JEE Main 2025 April 4 shift 2

In the decay chain $P \to Q \to R$, the mass of $P$ follows a simple exponential decay:
$N_P(t) = N_0 e^{-\lambda_P t}$
For the radioactive intermediate $Q$, the rate of change is given by the balance equation:
$\frac{dN_Q}{dt} = \lambda_P N_P - \lambda_Q N_Q$
Initially $N_Q(0) = 0$, so $N_Q$ increases, reaches a maximum when production equals decay, and then decreases as $P$ is depleted. The stable final product $R$ accumulates over time at a rate proportional to the amount of $Q$:
$\frac{dN_R}{dt} = \lambda_Q N_Q$
Thus, $N_R$ increases monotonically until all of $P$ and $Q$ have converted into $R$. Only graph B correctly represents the exponential decay of $P$, the peak for $Q$, and the monotonic increase of $R$.

Let the mains supply voltage be $V$ (same for every bulb). Each identical bulb is rated to consume power $p$ when connected alone across the mains. For a resistor-type load the power relation is $p = \frac{V^{2}}{R} \; -(1)$ where $R$ is the resistance of one bulb. From $(1)$, the resistance of each bulb is $R = \frac{V^{2}}{p} \; -(2)$ When the $n$ bulbs are connected \in SERIES, their resistances add up: $R_{\text{series}} = nR = n \left(\frac{V^{2}}{p}\right) = \frac{nV^{2}}{p} \; -(3)$ The current drawn from the mains is given by Ohm's law: $I = \frac{V}{R_{\text{series}}} = \frac{V}{\dfrac{nV^{2}}{p}} = \frac{p}{nV} \; -(4)$ The total power taken from the mains by the series combination is $P_{\text{total}} = I^{2} R_{\text{series}}$ Substituting from $(4)$ and $(3)$: $P_{\text{total}} = \left(\frac{p}{nV}\right)^{2} \times \frac{nV^{2}}{p}$ $\quad = \frac{p^{2}}{n^{2}V^{2}} \times \frac{nV^{2}}{p}$ $\quad = \frac{p}{n}$ Hence the total power drawn by the $n$ bulbs \in series is $\dfrac{p}{n}$. Option C is correct.

First convert the given dimensions to SI units: length $l = 9\text{ cm} = 0.09\text{ m}$, width $d = 4\text{ cm} = 0.04\text{ m}$. The initial area of the rectangular sheet is $A_0 = l \times d = 0.09 \times 0.04 = 0.0036\ \text{m}^2$ $-(1)$ Heat supplied to the sheet is $Q = 8.1 \times 10^{2}\ \text{J}$. Using the relation $Q = m\,C_v\,\Delta T$, the rise \in temperature is $\Delta T = \frac{Q}{m\,C_v} = \frac{8.1 \times 10^{2}}{0.1 \times 900} = 9\ \text{K}$ $-(2)$ For an isotropic solid, the coefficient of superficial (area) expansion is $\beta = 2\alpha$ $-(3)$ where $\alpha = 3.1 \times 10^{-5}\ \text{K}^{-1}$ is the linear expansion coefficient. From $(3)$, $\beta = 2 \times 3.1 \times 10^{-5} = 6.2 \times 10^{-5}\ \text{K}^{-1}$ $-(4)$ The fractional change \in area due to heating through $\Delta T$ is $\beta\,\Delta T$, so the absolute change \in area is $\Delta A = A_0 \,\beta \,\Delta T$ $-(5)$ Substituting from $(1)$, $(2)$ and $(4)$ into $(5)$: $\Delta A = 0.0036 \times 6.2 \times 10^{-5} \times 9$ Compute step by step: $6.2 \times 10^{-5} \times 9 = 5.58 \times 10^{-4}$; $0.0036 \times 5.58 \times 10^{-4} = 2.01 \times 10^{-6}\ \text{m}^2$. Rounding to two significant figures, $\Delta A \approx 2.0 \times 10^{-6}\ \text{m}^2$. Hence, the change \in area of the rectangular sheet is $2.0 \times 10^{-6}\ \text{m}^2$. Option A is correct.

