JEE Main 2026 Jan 22 shift 1

Consider a small gas element at distance r from the axis at end A. Because the tube rotates with angular speed ω\omegaω, that gas element needs centripetal force. Required centripetal force on volume element dV=A dr is $dm\omega ^2r$ where $dm=\rho Adr$ So force balance due to pressure difference gives $dP\cdot A=\rho Adr\omega ^2r$ Cancel A: $dP=\rho \omega ^2rdr$ For an ideal gas, $\rho=\frac{MP}{RT}$ (M is molar mass, one mole gas) Substitute: $dP=\frac{MP}{RT}\omega ^2rdr$ Rearrange: $\frac{dP}{P}=\frac{M}{RT}\omega ^2rdr$ Integrate from end A to end B: At $r=0,\quad P=P_A$ At $r=l,\quad P=P_B$ So $\int_{P_A}^{P_B}\frac{dP}{P}=\frac{M\omega^2}{RT}\int\ rdr$ $\ln\frac{P_B}{P_A}=\frac{M\omega^2}{RT}\cdot\frac{l^2}{2}l$ Which is same as $P_{B}=P_{A} exp(M \omega^{2}l^{2}/2RT)$

We have an equilateral prism, so the prism angle $A = 60^{\circ}$, and the refractive index $\mu = \sqrt{2}$. The emergent ray grazes along the second surface, which means the angle of emergence is $e = 90^{\circ}$. This means the angle of refraction at the second surface equals the critical angle $C$. At the critical angle, $\sin C = \frac{1}{\mu} = \frac{1}{\sqrt{2}}$ So, $C = 45^{\circ}$ This means the angle of refraction at the second surface is $r_2 = C = 45^{\circ}$. For a prism, the relation between the prism angle and the angles of refraction at the two surfaces is: $r_1 + r_2 = A$ where $r_1$ is the angle of refraction at the incident surface and $r_2$ is the angle of refraction at the emergent surface. So, $r_1 = A - r_2 = 60^{\circ} - 45^{\circ} = 15^{\circ}$ Hence, the angle of refraction at the incident surface is $15^{\circ}$, which corresponds to Option D.

$V_2 = 8V_1$, $T_2 = T_1/4$. For reversible: $PV^\gamma = const$... $TV^{\gamma-1} = const$. $(T_1/4)(8V_1)^{\gamma-1} = T_1 V_1^{\gamma-1}$. $8^{\gamma-1} = 4$. $2^{3(\gamma-1)} = 2^2$. $\gamma = 5/3$ (monatomic). He is monatomic. The answer is Option 2 : He.

We have a thin convex lens of focal length $f_1 = 5$ cm and a thin concave lens of focal length $f_2 = -4$ cm. An object is placed 10 cm before the convex lens. When the lenses are \in contact with no gap, the combined focal length is given by $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{5} + \frac{1}{-4} = \frac{4 - 5}{20} = -\frac{1}{20}$ So $F = -20$ cm (the combination acts as a diverging lens). Applying the thin lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{F}$ with $u = -10$ cm yields $\frac{1}{v} = \frac{1}{F} + \frac{1}{u} = -\frac{1}{20} + \frac{1}{-10} = -\frac{1}{20} - \frac{2}{20} = -\frac{3}{20}$ Thus $v = -\frac{20}{3}$ cm (virtual image on the same side as the object) and $m_1 = \frac{v}{u} = \frac{-20/3}{-10} = \frac{2}{3}$. With a 1 cm gap between the lenses, the convex lens first produces an image. For the convex lens with $u_1 = -10$ cm and $f_1 = 5$ cm, the lens equation gives $\frac{1}{v_1} = \frac{1}{f_1} + \frac{1}{u_1} = \frac{1}{5} + \frac{1}{-10} = \frac{2 - 1}{10} = \frac{1}{10}$ Hence $v_1 = 10$ cm (real image 10 cm to the right of the convex lens). The image from the convex lens acts as a virtual object for the concave lens, which is placed 1 cm to its right. The distance from the concave lens to this image is $10 - 1 = 9$ cm on the far side, so $u_2 = +9$ cm. Applying the lens formula with $f_2 = -4$ cm yields $\frac{1}{v_2} = \frac{1}{f_2} + \frac{1}{u_2} = \frac{1}{-4} + \frac{1}{9} = \frac{-9 + 4}{36} = -\frac{5}{36}$ Thus $v_2 = -\frac{36}{5}$ cm. The total magnification is the product of the magnifications of each lens: $m_2 = \frac{v_1}{u_1} \times \frac{v_2}{u_2} = \frac{10}{-10} \times \frac{-36/5}{9} = (-1)\left(-\frac{4}{5}\right) = \frac{4}{5}$ Therefore, the ratio of magnifications is $\left|\frac{m_1}{m_2}\right| = \left|\frac{2/3}{4/5}\right| = \left|\frac{2}{3} \times \frac{5}{4}\right| = \frac{10}{12} = \frac{5}{6}$ Hence, $\left|\frac{m_1}{m_2}\right| = \frac{5}{6}$.

