JEE Main 2025 23rd January Evening Shift Question Paper with Solutions

When water flows from a region of higher pressure to a region of lower pressure, the difference in pressure is converted into kinetic energy of the flowing water. According to Bernoulli's principle for horizontal flow (where gravitational effects can be ignored), this conversion can be expressed as:

$\frac{1}{2}\rho v^2 = P_1 - P_2$

Here,

$\rho$ is the density of water,

$v$ is the speed of the water,

$P_1$ is the initial pressure when the valve is closed,

$P_2$ is the pressure after the valve is opened.

By solving for $v$, we have:

$v = \sqrt{\frac{2(P_1 - P_2)}{\rho}}$

This equation shows that the speed of the water, $v$, is proportional to the square root of the pressure difference:

$v \propto \sqrt{P_1 - P_2}$

Thus, the correct relationship is given by the square root of the pressure difference.

The work done on an electric dipole in a uniform electric field when it is rotated from an angle $\theta_1$ to an angle $\theta_2$ is given by:

$W = -pE(\cos \theta_2 - \cos \theta_1)$

where:

$p = q \cdot d$ is the dipole moment,

$E$ is the electric field strength,

$\theta_1$ and $\theta_2$ are the initial and final angles between the dipole moment and the electric field.

Given:

Charge $q = 4 \, \mu \mathrm{C} = 4 \times 10^{-6} \, \mathrm{C}$

Distance $d = 18 \, \mathrm{cm} = 0.18 \, \mathrm{m}$

Electric field $E = 10^4 \, \mathrm{N/C}$

Initial angle $\theta_1 = 0^\circ$ (aligned with the field, equilibrium position)

Final angle $\theta_2 = 180^\circ$

First, calculate the dipole moment $p$:

$p = q \cdot d = 4 \times 10^{-6} \, \mathrm{C} \cdot 0.18 \, \mathrm{m} = 7.2 \times 10^{-7} \, \mathrm{Cm}$

Calculate the work done $W$:

$W = -pE(\cos 180^\circ - \cos 0^\circ)$

$W = -7.2 \times 10^{-7} \, \mathrm{Cm} \cdot 10^4 \, \mathrm{N/C} \cdot (-1 - 1)$

$W = 7.2 \times 10^{-7} \, \mathrm{Cm} \cdot 10^4 \, \mathrm{N/C} \cdot 2$

$W = 1.44 \times 10^{-2} \, \mathrm{J} = 14.4 \, \mathrm{mJ}$

Therefore, the work done on the dipole in rotating it from the equilibrium position through $180^\circ$ is 14.4 mJ (Option D).

We start with the energy function:

$E(t)= \alpha^3 e^{-\beta t},$

where $\beta=0.3\, \text{s}^{-1}$.

Step 1. Take the Logarithm

Taking the logarithm of both sides:

$\ln E = 3 \ln \alpha - \beta t.$

Step 2. Differentiate to Find the Relative Error

Differentiate both sides:

$\frac{dE}{E} = 3\,\frac{d\alpha}{\alpha} - \beta\, dt.$

For maximum error estimation, we consider the absolute values and sum the contributions:

$\left|\frac{\Delta E}{E}\right| \approx 3\,\left|\frac{\Delta \alpha}{\alpha}\right| + \beta\, |\Delta t|.$

Step 3. Plug in the Given Errors

The percentage error in $\alpha$ is $1.2\%$ (i.e., $\Delta \alpha/\alpha = 0.012$).

The percentage error in $t$ is $1.6\%$, so for $t=5\,\text{s}$:

$\Delta t = 0.016 \times 5 = 0.08\, \text{s}.$

Now substitute these values:

$\left|\frac{\Delta E}{E}\right| \approx 3(0.012) + 0.3(0.08).$

Calculating each term:

$3(0.012) = 0.036$ (or $3.6\%$),

$0.3(0.08) = 0.024$ (or $2.4\%$).

Step 4. Compute the Total Maximum Percentage Error

Add the contributions:

$\left|\frac{\Delta E}{E}\right| \approx 0.036 + 0.024 = 0.06,$

which is $6\%$.