First recall the symbols: • Planck's constant $h$ \bullet Angular momentum of an electron \in Bohr orbit $L$ Checking Statement (I) : "The dimensions of Planck's constant and angular momentum are same." Write dimensions of Planck's constant. Planck's constant is energy $\times$ time. Energy has dimension $ML^{2}T^{-2}$, therefore $[h] = ML^{2}T^{-2}\times T = ML^{2}T^{-1}$ Write dimensions of angular momentum. For a particle of mass $m$ moving with speed $v$ \in a circle of radius $r$, $L = mvr$ Dimensions: $[L] = M(LT^{-1})L = ML^{2}T^{-1}$ Thus $[h] = [L] = ML^{2}T^{-1}$, so Statement (I) is correct. Checking Statement (II) : "In Bohr's model electron revolve around the nucleus only \in those orbits for which angular momentum is integral multiple of Planck's constant." Bohr's postulate states $mvr = n\frac{h}{2\pi},\qquad n = 1,2,3,\dots$ $-(1)$ Equation $-(1)$ shows that angular momentum is an integral multiple of $\dfrac{h}{2\pi}$ (i.e. $\hbar$), not of $h$ itself. Hence Statement (II) is incorrect. Conclusion : Statement (I) is correct, Statement (II) is incorrect. This matches Option C. Therefore, the most appropriate answer is Option C .

The cylindrical rod is subjected to a horizontal (shear) force at its upper end while the lower end is fixed. Such a loading produces a uniform shear stress over the circular cross-section. Step 1 : Shear stress on the cross-section Shear stress $\tau$ is force per unit area: $\tau = \dfrac{F}{A}$ The cross-sectional area is $A = \pi r^{2}$. Given radius $r = 4\text{ cm} = 0.04\text{ m}$, so $A = \pi (0.04)^{2} = \pi \times 0.0016 = 0.0016\,\pi\;\text{m}^{2}$. With the applied force $F = 10^{5}\text{ N}$, $\tau = \dfrac{10^{5}}{0.0016\,\pi} = \dfrac{10^{5}\times625}{\pi} = \dfrac{6.25\times10^{7}}{\pi}\;\text{N m}^{-2}$ $-(1)$ Step 2 : Shear strain produced For an elastic material, shear strain $\gamma$ is related to shear stress by the shear modulus $G$: $\gamma = \dfrac{\tau}{G}$. Here $G = 10^{10}\;\text{N m}^{-2}$. Substitute $\tau$ from $(1)$: $\gamma = \dfrac{6.25\times10^{7}/\pi}{10^{10}} = \dfrac{6.25}{\pi\times10^{3}} = \dfrac{0.00625}{\pi}$. Step 3 : Relation between shear strain and angular displacement For a small angular displacement of the top relative to the fixed bottom, shear strain equals the angle \in radians: $\theta \approx \gamma$. Therefore, $\theta = \dfrac{0.00625}{\pi} = \dfrac{1}{160\,\pi}\;\text{radian}$. Result Angular displacement $\theta = \dfrac{1}{160\pi}\;\text{rad}$. Hence, the correct option is Option A .

Given: R₁ = R₂ = R₃ = 5 Ω R₄ = 10 Ω Check option A : Top branch: R₁ + R₂ = 5 + 5 = 10 Ω Bottom branch: R₃ + R₄ = 5 + 10 = 15 Ω These two branches are in parallel: $R_{eq}=(10\times15)/(10+15)$ = 150 / 25 = 6 Ω So this exactly matches the required resistance.

Arc AB is a quarter circle subtending $90^{\circ}$ and its electric field at center is given as E Now consider arc ABC. It consists of two quarter arcs: arc AB arc BC Each subtends $90^{\circ}$, so each produces field of magnitude E at center. For uniformly charged arc, field at center lies along bisector of the arc. For arc AB, field is along bisector of first quadrant diagonal toward southwest. For arc BC, field is along bisector of fourth quadrant diagonal toward northwest. These two field vectors are perpendicular because the bisectors differ by $90^{\circ}$ So resultant field due to arc ABC is $\sqrt{E^{2+}E^2}$ $=\sqrt{2}E$

