JEE Main 2025 3rd April Morning Shift Question Paper with Solutions

Melting takes place isothermally at $T = 273\;\text{K}$ and constant atmospheric pressure $P = 1\;\text{atm}$ (≈ $1.013\times10^{5}\;\text{Pa}$).

At constant pressure the heat absorbed equals the molar enthalpy of fusion:
$q_{p}= \Delta H_{fus} \;\;(\text{positive})$

The first-law relation connecting enthalpy and internal energy is
$\Delta H = \Delta U + P\,\Delta V \quad -(1)$

Rearranging, the change in internal energy is
$\Delta U = \Delta H - P\,\Delta V \quad -(2)$

When ice converts to water its volume decreases because water is denser than ice:
density of ice ≈ $0.917\;\text{g cm}^{-3}$, density of water ≈ $1.00\;\text{g cm}^{-3}$.
Hence $\Delta V \lt 0$ (final volume < initial volume).

Work done by the system on the atmosphere is
$w = P\,\Delta V$.
Since $\Delta V \lt 0$, $w$ is negative: the system does negative work, that is, the atmosphere does positive work on the system.

Substituting $\Delta V \lt 0$ in $(2)$:
$\Delta U = \Delta H - P\,(\text{negative}) = \Delta H + (\text{positive term})$.
Therefore $\Delta U \gt 0$; the internal energy increases, it certainly does not remain unchanged or decrease.

Summarising:
• Internal energy increases ⇒ Options A and C are incorrect.
• Work done on the system is positive ⇒ Option B is incorrect, Option D is correct.

Hence the correct choice is Option D: Positive work is done on the ice-water system by the atmosphere.

The tank is completely filled, so the vertical distance between the top free surface and the hole is the full column above the hole.

Depth of hole below the top surface:
$h = 1.6\,\text{m} - 0.9\,\text{m} = 0.7\,\text{m}$

1. Pressure at the hole due to the water column alone (hydrostatic law $P = \rho g h$):
Density of water $\rho = 1000\,\text{kg/m}^3$, acceleration due to gravity $g = 10\,\text{m/s}^2$.
$P_{\text{water}} = \rho g h = 1000 \times 10 \times 0.7 = 7000\,\text{Pa}$

2. Extra pressure caused by the 50 kg load placed on the water surface:
Area of the tank’s cross-section $A = 0.5\,\text{m}^2$.
Force of the load $F = mg = 50 \times 10 = 500\,\text{N}$.
Extra pressure $P_{\text{load}} = \dfrac{F}{A} = \dfrac{500}{0.5} = 1000\,\text{Pa}$

3. Total gauge pressure at the level of the hole:
$P_{\text{total}} = P_{\text{water}} + P_{\text{load}} = 7000 + 1000 = 8000\,\text{Pa}$

4. Speed of efflux (Torricelli’s theorem with an added surface pressure):
For a small orifice the velocity is obtained from energy balance
$v = \sqrt{\dfrac{2 P_{\text{total}}}{\rho}}$

Substituting values:
$v = \sqrt{\dfrac{2 \times 8000}{1000}} = \sqrt{16} = 4\,\text{m/s}$

Therefore, the velocity of water emerging from the hole is 4 m/s.
Option D.

The output $Y$ is directly dependent only on input $A$. The Boolean expression is simply $Y = A$

Option A:
$Y = (A + B) \cdot B$

$Y = A \cdot B + B \cdot B$

Since $B \cdot B = B$:

$Y = A \cdot B + B$

$Y = B \cdot (A + 1)$

Since $A + 1 = 1$:

$Y = B \cdot 1$

Option B:
$Y = A \cdot (A + B)$
$Y = A \cdot A + A \cdot B$
$Y = A + A \cdot B$
$Y = A \cdot (1 + B)$
$Y = A \cdot 1$
$Y = A$

Option C:
$Y = A \cdot (\overline{A + B})$
$Y = A \cdot (\overline{A} \cdot \overline{B})$
$Y = (A \cdot \overline{A}) \cdot \overline{B}$
Since $A \cdot \overline{A} = 0$:
$Y = 0 \cdot \overline{B}$
$Y = 0$

Option D:
$Y = \overline{A} \cdot (A + B)$
$Y = \overline{A} \cdot A + \overline{A} \cdot B$
$Y = 0 + \overline{A} \cdot B$
$Y = \overline{A} \cdot B$

The source emits light equally in all directions, so the energy spreads over the surface of a sphere of radius $r$.

