JEE Main 2026 4th April Morning Shift Question Paper with Solutions

The linear distance moved by the screw in one complete rotation is called the pitch.

Five complete rotations produce a linear advance of $2.5 \text{ mm}$.
Therefore, pitch $= \frac{2.5 \text{ mm}}{5} = 0.5 \text{ mm}$.

The least count (LC) is defined as
$LC = \frac{\text{pitch}}{\text{number of divisions on the circular scale}}$.

Here the circular scale has $100$ divisions, so
$LC = \frac{0.5 \text{ mm}}{100} = 0.005 \text{ mm}$.

Writing $0.005$ in scientific notation:
$0.005 \text{ mm} = 5 \times 10^{-3} \text{ mm}$.

Hence, the least count of the screw gauge is $5 \times 10^{-3} \text{ mm}$.
Option D which is: $5 \times 10^{-3}$.

The bulk modulus (incompressibility) of a fluid is defined as

$K = -\frac{\Delta P}{\Delta V/V}$ $-(1)$

where
  $\Delta P$ = rise in pressure,
  $\Delta V$ = change in volume (negative for compression),
  $\Delta V/V$ = fractional change in volume.

We are interested only in the magnitude of the fractional change, so from $-(1)$

$\left|\frac{\Delta V}{V}\right| = \frac{\Delta P}{K}$ $-(2)$

Given data:
  $\Delta P = 6.3 \times 10^{7}\,\text{N/m}^2$
  $K = 2.1 \times 10^{9}\,\text{N/m}^2$

Substituting in $-(2)$:

$\left|\frac{\Delta V}{V}\right| = \frac{6.3 \times 10^{7}}{2.1 \times 10^{9}}$

Divide the numbers and powers of 10 separately:

$\frac{6.3}{2.1} = 3 \quad\text{and}\quad 10^{7}/10^{9} = 10^{-2} = 0.01$

Hence

$\left|\frac{\Delta V}{V}\right| = 3 \times 0.01 = 0.03$

The fractional decrease in volume is $0.03$, i.e. a $3\%$ decrease.

Therefore, the percentage decrease in the volume of water is $3\%$.

Option B which is: $3\%$

Let the length of the incline (measured along the plane) be $s$. The block is released from rest from the top in both cases.

Case 1: Smooth plane
Angle of the plane, $\theta = 45^{\circ}$.
Net acceleration down the plane is only the component of gravity:$a_s = g\sin\theta$

The time taken to travel the distance $s$ starting from rest is obtained from $s = \tfrac12 a t^2$, so
$t_s = \sqrt{\frac{2s}{a_s}}$

Case 2: Rough plane (kinetic coefficient $\mu_k = \mu$)
The force of kinetic friction acts up the plane with magnitude $\mu N = \mu mg\cos\theta$.
Net acceleration down the plane:$a_r = g\sin\theta - \mu g\cos\theta$
Time taken:$t_r = \sqrt{\frac{2s}{a_r}}$

The problem states that $t_r$ is 50 % larger than $t_s$:$t_r = 1.5\,t_s$

Taking the ratio of the two times:

$\frac{t_r}{t_s} = \sqrt{\frac{2s}{a_r}}\;\Big/\;\sqrt{\frac{2s}{a_s}} = \sqrt{\frac{a_s}{a_r}}$

Given $\dfrac{t_r}{t_s}=1.5$, so$1.5=\sqrt{\frac{a_s}{a_r}}$

Squaring both sides:$\frac{a_s}{a_r}=1.5^2=2.25$
$\Rightarrow a_r=\frac{a_s}{2.25}=\frac{4}{9}\,a_s$

Now substitute the explicit expressions for the accelerations at $\theta=45^{\circ}$:
$a_s = g\sin45^{\circ}= \frac{g}{\sqrt2}$
$a_r = g\sin45^{\circ}-\mu g\cos45^{\circ} = \frac{g}{\sqrt2}\,(1-\mu) = a_s(1-\mu)$

Using $a_r = \dfrac{4}{9}a_s$:

$a_s(1-\mu)=\frac{4}{9}a_s$
Cancelling $a_s$ (non-zero):$1-\mu=\frac{4}{9}$
$\mu = 1-\frac{4}{9}=\frac{5}{9}$

Therefore the coefficient of kinetic friction is $\dfrac{5}{9}$.