Let the volume of the smaller vessel be $V$. The larger vessel has double this volume, so its volume is $2V$. Initial data Large vessel : $P_1 = 8\text{ kPa},\; T_1 = 1000\text{ K},\; V_1 = 2V$ Small vessel : $P_2 = 7\text{ kPa},\; T_2 = 500\text{ K},\; V_2 = V$ Using the ideal-gas equation $PV = nRT$, the initial number of moles \in each vessel is $n_{1i} = \frac{P_1 V_1}{R T_1} = \frac{8 \times 2V}{R \times 1000} = \frac{16V}{1000R} = 0.016\frac{V}{R}$ $n_{2i} = \frac{P_2 V_2}{R T_2} = \frac{7 \times V}{R \times 500} = \frac{7V}{500R} = 0.014\frac{V}{R}$ Total moles before connection $n_{\text{total}} = n_{1i} + n_{2i} = 0.016\frac{V}{R} + 0.014\frac{V}{R} = 0.030\frac{V}{R}$ After connecting the vessels, both are maintained at a common temperature $T_f = 600\text{ K}$. At steady state the common pressure is $P_f$. Moles \in the large vessel at this stage: $n_{1f} = \frac{P_f \, V_1}{R T_f} = \frac{P_f \times 2V}{R \times 600} = \frac{P_f V}{300R}$ Moles \in the small vessel: $n_{2f} = \frac{P_f \, V_2}{R T_f} = \frac{P_f \times V}{R \times 600} = \frac{P_f V}{600R}$ Total moles remain unchanged, so $n_{1f} + n_{2f} = n_{\text{total}}$ $\frac{P_f V}{300R} + \frac{P_f V}{600R} = 0.030\frac{V}{R}$ Combine the terms on the left: $P_f V \left(\frac{1}{300} + \frac{1}{600}\right)\!\bigg/\!R = P_f V \left(\frac{2 + 1}{600}\right)\!\bigg/\!R = \frac{P_f V}{200R}$ Equate this to the right side and solve for $P_f$: $\frac{P_f V}{200R} = 0.030\frac{V}{R} \;\;\Longrightarrow\;\; P_f = 0.030 \times 200 = 6\text{ kPa}$ Hence the common steady-state pressure \in both vessels is $6\text{ kPa}$. Option B is correct.

The object is initially at rest at a height $h = 3R$ above Earth's surface. Therefore its initial distance from Earth's centre is $r_0 = R + h = R + 3R = 4R$. To "never return to Earth" the object must just reach infinity with zero speed. Hence we apply conservation of mechanical energy between the initial point $r_0$ and infinity. Initial mechanical energy Kinetic: $K_i = \frac12 m v^2$ (where $v$ is the required minimum speed). Gravitational potential: $U_i = -\frac{GMm}{r_0} = -\frac{GMm}{4R}$. Final mechanical energy at infinity Kinetic: $K_f = 0$ (minimum condition). Potential: $U_f = 0$ (by convention). Conservation of energy gives $K_i + U_i = K_f + U_f$ $\frac12 m v^2 - \frac{GMm}{4R} = 0$. Solve for $v$: $\frac12 m v^2 = \frac{GMm}{4R}$ $v^2 = \frac{2GM}{4R} = \frac{GM}{2R}$ $v = \sqrt{\frac{GM}{2R}}$. The minimum speed required is $\sqrt{\dfrac{GM}{2R}}$, which corresponds to Option A.

Initially $C_1=C_2=C_3=5\mu F$ Dielectric constant 4 is inserted \in $C_1$​, so $C_1'=4(5)=20\mu F$ From circuit: Top branch: $C_1\ ​and\ C_2$ are \in series Bottom branch:$C_3$ alone These two branches are \in parallel. Top branch equivalent $C_{top}=\frac{C_1'C_2}{C_1'+C_2}$ $=\frac{20\cdot5}{20+5}$ $\frac{100}{25}=4\mu F$ Total equivalent Parallel combination: $C_{eq}=C_{top}+C_3$ $=4+5$ $=9\mu F$