Energy flux (intensity) at distance $r$ is given by the inverse-square law:
$I = \frac{P}{4 \pi r^{2}}$ where $P$ is the power of the source.

Substitute $P = 450 \text{ W}$ and $r = 2 \text{ m}$:
$I = \frac{450}{4 \pi (2)^{2}} = \frac{450}{16 \pi}$ W m$^{-2}$.

Numerical value:
$I = \frac{450}{50.265} \approx 8.96 \text{ W m}^{-2}$.

For a perfectly reflecting surface, radiation pressure is twice that for a perfectly absorbing surface. The formula is
$P_{\text{rad}} = \frac{2I}{c}$ where $c = 3 \times 10^{8} \text{ m s}^{-1}$.

Insert the value of $I$:
$P_{\text{rad}} = \frac{2 \times 8.96}{3 \times 10^{8}}$ Pa
$= \frac{17.92}{3 \times 10^{8}} \text{ Pa}$.

Simplify:
$P_{\text{rad}} \approx 5.97 \times 10^{-8} \text{ Pa} \approx 6 \times 10^{-8} \text{ Pa}$.

Therefore, the radiation pressure on the perfectly reflecting surface is $6 \times 10^{-8}$ Pascals.

Correct option: Option C.

$R_{total} = \rho \frac{L}{A}$

$R_{total} = (2 \times 10^{-6}) \times \frac{25}{5 \times 10^{-6}}$

$R_{total} = 2 \times \frac{25}{5}$

$R_{total} = 2 \times 5$

$R_{total} = 10\ \Omega$

The wire is now bent into a complete circle. When calculating the resistance between two diametrically opposite points, these points divide the circular wire into two equal semicircular halves.

The resistance of each half will be exactly half of the total resistance:

$R_1 = R_2 = \frac{R_{total}}{2} = \frac{10}{2} = 5\ \Omega$

These two halves act as two resistors connected in parallel between the diametrically opposite points.

The equivalent resistance ($R_{eq}$) for two resistors in parallel is given by:

$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$

$R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2}$

$R_{eq} = \frac{5 \times 5}{5 + 5}$

$R_{eq} = \frac{25}{10}$

$R_{eq} = 2.5\ \Omega$

Treat both blocks as one system (if no slipping occurs).

Total mass

M+m

Restoring force by spring:

F=−kx

So equation of motion is

$(M+m)a=-kx$

Thus

$a=-\frac{kx}{M+m}$

So (B) is correct.

For SHM,

$\omega=\sqrt{\frac{k}{M+m}}$

Hence

$T=2\pi\sqrt{\frac{M+m}{k}}$

So (A) is correct.

For upper block m, only horizontal force is friction.

It must provide acceleration

$a=\frac{kx}{M+m}$

Therefore friction needed is

$f=ma$

$f=\frac{mkx}{M+m}$

So (C) is wrong 

For no slipping,

required friction must not exceed maximum static friction:

$\frac{mkA}{M+m}\le\mu mg$

Cancel m:

$\frac{kA}{M+m}\le\mu g$

Thus maximum amplitude

$A_{\max}=\frac{\mu(M+m)g}{k}$

So (D) is correct.

Now (E):

Maximum possible friction between blocks is

$f_{\max}=\mu N$

Normal reaction is only due to upper block’s weight:

N=mg

So

$f_{\max}=\mu mg$

So (E) is false.

Let’s check each graph using basic rules of 1D motion.

