JEE Main 2025 7th April Morning Shift Question Paper with Solutions

We are given that two harmonic waves moving in the same direction superimpose to form:

where $t$ is in seconds.

Step 1: Decompose using the product-to-sum identity

We use the trigonometric identity:

Applying this with $A = 50.5t$ and $B = 1.5t$:

Step 2: Identify the two component waves

The superposition consists of two harmonic waves with angular frequencies:

$\omega_1 = 49 \text{ rad/s}$ and $\omega_2 = 52 \text{ rad/s}$

Step 3: Find the beat frequency and beat period

The beat angular frequency is:

The beat period is:

Rounding to the nearest integer: $T_{beat} \approx 2 \text{ s}$.

Hence, the correct answer is Option D.

To find the amplitude of the resultant wave for the given electric field components, we use the formula for the resultant amplitude of two combining waves. These components are:

${𝐸}_1={𝐸}_0\sin 𝜔𝑡$

${𝐸}_2={𝐸}_0\sin (𝜔𝑡+\frac{𝜋}{3})$

The general formula for the amplitude $𝐸$ of two superimposed waves ${𝐸}_1$ and ${𝐸}_2$ with a phase difference is:

$𝐸=\sqrt{\surd {𝐸}_2^1+{𝐸}_2^2+2{𝐸}_1{𝐸}_2\cos 𝜙}$

where $𝜙$ is the phase difference between the two waves. In this scenario, $𝜙=\frac{𝜋}{3}$.

Substituting the given expressions and phase difference into the formula, we get:

$𝐸=\sqrt{\surd ({𝐸}_0{)}^2+({𝐸}_0{)}^2+2\left({𝐸}_0\right)\left({𝐸}_0\right)\cos \left(\frac{𝜋}{3}\right)}$

Given that $\cos \left(\frac{𝜋}{3}\right)=\frac{1}{2}$, this simplifies to:

$𝐸=\sqrt{\surd 2{𝐸}_2^0+{𝐸}_2^0}=\sqrt{\surd 3}{𝐸}_0$

Thus, the amplitude of the resultant wave is $\sqrt{\surd 3}{𝐸}_0\approx 1.73{𝐸}_0$.

All edges have equal resistance. Total wire resistance = R.
Number of edges in a tetrahedron = 6

So resistance of each edge:

$r=\frac{R}{6}$

Now find equivalent between two vertices (say A and B).

By symmetry, the other two vertices are at same potential → no current flows between them.

So that edge can be ignored.

Now circuit reduces to:

• Direct edge A-B: r
• Two identical paths A → (top nodes) → B
  Each path has 2 resistors in series → 2r

So we have 3 parallel paths:

$R_{eq}=r\parallel2r\parallel2r$

$R_{eq}=\frac{r}{2}$

Substitute $r=R/6$

$R_{eq}=\frac{R}{12}$

When a charged particle ($q, m$) enters a uniform magnetic field ($B$) with velocity ($v$) perpendicular to the field, it executes uniform circular motion.

Radius ($r$), $r = \frac{mv}{qB}$

Region 2:

The particle enters Region 2 with magnetic field $B_2$. It describes a semicircular arc and returns to the interface.

Radius in Region 2: $r_2 = \frac{mv}{qB_2}$

Displacement along the interface: $d_2 = 2r_2 = \frac{2mv}{qB_2}$

Assuming a positive charge moving downward, the magnetic force ($\vec{F} = q\vec{v} \times \vec{B}$) directed into the page causes the particle to curve to the right.

Region 1:

Upon returning to the interface, the particle enters Region 1 with magnetic field $B_1$. Because the velocity vector is now reversed (pointing upward), the force direction also reverses.

Radius in Region 1: $r_1 = \frac{mv}{qB_1}$

Displacement along the interface: $d_1 = 2r_1 = \frac{2mv}{qB_1}$

The particle now curves to the left.