First, remember: Slope of x-t graph = velocity Straight line → constant velocity Horizontal line → zero velocity Average velocity = (final position − initial position) / time (A) Average velocity from 0 to 3 s From the graph: $At\ t=0,x=0$ $At\ t=3,x=5$ So, $v_{avg}=\frac{5-0}{3}=\frac{5}{3}\approx1.67\text{ m/s}$ Given statement says $10m/s,$ which is clearly wrong. So, (A) is false (B) Average velocity from 3 to 5 s From graph: $At\ t=3,x=5$ $At\ t=5,x=5$ So displacement does not change $v_{avg}=\frac{5-5}{5-3}=0$ This means the object is at rest. So, (B) is true (C) Instantaneous velocity at t=2 s At t=2, the graph lies on the straight line between: $(1,-5)and(3,5)$ Slope (velocity) of this line: $v=\frac{5-(-5)}{3-1}=\frac{10}{2}=5\text{ m/s}$ Since it's a straight line, velocity is constant \in that region. So, (C) is true (D) Compare velocities Average velocity from 5 to 7: $At\ t=5,x=5$ $At t=7,x=0$ $v_{avg}=\frac{0-5}{7-5}=\frac{-5}{2}=-2.5$ Instantaneous velocity at 6.5 s: This lies between t=6 and t=7, where graph is a straight line from: (6,10)to (7,0) $v=\frac{0-10}{7-6}=-10$ Clearly: $-2.5\ne -10$ So, (D) is false (E) Average velocity from 0 to 9 s At t=0, x=0 At t=9, x=0 $v_{avg}=\frac{0-0}{9-0}=0$ So, (E) is true Final Answer: (B), (C), (E) only

For a wheel that rolls without slipping, the centre of mass (CM) moves with linear speed $v_{CM}$ and the wheel simultaneously rotates with angular speed $\omega$ such that $v_{CM}=R\omega$, where $R$ is the radius. The linear velocity of any point on the rim is the vector \sum of the translational velocity $v_{CM}$ (same for every point, directed horizontally) and the tangential velocity due to rotation $\omega R$ (perpendicular to the radius through that point). The highest point on the rim has its rotational velocity \in the same horizontal direction as $v_{CM}$, so the speeds add: $v_{\text{top}} = v_{CM}+R\omega = v_{CM}+v_{CM}=2v_{CM}$. Given $v_{\text{top}} = 8 \text{ m/s}$, we get $2v_{CM}=8 \implies v_{CM}=4 \text{ m/s}$. Consider a point on the rim that lies at the same vertical level as the centre (rightmost point if the wheel rolls to the right). \bullet Translational velocity: $4 \text{ m/s}$ to the right. \bullet Rotational (tangential) velocity: $\omega R = v_{CM}=4 \text{ m/s}$ vertically downward. These two perpendicular components give the resultant speed $v_{\text{side}}=\sqrt{(4)^{2+}(4)^2}=4\sqrt{2} \text{ m/s}$. Therefore, the speed of the particle on the rim at the same level as the centre is $4\sqrt{2} \text{ m/s}$. Hence, Option A is correct.

The main scale of the travelling microscope has $300$ equal divisions spread over a length of $15 \text{ cm}$. Length of one main scale division (MSD) is therefore $\text{MSD} = \frac{15 \text{ cm}}{300} = 0.05 \text{ cm}$ The vernier scale has $25$ divisions that coincide with $24$ divisions of the main scale. Thus, length of the entire vernier scale is $25 \text{ VSD} = 24 \text{ MSD}$ Using $\text{MSD} = 0.05 \text{ cm}$, $25 \text{ VSD} = 24 \times 0.05 \text{ cm} = 1.2 \text{ cm}$ Length of one vernier scale division (VSD) is then $\text{VSD} = \frac{1.2 \text{ cm}}{25} = 0.048 \text{ cm}$ Formula for least count (LC) of a vernier instrument: $\text{LC} = \text{MSD} - \text{VSD}$ Substituting the values, $\text{LC} = 0.05 \text{ cm} - 0.048 \text{ cm} = 0.002 \text{ cm}$ Hence, the least count of the travelling microscope is $0.002 \text{ cm}$. Option B is correct.