Option C which is: $\dfrac{5}{9}$

The reaction is
$\;^3X + \;^3X + \;^4Y \;\longrightarrow\; ^{10}Z$

First write the total binding energy (BE) of each nucleus.
Binding energy of a nucleus = (binding energy per nucleon) $\times$ (mass number).

For mass number $3$:
Total BE = $5.6\;\text{MeV per nucleon}\times 3 = 16.8\;\text{MeV}$.
There are two such nuclei, so initial contribution = $2 \times 16.8 = 33.6\;\text{MeV}$.

For mass number $4$:
Total BE = $7.4\;\text{MeV per nucleon}\times 4 = 29.6\;\text{MeV}$.

Therefore, total initial binding energy is
$E_{\text{initial}} = 33.6 + 29.6 = 63.2\;\text{MeV}$.

For the product nucleus of mass number $10$:
Total BE = $6.1\;\text{MeV per nucleon}\times 10 = 61.0\;\text{MeV}$.

The change in binding energy (final minus initial) is
$\Delta E = E_{\text{final}} - E_{\text{initial}} = 61.0 - 63.2 = -2.2\;\text{MeV}.$

A negative value means the final nucleus is less tightly bound; energy of $2.2\;\text{MeV}$ must be supplied, which equals the increase in mass-energy of the system:
$\Delta M c^2 = 2.2\;\text{MeV}.$

Hence, the required value is $2.2\;\text{MeV}$.
Option C which is: $2.2$

Let the original solid sphere have uniform density $\rho$. Hence

$M=\rho \left(\frac{4}{3}\pi R^{3}\right) \qquad -(1)$

Step 1: Determine the radius $r$ of the smaller sphere
The smaller part has mass $\dfrac{M}{8}$ and is recast into a solid sphere of radius $r$:

$\frac{M}{8}= \rho \left(\frac{4}{3}\pi r^{3}\right) \qquad -(2)$

Divide $(2)$ by $(1)$:

$\frac{1}{8}= \frac{r^{3}}{R^{3}} \;\;\Longrightarrow\;\; r^{3}= \frac{R^{3}}{8} \;\;\Longrightarrow\;\; r=\frac{R}{2}$

Step 2: Determine the thickness $t$ of the disc
The remaining mass $\dfrac{7M}{8}$ is cast into a uniform circular disc of radius $2R$ and thickness $t$. Volume of the disc = $\pi (2R)^{2}\,t = 4\pi R^{2}t$, so

$\frac{7M}{8}= \rho \bigl(4\pi R^{2}t\bigr) \qquad -(3)$

Substitute $M$ from $(1)$ into $(3)$:

$\frac{7}{8}\,\rho\!\left(\frac{4}{3}\pi R^{3}\right)= \rho \left(4\pi R^{2}t\right)$

Cancel $\rho\pi$ and simplify:

$\frac{7R^{3}}{6}=4R^{2}t \;\;\Longrightarrow\;\; t=\frac{7R}{24}$

Step 3: Moment of inertia of the small sphere
For a solid sphere about any diameter, $I=\dfrac{2}{5}mr^{2}$. Here $m=\dfrac{M}{8}$ and $r=\dfrac{R}{2}$, hence

$I_{1}= \frac{2}{5}\left(\frac{M}{8}\right)\left(\frac{R}{2}\right)^{2} =\frac{2}{5}\cdot\frac{M}{8}\cdot\frac{R^{2}}{4} =\frac{M R^{2}}{80}$

Step 4: Moment of inertia of the disc about its diameter
For a uniform disc of mass $m$ and radius $a$, $I=\frac{1}{4} m a^{2}$ about any diameter lying in the plane of the disc. Here $m=\dfrac{7M}{8}$ and $a=2R$, hence

$I_{2}= \frac{1}{4}\left(\frac{7M}{8}\right)(2R)^{2} =\frac{1}{4}\cdot\frac{7M}{8}\cdot 4R^{2} =\frac{7M R^{2}}{8}$

Step 5: Required ratio

$\frac{I_{2}}{I_{1}} =\frac{\dfrac{7M R^{2}}{8}}{\dfrac{M R^{2}}{80}} =\frac{7}{8}\times 80 =70$

Therefore, $\dfrac{I_{2}}{I_{1}}=70$.