Estimate the diameter of a nucleus (Z = 79) from alpha-particle scattering data. When an alpha particle (charge $2e$, kinetic energy $KE$) approaches a nucleus head-on, all its kinetic energy converts to electrostatic potential energy at the distance of closest approach ($r_0$): $KE = \frac{1}{4\pi\epsilon_0} \cdot \frac{(Ze)(2e)}{r_0}$ The diameter $d_0 = 2r_0$. Solving for $r_0$ gives $r_0 = \frac{1}{4\pi\epsilon_0} \cdot \frac{2Ze^2}{KE}$ With $KE = 7.9$ MeV $= 7.9 \times 10^6 \times 1.6 \times 10^{-19}$ J $= 12.64 \times 10^{-13}$ J $= 1.264 \times 10^{-12}$ J, we substitute numerical values into the expression for $r_0$: $r_0 = 9 \times 10^9 \times \frac{2 \times 79 \times (1.6 \times 10^{-19})^2}{1.264 \times 10^{-12}}$ The numerator is $2 \times 79 = 158$ and $(1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38}$, so $158 \times 2.56 \times 10^{-38} = 404.48 \times 10^{-38} = 4.0448 \times 10^{-36}$ and $9 \times 10^9 \times 4.0448 \times 10^{-36} = 36.4 \times 10^{-27} = 3.64 \times 10^{-26}$ giving $r_0 = \frac{3.64 \times 10^{-26}}{1.264 \times 10^{-12}} = 2.88 \times 10^{-14}$ m. Therefore the diameter is $d_0 = 2r_0 = 2 \times 2.88 \times 10^{-14} = 5.76 \times 10^{-14}$ m, which corresponds to Option C: $5.76 \times 10^{-14}$ m.

We use the relation between electric field and potential: $\overrightarrow{E} = -\nabla V\,.$ For a two-dimensional field $\overrightarrow{E}=E_x\,\widehat{i}+E_y\,\widehat{j}$, this gives $E_x = -\frac{\partial V}{\partial x},\quad E_y = -\frac{\partial V}{\partial y}\,.$ Given $E_x = A\,x$ and $E_y = B\,y$, we write: $-\frac{\partial V}{\partial x} = A\,x\quad\Longrightarrow\quad \frac{\partial V}{\partial x} = -A\,x \quad -(1)$ Integrate $(1)$ with respect to $x$ (treating $y$ as constant): $V(x,y) = -A\int x\,dx + f(y) = -\frac{A\,x^2}{2} + f(y)\quad -(2)$ Similarly, from $E_y$: $-\frac{\partial V}{\partial y} = B\,y\quad\Longrightarrow\quad \frac{\partial V}{\partial y} = -B\,y \quad -(3)$ Differentiate expression $(2)$ with respect to $y$: $\frac{\partial V}{\partial y} = \frac{d}{dy}\Bigl(-\frac{A\,x^2}{2} + f(y)\Bigr) = f'(y)\quad -(4)$ Equate $(3)$ and $(4)$: $f'(y) = -B\,y$ Integrate with respect to $y$: $f(y) = -B\int y\,dy + C = -\frac{B\,y^2}{2} + C\quad -(5)$ Substitute $(5)$ into $(2)$ to obtain the general potential: $V(x,y) = -\frac{A\,x^2}{2} - \frac{B\,y^2}{2} + C\quad -(6)$ Use the condition $V(10,20)=500\,$V \in $(6)$: $500 = -\frac{A\,(10)^2}{2} - \frac{B\,(20)^2}{2} + C\,.$ Substitute $A=10\,$V/m2 and $B=5\,$V/m2: $500 = -\frac{10\times100}{2} - \frac{5\times400}{2} + C = -500 -1000 + C\,.$ Thus, $C = 500 + 1500 = 2000\quad -(7)$ The potential at the origin $(0,0)$ is: $V(0,0) = -\frac{A\cdot0^2}{2} - \frac{B\cdot0^2}{2} + C = C = 2000\;\text{V}\,.$ Answer: 2000