The forces on the block are: • weight $mg$ downward \bullet normal reaction $N$ upward \bullet applied force $\mathbf{F}$ making $45^{\circ}$ with the horizontal \bullet kinetic friction $f_k$ opposite to the motion along the surface. Resolve the applied force: Horizontal component $F_x = F\cos 45^{\circ} = \dfrac{F}{\sqrt{2}}$ Vertical component $F_y = F\sin 45^{\circ} = \dfrac{F}{\sqrt{2}}$ Vertical equilibrium (no vertical motion): $N + F_y = mg$ so $N = mg - \frac{F}{\sqrt{2}} \quad -(1)$ Friction coefficient is $\mu = 0.25$, hence $f_k = \mu N = \mu\left( mg - \frac{F}{\sqrt{2}} \right) \quad -(2)$ The block is pulled steadily (horizontal acceleration zero). Therefore horizontal forces balance: $F_x = f_k$ $\frac{F}{\sqrt{2}} = \mu\left( mg - \frac{F}{\sqrt{2}} \right) \quad -(3)$ Solve equation $-(3)$ for $F$: $\frac{F}{\sqrt{2}} + \mu\frac{F}{\sqrt{2}} = \mu mg$ $F(1+\mu)\frac{1}{\sqrt{2}} = \mu mg$ $F = \frac{\mu mg\sqrt{2}}{1+\mu}$ Insert the data $m = 25\ \text{kg},\; g = 9.8\ \text{m s}^{-2},\; \mu = 0.25$: $F = \frac{0.25 \times 25 \times 9.8 \times 1.414}{1 + 0.25}$ $F \approx \frac{86.6}{1.25} \approx 69.3\ \text{N}$ Horizontal component of the applied force: $F_x = \frac{F}{\sqrt{2}} \approx \frac{69.3}{1.414} \approx 49.0\ \text{N}$ Work done by the external force over the displacement $s = 5\ \text{m}$: $W = F_x\,s = 49.0 \times 5 \approx 245\ \text{J}$ Therefore, the work done by the external force is $245\ \text{J}$. Correct option: Option C

Let the incident beam on $P_1$ be un-polarised with intensity $I_0$. Step 1 (Polariser $P_1$) An ideal polariser transmits one-half of the incident intensity. Hence, after $P_1$ the light is plane-polarised with $I_1 = \frac{I_0}{2}$ $-(1)$ Step 2 (Geometry of the three axes) $P_1$ and $P_2$ are crossed, so the angle between their transmission axes is $90^\circ$ (i.e. $\frac{\pi}{2}$). Insert $P_3$ between them such that the angle between $P_2$ and $P_3$ is $\theta$. Therefore, the angle between $P_1$ and $P_3$ is $\frac{\pi}{2} - \theta$. Step 3 (Intensity after $P_3$) Using Malus' law (transmitted intensity $I = I_{\text{\in}}\cos^2\alpha$, where $\alpha$ is the angle between the light's polarisation and the polariser's axis), we have $I_2 = I_1 \cos^2\!\left(\frac{\pi}{2}-\theta\right) = \frac{I_0}{2}\,\sin^2\theta$ $-(2)$ Step 4 (Intensity after $P_2$) Now the light emerging from $P_3$ makes an angle $\theta$ with the axis of $P_2$, so again applying Malus' law, $I_3 = I_2 \cos^2\theta = \frac{I_0}{2}\,\sin^2\theta\,\cos^2\theta$ Combine the trigonometric factors: $\sin^2\theta\,\cos^2\theta = \left(\sin\theta\cos\theta\right)^2 = \left(\tfrac{1}{2}\sin2\theta\right)^2 = \frac{1}{4}\sin^2 2\theta$ Hence $I_3 = \frac{I_0}{2}\times \frac{1}{4}\sin^2 2\theta = \frac{I_0}{8}\,\sin^2 2\theta$ $-(3)$ Step 5 (Maximising the transmitted intensity) The factor $\sin^2 2\theta$ attains its maximum value of $1$ when $2\theta = \frac{\pi}{2},\,\frac{3\pi}{2},\ldots$ Taking the first positive solution inside $0 \lt \theta \lt \frac{\pi}{2}$ gives $\theta = \frac{\pi}{4}$. Conclusion For maximum transmitted intensity through the three polarisers, the angle between $P_2$ and $P_3$ must be $\frac{\pi}{4}$. Therefore, Option A is correct.