Net displacement ($S$): $S = |d_1 - d_2|$

$S = 2r_1 - 2r_2 = \frac{2mv}{qB_1} - \frac{2mv}{qB_2}$

$S = \frac{mv}{qB_1} \left( 1 - \frac{B_1}{B_2} \right) \times 2$

To identify the physical quantity represented by $\epsilon_0\dfrac{d\Phi_E}{dt}$ we must compare its dimensions with those of the given options.

Step 1: Dimension of electric flux $\Phi_E$
Electric flux is defined as $\Phi_E=\displaystyle\oint \mathbf E\cdot d\mathbf A$.
• Dimension of electric field $\mathbf E$:
  $\mathbf E=\dfrac{\text{Force}}{\text{Charge}}=\dfrac{MLT^{-2}}{IT}=MLT^{-3}I^{-1}$.
• Area has dimension $L^2$.
Hence,
$[\Phi_E]=MLT^{-3}I^{-1}\times L^2=ML^{3}T^{-3}I^{-1}$.

Step 2: Dimension of $\dfrac{d\Phi_E}{dt}$
Time derivative multiplies the time dimension by $T^{-1}$:
$\Bigl[\dfrac{d\Phi_E}{dt}\Bigr]=ML^{3}T^{-3}I^{-1}\times T^{-1}=ML^{3}T^{-4}I^{-1}.$

Step 3: Dimension of $\epsilon_0$
From Coulomb’s law $F=\dfrac{1}{4\pi\epsilon_0}\dfrac{q_1q_2}{r^2}$ we get
$\epsilon_0=\dfrac{q^2}{Fr^{2}}.$
• $[q]=IT,$ so $[q^2]=I^{2}T^{2}$.
• $[F]=MLT^{-2},\quad [r^2]=L^{2}.$
Therefore,
$[\epsilon_0]=\dfrac{I^{2}T^{2}}{MLT^{-2}\times L^{2}}=M^{-1}L^{-3}T^{4}I^{2}.$

Step 4: Dimension of $\epsilon_0\dfrac{d\Phi_E}{dt}$
Multiply the results of Steps 2 and 3:
$[\epsilon_0]\,[d\Phi_E/dt]=(M^{-1}L^{-3}T^{4}I^{2})(ML^{3}T^{-4}I^{-1})=I.$

The final dimension is that of electric current. Thus $\epsilon_0\dfrac{d\Phi_E}{dt}$ represents a quantity having the dimensions of electric current.

Hence the correct choice is Option D: Electric current.

The bent rod can be treated as two uniform rods of lengths $3L$ and $2L$ joined at the corner.

For the horizontal rod, its centre of mass is at  $x_1=L,\qquad y_1=0$

For the vertical rod, its centre of mass is at  $x_2=0,\qquad y_2=\frac{3L}{2}$

Since the rod is uniform, mass is proportional to length.

Thus,  $x_{cm}=\frac{(2L)(L)+(3L)(0)}{5L}$

$x_{cm}=\frac{2L}{5}$

Similarly,  $y_{cm}=\frac{(2L)(0)+(3L)\left(\frac{3L}{2}\right)}{5L}$

$y_{cm}=\frac{9L}{10}$

Given,  $L=10\ \mathrm{cm}$

Therefore,  $x_{cm}=4\ \mathrm{cm}$

$y_{cm}=9\ \mathrm{cm}$

Hence, the position vector of the centre of mass is  ${4\hat{i}+9\hat{j}\ \mathrm{cm}}$

Hence, the correct option is D.

The magnetic field inside a long, closely-wound solenoid is

$B = \mu H$

where $\mu$ is the permeability of the medium and $H$ is the magnetic field strength produced by the current.

In vacuum (or air) $\mu = \mu_0$, so the field is

$B_0 = \mu_0 H$ $-(1)$

When the core is filled by a material of magnetic susceptibility $\chi$, its permeability becomes

$\mu = \mu_0(1 + \chi)$

Therefore the new magnetic field is

$B = \mu H = \mu_0(1 + \chi)H$ $-(2)$

Divide $(2)$ by $(1)$ to find the ratio of the two fields:

$\frac{B}{B_0} = \frac{\mu_0(1 + \chi)H}{\mu_0 H} = 1 + \chi$

The fractional (i.e. relative) increase in the field is

$\frac{B - B_0}{B_0} = (1 + \chi) - 1 = \chi$

Hence the percentage increase is

$\chi \times 100\%$

For magnesium $\chi_{mg} = 1.2 \times 10^{-5}$, so

Percentage increase $= 1.2 \times 10^{-5} \times 100\% = 1.2 \times 10^{-3}\%$

Option A expresses this value as $\dfrac{6}{5} \times 10^{-3}\% = 1.2 \times 10^{-3}\%$, which matches our result.