Key idea: for any physical motion, at a given time there must be a unique value of position/velocity/etc. Also, graphs must satisfy physical definitions (like velocity = dx/dt, distance can’t decrease, etc.)

(A) Phase vs Time (straight line)
Phase increasing linearly with time is possible (like uniform motion in SHM phase representation).
No contradiction → valid.

(B) Velocity vs Displacement (circle)
This can represent motion like SHM where

$v^2\propto(A^2-x^2)$

which gives a circular/elliptic relation.
For each displacement, two velocities (forward/backward) are allowed → physically valid.

(C) Velocity vs Time (circle)
This is not possible.
At a given time, velocity must have a single value.
But a circle gives two velocities for the same time → violates definition.
So invalid.

(D) Total distance vs Time
Total distance must:

• always increase (never decrease)
• can be constant temporarily (if particle stops)

The graph shown is non-decreasing → valid.

Final answer:
A, B and D only
(C is not possible because it gives multiple velocities at the same time)

First configuration

Two dielectrics fill half the thickness each, so they are in series.

Each slab has thickness

$\frac{d}{2}$

and full area AAA.

Capacitances:

$C_a=\frac{\varepsilon_1A}{d/2}$

$=\frac{2\varepsilon_1A}{d}$

$C_b=\frac{\varepsilon_2A}{d/2}=\frac{2\varepsilon_2A}{d}$

Series combination:

$\frac{1}{C_1}=\frac{1}{C_a}+\frac{1}{C_b}$$\frac{1}{C_1}=\frac{d}{2A}\left(\frac{1}{\varepsilon_1}+\frac{1}{\varepsilon_2}\right)$

Thus

$C_1=\frac{2A\varepsilon_1\varepsilon_2}{d(\varepsilon_1+\varepsilon_2)}$

Second configuration

Each dielectric occupies half area

$\frac{A}{2}$

and full thickness ddd, so they are in parallel.

Capacitances:

$C'_1=\frac{\varepsilon_1(A/2)}{d}$

$C'_2=\frac{\varepsilon_2(A/2)}{d}$

Hence

$C_2=C'_1+C'_2$$C_2=\frac{A}{2d}(\varepsilon_1+\varepsilon_2)$

Now ratio:

$\frac{C_1}{C_2}=\frac{\frac{2A\varepsilon_1\varepsilon_2}{d(\varepsilon_1+\varepsilon_2)}}{\frac{A}{2d}(\varepsilon_1+\varepsilon_2)}$

Simplifying,

$\frac{C_1}{C_2}=\frac{4\varepsilon_1\varepsilon_2}{(\varepsilon_1+\varepsilon_2)^2}$

So correct answer is

$\frac{4\varepsilon_1\varepsilon_2}{(\varepsilon_1+\varepsilon_2)^2}$

The four physical quantities are gravitational constant $G$, gravitational potential energy $U$, gravitational potential $V$ and acceleration due to gravity $g$. We first write the defining equations for each quantity and then extract their dimensional formulas.

Case A : Gravitational constant  $G$
Newton’s law gives the gravitational force between two masses $m_1$ and $m_2$ separated by distance $r$ as
$F = \dfrac{G\,m_1\,m_2}{r^{2}}$
Re-arranging,
$G = \dfrac{F\,r^{2}}{m_1\,m_2}$
Force has dimensions $[MLT^{-2}]$. Hence
$[G] = \dfrac{[MLT^{-2}]\,L^{2}}{M\,M} = [M^{-1}L^{3}T^{-2}]$

Case B : Gravitational potential energy  $U$
Potential energy near Earth’s surface is $U = mgh$.
Mass $m$ has $[M]$, $g$ has $[LT^{-2}]$ and height $h$ has $[L]$. Therefore
$[U] = [M]\,[LT^{-2}]\,[L] = [ML^{2}T^{-2}]$

Case C : Gravitational potential  $V$
Potential is energy per unit mass: $V = \dfrac{U}{m}$.
Divide the dimensions of $U$ obtained above by $[M]$:
$[V] = \dfrac{[ML^{2}T^{-2}]}{[M]} = [L^{2}T^{-2}]$

