JEE Main 2025 24th January Morning Shift Question Paper with Solutions

$I_{\text{rms}} = \sqrt{\frac{I_A^2 + I_B^2}{2}}$

To determine the root mean square (r.m.s) value of the alternating current given by

$I(t) = I_A \sin (\omega t) + I_B \cos (\omega t),$

we start by squaring the current:

$I(t)^2 = I_A^2 \sin^2 (\omega t) + I_B^2 \cos^2 (\omega t) + 2 I_A I_B \sin (\omega t) \cos (\omega t).$

The r.m.s value is defined as the square root of the average of this squared current over one complete period. When averaging over a full cycle, we utilize the known averages:

$\langle \sin^2 (\omega t) \rangle = \frac{1}{2}, \quad \langle \cos^2 (\omega t) \rangle = \frac{1}{2}, \quad \langle \sin (\omega t) \cos (\omega t) \rangle = 0.$

Thus, the time-averaged square of the current becomes:

$\langle I(t)^2 \rangle = I_A^2 \frac{1}{2} + I_B^2 \frac{1}{2} = \frac{I_A^2 + I_B^2}{2}.$

Taking the square root gives the r.m.s current:

$I_{\text{rms}} = \sqrt{\frac{I_A^2 + I_B^2}{2}}.$

This corresponds to Option B.

$P_{\text{bubble}} = P_{\text{atm}} + \rho g h + \frac{2S}{r}$

Given that the pressure inside the bubble is $P_{\text{atm}} + 2100\,\text{N/m}^2$ and the liquid pressure at a depth of 20 cm is

$P_{\text{liquid}} = P_{\text{atm}} + \rho g h = P_{\text{atm}} + 1000 \times 10 \times 0.20 = P_{\text{atm}} + 2000\,\text{N/m}^2,$

the pressure difference due solely to surface tension is

$\Delta P = \Bigl(P_{\text{bombubble}} - P_{\text{liquid}}\Bigr) = \bigl(P_{\text{atm}}+2100\bigr) - \bigl(P_{\text{atm}}+2000\bigr) = 100\,\text{N/m}^2.$

The Laplace pressure difference for a bubble (which has one interface) is given by

$\Delta P = \frac{2S}{r}.$

The given bubble radius is 0.1 cm, which in SI units is

$r = 0.1\,\text{cm} = 0.001\,\text{m}.$

Substituting the known values into the Laplace equation gives

$\frac{2S}{0.001} = 100.$

Solving for $S$:

$2S = 100 \times 0.001 = 0.1,$

$S = \frac{0.1}{2} = 0.05\,\text{N/m}.$

Thus, the surface tension of the liquid is

$\boxed{0.05\,\text{N/m}}.$

$u = \frac{1}{2}\varepsilon_0 E^2$

The energy density of an electric field in free space is given by the expression above. For a parallel plate capacitor with plate area $A$ and separation $d$, the volume between the plates is

$V = Ad.$

Multiplying the energy density by the volume gives the total energy stored:

$U = u \times V = \frac{1}{2}\varepsilon_0 E^2 \times Ad = \frac{1}{2}\varepsilon_0 E^2 Ad.$

Thus, the potential energy stored in the capacitor is

$\frac{1}{2}\varepsilon_0 E^2 Ad.$

This corresponds to the option:

$\text{Option D: } \frac{1}{2}\varepsilon_0 E^2 Ad.$

$N \cos\theta - \mu N \sin\theta = mg$

$N \sin\theta + \mu N \cos\theta = \frac{mv_0^2}{r}$

Dividing the second equation by the first gives

$\frac{v_0^2}{r} = \frac{g\left(\sin\theta + \mu \cos\theta\right)}{\cos\theta - \mu \sin\theta}.$

Multiplying both sides by $\cos\theta - \mu\sin\theta$ leads to

$\frac{v_0^2}{r}(\cos\theta - \mu \sin\theta) = g(\sin\theta + \mu \cos\theta).$

Expanding and grouping the terms with $\mu$:

$\frac{v_0^2}{r}\cos\theta - g\sin\theta = \mu\left(\frac{v_0^2}{r}\sin\theta + g\cos\theta\right).$

Thus, solving for $\mu$:

$\mu = \frac{\frac{v_0^2}{r}\cos\theta - g\sin\theta}{\frac{v_0^2}{r}\sin\theta + g\cos\theta}.$

Multiplying the numerator and the denominator by $r$ gives

$\mu = \frac{v_0^2\cos\theta - rg\sin\theta}{v_0^2\sin\theta + rg\cos\theta}.$

Dividing both the numerator and denominator by $\cos\theta$ (assuming $\cos\theta \neq 0$):

$\mu = \frac{v_0^2 - rg\tan\theta}{v_0^2\tan\theta + rg}.$

This result matches the expression

$\mu=\frac{v_0^2 - rg\,\tan\theta}{rg + v_0^2\,\tan\theta}.$

Thus, the correct answer is:

Option C

$d \sin \theta = m \lambda$

For two wavelengths, $\lambda_1 = 600\,\text{nm}$ and $\lambda_2 = 480\,\text{nm}$, the bright fringes coincide when the conditions

$m \lambda_2 = n \lambda_1$

are simultaneously satisfied for integers $m$ and $n$, representing the bright fringe order for each wavelength.

Rearranging the equation gives:

$\frac{m}{n} = \frac{\lambda_1}{\lambda_2} = \frac{600}{480} = \frac{5}{4}$

Thus, the smallest positive integers that satisfy this ratio are:

$m = 5 \quad \text{and} \quad n = 4.$

This means the first coincidence of the bright fringes occurs at the 5th bright fringe for the 480 nm light.

The work done by a force is given by the integral of the force over the displacement. The force is given by:

$F(x) = \alpha + \beta x^2$

To find the work done when the object is displaced from $x = 0$ to $x = 1$, we compute:

$W = \int_{0}^{1} (\alpha + \beta x^2) \, dx$

Substituting $\alpha = 1 \, \text{N}$:

$W = \int_{0}^{1} (1 + \beta x^2) \, dx = \int_{0}^{1} 1 \, dx + \int_{0}^{1} \beta x^2 \, dx$

Calculating each integral separately:

$\int_{0}^{1} 1 \, dx = [x]_{0}^{1} = 1$

$\int_{0}^{1} \beta x^2 \, dx = \beta \left[ \frac{x^3}{3} \right]_{0}^{1} = \beta \left[\frac{1^3}{3} - \frac{0^3}{3}\right] = \frac{\beta}{3}$

Thus, the total work done is:

$W = 1 + \frac{\beta}{3}$

We are given that the work done is 5 J, so:

$1 + \frac{\beta}{3} = 5$

Subtract 1 from both sides:

$\frac{\beta}{3} = 4$

Multiply both sides by 3 to solve for $\beta$:

$\beta = 12 \, \text{N/m}^2$

Therefore, the value of $\beta$ is $\boxed{12 \, \text{N/m}^2}$. Thus, Option C $12 \, \text{N/m}^2$ is the correct answer.

The acceleration of the center of mass for a rolling object down an incline is given by:

$a = \frac{g \sin\theta}{1 + \frac{I}{m r^2}}$

For a uniform solid cylinder, the moment of inertia about its central axis is:

$I = \frac{1}{2} m r^2$

Substituting this into the expression for acceleration:

$a = \frac{g \sin\theta}{1 + \frac{\frac{1}{2} m r^2}{m r^2}} = \frac{g \sin\theta}{1 + \frac{1}{2}} = \frac{g \sin\theta}{\frac{3}{2}} = \frac{2}{3} g \sin\theta$

For an incline of $45^\circ$, we have:

$\sin 45^\circ = \frac{\sqrt{2}}{2}$

Thus, the acceleration becomes:

$a = \frac{2}{3} g \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2} g}{3}$

Therefore, the linear acceleration of the cylinder's axis is:

$\frac{\sqrt{2} g}{3}$

$\text{Each side of the triangle has a resistance of } R_{\text{side}} = \frac{9\,\Omega}{3} = 3\,\Omega.$

When measuring the equivalent resistance between any two vertices, there are two paths:

A direct path along one side with resistance:

$R_1 = 3\,\Omega.$

An indirect path passing through the other two sides in series:

$R_2 = 3\,\Omega + 3\,\Omega = 6\,\Omega.$

These two paths are in parallel, so the equivalent resistance is calculated by:

$\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{3\,\Omega} + \frac{1}{6\,\Omega} = \frac{2}{6\,\Omega} + \frac{1}{6\,\Omega} = \frac{3}{6\,\Omega}.$

Thus,

$R_{\text{eq}} = \frac{6\,\Omega}{3} = 2\,\Omega.$

The equivalent resistance across any two vertices of the equilateral triangle is therefore:

$2\,\Omega.$

$P_1 = 2.5\,D,\quad f_1 = \frac{1}{P_1} = \frac{1}{2.5} = 0.4\,m$

After an increase in optical power by 0.1 D:

$P_2 = 2.5\,D + 0.1\,D = 2.6\,D,\quad f_2 = \frac{1}{P_2} = \frac{1}{2.6} \approx 0.3846\,m$

The change in focal length is:

$\Delta f = f_1 - f_2 = 0.4\,m - 0.3846\,m \approx 0.0154\,m$

The relative decrease in focal length is then calculated as:

$\text{Relative decrease} = \frac{\Delta f}{f_1} = \frac{0.0154}{0.4} \approx 0.0385 \approx 0.04$

Thus, the relative decrease in focal length is approximately 0.04.

$\vec{v_0} = \frac{v_0}{\sqrt{2}}\,\hat{i} + \frac{v_0}{\sqrt{2}}\,\hat{j}.$

At maximum height, the vertical component of the velocity becomes zero. Setting the vertical velocity to zero:

$\frac{v_0}{\sqrt{2}} - g\,t = 0 \quad \Longrightarrow \quad t = \frac{v_0}{g\sqrt{2}}.$

The coordinates at this time are:

$x = \frac{v_0}{\sqrt{2}}\,t = \frac{v_0}{\sqrt{2}} \cdot \frac{v_0}{g\sqrt{2}} = \frac{v_0^2}{2g},$

$y = \frac{v_0}{\sqrt{2}}\,t - \frac{1}{2}g\,t^2 = \frac{v_0^2}{2g} - \frac{1}{2}g\,\frac{v_0^2}{2g^2} = \frac{v_0^2}{2g} - \frac{v_0^2}{4g} = \frac{v_0^2}{4g}.$

Thus, the position vector at maximum height is:

$\vec{r} = \frac{v_0^2}{2g}\,\hat{i} + \frac{v_0^2}{4g}\,\hat{j}.$

At maximum height, the only nonzero component of velocity is the horizontal one:

$\vec{v} = \frac{v_0}{\sqrt{2}}\,\hat{i}.$

The angular momentum about the origin is given by:

$\vec{L} = m\,\vec{r} \times \vec{v}.$

Writing out the cross product:

$\vec{r} \times \vec{v} = \left(\frac{v_0^2}{2g}\,\hat{i} + \frac{v_0^2}{4g}\,\hat{j}\right) \times \frac{v_0}{\sqrt{2}}\,\hat{i}.$

Since the cross product of $\hat{i}$ with itself is zero and:

$\hat{j} \times \hat{i} = -\hat{k},$

we have:

$\vec{r} \times \vec{v} = \frac{v_0^2}{4g} \cdot \frac{v_0}{\sqrt{2}}\, (-\hat{k}) = -\frac{v_0^3}{4\sqrt{2}g}\,\hat{k}.$

Thus, the angular momentum is:

$\vec{L} = -\frac{m\,v_0^3}{4\sqrt{2}g}\,\hat{k}.$

This means the magnitude is:

$\frac{m\,v_0^3}{4\sqrt{2}g},$

and its direction is along the negative $\hat{k}$ (or negative $z$-axis).