Thus, the maximum percentage error in the energy at $t = 5\,\text{s}$ is:

$\boxed{6\%}$

Answer: Option A (6%).

$\frac{dV}{dt} = \frac{I}{C}$

Given that the displacement current is $I = 0.25 \times 10^{-3} \, \text{A}$ and the capacitance is $C = 2.5 \times 10^{-6} \, \text{F},$ the rate of change of the potential difference is calculated as follows:

$\frac{dV}{dt} = \frac{0.25 \times 10^{-3}}{2.5 \times 10^{-6}} = \frac{0.25}{2.5} \times \frac{10^{-3}}{10^{-6}} = 0.1 \times 10^3 = 100 \, \text{V/s}.$

Thus, the magnitude of the rate of change of the potential difference is $100 \, \text{V/s}$.

Let's analyze the wave equation step by step.

The given wave is:

$y(x, t) = 4.0 \sin \left[20 \times 10^{-3} x + 600 t\right] \text{ mm}.$

First, simplify the coefficient of $x$:

$20 \times 10^{-3} = 0.02 \, \text{mm}^{-1}.$

So the equation becomes:

$y(x, t) = 4.0 \sin\left(0.02x + 600t\right) \text{ mm}.$

A standard form for a travelling wave is:

$y(x, t) = A \sin(kx - \omega t)$

which represents a wave moving in the positive $x$-direction with speed $v = \frac{\omega}{k}$.

Notice that our wave equation has the form:

$\sin(0.02x + 600t)$

The positive sign in front of $600t$ means we can rewrite the phase as:

$0.02x + 600t = 0.02x - (-600t),$

which indicates that the angular frequency $\omega$ in the standard form is effectively $-600$.

The velocity $v$ of a wave is determined from the phase (for a constant phase, $\phi =$ constant):

$k x + \omega t = \text{constant}.$

Differentiating with respect to $t$:

$k \frac{dx}{dt} + \omega = 0,$

which gives:

$\frac{dx}{dt} = -\frac{\omega}{k}.$

Substituting the values:

$k = 0.02 \, \text{mm}^{-1}$

$\omega = 600 \, \text{s}^{-1}$

We have:

$v = -\frac{600}{0.02} = -30000 \text{ mm/s}.$

Convert the velocity from mm/s to m/s:

$-30000 \, \text{mm/s} = -30 \, \text{m/s}.$

Thus, the velocity of the wave is $-30 \, \text{m/s}$.

The correct answer is Option D.

Let's break down the problem step by step:

The ball is projected with kinetic energy

$\mathrm{KE} = \frac{1}{2} m v^2,$

where $v$ is the initial speed.

The ball is launched at an angle of $60^\circ$. So, its initial velocity components are:

Horizontal: $v_x = v \cos(60^\circ) = \frac{v}{2}$

Vertical: $v_y = v \sin(60^\circ)$

At the highest point of its flight, the vertical component of the velocity becomes zero (i.e., $v_y = 0$), but the horizontal component remains unchanged.

Therefore, the kinetic energy at the highest point is only due to the horizontal velocity:

$\mathrm{KE}_{\text{top}} = \frac{1}{2} m v_x^2$

Substitute $v_x = \frac{v}{2}$:

$\mathrm{KE}_{\text{top}} = \frac{1}{2} m \left(\frac{v}{2}\right)^2 = \frac{1}{2} m \frac{v^2}{4} = \frac{1}{8} m v^2$

Notice that the initial kinetic energy is

$\mathrm{KE} = \frac{1}{2} m v^2.$

So we can write:

$\mathrm{KE}_{\text{top}} = \frac{1}{8} m v^2 = \frac{1}{4} \left(\frac{1}{2} m v^2\right) = \frac{1}{4} \mathrm{KE}$

Thus, the kinetic energy at the highest point is $\frac{\mathrm{KE}}{4}$.

Answer: Option C.

The correct matching is:

(A) Permeability of free space has dimensions

$[MLT^{-2}A^{-2}]$

which corresponds to entry (III).