Case D : Acceleration due to gravity  $g$
Acceleration is rate of change of velocity: $g = \dfrac{dv}{dt}$.
Velocity $v$ has $[LT^{-1}]$, so acceleration has
$[g] = \dfrac{[LT^{-1}]}{[T]} = [LT^{-2}]$

Now compare with LIST-II:
I. $[LT^{-2}]$  II. $[L^{2}T^{-2}]$  III. $[ML^{2}T^{-2}]$  IV. $[M^{-1}L^{3}T^{-2}]$

Matching each quantity with its dimensional formula:
A. $G$ ⟶ IV  B. $U$ ⟶ III  C. $V$ ⟶ II  D. $g$ ⟶ I

Thus the correct set is A-IV, B-III, C-II, D-I, which corresponds to Option A.

The force $F$ is applied at the highest point of the sphere. Taking torque about the point of contact with the ground, the friction force produces no torque.

The perpendicular distance of the force from the point of contact is $2r$. Therefore,  $F(2r)=I_P\alpha$

Using the parallel-axis theorem,  $I_P=I+mr^2$

For a solid sphere,  $I=\frac{2}{5}mr^2$

Therefore,  $I_P=\frac{2}{5}mr^2+mr^2=\frac{7}{5}mr^2$

Using the rolling condition,  $\alpha=\frac{a}{r}$

Hence,  $F(2r)=\frac{7}{5}mr^2\frac{a}{r}$

$2F=\frac{7}{5}ma$

Therefore,  $a=\frac{10F}{7m}$

Substituting $F=49\,N$ and $m=20\,kg$,

$a=\frac{10\times49}{7\times20}$

$a=3.5\,m/s^2$

Therefore,  ${a=3.5\,m/s^2}$

Hence, the correct option is A.

Since the gas is heated isothermally,

$ΔU=0$

(for ideal gas internal energy depends only on temperature)

So heat supplied by filament equals work done by gas:

$Q=W$

Total work done by gas has two parts:

1. Work done against atmosphere/piston weight
2. Work done in stretching spring

At equilibrium initially,

$P_0A=Mg$

where A is piston area.

Gas expands from height $L_0$ to $L_1$​

Change in volume:

$\Delta V=A(L_1-L_0)$

Work against piston weight:

$W_1=Mg(L_1-L_0)$

Spring force law:

$F=kx^3$

Initially spring is natural length, so initial extension

x=0

Final extension:

$x=L_1-L_0$

Work stored in spring:

$W_2​=\int kx^3dx\ \ from\ 0\ to\ L_{1\ }-L_0$

$\frac{k}{4}(L_1-L_0)^4$

Also because process is reversible isothermal,

$W=nRT\ln\frac{V_1}{V_0}$

with

$V_1=AL_1,\quad V_0=AL_0$

so

$W_3=nRT\ln\frac{L_1}{L_0}$

This is the expansion work of ideal gas.

Hence total energy supplied by filament

$Q=nRT\ln\frac{L_1}{L_0}+\frac{k}{4}(L_1-L_0)^4+Mg(L_1-L_0)$

For an adiabatic process involving an ideal gas, the relation between temperature and volume is
$T\,V^{\gamma-1}= \text{constant}$

Given data:
Initial volume $V_1 = 800\;\text{cm}^3$
Final volume $V_2 = 200\;\text{cm}^3$
Initial temperature $T_1 = 27^\circ\text{C} = 27 + 273 = 300\;\text{K}$
Specific-heat ratio $\gamma = 1.5$

Applying the adiabatic relation for the final temperature $T_2$:
$T_1\,V_1^{\gamma-1}=T_2\,V_2^{\gamma-1}$

Solve for $T_2$:
$T_2 = T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}$

Compute the volume ratio and exponent:
$\frac{V_1}{V_2} = \frac{800}{200} = 4$
$\gamma - 1 = 1.5 - 1 = 0.5$

Therefore,
$T_2 = 300\,\text{K}\;\times\;4^{0.5} = 300\,\text{K}\;\times\;2 = 600\,\text{K}$

The change in temperature is
$\Delta T = T_2 - T_1 = 600\,\text{K} - 300\,\text{K} = 300\,\text{K}$

Hence, the temperature rises by $300\;\text{K}$.
Option D is correct.