Option B which is: $70$

Let the common initial speed be $u$ and the acceleration due to gravity be $g$.

The horizontal range of a projectile launched with angle $\theta$ is $R = \dfrac{u^{2}\sin 2\theta}{g}$.

Case 1: $\theta_1 = 15^{\circ}$
$R_1 = \dfrac{u^{2}\sin 2(15^{\circ})}{g} = \dfrac{u^{2}\sin 30^{\circ}}{g}$

Case 2: $\theta_2 = 30^{\circ}$
$R_2 = \dfrac{u^{2}\sin 2(30^{\circ})}{g} = \dfrac{u^{2}\sin 60^{\circ}}{g}$

Take the ratio $\dfrac{R_1}{R_2}$:

$\dfrac{R_1}{R_2} = \dfrac{\sin 30^{\circ}}{\sin 60^{\circ}}$

Use the standard trigonometric values $\sin 30^{\circ} = \dfrac{1}{2}$ and $\sin 60^{\circ} = \dfrac{\sqrt{3}}{2}$:

$\dfrac{R_1}{R_2} = \dfrac{\tfrac{1}{2}}{\tfrac{\sqrt{3}}{2}} = \dfrac{1}{\sqrt{3}}$

The problem states $R_1 : R_2 = 1 : x$, hence

$x = \sqrt{3}$

Therefore, the required value is $\sqrt{3}$.

Option B which is: $\sqrt{3}$

The photoelectric equation relates the maximum kinetic energy of the emitted electrons to the stopping potential:

$\text{K.E.}_{\max} = eV_0 = h\nu - \phi$

where
  • $h$ is Planck’s constant,
  • $\nu$ is the frequency of the incident radiation,
  • $\phi$ is the work function of the metal,
  • $e$ is the electronic charge.

Re-writing it in the form of a straight line:

$V_0 = \frac{h}{e}\,\nu - \frac{\phi}{e} \qquad -(1)$

Equation $(1)$ shows that a plot of $V_0$ (vertical axis) versus $\nu$ (horizontal axis) is a straight line with
  • slope $\frac{h}{e}$ (same for every metal),
  • intercept on the $V_0$-axis equal to $-\frac{\phi}{e}$, and
  • intercept on the $\nu$-axis (threshold frequency) $\nu_0 = \frac{\phi}{h}$.

For a fixed incident wavelength (that is, a fixed frequency $\nu$ common to all three metals), the maximum kinetic energy is directly proportional to the stopping potential $V_0$. Hence, the metal that exhibits the largest $V_0$ at that chosen $\nu$ will emit the electrons with the greatest kinetic energy.

Because the slope is identical for all three lines, the line lying highest on the graph at any given $\nu$ is the one whose $\nu$-axis intercept $\nu_0$ is the smallest (lowest threshold frequency). A smaller $\nu_0$ means a smaller work function $\phi$, so the electrons require less energy to be liberated and therefore retain more kinetic energy.

Among the three plotted lines, metal $X_1$ has the lowest threshold frequency (its line meets the $\nu$-axis farthest to the left). Consequently, for the same incident frequency (or wavelength) it gives the largest stopping potential $V_0$ and, therefore, the greatest maximum kinetic energy of the emitted electrons.

Hence, the required metal is:

Option A which is: $X_1$

For a single-slit Fraunhofer diffraction pattern the angular position of the $m^{\text{th}}$ minimum is given by the condition

$a \sin\theta_m = m\lambda, \qquad m = 1,2,3,\dots$

Here $a$ is the slit width and $\lambda$ is the wavelength of the light.