$B = \frac{2 \sqrt{2}\,\mu_0 I}{\pi L}$

For a square loop of side length

$L = \frac{1}{\sqrt{2}} \text{ m},$

the perpendicular distance from the center to any side is

$d = \frac{L}{2} = \frac{1}{2\sqrt{2}} \text{ m}.$

For a finite straight wire, the magnetic field at a point at distance $d$ from the wire is given by the Biot–Savart expression

$B_{\text{side}} = \frac{\mu_0 I}{4\pi d} \left(\sin\theta_1 + \sin\theta_2\right)$

where $\theta_1$ and $\theta_2$ are the angles between the extended wire and the line joining the ends of the wire with the observation point. For one side of the square, by symmetry, the midpoint of the side and the center of the square give

$\tan\theta = \frac{L/2}{L/2} = 1 \quad\Rightarrow\quad \theta = 45^\circ.$

Thus,

$\sin\theta_1 = \sin\theta_2 = \sin 45^\circ = \frac{1}{\sqrt{2}}.$

Then the field due to one side is

$B_{\text{side}} = \frac{\mu_0 I}{4\pi \left(\frac{L}{2}\right)} \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) = \frac{\mu_0 I}{4\pi \left(\frac{L}{2}\right)} \left(\frac{2}{\sqrt{2}}\right) = \frac{\mu_0 I \sqrt{2}}{2\pi L}.$

Since there are 4 sides with contributions that add vectorially in the same direction at the center, the total magnetic field is

$B = 4 \cdot B_{\text{side}} = \frac{4 \mu_0 I \sqrt{2}}{2\pi L} = \frac{2 \sqrt{2}\,\mu_0 I}{\pi L}.$

Substitute

$L = \frac{1}{\sqrt{2}},$

to obtain

$B = \frac{2 \sqrt{2}\,\mu_0 I}{\pi \left(\frac{1}{\sqrt{2}}\right)} = \frac{2 \sqrt{2}\,\mu_0 I \sqrt{2}}{\pi} = \frac{4\,\mu_0 I}{\pi}.$

Using the provided values

$I = 5 \text{ A} \quad \text{and} \quad \mu_0 = 4\pi \times 10^{-7} \text{ T\,m/A},$

the magnetic field becomes

$B = \frac{4 \times \left(4\pi \times 10^{-7}\right) \times 5}{\pi} = \frac{80\pi \times 10^{-7}}{\pi} = 80 \times 10^{-7} \text{ T} = 8 \times 10^{-6} \text{ T}.$

Thus, the value of $p$ is

$p = 8.$

$\frac{1}{f} = \left(\frac{n}{n'} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

For a thin plano-convex lens, one surface is flat, meaning $R_2 = \infty$ and $\frac{1}{R_2} = 0.$ The formula reduces to:

$\frac{1}{f} = \left(\frac{n}{n'} - 1\right)\frac{1}{R_1}$

Thus, the focal length is:

$f = \frac{R_1}{\left(\frac{n}{n'} - 1\right)}$

For the First Lens (in Air)

Refractive index of lens, $n = 1.5$

Surrounding medium (air), $n' = 1$

Radius of curvature, $R_1 = 2 \text{ cm}$

Substitute into the formula:

$f_1 = \frac{2}{\left(\frac{1.5}{1} - 1\right)} = \frac{2}{(1.5 - 1)} = \frac{2}{0.5} = 4 \text{ cm}$

For the Second Lens (in Liquid)