First recognise the four classes given in LIST-II.

I Chemical reaction - involves only electrons, no change in nucleus, energy of the order of $\text{kJ mol}^{-1}$.

II Fusion with $+$ve Q value - two light nuclei combine, total mass decreases, energy is released ($Q \gt 0$).

III Fission - a heavy nucleus splits into two medium-mass nuclei and a few neutrons, energy is released ($Q \gt 0$).

IV Fusion with $-$ve Q value - two nuclei combine but the product mass is larger, energy is absorbed ($Q \lt 0$).

Case A: $^{1}_{0}n + {}^{235}_{92}U \rightarrow {}^{140}_{54}Xe + {}^{94}_{38}Sr + 2\,^{1}_{0}n$
A single heavy nucleus $^{235}U$ breaks into two medium nuclei plus neutrons. According to the definition, this is nuclear fission. Hence A → III.

Case B: $2H_2 + O_2 \rightarrow 2H_2O$
Only the electronic configuration of atoms changes; nuclei remain unchanged. This is an ordinary chemical reaction. Hence B → I.

Case C: $^{2}_{1}H + {}^{2}_{1}H \rightarrow {}^{3}_{2}He + {}^{1}_{0}n$
Two deuterons fuse to form a helium-3 nucleus and a neutron.
Mass of reactants: $M_r = 2 \times 2.014102\,u = 4.028204\,u$.
Mass of products: $M_p = 3.016029\,u + 1.008665\,u = 4.024694\,u$.
$\Delta M = M_r - M_p = 0.003510\,u \gt 0$, so $Q = \Delta M c^2 \approx 3.27\ \text{MeV} \gt 0$.
Energy is released; it is a fusion reaction with positive Q value. Hence C → II.

Case D: $^{1}_{1}H + {}^{3}_{1}H \rightarrow {}^{2}_{1}H + {}^{2}_{1}H$
Mass of reactants: $1.007825\,u + 3.016049\,u = 4.023874\,u$.
Mass of products: $2 \times 2.014102\,u = 4.028204\,u$.
$\Delta M = M_r - M_p = -0.004330\,u \lt 0$, so $Q = \Delta M c^2 \approx -4.03\ \text{MeV} \lt 0$.
Energy is absorbed; it is a fusion reaction with negative Q value. Hence D → IV.

Thus the correct matching is
A - III, B - I, C - II, D - IV.

The option representing this pattern is Option B.

Final Answer: Option B

The shell is a thin, uniformly charged spherical shell of radius $R = 10 \text{ cm} = 0.10 \text{ m}$.
The given electrostatic potential on its surface is $V(R)=120 \text{ V}$.

Key electrostatics results for a uniformly charged thin spherical shell
1. For any point inside the shell ($r \lt R$): the electric field is zero, $E=0$, and the electrostatic potential is the same everywhere as on the surface. Hence $V(r)=V(R)$.
2. For any point outside the shell ($r \gt R$): the shell behaves as if all its charge were concentrated at its centre. Therefore $V(r)=\dfrac{kQ}{r}$, where $Q$ is the total charge and $k=\dfrac{1}{4\pi\varepsilon_0}$.

We use these two facts to find the required potentials.

Case 1: At the centre of the shell ( $r = 0$ )
Since $r \lt R$, the potential equals the surface potential: $V(0)=V(R)=120 \text{ V}$.

Case 2: At an interior point $r = 5 \text{ cm} = 0.05 \text{ m}$
Again $r \lt R$, so $V(5\text{ cm})=V(R)=120 \text{ V}$.