When the screen is at a large distance $D$, the linear distance $y_m$ of the $m^{\text{th}}$ minimum from the central axis is obtained using the small-angle approximation $\sin\theta_m \approx \tan\theta_m \approx \theta_m$:

$y_m = D \tan\theta_m \approx D \theta_m = D \sin\theta_m = D \left(\frac{m\lambda}{a}\right) = \frac{m\lambda D}{a}$

Thus,
for the first minimum $\bigl(m = 1\bigr): \; y_1 = \frac{\lambda D}{a}$
for the third minimum $\bigl(m = 3\bigr): \; y_3 = \frac{3\lambda D}{a}$

The required linear separation between the first and third minima is

$y_3 - y_1 = \frac{3\lambda D}{a} - \frac{\lambda D}{a} = \frac{2\lambda D}{a}$

Hence the correct choice is:
Option C which is: $\frac{2D\lambda}{a}$

The extension of a uniform string under a steady load is obtained from
$\Delta L = \dfrac{F\,L}{A\,Y}$
where $F$ is the tension (equal to the load’s weight), $L$ is the original length, $A$ is the cross-sectional area and $Y$ is Young’s modulus.

Data for the two strings
String A: $L_1 = 0.314\ \text{m},\quad Y_1 = 2 \times 10^{10}\ \text{N m}^{-2}$
String B: $L_2 = 2L_1 = 0.628\ \text{m},\quad Y_2 = 2Y_1 = 4 \times 10^{10}\ \text{N m}^{-2}$
Common radius: $r = 0.2\ \text{mm} = 0.2 \times 10^{-3}\ \text{m}$

Cross-sectional area of either string:
$A = \pi r^{2} = \pi\,(0.2 \times 10^{-3})^{2}$ $A = \pi \times 0.04 \times 10^{-6} = 4\pi \times 10^{-8}\ \text{m}^{2}$

Load attached at the lower end:
$m = 0.8\ \text{kg},\quad g = 10\ \text{m s}^{-2}$ $F = mg = 0.8 \times 10 = 8\ \text{N}$

Extension of String A
$\Delta L_1 = \dfrac{F\,L_1}{A\,Y_1} = \dfrac{8 \times 0.314}{(4\pi \times 10^{-8})\,(2 \times 10^{10})}$ Denominator: $4\pi \times 10^{-8} \times 2 \times 10^{10} = 8\pi \times 10^{2} = 800\pi$
Hence
$\Delta L_1 = \dfrac{2.512}{800\pi}\ \text{m} \approx 0.000999\ \text{m} = 0.999\ \text{mm}$

Extension of String B
$\Delta L_2 = \dfrac{F\,L_2}{A\,Y_2} = \dfrac{8 \times 0.628}{(4\pi \times 10^{-8})\,(4 \times 10^{10})}$ Denominator: $4\pi \times 10^{-8} \times 4 \times 10^{10} = 16\pi \times 10^{2} = 1600\pi$
So
$\Delta L_2 = \dfrac{5.024}{1600\pi}\ \text{m} \approx 0.000999\ \text{m} = 0.999\ \text{mm}$

Total extension of the series combination
$\Delta L_{\text{total}} = \Delta L_1 + \Delta L_2 \approx 0.999\ \text{mm} + 0.999\ \text{mm} \approx 1.998\ \text{mm}$

Rounded to the nearest millimetre, the net change in length is $2\ \text{mm}$.

Option B which is: $2$

Assume both gases behave ideally. For an ideal gas the relation is $PV = nRT$, where $n$ is the number of moles and $R$ is the universal gas constant.

For gas 1 before mixing,

$P_1V_1 = n_1RT_1 \; \Rightarrow \; n_1 = \frac{P_1V_1}{RT_1} \; -(1)$

For gas 2 before mixing,

$P_2V_2 = n_2RT_2 \; \Rightarrow \; n_2 = \frac{P_2V_2}{RT_2} \; -(2)$

After mixing the two samples, the total number of moles is the sum of $n_1$ and $n_2$:

$n = n_1 + n_2 = \frac{P_1V_1}{RT_1} + \frac{P_2V_2}{RT_2} \; -(3)$

The mixture finally occupies volume $V$ at pressure $P$ and temperature $T$, so once again using the ideal-gas equation,

$PV = nRT \; -(4)$

Insert $n$ from $(3)$ into $(4)$:

$PV = \left( \frac{P_1V_1}{RT_1} + \frac{P_2V_2}{RT_2} \right) RT$

The gas constant $R$ cancels from numerator and denominator, giving

$PV = T\left( \frac{P_1V_1}{T_1} + \frac{P_2V_2}{T_2} \right) \; -(5)$

Solve $(5)$ for the final temperature $T$:

$T = \frac{PV}{\dfrac{P_1V_1}{T_1} + \dfrac{P_2V_2}{T_2}}$

Multiply numerator and denominator by $T_1T_2$ to simplify:

$T = \frac{T_1T_2PV}{T_2P_1V_1 + T_1P_2V_2}$

Thus, the temperature of the mixture is

$\boxed{\displaystyle T = \frac{T_1T_2PV}{T_2P_1V_1 + T_1P_2V_2}}$

Option B is correct.

$P = \frac{P_0}{1 + \frac{V_0^2}{V^2}} = \frac{P_0 V^2}{V^2 + V_0^2}$

For sample A at $V_A = V_0$: $P = \frac{P_0 V_0^2}{V_0^2 + V_0^2} = \frac{P_0}{2}$

For sample B at $V_B = 3V_0$: $P = \frac{P_0 (3V_0)^2}{(3V_0)^2 + V_0^2} = \frac{9P_0 V_0^2}{9V_0^2 + V_0^2} = \frac{9}{10}P_0$

$T_A = \frac{P_A V_A}{2R} = \frac{\left(\frac{P_0}{2}\right) V_0}{2R} = \frac{P_0 V_0}{4R}$

$T_B = \frac{P_B V_B}{2R} = \frac{\left(\frac{9}{10}P_0\right) (3V_0)}{2R} = \frac{27P_0 V_0}{20R}$

$T_B - T_A = \frac{27P_0 V_0 - 5P_0 V_0}{20R} = \frac{22P_0 V_0}{20R} = \frac{11P_0 V_0}{10R}$

A voltmeter is basically a galvanometer designed to draw as little current as possible from the circuit under test. Its scale is calibrated such that the full-scale deflection occurs when the potential difference across its own (internal) resistance is equal to the rated range.

Given data for the original voltmeter:
internal resistance $R_v = x \,\Omega$
maximum readable voltage (full-scale) $V_{\text{old}} = 20 \text{ V}$

We want to extend the range to a new value
$V_{\text{new}} = 30 \text{ V}$

Method: Add a resistance in series (called a multiplier resistance) so that the larger voltage is divided between the voltmeter and the added resistor, while the same full-scale current still flows through the meter.

Let the required series resistance be $R_s$. At full-scale current $I_{\text{fs}}$ we have:

For the original meter alone:
$I_{\text{fs}} = \frac{V_{\text{old}}}{R_v} = \frac{20}{x} \quad -(1)$

After adding $R_s$ in series, the same current $I_{\text{fs}}$ must flow when the applied voltage is $V_{\text{new}} = 30 \text{ V}$:

$V_{\text{new}} = I_{\text{fs}}\,(R_v + R_s) \quad -(2)$

Substitute $I_{\text{fs}}$ from $(1)$ into $(2)$:

$30 = \frac{20}{x}\,(R_v + R_s)$

But $R_v = x$, so

$30 = \frac{20}{x}\,(x + R_s) \; \Longrightarrow \; 30 = 20\left(1 + \frac{R_s}{x}\right)$

Divide both sides by 20:

$\frac{30}{20} = 1 + \frac{R_s}{x} \;\; \Longrightarrow \;\; \frac{3}{2} = 1 + \frac{R_s}{x}$

Hence

$\frac{R_s}{x} = \frac{3}{2} - 1 = \frac{1}{2}$

$\Rightarrow \; R_s = \frac{x}{2}$

Therefore, connecting a series resistor of $\dfrac{x}{2}\,\Omega$ extends the voltmeter range from 20 V to 30 V.

Adding any resistor in parallel would reduce, not increase, the effective resistance and hence the measurable voltage, so options B and D are unsuitable. Option C gives too high a value, leading to a 40-V range. Thus the correct choice is:

Option A which is: connect resistor of $\frac{x}{2} \; \Omega$ in series.

The input $A$ passes through a NOT gate, resulting in $\overline{A}$.