We need to determine when induced current in $C_2$ flows clockwise. Three coils on a common axis-$C_1$ on the left, $C_2$ \in the middle, and $C_3$ on the right-carry currents of magnitude $I$, with $C_1$ carrying current anticlockwise as viewed from the left and $C_3$ carrying current clockwise as viewed from the left. By the right-hand rule, the anticlockwise current \in $C_1$ produces a magnetic field pointing to the right through $C_2$, while the clockwise current \in $C_3$ produces a magnetic field pointing to the left through $C_2$. Since $C_2$ is equidistant from both coils, the net magnetic flux through $C_2$ is initially zero. When $C_1$ moves toward $C_2$, the rightward flux through $C_2$ increases. At the same time, when $C_3$ moves away from $C_2$, the leftward flux through $C_2$ decreases, which is equivalent to a further increase \in rightward flux. Hence the net rightward flux through $C_2$ increases. By Lenz's law, the induced current \in $C_2$ must oppose this increase \in rightward flux by producing leftward flux of its own. A clockwise current (viewed from the left) generates a leftward magnetic field inside $C_2$, so the induced current flows clockwise, matching Option B. Therefore, the answer is Option B.

Using components: Magnitude due to each charge, $E_0=\frac{1}{4\pi\ \epsilon_o}.\left(\frac{Q}{R^2}\right)$ Fields from top and bottom charges cancel. Now remaining four charges: At $30^{\circ}$, $\vec{E}_1=-E_0\left(\frac{\sqrt{3}}{2}i+\frac{1}{2}j\right)$ At $-30^{\circ}$, $\vec{E}_2=-E_0\left(\frac{\sqrt{3}}{2}i-\frac{1}{2}j\right)$ At $150^{\circ}$ (negative charge), $\vec{E}_3=E_0\left(-\frac{\sqrt{3}}{2}i+\frac{1}{2}j\right)$ At $-150^{\circ}$, $\vec{E}_4=E_0\left(\frac{\sqrt{3}}{2}i+\frac{1}{2}j\right)$ Adding x-components: $E_x=-\frac{\sqrt{3}}{2}E_0-\frac{\sqrt{3}}{2}E_0-\frac{\sqrt{3}}{2}E_0+\frac{\sqrt{3}}{2}E_0$ $=-\sqrt{3}E_0$ Adding y-components: $E_y=-\frac{1}{2}E_0+\frac{1}{2}E_0+\frac{1}{2}E_0+\frac{1}{2}E_0$ =$E_0$ Therefore, $E=-\sqrt{\ 3}E_0i\ \ +\ E_0j$ Substitute $E_0$​: $-\frac{Q}{4\pi\epsilon_{o}R^{2}}\left(\sqrt{3}\widehat{i}- \widehat{j} \right)$

We have two solid spheres in contact: Sphere A (mass $m_1 = 5$ kg, radius $r_1 = 0.1$ m) and Sphere B (mass $m_2 = 10$ kg, radius $r_2 = 0.2$ m). We need the moment of inertia about a tangent at the point of contact. Let the point of contact be T. The tangent at T is perpendicular to the line joining the centres. The centre of Sphere A is at distance $r_1 = 0.1$ m from T, and the centre of Sphere B is at distance $r_2 = 0.2$ m from T (on the other side). For a solid sphere, the moment of inertia about its own centre is $I_{cm} = \dfrac{2}{5}mr^2$. Using the parallel axis theorem, the moment of inertia about a line at distance $d$ from the centre is $I = I_{cm} + md^2$. For Sphere A (distance from centre to tangent = $r_1$): $I_A = \frac{2}{5}m_1 r_1^2 + m_1 r_1^2 = \frac{7}{5}m_1 r_1^2$ $I_A = \frac{7}{5} \times 5 \times (0.1)^2 = \frac{7}{5} \times 5 \times 0.01 = 7 \times 0.01 = 0.07 \text{ kg.m}^2$ For Sphere B (distance from centre to tangent = $r_2$): $I_B = \frac{2}{5}m_2 r_2^2 + m_2 r_2^2 = \frac{7}{5}m_2 r_2^2$ $I_B = \frac{7}{5} \times 10 \times (0.2)^2 = \frac{7}{5} \times 10 \times 0.04 = 14 \times 0.04 = 0.56 \text{ kg.m}^2$ Thus the total moment of inertia is $I = I_A + I_B = 0.07 + 0.56 = 0.63 \text{ kg.m}^2$ The answer is $\boxed{0.63}$ kg.m$^2$, which corresponds to Option 4.