Therefore, the correct choice is Option A.

The focal length of a thin lens in a surrounding medium is obtained from the Lens-Maker’s formula:

$\frac{1}{f} = \left(\frac{\mu_{\text{lens}}}{\mu_{\text{medium}}} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \quad -(1)$

For a given lens the radii $R_1$ and $R_2$ do not change; only the refractive-index term varies when the medium is changed.

Case 1: Lens in air
Here $\mu_{\text{medium}} = 1$, $\mu_{\text{lens}} = 1.6$ and the focal length is given as $f_1 = 12\ \text{cm}$.

Substituting in $(1)$:

$\frac{1}{12} = \left(\frac{1.6}{1} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ $\Rightarrow \frac{1}{12} = (1.6 - 1)K$ where $K = \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$.

Hence $K = \frac{1}{12 \times 0.6} = \frac{1}{7.2}\quad -(2)$

Case 2: Lens in water
Now $\mu_{\text{medium}} = 1.28$, while $\mu_{\text{lens}} = 1.6$ remains the same. Let the new focal length be $f_2$.

Using $(1)$ again:

$\frac{1}{f_2} = \left(\frac{1.6}{1.28} - 1\right)K$ First evaluate the refractive-index factor: $\frac{1.6}{1.28} = 1.25 \; \Rightarrow \; 1.25 - 1 = 0.25$.

Therefore $\frac{1}{f_2} = 0.25 \times K$ Insert $K$ from $(2)$: $\frac{1}{f_2} = 0.25 \times \frac{1}{7.2} = \frac{0.25}{7.2}$ $\frac{1}{f_2} = 0.034722\dots$ Hence $f_2 = \frac{1}{0.034722\dots} = 28.8\ \text{cm}$.

Convert to millimetres: $28.8\ \text{cm} = 28.8 \times 10 = 288\ \text{mm}$.

Thus, the focal length of the lens in water is $288\ \text{mm}$.

Option B is correct.

The given current has two parts:
a constant (dc) part : $I_{dc}=5\sqrt{2}$ Amp
a sinusoidal (ac) part : $i_{ac}=10\cos\!\left(650\pi t+\frac{\pi}{6}\right)$ Amp

For any quantity that is the sum of a dc value and a sinusoidal value, the mean-square over one full cycle is the sum of the individual mean-squares, because the average of the product of the dc term with the ac term over a complete cycle is zero.

Therefore, the r.m.s. value of the total current is given by
$I_{rms}=\sqrt{I_{dc}^{\,2}+I_{ac,rms}^{\,2}}$

First find the r.m.s. of the ac component. If the peak (amplitude) of the cosine term is $I_0$, then
$I_{ac,rms}=\frac{I_0}{\sqrt{2}}$

Here $I_0=10$ Amp, so
$I_{ac,rms}=\frac{10}{\sqrt{2}}=5\sqrt{2}\text{ Amp}$

Now substitute into the overall r.m.s. formula:
$I_{rms}=\sqrt{\left(5\sqrt{2}\right)^{2}+\left(5\sqrt{2}\right)^{2}}$

Calculate the squares:
$\left(5\sqrt{2}\right)^{2}=25\times2=50$

Hence
$I_{rms}=\sqrt{50+50}= \sqrt{100}=10\text{ Amp}$

So, the r.m.s. value of the current is 10 Amp.

Answer: Option C (10 Amp)

The equivalent focal length of two thin lenses separated by a distance is found with the lens-combination formula
$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1\,f_2}$
where
$f_1, f_2$ are the focal lengths of the individual lenses (signs taken from the Cartesian convention) and $d$ is the separation between their optical centres.