(B) Magnetic field has dimensions

$[MT^{-2}A^{-1}]$

which corresponds to entry (II).

(C) Magnetic moment in SI units is expressed in Ampere-square meters, i.e.,

$[L^2A]$

which corresponds to entry (IV).

(D) Torsional constant (torque per unit angular displacement) has dimensions

$[ML^2T^{-2}]$

which corresponds to entry (I).

Thus, the correct answer is:

Option A: (A)-(III), (B)-(II), (C)-(IV), (D)-(I)

Let's break down the problem step by step:

In the photoelectric effect, the energy of a photon is given by:

$h\nu = \phi + K_e$

where:

$\phi$ is the work function of the metal,

$K_e$ is the kinetic energy of the ejected electron.

The kinetic energy of the ejected electrons is provided by the stopping potential:

$K_e = eV$

Given that the stopping potential is $V = 2 \text{ V}$, we have:

$K_e = 2 \text{ eV}$

Now, we can calculate the photon energy:

$h\nu = \phi + K_e = 2.14 \text{ eV} + 2 \text{ eV} = 4.14 \text{ eV}$

The relation between the photon's energy and its wavelength is:

$h\nu = \frac{hc}{\lambda}$

Rearranging for $\lambda$:

$\lambda = \frac{hc}{h\nu}$

Given that $hc = 1242 \text{ eV·nm}$, substitute the values:

$\lambda = \frac{1242 \text{ eV·nm}}{4.14 \text{ eV}}$

Calculate the wavelength:

$\lambda \approx \frac{1242}{4.14} \approx 300 \text{ nm}$

Thus, the wavelength of the incident electromagnetic wave is approximately 300 nm.

The correct answer is Option C: 300 nm.

To find the work done by an ideal gas along the path $ABCD$ on the given $P-V$ diagram, we calculate the work done on each segment individually and then sum them up.

Segment $AB$: The work is given by $\text{w}_{AB} = P_0 \times V_0$, contributing $P_0 V_0$.

Segment $BC$: No volume change occurs along this path, resulting in zero work: $\text{w}_{BC} = 0$.

Segment $CD$: The gas compresses, performing negative work: $\text{w}_{CD} = -(2P_0 \times 2V_0) = -4P_0 V_0$.

Adding the work done in each segment:

\begin{align*} \text{w}_{ABCD} &= \text{w}_{AB} + \text{w}_{BC} + \text{w}_{CD} \\ &= P_0 V_0 + 0 - 4P_0 V_0 \\ &= -3P_0 V_0 \end{align*}

Thus, the total work done by the gas along the path $ABCD$ is $-3P_0 V_0$.

$\omega = \frac{1}{\sqrt{LC}}$

For a series LCR circuit, the maximum current occurs at resonance, where the inductive reactance equals the capacitive reactance. Given the values:

Inductance: $L = 100 \, \text{mH} = 0.1 \, \text{H}$

Capacitance: $C = 25 \, \text{nF} = 25 \times 10^{-9} \, \text{F}$

we first calculate the product $LC$:

$LC = 0.1 \times 25 \times 10^{-9} = 2.5 \times 10^{-9}$

Next, compute the square root of the product:

$\sqrt{LC} = \sqrt{2.5 \times 10^{-9}}$

Recognize that:

$\sqrt{2.5 \times 10^{-9}} = \sqrt{2.5} \times \sqrt{10^{-9}} \approx 1.581 \times 10^{-4.5}$

Since $10^{-4.5} = 3.162 \times 10^{-5}$, we have:

$\sqrt{LC} \approx 1.581 \times 3.162 \times 10^{-5} \approx 5.0 \times 10^{-5}$

Now, the angular frequency at resonance becomes:

$\omega = \frac{1}{5.0 \times 10^{-5}} = 2.0 \times 10^{4} \, \text{radians/s}$

Thus, the angular frequency of the AC source for maximum current in the circuit is:

$\omega = 2 \times 10^4 \, \text{radians/s}$

Given:

Galvanometer resistance:

$R_g = 30 \, \Omega$

Full-scale deflection current:

$I_g = 20 \, mA = 0.02 \, A$

Maximum current to be measured:

$I = 3 \, A$

When using the galvanometer to measure 3 A, the extra current passing through the shunt resistor ($R_s$) is:

$I_s = I - I_g = 3 - 0.02 = 2.98 \, A.$

Since the galvanometer and the shunt resistor are connected in parallel, their voltage drops must be equal. Therefore:

$I_g R_g = I_s R_s.$

Substitute the given values:

$0.02 \times 30 = 2.98 \times R_s \quad \Rightarrow \quad R_s = \frac{0.6}{2.98} \approx 0.2013 \, \Omega.$

The shunt resistance is represented as:

$R_s = \frac{30}{X} \, \Omega.$

Equate the two expressions:

$\frac{30}{X} = 0.2013.$

Solving for $X$:

$X = \frac{30}{0.2013} \approx 149.$

Thus, the value of $X$ is approximately $149.$

For an electromagnetic wave in free space, the magnitudes of the electric and magnetic fields are related by the speed of light, $c \approx 3 \times 10^8 \, \mathrm{m/s},$ through the equation

$E = cB.$

Here's how to determine the magnetic field:

We are given the electric field magnitude: $E=9.3 \, \mathrm{V/m}.$

Use the relationship to find the magnetic field magnitude:

$B = \frac{E}{c} = \frac{9.3}{3 \times 10^8} \, \mathrm{T}.$

Calculating the value gives:

$B \approx 3.1 \times 10^{-8} \, \mathrm{T}.$

Since the electric field is along the $y$ direction and the wave propagates along the $+x$ direction, by the right-hand rule, the magnetic field is oriented along the $z$ direction.

Thus, the magnetic field vector is

$\mathrm{B}_z = 3.1 \times 10^{-8} \, \mathrm{T},$

which corresponds to Option C.

To find the difference between the pressure inside the air bubble and the atmospheric pressure, we use the fact that at depth $h$ in a liquid of density $\rho$, the external (hydrostatic) pressure exceeds atmospheric by $\rho g h$. Additionally, for a spherical bubble in a liquid, the internal pressure exceeds the external pressure by $\frac{2T}{r}$, where $T$ is the surface tension and $r$ is the radius of the bubble.

Hence, the internal pressure of the bubble is

$P_\text{inside} = P_\text{atm} + \rho g h + \frac{2T}{r}.$

Therefore, the difference between the bubble's internal pressure and the atmospheric pressure is

$\Delta P = P_\text{inside} - P_\text{atm} = \rho g h + \frac{2T}{r}.$

Substituting the given values:

$\rho = 10^3 \,\text{kg/m}^3$

$g = 10 \,\text{m/s}^2$

$h = 0.20 \,\text{m}$

$T = 0.095 \,\text{J/m}^2$

$r = 1.0 \times 10^{-3} \,\text{m}$

we compute each term:

Hydrostatic term:

$\rho g h = (10^3 \,\text{kg/m}^3) \cdot (10 \,\text{m/s}^2) \cdot (0.20 \,\text{m}) = 2000 \,\text{N/m}^2.$

Surface tension term:

$\frac{2T}{r} = \frac{2 \times 0.095 \,\text{J/m}^2}{1.0 \times 10^{-3} \,\text{m}} = \frac{2 \times 0.095}{10^{-3}} = 2 \times 95 = 190 \,\text{N/m}^2.$

Hence,

$\Delta P = 2000 \,\text{N/m}^2 + 190 \,\text{N/m}^2 = 2190 \,\text{N/m}^2.$

Thus, the required pressure difference between the inside of the bubble and the atmospheric pressure is

$\boxed{2190 \,\text{N/m}^2.}$

Moment of Inertia and Angular Motion

For a solid circular disk, the moment of inertia is given by

$I = \frac{1}{2}MR^2.$

The angular position is defined as

$\theta(t) = 5t^2 - 8t.$

Angular Velocity and Acceleration

Differentiate with respect to time to obtain the angular velocity:

$\omega(t) = \frac{d\theta}{dt} = 10t - 8.$

Differentiating again, the angular acceleration is:

$\alpha(t) = \frac{d\omega}{dt} = 10.$

At time $t = 2 \, \text{s}$:

Angular velocity:

$\omega(2) = 10(2) - 8 = 12 \, \text{rad/s}.$

Angular acceleration:

$\alpha(2) = 10 \, \text{rad/s}^2.$

Torque Calculation

The torque applied by the external force is related to the moment of inertia and angular acceleration:

$\tau = I \alpha = \frac{1}{2}MR^2 \times 10 = 5MR^2.$

Power Delivered

Power delivered by a torque is given by:

$P = \tau \omega.$

At $t = 2 \, \text{s}$, substitute the values:

$P = 5MR^2 \times 12 = 60MR^2.$

Thus, the power delivered by the applied torque at $t = 2 \, \text{s}$ is:

$\boxed{60MR^2}.$

Let the amplitude from the slit of width $d$ be proportional to $E$ and from the slit of width $xd$ be proportional to $xE$ (with $x>1$, as is evident from the given options).

For two coherent waves with amplitudes $E_1$ and $E_2$, the resultant intensity when added in phase (maximum) and out of phase (minimum) is given by

$I_{\max} \propto (E_1+E_2)^2 \quad \text{and} \quad I_{\min} \propto (E_2-E_1)^2.$

Here, setting

$E_1 = E$ (from the narrower slit of width $d$) and

$E_2 = xE$ (from the wider slit of width $xd$),

we have

$I_{\max} \propto (E+xE)^2 = E^2 (1+x)^2,$

and since $x > 1$ the subtraction in the destructive case gives

$I_{\min} \propto (xE-E)^2 = E^2 (x-1)^2.$

The ratio of maximum to minimum intensity is therefore

$\frac{I_{\max}}{I_{\min}} = \frac{(1+x)^2}{(x-1)^2}.$

We are given that

$\frac{(1+x)^2}{(x-1)^2} = \frac{9}{4}.$

Taking the square root of both sides (noting that all quantities are positive) leads to

$\frac{1+x}{x-1} = \frac{3}{2}.$

To solve for $x$, cross-multiply:

$2(1+x) = 3(x-1).$

Expanding both sides:

$2 + 2x = 3x - 3.$

Rearrange the equation to isolate $x$:

$2 + 2x + 3 = 3x \quad \Longrightarrow \quad 5 + 2x = 3x,$

which implies

$x = 5.$

Thus, the value of $x$ is

$\boxed{5}.$

$\textbf{Step 1: Symmetry at Minimum Deviation}$

In a prism, when the deviation is minimum, the path of light is symmetric. This means that the angle of incidence ($i$) is equal to the angle of emergence ($i'$), and the light inside the prism makes equal angles with the prism faces. If the prism angle is $A$, then the refracted angle at each interface is:

$r = \frac{A}{2}$

$\textbf{Step 2: Relating Deviation to the Angles}$

The formula for the deviation ($D$) in a prism is given by:

$D = i + i' - A$

At minimum deviation, $i = i'$, so:

$D_{\text{min}} = 2i - A$

The special condition given is that the minimum deviation is equal to the prism angle:

$D_{\text{min}} = A$

Thus:

$A = 2i - A \quad \Longrightarrow \quad 2i = 2A \quad \Longrightarrow \quad i = A$

$\textbf{Step 3: Applying Snell's Law}$

At the first surface, Snell's law gives:

$\sin i = \mu \sin r$

Substitute the values $i = A$ and $r = \frac{A}{2}$:

$\sin A = \mu \sin\frac{A}{2}$

Given that the refractive index $\mu = \sqrt{3}$, we have:

$\sin A = \sqrt{3} \sin\frac{A}{2}$

$\textbf{Step 4: Solving the Equation}$

Recall the double-angle formula for sine:

$\sin A = 2 \sin\frac{A}{2} \cos\frac{A}{2}$

Substitute this into the previous equation:

$2 \sin\frac{A}{2} \cos\frac{A}{2} = \sqrt{3} \sin\frac{A}{2}$

Assuming $\sin\frac{A}{2} \neq 0$, we can divide both sides by $\sin\frac{A}{2}$:

$2 \cos\frac{A}{2} = \sqrt{3}$

Solve for $\cos\frac{A}{2}$:

$\cos\frac{A}{2} = \frac{\sqrt{3}}{2}$

Since:

$\cos 30^\circ = \frac{\sqrt{3}}{2}$

It follows:

$\frac{A}{2} = 30^\circ \quad \Longrightarrow \quad A = 60^\circ$

$\textbf{Final Answer:}$

The angle of the prism is $60^\circ$.