The circuit consists of two NOT gates (NOR gates with inputs tied) acting on $A$ and $B$, followed by a NAND gate. The top output is $Y_1 = \text{NAND}(\bar{A}, \bar{B}) = \overline{\bar{A} \cdot \bar{B}} = A + B$. The bottom NOR gate provides $Y_2 = \overline{C + D}$. The LED glows when there is a potential difference between the outputs, specifically when $Y_1 = 1$ and $Y_2 = 0$.

For option D ($A=1, B=1, C=0, D=1$):

1. $Y_1 = 1 + 1 = 1$.
2. $Y_2 = \overline{0 + 1} = \overline{1} = 0$.
   Since $Y_1=1$ and $Y_2=0$, the LED terminal potential difference allows it to glow.

Spring constant: $F=kx\Rightarrow k=\frac{F}{x}$ $[k]=\frac{MLT^{-2}}{L}=[MT^{-2}]$ So, (A) matches with (II) Thermal conductivity: From Fourier's law: $\frac{Q}{t}=kA\frac{\Delta T}{L}$ $k=\frac{Q}{t}\cdot\frac{L}{A\Delta T}$ $[Q]=ML^2T^{-2}$ $[k]=\frac{ML^2T^{-2}}{T}\cdot\frac{L}{L^2\cdot K}=[MLT^{-3}K^{-1}]$ So, (B) matches with (IV) Boltzmann constant: $E=k_BT$ $\left[K_B\right]=\frac{E}{T}=\frac{ML^2T^{-2}}{K}=[ML^2T^{-2}K^{-1}]$ So, (C) matches with (I) Inductive reactance: $X_L=\omega LXL​=\omega L$ $[X_L]=[\text{resistance}]=[ML^2T^{-3}A^{-2}]$ So, (D) matches with (III) Final matching: (A)−(II), (B)−(IV), (C)−(I), (D)−(III)

Let the distance from the center of the square to each corner be $r$. The magnitude of the gravitational force from a mass $m$ on a test mass $m_0$ at the center is $F = \frac{G m m_0}{r^2}$. Let $k = \frac{G M m_0}{r^2}$. So the force from a mass $nM$ has magnitude $nk$. In the given figure: Diagonal 1 (connecting $M$ and $3M$): The masses pull \in opposite directions. The net force along this diagonal is $3k - 1k = 2k$ (acting towards the $3M$ mass). Diagonal 2 (connecting $2M$ and $4M$): The net force along this diagonal is $4k - 2k = 2k$ (acting towards the $4M$ mass). Since the diagonals of a square are perpendicular: $F_1 = \sqrt{(2k)^2 + (2k)^2} = \sqrt{8}k = 2\sqrt{2}k$ When the positions of $3M$ and $4M$ are interchanged: The $4M$ mass is now at the Top-Right corner (paired with $M$). The $3M$ mass is now at the Top-Left corner (paired with $2M$). Diagonal 1 (connecting $M$ and $4M$): Net force $= 4k - 1k = 3k$. Diagonal 2 (connecting $2M$ and $3M$): Net force $= 3k - 2k = 1k$. $F_2 = \sqrt{(3k)^2 + (1k)^2} = \sqrt{9k^2 + k^2} = \sqrt{10}k$ $\frac{F_1}{F_2} = \frac{\sqrt{8}k}{\sqrt{10}k} = \sqrt{\frac{8}{10}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$ $\frac{2}{\sqrt{5}} = \frac{\alpha}{\sqrt{5}} \implies \alpha = 2$

To find the minimum frequency of the photon, we first calculate the mass defect ($\Delta m$) required to produce 4 $\alpha$-particles:
$\Delta m = (4 \times 4.002 \text{ amu}) - 15.348 \text{ amu} = 16.008 - 15.348 = 0.66 \text{ amu}$
Converting the mass defect into kilograms:
$\Delta m = 0.66 \times 1.66 \times 10^{-27} \text{ kg} = 1.0956 \times 10^{-27} \text{ kg}$
The energy required is given by $E = \Delta m c^2$:
$E = (1.0956 \times 10^{-27} \text{ kg}) \times (3 \times 10^8 \text{ m/s})^2 = 9.8604 \times 10^{-11} \text{ J}$
Using the relation $E = hf$, we find the frequency:
$f = \frac{E}{h} = \frac{9.8604 \times 10^{-11} \text{ J}}{6.6 \times 10^{-34} \text{ J}\cdot\text{s}} \approx 1.494 \times 10^{23} \text{ Hz}$
Given the options provided, the magnitude shift in the target answer implies an adjustment in units or constants usually associated with such problems, leading to:
$f = 14.94 \times 10^{19} \text{ kHz}$

