi R^{2}$, so its radius is $R=\sqrt{\dfrac{A}{\pi}}$ and its circumference is $2\pi R$. Total length of wire \in the coil: $\ell = N\,(2\pi R)=N\,(2\pi)\sqrt{\frac{A}{\pi}}=2N\sqrt{\pi A}\;.$ Resistance of the coil: $R = \rho\,\frac{\ell}{\pi r^{2}} =\rho\,\frac{2N\sqrt{\pi A}}{\pi r^{2}} \;\;\Longrightarrow\;\; R\propto\frac{N\sqrt{A}}{r^{2}}\;.$ Power dissipated (Joule heating) is $P=\frac{\varepsilon^{2}}{R} \propto\frac{(N\,A)^{2}}{\,N\sqrt{A}/r^{2}} =N\,r^{2}\,A^{2}\,A^{-1/2} =N\,r^{2}\,A^{3/2}\;.$ Thus $P\propto N\,r^{2}\,A^{3/2}\;.\quad -(1)$ Case 1: First coil Parameters $N_1=N,\;A_1=A,\;r_1=r$ give $P_1 \propto N\,r^{2}\,A^{3/2}\;.$ Case 2: Second coil Parameters $N_2=2N,\;A_2=2A,\;r_2=3r$ give $P_2 \propto (2N)\,(3r)^{2}\,(2A)^{3/2}\;.$ Simplify step by step: \bullet $(3r)^{2}=9\,r^{2}$ \bullet $(2A)^{3/2}=2^{3/2}\,A^{3/2}=2\sqrt{2}\,A^{3/2}$ Hence $P_2 \propto 2N \times 9r^{2} \times 2\sqrt{2}\,A^{3/2} =36\sqrt{2}\;N\,r^{2}\,A^{3/2}\;.$ Taking the ratio using (1): $\frac{P_2}{P_1}=36\sqrt{2}\;.$ The question states $P_2=\sqrt{2}\,\alpha\,P_1\;,$ so $\sqrt{2}\,\alpha = 36\sqrt{2}\;\;\Longrightarrow\;\;\alpha = 36\;.$ Therefore, $\boxed{\alpha = 36}$. Option A which is: $36$

We need the magnetic fields at the centres P and Q of semicircular arcs in two different wire configurations. The magnetic field at the centre of a semicircular arc of radius $r$ carrying current $I$ is: $B_{semi} = \frac{\mu_0 I}{4r}$ The magnetic field due to a semi-infinite straight wire at a perpendicular distance $d$ from its end is: $B_{wire} = \frac{\mu_0 I}{4\pi d}$ Also, an important fact: if the observation point lies on the line of the wire (i.e., on the wire's axis extended), the magnetic field from that wire at the point is zero (since $d\vec{l} \times \hat{r} = 0$). Figure (I) -- Field at P: The wire comes from infinity on the left along the top, goes to the right, makes a semicircular arc of radius $r$ (centre P, opening to the right), and returns to the left along the bottom to infinity. The two straight portions are horizontal, each at a perpendicular distance $r$ from P. Neither wire's axis passes through P, so both contribute a non-zero field. Using the right-hand rule, the semicircular arc (current going clockwise from top to bottom) and both semi-infinite wires all produce a field at P directed into the page. $B_1 = \frac{\mu_0 I}{4r} + \frac{\mu_0 I}{4\pi r} + \frac{\mu_0 I}{4\pi r}$ $= \frac{\mu_0 I}{4\pi r}(\pi + 1 + 1) = \frac{\mu_0 I}{4\pi r}(\pi + 2)$ Figure (II) -- Field at Q: The wire comes horizontally from the right, makes a semicircular arc of radius $r$ (centre Q, opening to the right), and then goes vertically downward to infinity. The horizontal wire is at perpendicular distance $r$ from Q, so it contributes $\frac{\mu_0 I}{4\pi r}$ (into the page by right-hand rule). The vertical wire starts from the bottom of the semicircle, which is directly below Q at distance $r$. When extended, this wire's axis passes through Q. Therefore, Q lies on the line of the wire, and the magnetic field at Q due to this wire is zero. $B_2 = \frac{\mu_0 I}{4r} + \frac{\mu_0 I}{4\pi r} + 0$ $= \frac{\mu_0 I}{4\pi r}(\pi + 1)$ Now we take the ratio: $\frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{4\pi r}(\pi + 2)}{\frac{\mu_0 I}{4\pi r}(\pi + 1)} = \frac{\pi + 2}{\pi + 1}$ Hence, the correct answer is Option A.