 This signal ($\overline{A}$) and the input $B$ are fed into a NAND gate.

Therefore, the boolean expression for the output $Y$ is:

$Y = \overline{\overline{A} \cdot B}$

According to De Morgan's Law, $\overline{X \cdot Y} = \overline{X} + \overline{Y}$. Applying this to our expression:

$Y = \overline{\overline{A}} + \overline{B}$

Since the double complement of a variable is the variable itself ($\overline{\overline{A}} = A$):

$Y = A + \overline{B}$

This means the output $Y$ is the bitwise OR operation of $A$ and the 1's complement of $B$.

Given the inputs:

$A = 1101$

$B = 1010$

First, find the 1's complement (bitwise NOT) of $B$:

$\overline{B} = 0101$

Next, perform the bitwise OR addition ($+$) of $A$ and $\overline{B}$:

Y= 1101 

Let the concave mirror be taken as the origin of the coordinate system and the principal axis be the positive x-axis extending to the left of the mirror (Cartesian sign convention).
Hence every object distance is negative and every real image distance is also negative. We shall finally use magnitudes for the required lengths.

Focal length of the concave mirror: $f = -10\ \text{cm}$

The rod of length 10 cm is kept along the principal axis so that its nearer end $A$ is 20 cm from the pole of the mirror and its farther end $B$ is 30 cm from the pole.
Therefore, using sign convention,

$u_A = -20\ \text{cm}, \qquad u_B = -30\ \text{cm}$

For each end we apply the mirror formula
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \qquad -(1)$

Image of end A

Substituting $u = u_A = -20\ \text{cm}$ in (1):
$\frac{1}{v_A} + \frac{1}{-20} = \frac{1}{-10}$
$\frac{1}{v_A} = \frac{1}{-10} + \frac{1}{20} = \frac{-2 + 1}{20} = -\frac{1}{20}$
$v_A = -20\ \text{cm}$

Image of end B

Substituting $u = u_B = -30\ \text{cm}$ in (1):
$\frac{1}{v_B} + \frac{1}{-30} = \frac{1}{-10}$
$\frac{1}{v_B} = \frac{1}{-10} + \frac{1}{30} = \frac{-3 + 1}{30} = -\frac{2}{30} = -\frac{1}{15}$
$v_B = -15\ \text{cm}$

Both images are real and lie on the principal axis in front of the mirror. Their separation (length of the image of the rod) is the difference in the magnitudes of $v_A$ and $v_B$:

$\text{Image length} = |v_A| - |v_B| = 20\ \text{cm} - 15\ \text{cm} = 5\ \text{cm}$

Hence the length of the image is 5 cm.

Option B which is: $5$

For a parallel-plate air capacitor of plate area $A$ and separation $x$, the capacitance is given by
$C = \frac{\varepsilon_0 A}{x}\quad -(1)$

The capacitor remains connected to the battery, so the potential difference across the plates is a constant $V$.

Electrostatic energy stored in a capacitor at constant voltage is
$U = \tfrac12 C V^{2}\quad -(2)$

Substituting $C$ from $(1)$ into $(2)$:
$U = \tfrac12 \varepsilon_{0} A \frac{V^{2}}{x}\quad -(3)$

Differentiate $(3)$ with respect to time to find the rate of change of energy:
$\frac{dU}{dt} = \tfrac12 \varepsilon_{0} A V^{2}\,\frac{d}{dt}\!\left(\frac{1}{x}\right)$
Using $\dfrac{d}{dt}\!\left(\frac{1}{x}\right)= -\dfrac1{x^{2}}\dfrac{dx}{dt}$, we get
$\frac{dU}{dt}= -\tfrac12 \varepsilon_{0} A V^{2}\,\frac{1}{x^{2}}\frac{dx}{dt}\quad -(4)$

The plates are being pulled apart at a uniform speed $v$, so $\dfrac{dx}{dt}=v$ (a positive constant). Substituting this into $(4)$:
$\frac{dU}{dt}= -\tfrac12 \varepsilon_{0} A V^{2}\,v\,\frac{1}{x^{2}}$

The magnitude of the time rate of change of electrostatic energy is therefore proportional to $x^{-2}$. Hence
$\left|\frac{dU}{dt}\right|\propto x^{-2}$ and the required exponent is $\alpha=-2$.