We can model the thermal conduction network as an electrical circuit where thermal resistance is $R = \frac{L}{KA}$. Since dimensions are equal and $K_x = 3K_y$, the resistance of rod $y$ is three \times that of rod $x$. Let $R_x = R$ and $R_y = 3R$. 1. Equivalent Resistance of the Network The middle section consists of two parallel paths between junctions $B$ and $E$: Path BCE: $R_y + R_y = 3R + 3R = 6R$ Path BDE: $R_x + R_x = R + R = 2R$ Equivalent resistance $R_{BE} = \frac{6R \times 2R}{6R + 2R} = 1.5R$. The total resistance of the series system $A \to B \to E \to F$ is: $R_t​​=R+1.5R+3R=5.5R$ 2. Temperature at Junction B ($T_B$) The temperature drop across rod $AB$ is proportional to its resistance relative to the total: $100 - T_B = \frac{R_{AB}}{R_{total}}(100 - 40)$ $100 - T_B = \frac{R}{5.5R}(60) \approx 10.9^\circ\text{C}$ $T_B \approx 89.1^\circ\text{C}$ 3. Temperature at Junction E ($T_E$) Similarly, the temperature at $E$ can be found by calculating the drop from $A$ to $E$: $T_E = 100 - \frac{R_{AB} + R_{BE}}{R_{total}}(60)$ $T_E = 100 - \frac{2.5R}{5.5R}(60) \approx 100 - 27.3^\circ\text{C}$ $T_E \approx 72.7^\circ\text{C}$ Rounding to the nearest whole numbers, we get $89^\circ\text{C}$ and $73^\circ\text{C}$.

step 1: initial balance Null point at 40 cm from P, so R₁ / R₂ = 40 / (100 − 40) = 40 / 60 = 2 / 3 So, R₁ = (2/3) R₂ ... (1) step 2: after adding 16 Ω in parallel with R₂ New null point = 50 cm → perfectly centered So, R₁ / (R₂ || 16) = 50 / 50 = 1 That means: R₁ = R₂ || 16 Equivalent of parallel: R₂ || 16 = (16R₂) / (R₂ + 16) So, R₁ = (16R₂) / (R₂ + 16) ... (2) step 3: substitute (1) into (2) (2/3)R₂ = (16R₂) / (R₂ + 16) Cancel R₂: 2/3 = 16 / (R₂ + 16) Cross multiply: 2(R₂ + 16) = 48 2R₂ + 32 = 48 2R₂ = 16 R₂ = 8 Ω step 4: find R₁ R₁ = (2/3) × 8 = 16/3 Ω

We need to find the initial speed of a projectile thrown at $60°$ to the horizontal. The angle of projection is $\theta = 60°$, and at some point the speed is $20$ m/s when the direction of motion is $45°$ with the horizontal. The horizontal component of velocity remains constant throughout the motion, given by $v_x = u\cos 60° = \frac{u}{2}$. At the point where the direction is $45°$, the horizontal component is also $v_x = 20\cos 45° = \frac{20}{\sqrt{2}} = 10\sqrt{2}$. Equating the two expressions for the horizontal component gives $\frac{u}{2} = 10\sqrt{2}$, from which it follows that $u = 20\sqrt{2}$. Therefore, the initial speed is $20\sqrt{2}$ m/s, which matches Option D.

At terminal velocity ($v_t$), the gravitational force is balanced by the magnetic force.

1. Induced emf in the rod: $e = Blv_t$
2. Induced current: $I = \frac{e}{R} = \frac{Blv_t}{R}$
3. Magnetic force acting upwards: $F_m = BIl = \frac{B^2l^2v_t}{R}$

Equating forces:
$mg = \frac{B^2l^2v_t}{R}$
$v_t = \frac{mgR}{B^2l^2}$

Hence, option B is correct.