All three quantities are first expressed in the same unit. Working in centimetres:
$f_1 = 30\; \text{cm},\; f_2 = 10\; \text{cm},\; d = 10\; \text{cm}$

Substituting into the formula gives
$\frac{1}{F} = \frac{1}{30} + \frac{1}{10} - \frac{10}{30 \times 10}$

Simplify each term:
$\frac{1}{30} = 0.0333\ldots,\quad \frac{1}{10} = 0.1,\quad \frac{10}{30 \times 10} = \frac{10}{300} = 0.0333\ldots$

Add and subtract term by term:
$\frac{1}{F} = 0.0333\ldots + 0.1 - 0.0333\ldots = 0.1$

Thus
$F = \frac{1}{0.1} = 10\; \text{cm}$

Convert the focal length to metres for power calculation:
$F = 10\; \text{cm} = 0.10\; \text{m}$

Lens power is defined as
$P = \frac{1}{F\;(\text{in metres})}$
so
$P = \frac{1}{0.10} = 10\; \text{dioptre}$

Therefore, the power of the lens combination is $10\; \text{D}$, which corresponds to Option D.

Using voltage division to test if Zener breakdown occurs: $V_{\text{open}} = V_s \left(\frac{R_L}{R_s + R_L}\right)$

Given: $V_s = 12\text{ V}$, $R_s = 100\ \Omega$, $R_L = 400\ \Omega$

$V_{\text{open}} = 12 \times \left(\frac{400}{100 + 400}\right) = 12 \times \frac{4}{5} = 9.6\text{ V}$

Since $V_{\text{open}} > V_Z$ ($9.6\text{ V} > 4\text{ V}$), the Zener diode operates in the breakdown region:

$V_L = V_Z = 4\text{ V}$

Using Ohm's law for the load branch containing the ammeter:

$I_A = \frac{V_L}{R_L} = \frac{4}{400} = 0.01\text{ A} = 10\text{ mA}$

Let the common initial speed of the two projectiles be $u$ and let the acceleration due to gravity be $g$ (acting downward).

Case 1: Projection angle $\theta_1 = 45^\circ + \alpha$
Time of flight formula for a projectile launched from ground level is
$T = \frac{2u \sin\theta}{g}$
Thus, $T_1 = \frac{2u \sin (45^\circ + \alpha)}{g}$ $-(1)$

Case 2: Projection angle $\theta_2 = 45^\circ - \alpha$
Similarly,
$T_2 = \frac{2u \sin (45^\circ - \alpha)}{g}$ $-(2)$

We require the ratio $\dfrac{T_1}{T_2}$. Using $(1)$ and $(2)$, the common factors $\frac{2u}{g}$ cancel out:

$\frac{T_1}{T_2} = \frac{\sin (45^\circ + \alpha)}{\sin (45^\circ - \alpha)}$ $-(3)$

Apply the sine addition-subtraction identities:
$\sin (45^\circ \pm \alpha) = \sin 45^\circ \cos\alpha \pm \cos 45^\circ \sin\alpha$

Since $\sin 45^\circ = \cos 45^\circ = \frac{\sqrt{2}}{2}$, we obtain
$\sin (45^\circ + \alpha) = \frac{\sqrt{2}}{2}(\cos\alpha + \sin\alpha)$
$\sin (45^\circ - \alpha) = \frac{\sqrt{2}}{2}(\cos\alpha - \sin\alpha)$

Substitute these into $(3)$:

$\frac{T_1}{T_2} \;=\; \frac{\frac{\sqrt{2}}{2}(\cos\alpha + \sin\alpha)}{\frac{\sqrt{2}}{2}(\cos\alpha - \sin\alpha)}$

The factor $\frac{\sqrt{2}}{2}$ cancels out, giving

$\frac{T_1}{T_2} = \frac{\cos\alpha + \sin\alpha}{\cos\alpha - \sin\alpha}$ $-(4)$

To convert the expression to a form involving $\tan\alpha$, divide numerator and denominator of $(4)$ by $\cos\alpha$:

$\frac{T_1}{T_2} = \frac{1 + \tan\alpha}{1 - \tan\alpha}$

Hence, the required ratio of times of flight is $\dfrac{1 + \tan\alpha}{1 - \tan\alpha}$, which corresponds to Option D.