Case 3: At an exterior point $r = 15 \text{ cm} = 0.15 \text{ m}$
Here $r \gt R$, so use the point-charge formula.

The charge $Q$ is not given directly, but we can eliminate it using the known surface potential:
$V(R)=\dfrac{kQ}{R}=120 \text{ V} \quad -(1)$

For $r = 0.15 \text{ m}$, $V(15\text{ cm})=\dfrac{kQ}{r}$

Divide this by equation $(1)$ to remove $Q$:
$\dfrac{V(15\text{ cm})}{120}=\dfrac{\dfrac{kQ}{r}}{\dfrac{kQ}{R}}=\dfrac{R}{r}$

Hence
$V(15\text{ cm})=120 \times \dfrac{R}{r}=120 \times \dfrac{0.10}{0.15}=120 \times \dfrac{2}{3}=80 \text{ V}$.

Summary of potentials
Centre (0 cm): $120 \text{ V}$
Inside at 5 cm: $120 \text{ V}$
Outside at 15 cm: $80 \text{ V}$

These values match Option A: 120 V, 120 V, 80 V.

We are given that the work function of a metal is $\phi = 3 \, eV$. We need to find which color of visible light can cause emission of photoelectrons.

For photoelectron emission to occur, the energy of the incident photon must be at least equal to the work function of the metal:
$E \geq \phi$

The energy of a photon is related to its wavelength by:
$E = \frac{hc}{\lambda}$

where $hc = 1240 \, eV \cdot nm$.

For the minimum photon energy of $3 \, eV$, the maximum wavelength is:
$\lambda_{max} = \frac{hc}{\phi} = \frac{1240}{3} \approx 413 \, nm$

So, only light with wavelength $\lambda \leq 413 \, nm$ can cause photoemission.

Now let us check which color of visible light satisfies this condition:

- Red light: $\lambda \approx 620 - 750 \, nm$ (energy $\approx 1.65 - 2.0 \, eV$) — Not sufficient
- Yellow light: $\lambda \approx 570 - 590 \, nm$ (energy $\approx 2.1 - 2.2 \, eV$) — Not sufficient
- Green light: $\lambda \approx 495 - 570 \, nm$ (energy $\approx 2.2 - 2.5 \, eV$) — Not sufficient
- Blue light: $\lambda \approx 400 - 495 \, nm$ (energy $\approx 2.5 - 3.1 \, eV$) — Sufficient at shorter wavelengths

Blue light at its shorter wavelength end (around $400 - 413 \, nm$) has energy of about $3.0 - 3.1 \, eV$, which is equal to or greater than the work function of $3 \, eV$. Therefore, blue light can cause photoelectron emission.

Red, yellow, and green light all have energies less than $3 \, eV$ and cannot cause photoemission from this metal.

Hence, the correct answer is Option B.

Let the particle be released from rest at height $S$ above the earth’s surface. Choose the earth’s surface as the reference level of zero potential energy.

Initial potential energy: $U_i = mgS$
Initial kinetic energy: $K_i = 0$ (because it is released from rest)

Total mechanical energy remains constant, so for any later instant
$K + U = mgS$  $-(1)$

At some height $h$ (measured from the surface) the kinetic energy is given to be three times the potential energy:
$K = 3U$  $-(2)$

Let the potential energy at this instant be $U = mgh$. Using $-(2)$, the kinetic energy at the same instant is
$K = 3mgh$.

Substitute these values of $K$ and $U$ into the energy conservation equation $-(1)$:
$3mgh + mgh = mgS$
$4mgh = mgS$
$h = \frac{S}{4}$.

Now compute the speed. Kinetic energy at height $h$ is also $K = \frac12 mv^2$. Equate this with $3mgh$:

$\frac12 mv^2 = 3mgh$
$v^2 = 6gh$
Insert $h = \frac{S}{4}$:
$v^2 = 6g\left(\frac{S}{4}\right) = \frac{3gS}{2}$
$v = \sqrt{\frac{3gS}{2}}$.