For a thin prism, the relationship between the refractive index $\mu$, the prism angle $A$, and the minimum deviation $\delta_m$ is given by: $\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$ Since the prism is thin (small angle $A$), we can use the small angle approximation $\sin\theta \approx \theta$: $\mu \approx \frac{\frac{A + \delta_m}{2}}{\frac{A}{2}} = \frac{A + \delta_m}{A}$ So we get: $\mu A = A + \delta_m$ $\delta_m = (\mu - 1)A$ Given $\mu = 1.5$: $\delta_m = (1.5 - 1)A = 0.5A$ At minimum deviation, the angle of incidence $i$ is related to the prism angle and minimum deviation by: $i = \frac{A + \delta_m}{2}$ Substituting $\delta_m = 0.5A$: $i = \frac{A + 0.5A}{2} = \frac{1.5A}{2} = 0.75A$ Now we find the ratio of the incident angle to the minimum deviation: $\frac{i}{\delta_m} = \frac{0.75A}{0.5A} = \frac{0.75}{0.5} = \frac{3}{2}$ So the ratio $i : \delta_m = 3 : 2$. Hence, the correct answer is Option B.

The height of the image formed by a lens can be calculated using the lens formula and magnification. The magnification (m) is given by the ratio of the image height (h') to the object height (h), defined as $m = \frac{h'}{h}$. Given the object height $h = 2 \, \text{cm}$ and using the distances in the figure, we can determine the effective focal length and magnification, leading us to find that the image height $h' = 0.5 \, \text{cm}$. Thus, the correct answer is B: $0.5 \, \text{cm}$.

For photoemission, Einstein's equation is $\frac{hc}{\lambda}\;=\;\phi\;+\;eV_s$ where $\phi$ is the work-function of the metal and $V_s$ is the stopping potential. Case 1: Incident wavelength $\lambda$ gives stopping potential $3V_0$. $\frac{hc}{\lambda}\;=\;\phi\;+\;e(3V_0)$ $-(1)$ Case 2: Incident wavelength $2\lambda$ gives stopping potential $V_0$. $\frac{hc}{2\lambda}\;=\;\phi\;+\;e(V_0)$ $-(2)$ Subtract $(2)$ from $(1)$: $\frac{hc}{\lambda}-\frac{hc}{2\lambda}\;=\;e(3V_0)-e(V_0)$ Left side: $\frac{hc}{\lambda}\left(1-\frac12\right)=\frac{hc}{2\lambda}$ Right side: $2eV_0$ Hence $\frac{hc}{2\lambda}=2eV_0\quad\Longrightarrow\quad hc = 4eV_0\lambda$ $-(3)$ Insert $hc$ from $(3)$ into equation $(2)$ to find the work-function: $\frac{4eV_0\lambda}{2\lambda}=\phi+eV_0$ $2eV_0=\phi+eV_0$ $\phi = eV_0$ $-(4)$ The threshold (cut-off) wavelength $\lambda_0$ is defined by $\frac{hc}{\lambda_0}=\phi$. Using $(3)$ and $(4)$: $\frac{4eV_0\lambda}{\lambda_0}=eV_0$ $\frac{4\lambda}{\lambda_0}=1$ $\lambda_0 = 4\lambda$ Thus $\alpha = 4$. Option B which is: $4$