Refractive index of lens, $n = 1.5$

Surrounding medium (liquid), $n' = 1.2$

Radius of curvature, $R_1 = 3 \text{ cm}$

Substitute into the formula:

$f_2 = \frac{3}{\left(\frac{1.5}{1.2} - 1\right)}$

First, calculate the term inside the parentheses:

$\frac{1.5}{1.2} = 1.25 \quad \Rightarrow \quad 1.25 - 1 = 0.25$

Now, compute $f_2$:

$f_2 = \frac{3}{0.25} = 12 \text{ cm}$

Ratio of the Focal Lengths

$\frac{f_1}{f_2} = \frac{4}{12} = \frac{1}{3}$

Thus, the ratio of $f_1$ to $f_2$ is:

$1 : 3$

Statement A: The junction area of solar cell is made very narrow compared to a photo diode.

• Let's think about the purpose of a solar cell. Its main job is to collect as much sunlight as possible to generate electricity.

• To collect more light, it needs a larger surface area. Therefore, the $𝑝-𝑛$ junction of a solar cell is made with a very large area (typically a few square centimeters).

• This statement says the area is narrow, which is the opposite. So, statement A is Incorrect.

Statement B: Solar cells are not connected with any external bias.

• Remember, a solar cell is like a small battery. It generates an electromotive force (e.m.f.) or voltage when light shines on it. It gives power; it doesn't take power from an external battery to work.

• We don't apply any external bias to it. We simply connect a load, like a bulb, across it to use the electricity it produces.

• So, statement B is Correct.

Statement C: LED is made of lightly doped p-n junction.

• An LED (Light Emitting Diode) works by the recombination of electrons and holes at the junction. To get bright light, we need a very high rate of recombination.

• A high rate of recombination is only possible if there are a large number of electrons on the n-side and holes on the p-side.

• To achieve this high concentration of charge carriers, the $𝑝-𝑛$ junction of an LED is heavily doped, not lightly doped.

• So, statement C is Incorrect.

Statement D: Increase of forward current results in continuous increase of LED light intensity.

• It is true that the intensity of light from an LED increases as we increase the forward current. More current means more electrons and holes are crossing the junction and recombining per second, which produces more light.

• However, we cannot increase the current indefinitely. Every LED has a maximum current rating mentioned by the manufacturer. If we pass a current higher than this safe limit, the LED will overheat and get permanently damaged.

• So, the increase is not "continuous" without any limit.

• Therefore, this statement is Incorrect.

Statement E: LEDs have to be connected in forward bias for emission of light.

• This is the basic condition for an LED to work. When we connect it in forward bias, the potential barrier at the junction is lowered.

• This allows the majority carriers to flow across the junction and recombine. This recombination process releases energy in the form of light (photons).

• In reverse bias, the barrier height increases, very little current flows, and no light is emitted.

• So, statement E is Correct.

Final Conclusion:

Based on our step-by-step analysis, the only correct statements are B and E.

Looking at the options provided:

Option A: A, C, E Only

Option B: B, E Only

Option C: A, C Only

Option D: B, D, E Only

Thus, the correct choice is Option B.

$32.3 \quad (3\ \text{sig figs}), \quad 1125 \quad (4\ \text{sig figs}), \quad 27.4 \quad (3\ \text{sig figs})$

In multiplication and division, the final answer must be reported with the same number of significant figures as the factor with the fewest significant figures, which here is 3.

First, calculate the unrounded result:

$y = \frac{32.3 \times 1125}{27.4} \approx \frac{36337.5}{27.4} \approx 1326.186$

Expressing $1326.186$ in scientific notation:

$1326.186 \approx 1.326186 \times 10^3$

Rounding to 3 significant figures:

The first three significant digits are 1, 3, and 2.

Considering the fourth digit (6) causes the last significant digit to round up.