Therefore, the height of the particle above the earth’s surface and its speed at that instant are
$h = \frac{S}{4}, \qquad v = \sqrt{\frac{3gS}{2}}$.

Hence, the correct option is Option D.

Write the three measured masses with their decimal places aligned:
$$\begin{array}{r} 435.42 \\[2pt] 226.3 \\[2pt] \;0.125\\ \hline \end{array}$$

Rule for addition/subtraction with significant figures:
The final result must be rounded off to the same number of decimal places as the quantity having the least number of decimal places.

Count decimal places in each measurement:
435.42 g  → 2 decimal places
226.3 g  → 1 decimal place (least among the three)
0.125 g  → 3 decimal places

Perform the addition without rounding first:
$435.42 + 226.3 + 0.125 = 661.845$ g

The least number of decimal places is 1, so retain only one digit after the decimal and round the result accordingly:
661.845 g  → 661.8 g (since the next digit 4 is less than 5)

Therefore, the correct sum expressed with proper significant figures is $661.8$ g.

Hence, the required option is Option C.

The thin lens is placed in air, so we use the lens-maker’s formula

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

where
$f$ = focal length of the lens,
$n$ = refractive index of lens material with respect to air,
$R_1$ = radius of curvature of the first surface (the surface that light meets first),
$R_2$ = radius of curvature of the second surface.

Sign convention (Cartesian):
• Distances measured to the right of the surface are positive.
• For a double-convex lens, the first surface bulges toward the left (center of curvature on the right), so $R_1 \gt 0$.
• The second surface bulges toward the right (center of curvature on the left), so $R_2 \lt 0$.

Given magnitudes:
$|R_1| = 10 \text{ cm},\; |R_2| = 15 \text{ cm},\; f = 12 \text{ cm}$

Applying the signs:
$R_1 = +10 \text{ cm},\; R_2 = -15 \text{ cm}$

Substitute into the lens-maker’s formula:

$\frac{1}{12} = (n-1)\left(\frac{1}{10} - \frac{1}{-15}\right)$

Simplify the bracket first:

$\frac{1}{10} - \left(-\frac{1}{15}\right) = \frac{1}{10} + \frac{1}{15}$

Take the LCM (30):

$\frac{1}{10} + \frac{1}{15} = \frac{3}{30} + \frac{2}{30} = \frac{5}{30} = \frac{1}{6}$

Thus

$\frac{1}{12} = (n-1)\left(\frac{1}{6}\right)$

Cross-multiply:

$(n-1) = \frac{1}{12} \times 6 = \frac{6}{12} = 0.5$

Add 1 on both sides:

$n = 1 + 0.5 = 1.5$

Therefore, the refractive index of the lens material is $1.5$.

Option C is correct.

given
angle φ is measured from vertical

so convert to usual projectile angle θ (from horizontal):

θ = 90° − φ

maximum height formula:

hₘ = (u² sin²θ) / (2g)

substitute θ:

hₘ = (u² sin²(90° − φ)) / (2g)

use identity:

sin(90° − φ) = cosφ

so,

hₘ = (u² cos²φ) / (2g)

so relation is:

hₘ ∝ cos²φ

now understand behavior of cos²φ

when φ = 0°
cosφ = 1 → hₘ is maximum

when φ increases
cosφ decreases → hₘ decreases

when φ = 90°
cosφ = 0 → hₘ = 0

so graph must:

start at maximum value at φ = 0
decrease continuously
reach zero at φ = 90°

shape:

cos²φ decreases slowly at first, then faster
so curve is downward bending (not straight)

check options:

A → increasing → wrong
B → goes up then down → wrong
C → starts high, smoothly decreases to zero → correct
D → decreases too sharply like exponential → wrong

final answer: C

For a prism of refracting angle $A$ made of material having refractive index $n$, when the deviation is minimum the ray of light follows a symmetrical path.