To determine the current through the battery in this circuit, we need to consider the equivalent resistance and the biasing of the components involved.

Both components in the circuit are forward-biased. Given this, the equivalent resistance ($R_{\text{eq}}$) is $10 \, \Omega$.

The current ($i$) through the circuit can be calculated using Ohm's law:

$i = \frac{V}{R} = \frac{5 \, \text{V}}{10 \, \Omega} = 0.5 \, \text{A}$

Thus, the current through the battery is $0.5 \, \text{A}$.

Let's analyze both statements:

Assertion (A): “The binding energy per nucleon is found to be practically independent of the atomic number A, for nuclei with mass numbers between 30 and 170.”

In fact, for nuclei in this mass range, the binding energy per nucleon is nearly constant (around 7–8 MeV per nucleon). This is because the attractive nuclear force saturates, leading to a relatively uniform binding energy per nucleon in medium to heavy nuclei.

Therefore, Assertion (A) is true.

Reason (R): “Nuclear force is long range.”

The nuclear force (or strong force) that holds the nucleons (protons and neutrons) together is actually very short-ranged. Its effective range is approximately 1 to 3 femtometers (fm), after which the force rapidly decreases.

Hence, Reason (R) is false.

Since the assertion is correct and the reason is incorrect, the correct answer is:

Option A: (A) is true but (R) is false.

$r = R + \frac{R}{3} = \frac{4R}{3}$

For a circular orbit, the gravitational force provides the required centripetal force:

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

which simplifies to

$v = \sqrt{\frac{GM}{r}}.$

The angular momentum $L$ of the satellite is given by

$L = mvr.$

Substitute the values:

Satellite mass: $m = \frac{M}{2}$

Orbital radius: $r = \frac{4R}{3}$

Speed: $v = \sqrt{\frac{GM}{\frac{4R}{3}}} = \sqrt{\frac{3GM}{4R}}$

Thus,

$L = \frac{M}{2} \cdot \frac{4R}{3} \cdot \sqrt{\frac{3GM}{4R}}.$

Simplify the expression:

$L = \frac{M \cdot 4R}{6} \sqrt{\frac{3GM}{4R}} = \frac{2MR}{3} \sqrt{\frac{3GM}{4R}}.$

Notice that the square root can be combined as:

$\sqrt{\frac{3GM}{4R}} = \sqrt{\frac{3}{4}} \sqrt{\frac{GM}{R}}.$

Therefore,

$L = \frac{2MR}{3} \cdot \sqrt{\frac{3}{4}} \sqrt{\frac{GM}{R}} = \frac{2MR \sqrt{3}}{3 \cdot 2} \sqrt{\frac{GM}{R}} = \frac{M \sqrt{3}}{3} \sqrt{GMR},$

where we used the fact that $R \sqrt{\frac{1}{R}} = \sqrt{R}$ to form $\sqrt{GMR}$.

So the final expression is:

$L = \frac{M}{\sqrt{3}} \sqrt{GMR}.$

It is given that the angular momentum can also be expressed as:

$L = M \sqrt{\frac{GMR}{x}}.$

Equate the two expressions:

$\frac{M}{\sqrt{3}} \sqrt{GMR} = M \sqrt{\frac{GMR}{x}}.$

Cancel $M$ and $\sqrt{GMR}$ on both sides (assuming they are nonzero):

$\frac{1}{\sqrt{3}} = \sqrt{\frac{1}{x}}.$

Taking squares on both sides:

$\frac{1}{3} = \frac{1}{x}.$

Thus,

$x = 3.$

To determine the change in entropy when water is heated from temperature $T_1$ to $T_2$ (here, $T_\gamma$ is equivalent to $T_2$), we use the definition of entropy change for a reversible process:

$\Delta S = \int_{T_1}^{T_2} \frac{dQ}{T}$

Since the water is being heated slowly, the process is reversible, and the heat added is given by:

$dQ = mc\,dT$

where:

• $m$ is the mass of the water,

• $c$ is the specific heat capacity (given as $1\,\mathrm{J\,kg^{-1}\,K^{-1}}$).