Thus,

$1.326186 \times 10^3 \approx 1.33 \times 10^3$

In standard decimal form, this is written as:

$1330$

Hence, the final reported value is:

$y = 1330$

For an ideal gas, the equation of state is given by

$pV = nRT.$

If during a process the pressure increases linearly with temperature (i.e., $p \propto T$), then from the ideal gas law (with a fixed amount of gas, $n$, and with the gas constant, $R$), it follows that

$\frac{p}{T} = \frac{nR}{V}.$

Since $nR$ is constant, a linear relationship between $p$ and $T$ implies that $\frac{p}{T}$ remains constant throughout the process. This can only be true if the volume $V$ is constant. Hence, the process is isochoric.

For an isochoric (constant volume) process:

The work done by the gas is given by

$W = \int p\, dV,$

and since $dV = 0$, we have $W = 0.$ (Statement A is true.)

The internal energy change for an ideal gas depends only on temperature:

$\Delta U = nC_V\Delta T.$

If the temperature increases, then $\Delta U > 0.$ (Statement D is true.)

For an isochoric process, no expansion work is done, so any heat added goes entirely into increasing the internal energy. This means the heat added is not different from the change in internal energy (in fact, $Q = \Delta U$). (Statement B is false.)

Since the process is isochoric, the volume does not change (and certainly is not increased). (Statement C is false, and Statement E stating it is isochoric is true.)

Thus, the correct true statements are A, D, and E.

To solve the problem, we need to compare the increase in surface area when the big drop is split into small drops.

Step 1: Determine the radius of the small drops

For 27 small drops:

Total volume is conserved:

$27 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \implies r^3 = \frac{R^3}{27} \implies r = \frac{R}{3}.$

For 64 small drops:

Similarly,

$64 \times \frac{4}{3}\pi r'^3 = \frac{4}{3}\pi R^3 \implies r'^3 = \frac{R^3}{64} \implies r' = \frac{R}{4}.$

Step 2: Calculate the surface area before and after the break-up

Initial surface area of the big drop:

$A_{\text{initial}} = 4\pi R^2.$

For 27 drops:

Surface area of one small drop:

$4\pi\left(\frac{R}{3}\right)^2 = \frac{4\pi R^2}{9}.$

Total surface area:

$A_{27} = 27 \times \frac{4\pi R^2}{9} = 3 \times 4\pi R^2 = 12\pi R^2.$

Increase in surface area:

$\Delta A_{27} = 12\pi R^2 - 4\pi R^2 = 8\pi R^2.$

For 64 drops:

Surface area of one small drop:

$4\pi\left(\frac{R}{4}\right)^2 = 4\pi \frac{R^2}{16} = \frac{\pi R^2}{4}.$

Total surface area:

$A_{64} = 64 \times \frac{\pi R^2}{4} = 16\pi R^2.$

Increase in surface area:

$\Delta A_{64} = 16\pi R^2 - 4\pi R^2 = 12\pi R^2.$

Step 3: Relate work done to the change in surface area

Since the work done is proportional to the increase in surface area:

$\text{Work} \propto \Delta A.$

We know that breaking into 27 drops requires 10 J corresponding to an increase of $8\pi R^2$.

Thus, if $W$ is the work done,

$W_{27} = k \cdot 8\pi R^2 = 10\, \text{J},$

where $k$ is the proportionality constant.

For 64 drops:

$W_{64} = k \cdot 12\pi R^2.$

Dividing the two equations:

$\frac{W_{64}}{10} = \frac{12\pi R^2}{8\pi R^2} = \frac{12}{8} = \frac{3}{2}.$

Thus,

$W_{64} = 10 \times \frac{3}{2} = 15\, \text{J}.$

Final Answer: 15 J (Option D)

Total flux through square $=\frac{\mathrm{q}}{\epsilon_0}\left(\frac{1}{6}\right)$

Lets divide square is 8 equal parts.

Flux is same for each part.