Conditions at minimum deviation:
(i) Angle of incidence $i_1$ equals angle of emergence $i_2$.
(ii) Angle of refraction at first surface $r_1$ equals angle of refraction at second surface $r_2$, say $r$.
(iii) From the geometry of the prism, $r_1 + r_2 = A$, therefore at minimum deviation $2r = A \Rightarrow r = \dfrac{A}{2}$.

The deviation-incidence graph is a U-shaped curve with its lowest point at the minimum deviation $D_m$. Using Snell’s law at minimum deviation one writes

$n = \dfrac{\sin\left(\dfrac{A + D_m}{2}\right)}{\sin\left(\dfrac{A}{2}\right)}$ $-(1)$

Now each statement can be checked one by one.

Statement A: “The refracted ray inside prism becomes parallel to the base.”
In the isosceles triangle formed by the two refracting faces and the base, the internal ray makes the angle $r = A/2$ with each face. Hence it is parallel to the side opposite the apex, i.e. the base. So Statement A is TRUE.

Statement B: “Larger angle prisms provide smaller angle of minimum deviation.”
From equation $(1)$, keeping $n$ fixed, increase in $A$ increases the numerator $\sin\bigl(\tfrac{A+D_m}{2}\bigr)$ as well as the denominator $\sin\bigl(\tfrac{A}{2}\bigr)$, but the net result is that $D_m$ rises with $A$. A quick numerical check with $n = 1.50$ gives
for $A = 30^\circ,\; D_m \approx 15.7^\circ$;
for $A = 60^\circ,\; D_m \approx 37.2^\circ$.
Thus larger $A$ leads to larger, not smaller, $D_m$. So Statement B is FALSE.

Statement C: “Angle of incidence and angle of emergence become equal.”
This is the defining condition of minimum deviation (symmetrical path). Hence Statement C is TRUE.

Statement D: “There are always two sets of angle of incidence for which deviation will be the same except at minimum deviation setting.”
Because the deviation curve is symmetric about the minimum, every value of deviation greater than $D_m$ is obtained for two equal and opposite departures of $i$ from the symmetry point. Therefore Statement D is TRUE.

Statement E: “Angle of refraction becomes double of prism angle.”
We have already derived $r = \dfrac{A}{2}$, not $2A$. Hence Statement E is FALSE.

The TRUE statements are A, C, and D only.

Correct choice: Option A (A, C and D Only).

The only quantities that decide the collapse-time for three identical masses released from the corners of an equilateral triangle are

• the initial side length $a$,
• the mass of each sphere $m$,
• the universal constant of gravitation $G$.

Let the required time be $T$.  Assume a dependence of the form

$T \propto G^{\,p}\,m^{\,q}\,a^{\,r}$

Write dimensions (M ≡ mass, L ≡ length, T ≡ time):

$[T] = T^{1},\quad [G]=L^{3}M^{-1}T^{-2},\quad [m]=M^{1},\quad [a]=L^{1}$

Equating powers of the fundamental dimensions,

$L: \; 0 = 3p + r$ $M: \; 0 = -p + q$ $T: \; 1 = -2p$

Solving,

$p = -\frac12,\quad q = -\frac12,\quad r = \frac32$

Hence

$T \propto G^{-1/2}\,m^{-1/2}\,a^{3/2} \;=\;k\,\sqrt{\dfrac{a^{3}}{G\,m}}$

for some dimensionless constant $k$ that is the same for both experiments because the geometry is identical.

Initial situation: side $=a$, mass $=m$, time $=T = 4\,$s.

New situation: side $=2a$, mass $=2m$. Using the proportionality,

$\dfrac{T'}{T}=\dfrac{\sqrt{(2a)^{3}/(G\,2m)}}{\sqrt{a^{3}/(G\,m)}}$

$\qquad=\,\dfrac{2^{3/2}}{2^{1/2}} = 2$

Therefore $T' = 2T = 2 \times 4\text{ s} = 8\text{ s}$.

So the three spheres will collide after 8 seconds.