We need to evaluate two statements about fluid mechanics. Statement I: "Pressure of a fluid is exerted only on a solid surface in contact as the fluid-pressure does not exist everywhere in a still fluid." This is FALSE . By Pascal's law, pressure in a fluid at rest acts at every point within the fluid equally in all directions. Fluid pressure exists throughout the fluid volume, not just at solid surfaces. Statement II: "Excess potential energy of the molecules on the surface of a liquid, when compared to interior, results in surface tension." This is TRUE . Interior molecules are surrounded by neighbours on all sides (net force is zero), while surface molecules lack neighbours above, resulting in fewer attractive interactions and higher potential energy. The liquid minimizes this excess surface energy by minimizing its surface area - this tendency manifests as surface tension. The correct answer is Option (4) : Statement I is false but Statement II is true.

We need to find the escape velocity from planet B given the escape velocity from planet A is 10 km/s. Recall the escape velocity formula: $v_e = \sqrt{\frac{2GM}{R}} = \sqrt{\frac{2G \cdot \frac{4}{3}\pi R^3 \rho}{R}} = R\sqrt{\frac{8\pi G\rho}{3}}$ so $v_e \propto R\sqrt{\rho}$. For planet B, density $\rho_B = 0.1\rho_A$ and radius $R_B = 0.1R_A$. Therefore, $\frac{v_{eB}}{v_{eA}} = \frac{R_B\sqrt{\rho_B}}{R_A\sqrt{\rho_A}} = 0.1 \times \sqrt{0.1} = \frac{1}{10} \times \frac{1}{\sqrt{10}} = \frac{1}{10\sqrt{10}}$ It follows that $v_{eB} = \frac{10}{10\sqrt{10}} = \frac{1}{\sqrt{10}} \text{ km/s}$. Converting to m/s gives $v_{eB} = \frac{1000}{\sqrt{10}} = \frac{1000\sqrt{10}}{10} = 100\sqrt{10} \text{ m/s}$ The escape velocity from planet B is $100\sqrt{10}$ m/s, which matches Option B. Therefore, the answer is Option B.

Forces on bob in equilibrium: Weight downward mg Electric force horizontal qE Tension T along string balancing resultant of these two. Since equilibrium requires tension equal to resultant of perpendicular forces: $T=\sqrt{(mg)^{2+}(qE)^2}$

The electric field of a plane EM wave in a non-magnetic medium is $E_y = 20\sin(3 \times 10^6 x - 4.5 \times 10^{14} t)$ V/m. Find the dielectric constant. The wave number is $k = 3 \times 10^6$ rad/m and the angular frequency is $\omega = 4.5 \times 10^{14}$ rad/s. The phase velocity \in the medium is $v = \frac{\omega}{k} = \frac{4.5 \times 10^{14}}{3 \times 10^6} = 1.5 \times 10^8$ m/s. The refractive index is $n = \frac{c}{v} = \frac{3 \times 10^8}{1.5 \times 10^8} = 2$. For a non-magnetic medium ($\mu_r = 1$), we have $n = \sqrt{\epsilon_r}$, so $\epsilon_r = n^2 = 4$. The answer is 4 .

1. Max current $I_{max} = \frac{V}{r} = \frac{10}{10} = 1\text{ A}$.
2. Given current $I = \frac{I_{max}}{e} = \frac{1}{e}\text{ A}$.
3. Energy density $U_B = \frac{B^2}{2\mu_o} = \frac{1}{2}\mu_o n^2 I^2$.
4. Substituting values:
   $U_B = \frac{1}{2} (4\pi \times 10^{-7}) (10^4)^2 (\frac{1}{e})^2$
   $U_B = 2\pi \times 10^1 \times \frac{1}{e^2} = 20\pi \times \frac{1}{e^2}$
5. Comparing with $\alpha\pi \times \frac{1}{e^2}$, we get $\alpha = 20$.

A parallel beam of light (from infinity) in air is incident on a convex spherical glass surface (radius R = 50 cm, $n_{\text{glass}} = 1.5$). To find the distance $x$ from the center of curvature to the point of convergence, we apply the refraction formula at a spherical surface: $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$ where $n_1 = 1.0$ (air), $n_2 = 1.5$ (glass), $R = +50$ cm (convex surface, center of curvature on the glass side), and $u = -\infty$ (parallel beam). Substituting the values gives $\frac{1.5}{v} - \frac{1}{-\infty } = \frac{1.5 - 1}{50}$, so $\frac{1.5}{v} = \frac{0.5}{50} = 0.01$ and hence $v = \frac{1.5}{0.01} = 150$ cm. The image forms at 150 cm from the surface, inside the glass. The center of curvature is at 50 cm from the surface (inside the glass), while the image is at 150 cm from the surface. Therefore, $x = 150 - 50 = 100$ cm. The answer is 100 cm.