For an ideal gas molecule the ratio of specific heats is given by the kinetic-theory formula
$\gamma = \frac{C_p}{C_v} = 1 + \frac{2}{f}$
where $f$ is the total number of degrees of freedom of one molecule.

Case A : Triatomic rigid gas
A rigid triatomic molecule that is non-linear (e.g. $H_2O$) has
 • $3$ translational degrees + $3$ rotational degrees  $\Rightarrow f = 6$.
Therefore $\gamma = 1 + \frac{2}{6} = 1 + \frac13 = \frac43$.
So A matches List-II item III.

Case B : Diatomic non-rigid gas
A non-rigid diatomic molecule possesses vibration along the bond in addition to translation and rotation.
 • Translational : $3$, Rotational : $2$, Vibrational : $2$ (one kinetic + one potential)
Hence $f = 3 + 2 + 2 = 7$.
Thus $\gamma = 1 + \frac{2}{7} = \frac{9}{7}$.
So B matches List-II item IV.

Case C : Monoatomic gas
A monoatomic molecule can only translate.
 • $3$ translational degrees, no rotation or vibration  $\Rightarrow f = 3$.
Hence $\gamma = 1 + \frac{2}{3} = \frac{5}{3}$.
So C matches List-II item I.

Case D : Diatomic rigid gas
A rigid diatomic molecule has no vibrational contribution at ordinary temperatures.
 • Translational : $3$, Rotational : $2$  $\Rightarrow f = 5$.
Therefore $\gamma = 1 + \frac{2}{5} = \frac{7}{5}$.
So D matches List-II item II.

Collecting the correspondences:
A - III, B - IV, C - I, D - II

This is exactly Option A.

Final answer: Option A.

The block slides down a rough plane making an angle $\theta = 60^{\circ}$ with the horizontal.

For a body sliding down an inclined plane, the forces along (parallel to) the plane are:
  • Down the plane: component of weight, $mg\sin\theta$
  • Up the plane: kinetic friction, $f_k = \mu_k N$, where $N$ is the normal reaction.

First find the normal reaction. For an incline, $N = mg\cos\theta$ because no acceleration is perpendicular to the plane.

Therefore the kinetic friction magnitude is
$f_k = \mu_k N = \mu_k mg\cos\theta$.

Apply Newton’s second law along the plane (taking down the plane as positive).
Net force $= mg\sin\theta - f_k = ma$.

Substitute $f_k$ and the given acceleration $a = \dfrac{g}{2}$:
$mg\sin\theta - \mu_k mg\cos\theta = m\left(\dfrac{g}{2}\right)$.

Cancel the common factor $mg$ from every term:
$\sin\theta - \mu_k \cos\theta = \dfrac{1}{2}$ $-(1)$

Insert the trigonometric values for $\theta = 60^{\circ}$:
$\sin 60^{\circ} = \dfrac{\sqrt{3}}{2}, \qquad \cos 60^{\circ} = \dfrac{1}{2}$.

Equation $(1)$ becomes
$\dfrac{\sqrt{3}}{2} - \mu_k \left(\dfrac{1}{2}\right) = \dfrac{1}{2}$.

Multiply every term by 2 to remove the denominators:
$\sqrt{3} - \mu_k = 1$.

Solve for $\mu_k$:
$\mu_k = \sqrt{3} - 1$.

Hence the coefficient of kinetic friction is $\sqrt{3} - 1$, which matches Option A.

Answer: Option A

For a hydrogen-like ion, the energy of the electron in the $n^{\text{th}}$ Bohr orbit is

$E_n = -\,\frac{13.6\,Z^{2}}{n^{2}}\ \text{eV} \qquad -(1)$

where $Z$ is the atomic number of the ion.

The ground state corresponds to $n = 1$.

“2nd excitation energy state’’ means the electron is in the second excited level, i.e. $n = 3$ (first excited is $n = 2$, second excited is $n = 3$).