Option A which is: $-2$

The magnetic moment of a planar coil equals the algebraic sum of magnetic moments of its individual turns.
For one circular turn of radius $r$ carrying current $I$, magnetic moment $m = I \, A = I \, (\pi r^{2})$, directed perpendicular to the plane.

The coil here is a tightly wound flat spiral: the radius grows uniformly from $r_1 = 3 \text{ cm}$ to $r_2 = 6 \text{ cm}$ in $N = 200$ turns. If the turns are uniformly spaced in the radial direction, the number of turns per unit radius is$\frac{dn}{dr} = \frac{N}{\,r_2 - r_1\,}.$

Total magnetic moment $M$ is obtained by integrating the contribution $I \pi r^{2}\, dn$ from $r_1$ to $r_2$:$M = \int_{r_1}^{r_2} I \pi r^{2}\, dn = I \pi \int_{r_1}^{r_2} r^{2}\left(\frac{N}{r_2 - r_1}\right) dr.$

Carry out the integral:$M = \frac{I \pi N}{r_2 - r_1}\left[ \frac{r^{3}}{3} \right]_{r_1}^{r_2} = \frac{I \pi N}{3}\,\frac{r_2^{3} - r_1^{3}}{r_2 - r_1}.$

Factor the difference of cubes: $r_2^{3} - r_1^{3} = (r_2 - r_1)(r_2^{2} + r_1 r_2 + r_1^{2})$. The factor $r_2 - r_1$ cancels, giving the compact formula

$M = \frac{I \pi N}{3}\left(r_2^{2} + r_1 r_2 + r_1^{2}\right). \quad -(1)$

Insert the numerical data (convert cm to m):
$r_1 = 0.03 \text{ m}, \; r_2 = 0.06 \text{ m}, \; N = 200,\; I = 20 \text{ mA} = 0.02 \text{ A}.$ Compute the bracket in $(1)$:
$r_2^{2} = 0.06^{2} = 0.0036, \quad r_1 r_2 = 0.03 \times 0.06 = 0.0018, \quad r_1^{2} = 0.03^{2} = 0.0009.$ Sum $= 0.0036 + 0.0018 + 0.0009 = 0.0063.$ Dividing by 3: $\frac{0.0063}{3} = 0.0021.$ Now evaluate $M$:
$M = I \pi N \times 0.0021 = \bigl(0.02 \times 200\bigr)\pi \times 0.0021 = 4 \times \pi \times 0.0021 \approx 4 \times 3.1416 \times 0.0021 \approx 0.0264 \text{ A·m}^{2}.$

Expressing $M$ as $\alpha \times 10^{-2} \text{ A·m}^{2}$ gives $\alpha = 2.64.$ Hence

Option B which is: $2.64$

The time constant of an $RC$ network is given by
$\tau = R_{\text{eq}}\,C$

where
  • $C$ is the capacitance, and
  • $R_{\text{eq}}$ is the net resistance actually present in the current path that charges / discharges the capacitor.

Data from the problem:
  • Capacitor: $C = 20 \,\mu\text{F} = 20 \times 10^{-6}\,\text{F}$
  • Fixed resistor: $R = 100\,\Omega$
  • Two identical diodes whose forward (dynamic) resistance is $r_d = 10\,\Omega$ each.

Because the two diodes are connected with opposite orientation (as is usual in such clipping / rectifying circuits), only one diode is forward-biased at any instant. Hence, along the conduction path there is:

$R_{\text{eq}} = R + r_d = 100\,\Omega + 10\,\Omega = 110\,\Omega$

Therefore, the time constant becomes

$\tau = R_{\text{eq}}\,C = 110\,\Omega \times 20 \times 10^{-6}\,\text{F}$

$\tau = 2200 \times 10^{-6}\,\text{s} = 2.2 \times 10^{-3}\,\text{s}$

Comparing with $\tau = \alpha \times 10^{-3}\,\text{s}$, we get

$\alpha = 2.2$

Hence, the correct choice is
Option A which is: $2.2$