In an extrinsic semiconductor, one type of charge carrier is made the majority by adding a suitable impurity. If the dopant supplies extra electrons, the material becomes n-type. Step 1 : Nature of majority and minority carriers In an n-type semiconductor, electrons supplied by donor atoms are the majority carriers, while holes created by thermal generation are very few in number. Thus statement (A) "Holes are minority carriers" is TRUE . Step 2 : Kind of dopant required A silicon or germanium atom has four valence electrons. To donate an extra electron we must insert an element from group V (P, As, Sb, etc.) which has five valence electrons. Such an atom is called a pentavalent donor . Therefore statement (B) "The dopant is a pentavalent atom" is TRUE . Step 3 : Applying the mass-action law At thermal equilibrium, semiconductors (intrinsic or extrinsic) obey the mass-action law $n_e n_h = n_i^2 \quad -(1)$ where $n_i$ is the intrinsic carrier concentration. Hence the product $n_e n_h$ always equals, not differs from, $n_i^2$. \bullet Statement (C) "$n_e n_h \neq n_i^2$" contradicts equation $(1)$, so it is FALSE . \bullet Statement (D) "$n_e n_h \geq n_i^2$" is also inconsistent with $(1)$ (equality must hold), so it is FALSE . Step 4 : Origin of the minority holes In an n-type material, donor atoms produce extra electrons only. Holes arise solely from the thermal breaking of covalent bonds (electron-hole pair generation), exactly as in an intrinsic semiconductor. They are not created by the donor atoms. Therefore statement (E) "The holes are not generated due to the donors" is TRUE . Step 5 : Collecting the TRUE statements TRUE : (A), (B), (E) FALSE : (C), (D) Comparing with the given options, Option C lists exactly (A), (B), (E). Hence the correct answer is Option C .

The first law of thermodynamics for a quasi-static process is written as $\Delta Q = \Delta U + \Delta W$, where $\Delta W$ is the work done by the gas. For expansion/compression work at constant external pressure, $\Delta W = P\,\Delta V$, so the first law becomes $\Delta Q = \Delta U + P\,\Delta V$ $-(1)$. Now analyse each process \in List-I. Case A: Isobaric (constant pressure) Pressure stays constant, therefore $P = \text{constant}$. Substituting \in $(1)$ gives $\Delta Q = \Delta U + P\,\Delta V$. This is exactly the expression \in List-II (IV). Hence A \to IV. Case B: Isochoric (constant volume) Here $\Delta V = 0$, so no work is done: $\Delta W = P\,\Delta V = 0$. Using the first law: $\Delta Q = \Delta U + 0 \; \Rightarrow \; \Delta Q = \Delta U$. This matches List-II (II). Hence B \to II. Case C: Adiabatic (no heat exchange) By definition, $\Delta Q = 0$ for an adiabatic process. Thus the relation \in List-II (III) fits. Hence C \to III. Case D: Isothermal (constant temperature) For an ideal gas, internal energy depends only on temperature. Since temperature is constant, $\Delta U = 0$. First law then gives $\Delta Q = \Delta W$. This is relation (I) in List-II. Hence D → I. Collecting the matches: (A-IV), (B-II), (C-III), (D-I). The option with this sequence is Option C. Final Answer: Option C

The given equation is $x(t)=5\cos\left(628\,t+\frac{\pi}{2}\right)$. This is of standard form $x(t)=A\cos\left(\omega t+\phi\right)$, so the angular frequency is $\omega = 628\ \text{rad s}^{-1}$. Frequency $f$ is related to angular frequency by $\omega = 2\pi f$ $-(1)$. Substituting $\omega = 628$ rad s$^{-1}$ \in $(1)$, $f = \frac{\omega}{2\pi} = \frac{628}{2\pi}\ \text{Hz}$. Using $\pi \approx 3.14$, $2\pi \approx 6.28$, hence $f \approx \frac{628}{6.28} = 100\ \text{Hz}$. The relation between wave speed $v$, frequency $f$, and wavelength $\lambda$ is $v = f\lambda$ $-(2)$. Given velocity $v = 300\ \text{m s}^{-1}$, substituting $f = 100\ \text{Hz}$ \in $(2)$: $\lambda = \frac{v}{f} = \frac{300}{100}\ \text{m} = 3\ \text{m}$. Therefore, the wavelength of the wave is $3\ \text{m}$. Option B.