Using (1):

Energy of ground state ($n = 1$): $E_1 = -\,13.6\,Z^{2}\ \text{eV} \qquad -(2)$

Energy of 2nd excited state ($n = 3$): $E_3 = -\,\frac{13.6\,Z^{2}}{9}\ \text{eV} \qquad -(3)$

The energy difference between these two states is

$\Delta E = E_3 - E_1$

Substituting (2) and (3):

$\Delta E = -\,\frac{13.6\,Z^{2}}{9} - \bigl(-\,13.6\,Z^{2}\bigr)$

$\Delta E = 13.6\,Z^{2}\left(1 - \frac{1}{9}\right)$

$\Delta E = 13.6\,Z^{2}\left(\frac{8}{9}\right)$

$\Delta E = \frac{108.8\,Z^{2}}{9}\ \text{eV} \qquad -(4)$

According to the question, this difference equals $108.8\ \text{eV}$, so from (4)

$\frac{108.8\,Z^{2}}{9} = 108.8$

$\Rightarrow \frac{Z^{2}}{9} = 1$

$\Rightarrow Z^{2} = 9$

$\Rightarrow Z = 3$

Thus the atomic number of the ion is $Z = 3$, which corresponds to Option D.

The wavelengths of hydrogen spectral lines are obtained from the Rydberg‐formula,

$\frac{1}{\lambda}=R\left(\frac{1}{n_f^{2}}-\frac{1}{n_i^{2}}\right)$

where $R$ is the Rydberg constant, $n_f$ is the lower (final) level of the transition and $n_i$ is the upper (initial) level with $n_i \gt n_f$.

For any given series, the largest wavelength (smallest energy difference) comes from the transition having the smallest possible gap, that is $n_i = n_f+1$.

For the Lyman series, $n_f = 1$. Therefore take $n_i = 2$.

$\frac{1}{\lambda_L}=R\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)=R\left(1-\frac{1}{4}\right)=R\left(\frac{3}{4}\right)$

Hence $\lambda_L=\frac{4}{3R}$ $-(1)$

For the Balmer series, $n_f = 2$. Therefore take $n_i = 3$.

$\frac{1}{\lambda_B}=R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=R\left(\frac{1}{4}-\frac{1}{9}\right)=R\left(\frac{5}{36}\right)$

Hence $\lambda_B=\frac{36}{5R}$ $-(2)$

Taking the ratio of $(1)$ to $(2)$:

$\frac{\lambda_L}{\lambda_B}=\frac{\dfrac{4}{3R}}{\dfrac{36}{5R}}=\frac{4}{3R}\times\frac{5R}{36}=\frac{20}{108}=\frac{5}{27}$

Therefore, the required ratio is $5 : 27$.

Correct option: Option B (5 : 27)

$qvB = \frac{mv^2}{r}$

$r = \frac{mv}{qB}$

$E = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2E}{m}}$

$r = \frac{m}{qB} \sqrt{\frac{2E}{m}}$

$r = \frac{\sqrt{2mE}}{qB}$

$r \propto \sqrt{E}$

Graph D correctly represents the $r \propto \sqrt{E}$ curve.

Mass of the object, $m = 1000\text{ g} = 1\text{ kg}$.

Given time-dependent force
$\vec{F}(t) = 2t\,\hat{i} + 3t^{2}\,\hat{j}\; \text{N}$

Step 1 : Find acceleration.
Newton’s second law states $\vec{F} = m\vec{a}$, hence
$\vec{a}(t) = \frac{\vec{F}}{m} = 2t\,\hat{i} + 3t^{2}\,\hat{j}$ $-(1)$

Step 2 : Integrate acceleration to obtain velocity.
Assume the object starts from rest at $t = 0$, so $\vec{v}(0) = 0$.
Component-wise integration of $(1)$ gives

$v_x(t) = \int 2t\,dt = t^{2} + C_x$
$v_y(t) = \int 3t^{2}\,dt = t^{3} + C_y$

Using $\vec{v}(0)=0$ implies $C_x = 0$ and $C_y = 0$.
Therefore
$\vec{v}(t) = t^{2}\,\hat{i} + t^{3}\,\hat{j}$ $-(2)$