$\therefore$ Flux through shaded portion is $\frac{5}{8}$ (Total flux)

$=\frac{5}{8} \times \frac{\mathrm{q}}{\epsilon_0} \frac{1}{6}=\frac{5}{48} \frac{1}{\epsilon_0}$

$\therefore$ Required Ans. is 48

Derivation of the de Broglie wavelength after time $t$

The de Broglie wavelength is given by

$\lambda = \frac{h}{p},$

where $h$ is Planck's constant and $p$ is the momentum of the particle.

Initial Conditions:

Initially, the electron has a velocity

$\vec{v} = v_0\,\hat{i},$

so its momentum is

$p_0 = m v_0.$

The initial de Broglie wavelength is therefore

$\lambda_0 = \frac{h}{m v_0}.$

Effect of the Electric Field:

The electron enters an electric field

$\vec{E} = -E_0\,\hat{k}.$

For an electron with charge $-e$, the force is given by

$\vec{F} = -e\,\vec{E} = -e\left(-E_0\,\hat{k}\right) = e E_0\,\hat{k}.$

This force causes an acceleration in the $\hat{k}$ direction:

$a_z = \frac{F_z}{m} = \frac{e E_0}{m}.$

Momentum Components After Time $t$:

In the $\hat{i}$ direction, there is no force, so the momentum remains constant:

$p_x = m v_0.$

In the $\hat{k}$ direction, starting from rest, the momentum becomes:

$p_z = m v_z = m \left(a_z t\right) = e E_0 t.$

Total Momentum of the Electron:

The total momentum is the vector sum of the components:

$p = \sqrt{p_x^2 + p_z^2} = \sqrt{(m v_0)^2 + (e E_0 t)^2}.$

New de Broglie Wavelength:

Substituting the expression for the total momentum into the de Broglie relation gives:

$\lambda = \frac{h}{\sqrt{m^2 v_0^2 + e^2 E_0^2 t^2}}.$

Expressing this in terms of the initial de Broglie wavelength $\lambda_0 = \frac{h}{m v_0}$:

$\lambda = \frac{\lambda_0}{\sqrt{1+\frac{e^2 E_0^2 t^2}{m^2 v_0^2}}}.$

This final expression matches the option:

$\frac{\lambda_0}{\sqrt{1+\frac{e^2 E_0^2 t^2}{m^2 v_0^2}}}.$

The least count (L.C.) of a screw gauge is given by:

$\text{L.C.} = \frac{\text{Pitch}}{\text{Number of Divisions on Circular Scale}}$

Let the original pitch be $p$ and the original number of divisions be $n$. Then, the original least count is:

$\frac{p}{n} = 0.01 \, \text{mm}$

After modifications:

The pitch is increased by 75%, so the new pitch is:

$p_{\text{new}} = p + 0.75p = 1.75p$

The number of divisions is reduced by 50%, so the new number of divisions is:

$n_{\text{new}} = 0.5n$

The new least count is then:

$\text{L.C.}_{\text{new}} = \frac{p_{\text{new}}}{n_{\text{new}}} = \frac{1.75p}{0.5n} = \frac{1.75}{0.5} \times \frac{p}{n} = 3.5 \times \frac{p}{n}$

Substitute the original least count:

$\text{L.C.}_{\text{new}} = 3.5 \times 0.01 \, \text{mm} = 0.035 \, \text{mm}$

Expressed in the form $\times 10^{-3} \, \text{mm}$, this becomes:

$0.035 \, \text{mm} = 35 \times 10^{-3} \, \text{mm}$

Thus, the new least count is:

$35 \times 10^{-3} \, \text{mm}$

$\frac{1}{\lambda_{AC}} = \frac{E_A - E_C}{hc}, \quad \frac{1}{\lambda_{BC}} = \frac{E_B - E_C}{hc}$

Since the electron transitions from state A to state C in two steps (A to B and then B to C), the energy difference for the transition from A to B is given by