Substitute $dQ$ into the integral:

$\Delta S = \int_{T_1}^{T_2} \frac{mc\,dT}{T} = mc\, \int_{T_1}^{T_2} \frac{dT}{T}.$

Since $c = 1$, the equation simplifies to:

$\Delta S = m \int_{T_1}^{T_2} \frac{dT}{T}.$

Evaluating the integral, we have:

$\int_{T_1}^{T_2} \frac{dT}{T} = \ln{T}\Big|_{T_1}^{T_2} = \ln\left(\frac{T_2}{T_1}\right).$

Thus, the change in entropy is:

$\Delta S = m \ln\left(\frac{T_2}{T_1}\right).$

Comparing with the options provided, the correct answer is:

Option D: $m \ln\left(\frac{T_2}{T_1}\right).$

$\begin{aligned} & \mathrm{i}=\left(\frac{5}{25}\right) \\ & \mathrm{Q}=\mathrm{CV} \\ & \mathrm{Q}=\left(8 \times 10^{-6}\right)\left(\frac{5}{25} \times 10\right) \\ & \mathrm{Q}=\left(\frac{8 \times 5 \times 10^{-2}}{25}\right)=16 \mu \mathrm{C}\end{aligned}$

When a concave mirror is dipped in a liquid, its focal length is still determined by the geometry of the mirror. For a spherical concave mirror, the focal length is given by

$f = \frac{R}{2}$

where $R$ is the radius of curvature of the mirror. This derivation comes solely from geometrical considerations and the law of reflection, which states that the angle of incidence equals the angle of reflection. Since reflection is independent of the medium in which the mirror is immersed, the focal length remains unchanged.

Thus, even when the mirror is in a liquid of refractive index $\mu$, its focal length is still

$f.$

The correct answer is Option B: $f$.

We can solve this using Hooke's Law, which states:

$F = kx,$

where:

• $F$ is the force (tension),

• $k$ is the spring constant, and

• $x$ is the extension of the spring.

Follow these steps:

For the first scenario (extension $x_1$ under 5 N):

$5 = kx_1 \quad \Rightarrow \quad k = \frac{5}{x_1}.$

For the second scenario (extension $x_2$ under 7 N):

$7 = kx_2 \quad \Rightarrow \quad k = \frac{7}{x_2}.$

Equate the two expressions for $k$:

$\frac{5}{x_1} = \frac{7}{x_2} \quad \Rightarrow \quad \frac{x_1}{x_2} = \frac{5}{7} \quad \Rightarrow \quad x_2 = \frac{7}{5}x_1.$

To find the tension for the extension of $\left(5x_1 - 2x_2\right)$, use Hooke's Law again:

$\text{Tension, } F = k \left(5x_1 - 2x_2\right).$

Substitute $k = \frac{5}{x_1}$:

$F = \frac{5}{x_1} \left(5x_1 - 2x_2\right).$

Substitute the expression for $x_2$:

$F = \frac{5}{x_1} \left(5x_1 - 2\left(\frac{7}{5}x_1\right)\right) = \frac{5}{x_1} \left(5x_1 - \frac{14}{5}x_1\right).$

Simplify the expression inside the parentheses:

$5x_1 - \frac{14}{5}x_1 = \left(\frac{25}{5} - \frac{14}{5}\right)x_1 = \frac{11}{5}x_1.$

Now substitute back:

$F = \frac{5}{x_1} \cdot \frac{11}{5}x_1 = 11 \text{ N}.$