Step 3 : Calculate instantaneous power.
Power delivered by a force is the scalar (dot) product of force and velocity:
$P(t) = \vec{F}\!\cdot\!\vec{v}$

Using $\vec{F}(t)$ and $\vec{v}(t)$ from $(2)$:

$$\begin{aligned} P(t) &= (2t\,\hat{i} + 3t^{2}\,\hat{j}) \cdot (t^{2}\,\hat{i} + t^{3}\,\hat{j}) \\ &= 2t \cdot t^{2} + 3t^{2} \cdot t^{3} \\ &= 2t^{3} + 3t^{5} \end{aligned}$$

Step 4 : State the result.
The power generated by the force at time $t$ is
$P(t) = 2t^{3} + 3t^{5}\ \text{W}$.

Hence, the correct choice is Option D.

For a metallic wire stretched by a tensile force, the extension (elongation) produced is given by the Young’s modulus relation: $\Delta L = \dfrac{F\,L}{A\,Y}$ where
$F$ = applied force, $L$ = original length, $A$ = cross-sectional area, $Y$ = Young’s modulus.

The two wires A and B are made of the same material, so $Y$ is identical for both. The same force $F$ is applied to both wires, so $F$ also cancels when we take a ratio.

Hence the ratio of their extensions is determined only by their lengths and cross-sectional areas: $\dfrac{\Delta L_A}{\Delta L_B} \;=\; \dfrac{F\,L_A / (A_A\,Y)}{F\,L_B / (A_B\,Y)} = \dfrac{L_A}{L_B}\;\dfrac{A_B}{A_A}$

Step 1: Length ratio
Given $\dfrac{L_A}{L_B} = \dfrac{1}{3}$, so $L_A : L_B = 1 : 3$.

Step 2: Area ratio
For circular wires, the area $A = \dfrac{\pi d^{2}}{4}$, hence $A \propto d^{2}$. Given $\dfrac{d_A}{d_B} = 2$, so $\left(\dfrac{d_A}{d_B}\right)^2 = 2^{2} = 4$ leading to $\dfrac{A_A}{A_B} = 4$. Therefore $\dfrac{A_B}{A_A} = \dfrac{1}{4}$.

Step 3: Extension ratio
Substituting the two ratios into the expression for $\dfrac{\Delta L_A}{\Delta L_B}$: $\dfrac{\Delta L_A}{\Delta L_B} = \left(\dfrac{1}{3}\right)\left(\dfrac{1}{4}\right) = \dfrac{1}{12}$.

Thus $\Delta L_A : \Delta L_B = 1 : 12$.

The required ratio of elongations is 1 : 12, which corresponds to Option B.

Given:

• AB=30 cm
• AC=AD=40 cm (circular path centered at A, so $q_1$​ is at center)

Thus distance of $q_3$ from $q_1$​ stays constant during motion:

So potential energy due to $q_1$​ does not change.

Only contribution from $q_2$ changes.

Distances from $q_2$:

At C:

$r_{2c}=\sqrt{\ 30^2+40^2}=50cm$

At D:

Since AD=40 and AB=30,

BD=10 cm

so

$r_{2D}=10cm$

Change in potential energy:

$\Delta U=\frac{1}{4\pi\varepsilon_0}q_3q_2\left(\frac{1}{r_{2D}}-\frac{1}{r_{2C}}\right)$

Substitute:

$=\frac{1}{4\pi\varepsilon_0}q_3q_2\left(\frac{1}{10}-\frac{1}{50}\right)=\frac{1}{4\pi\varepsilon_0}q_3q_2\times\ \left(\frac{4}{50}\right)$
convert 50 cm to m
$=\frac{1}{4\pi\varepsilon_0}q_3q_2\times\ \left(\frac{4}{0.5}\right)$
final answer is 
$=\frac{8}{4\pi\varepsilon_0}q_3q_2$
now comparing it with the given equation 
$K=8q_2$