$E_A - E_B = (E_A - E_C) - (E_B - E_C)$

Expressing these energy differences in terms of wavelengths,

$\frac{hc}{\lambda_{AB}} = \frac{hc}{\lambda_{AC}} - \frac{hc}{\lambda_{BC}}$

Cancelling out $hc$, we obtain

$\frac{1}{\lambda_{AB}} = \frac{1}{\lambda_{AC}} - \frac{1}{\lambda_{BC}}$

Substitute the given values:

$\frac{1}{\lambda_{AB}} = \frac{1}{2000} - \frac{1}{6000}$

Finding a common denominator:

$\frac{1}{\lambda_{AB}} = \frac{3 - 1}{6000} = \frac{2}{6000} = \frac{1}{3000}$

Thus, the wavelength for the A to B transition is

$\lambda_{AB} = 3000 \, \mathring{A}.$

This matches Option D.

$\frac{1}{f} = \frac{2(n_2 - n_1)}{n_1 R}$

For a thin plano‑convex lens made of glass (refractive index $n_2 = 1.5$) immersed in a liquid (refractive index $n_1 = 1.2$), when the plane side is silver‐coated, a ray entering the curved surface from the liquid is refracted into the glass, reflects off the coated plane (thereby reversing its direction) and finally refracts back into the liquid at the same curved surface. In the paraxial approximation the net effect is equivalent to a mirror in the liquid with an effective focal length $f$.

A ray that is incident parallel to the optical axis (at height $h$) will, after this combined refraction–reflection–refraction process, emerge with an angular deviation corresponding to a mirror having a power given by:

$\frac{h}{f} = \frac{2(n_2 - n_1)}{n_1R}\, h\,.$

This gives the relation:

$\frac{1}{f} = \frac{2(n_2 - n_1)}{n_1 R}\,.$

Given that the effective focal length is $f = 0.2\,\text{m}$, substitute the values:

$\frac{1}{0.2} = \frac{2(1.5 - 1.2)}{1.2\,R}\,.$

Simplify the numerator:

$\frac{1}{0.2} = \frac{2(0.3)}{1.2\,R} = \frac{0.6}{1.2\,R}\,.$

Since

$\frac{0.6}{1.2} = 0.5,$

the equation becomes:

$5 = \frac{0.5}{R}\,.$

Solving for $R$:

$R = \frac{0.5}{5} = 0.10\,\text{m}\,.$

Thus, the radius of curvature of the curved surface of the lens is $0.10\,\text{m}$.

$T \propto R^{\frac{3}{2}}$

For a satellite in a circular orbit with radius $R$, the time period is given by

$T = 2\pi \sqrt{\frac{R^3}{GM}}.$

If the radius is increased to $1.03R$, the new time period $T'$ becomes

$T' = 2\pi \sqrt{\frac{(1.03R)^3}{GM}} = T \times (1.03)^{\frac{3}{2}}.$

For a small change, we can approximate

$(1.03)^{\frac{3}{2}} \approx 1 + \frac{3}{2} \times 0.03 = 1 + 0.045 = 1.045.$

This means the new time period is approximately 4.5% larger than the original time period.

Thus, the time period of the second satellite is larger by approximately 4.5%.

$Q = n\,C_P\,\Delta T \quad \text{and} \quad \Delta U = n\,C_V\,\Delta T$

For a monoatomic ideal gas, the molar specific heats are given by:

$C_V = \frac{3}{2}R \quad \text{and} \quad C_P = C_V + R = \frac{5}{2}R$

Thus, the ratio of the heat added to the change in internal energy is:

$\frac{Q}{\Delta U} = \frac{C_P}{C_V} = \frac{\frac{5}{2}R}{\frac{3}{2}R} = \frac{5}{3}$

Given that:

$\frac{E_1}{E_2} = \frac{x}{9}$

We equate the two ratios:

$\frac{5}{3} = \frac{x}{9}$

Multiplying both sides by 9:

$x = \frac{5}{3} \times 9 = 15$

Thus, the value of $x$ is 15.