The object is placed on the principal axis $30 \text{ cm}$ \in front of a convex mirror of focal length $f = +30 \text{ cm}$ (positive for a convex mirror \in the new Cartesian sign convention). Using the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$, with object distance $u = -30 \text{ cm}$, we get $\frac{1}{30} = \frac{1}{v} + \frac{1}{(-30)}$ $\frac{1}{v} = \frac{1}{30} + \frac{1}{30} = \frac{2}{30} = \frac{1}{15}$ $\Rightarrow \; v = +15 \text{ cm}$ The positive sign means the convex mirror forms a virtual image $I_2$ $15 \text{ cm}$ behind the mirror (on the right side of the pole). Now place a plane mirror between the object and the convex mirror. Let the distance between the two mirrors be $d \text{ cm}$. With the pole of the convex mirror as origin, the coordinates along the principal axis are: \bullet Pole of convex mirror: $x = 0$ \bullet Plane mirror surface: $x = -d$ \bullet Object: $x = -30 \text{ cm}$ The object is $30 - d \text{ cm}$ \in front of the plane mirror, so the plane mirror produces a virtual image $I_1$ the same distance behind it: Coordinate of $I_1$: $x(I_1) = (-d) + (30 - d) = 30 - 2d$ For the two images to coincide, $I_1$ and $I_2$ must lie at the same coordinate: $30 - 2d = +15$ $2d = 30 - 15 = 15$ $\therefore \; d = 7.5 \text{ cm}$ Hence, the distance between the convex mirror and the plane mirror must be $7.5 \text{ cm}$. Option B is correct.

For any physical quantity we can express its dimension in the form $[M^{P} L^{Q} T^{R} A^{S}]$, where $M$ stands for mass, $L$ for length, $T$ for time and $A$ for electric current. Electric dipole moment Definition : $p_{e}=q\,d$, where $q$ is charge and $d$ is the separation between the charges. Dimension of charge : $[q]=[A\,T]$ (current \times time). Therefore $[p_{e}] = [A\,T]\,[L] = [M^{0} L^{1} T^{1} A^{1}]$ Magnetic dipole moment For a current loop, $m_{m}=I\,A_{\text{loop}}$, where $I$ is current and $A_{\text{loop}}$ is the area of the loop. Area has dimension $[L^{2}]$, hence $[m_{m}] = [A]\,\$[L^{2}]\$ = [M^{0} L^{2} T^{0} A^{1}]$ Required ratio $\frac{p_{e}}{m_{m}}$ has dimension $\frac{[M^{0} L^{1} T^{1} A^{1}]\$}{[M^{0} L^{2} T^{0} A^{1}]\$}$ = $[M^{0} L^{1-2} T^{1-0} A^{1-1}]$ = $[M^{0} L^{-1} T^{1} A^{0}]$. Comparing with $[M^{P} L^{Q} T^{R} A^{S}]$: $P=0,\; Q=-1,\; R=1,\; S=0$. Hence the values of $P$ and $Q$ are 0 and −1 respectively, which corresponds to Option D.

When a charged particle is fired perpendicular to a uniform magnetic field, it traces a circular path. The time required for the particle to return to its original location for the first time is exactly the time period of this circular motion. The magnetic force provides the centripetal force: $qvB = \frac{mv^{2}}{r}$ This gives the radius $r = \frac{mv}{qB}$. The time period is: $T = \frac{2\pi r}{v} = \frac{2\pi m}{qB}$ Notice that the time period does not depend on the speed of the particle. Now, we are given: Charge: $q = 1.6 \, \mu C = 1.6 \times 10^{-6}$ C Mass: $m = 16 \, \mu g = 16 \times 10^{-6}$ kg Magnetic field: $B = 6.28$ T $\pi = 3.14$ Substituting into the time period formula: $T = \frac{2 \times 3.14 \times 16 \times 10^{-6}}{1.6 \times 10^{-6} \times 6.28}$ Let us compute the numerator: $2 \times 3.14 = 6.28$ $6.28 \times 16 \times 10^{-6} = 100.48 \times 10^{-6}$ Now the denominator: $1.6 \times 6.28 = 10.048$ $10.048 \times 10^{-6}$ So we get: $T = \frac{100.48 \times 10^{-6}}{10.048 \times 10^{-6}} = \frac{100.48}{10.048} = 10 \text{ s}$ So, the answer is $10$.

Let the initial mass of the solid sphere be $M$ and its initial radius be $R$. Moment of inertia of a solid sphere about any diameter is $I=\frac{2}{5}MR^{2}$ $-(1)$ When the sphere loses mass it keeps its shape, so it remains a uniform solid sphere of a smaller radius. If the radius becomes $\frac{R}{2}$, the new volume is $V'=\frac{4}{3}\pi\left(\frac{R}{2}\right)^{3}=\frac{1}{8}\left(\frac{4}{3}\pi R^{3}\right)$ With uniform density, mass is proportional to volume, hence $M'=\frac{1}{8}M$ $-(2)$ The new moment of inertia about the same diameter is $I'=\frac{2}{5}M'\left(\frac{R}{2}\right)^{2} =\frac{2}{5}\left(\frac{M}{8}\right)\left(\frac{R^{2}}{4}\right) =\frac{2}{5}MR^{2}\left(\frac{1}{32}\right) =\frac{I}{32}$ $-(3)$ The mass is lost symmetrically, so no external torque acts on the sphere. Therefore the angular momentum about the diameter remains constant: $I\omega_{1}=I'\omega_{2}$ $-(4)$ Substituting $I'=\frac{I}{32}$ from $(3)$ into $(4)$, we get $I\omega_{1}=\frac{I}{32}\,\omega_{2} \;\;\Longrightarrow\;\; \omega_{2}=32\,\omega_{1}$ Comparing with $\omega_{2}=x\omega_{1}$, we find $x=32$ Hence, the angular velocity of the sphere when its radius becomes $R/2$ is $32\omega_{1}$ and the required value of $x$ is 32 .

The speed of an electromagnetic wave in any medium is $v = \frac{1}{\sqrt{\mu \, \varepsilon}}$ where $\mu = \mu_r \mu_0$ and $\varepsilon = \varepsilon_r \varepsilon_0$. Putting these into the expression, the speed \in the medium becomes $v = \frac{1}{\sqrt{\mu_r \mu_0 \, \varepsilon_r \varepsilon_0}} = \frac{1}{\sqrt{\mu_r \varepsilon_r}\;\sqrt{\mu_0 \varepsilon_0}} = \frac{c}{\sqrt{\mu_r \varepsilon_r}}$ Here $c = \dfrac{1}{\sqrt{\mu_0 \varepsilon_0}}$ is the speed of light \in vacuum. Therefore the factor by which light travels faster \in vacuum than \in the given medium is $\frac{c}{v} = \sqrt{\mu_r \varepsilon_r}$ The medium has $\mu_r = \frac{10}{\pi}, \qquad \varepsilon_r = \frac{1}{0.0885}$ First find the product $\mu_r \varepsilon_r$: $\mu_r \varepsilon_r = \frac{10}{\pi}\;\times \;\frac{1}{0.0885} = \frac{10}{\pi \times 0.0885}$ Using $\pi \approx 3.1416$: $\pi \times 0.0885 \approx 3.1416 \times 0.0885 = 0.2780$ Hence $\mu_r \varepsilon_r \approx \frac{10}{0.2780} \approx 35.96$ Now take the square root: $\sqrt{35.96} \approx 5.996 \approx 6$ Thus $\frac{c}{v} \approx 6$ Therefore, the velocity of light in vacuum is greater than that in this medium by 6 times .

For a single slit of width $a$, the first diffraction minima occur at angles $\theta$ that satisfy $a \sin \theta = \pm \lambda \qquad -(1)$ Hence the central diffraction maximum extends from $-\theta_1$ to $+\theta_1$ where $\sin \theta_1 = \frac{\lambda}{a} \qquad -(2)$ For the Young's double-slit interference, bright fringes (maxima) are obtained at $d \sin \theta = m \lambda ,\; m = 0, \pm 1, \pm 2, \ldots \qquad -(3)$ The extreme interference maximum that can still lie inside the central diffraction maximum is obtained by equating the conditions $(2)$ and $(3)$: $d \sin \theta_1 = m_{\text{max}}\lambda$ $\Rightarrow d \left(\frac{\lambda}{a}\right) = m_{\text{max}}\lambda$ $\Rightarrow m_{\text{max}} = \frac{d}{a} \qquad -(4)$ The problem states that a total of 20 interference maxima are contained inside the central diffraction maximum. These 20 maxima are the bright fringes on both sides of the central bright fringe but excluding the central one itself. Therefore Number of maxima on one side $= \frac{20}{2}=10$ $\Rightarrow m_{\text{max}} = 10 \qquad -(5)$ Substituting $(5)$ into $(4)$ gives the required slit width: $a = \frac{d}{m_{\text{max}}} = \frac{1.5 \text{ mm}}{10} = 0.15 \text{ mm}$ Converting to centimetres, $0.15 \text{ mm} = 0.015 \text{ cm} = 15 \times 10^{-3} \text{ cm}$ Thus, the width of each slit is $15 \times 10^{-3}$ cm, so the required value of $x$ is $15$.