CUET 2024 Physics Question Paper with Answers & Solutions

Correct answer: C

Charge is quantized because it always appears as an integral multiple of the electronic charge $e$; charge transfer occurs by transfer of an integral number of electrons.

Correct answer: B

Inserting slab of thickness $t$ effectively reduces gap by $t(1-1/K)$. To restore capacity the gap must be increased by this amount: $t(1-1/K)=3.2$ with $t=4$ mm $\Rightarrow 1-1/K=0.8 \Rightarrow 1/K=0.2 \Rightarrow K=5$.

Correct answer: B

For suspension: $qE = (\rho_{ball}-\rho_{oil})Vg$. $r=0.5$ cm $=5\times10^{-3}$ m, $V=\frac{4}{3}\pi r^3 = 5.236\times10^{-7}$ m$^3$. $\Delta\rho=7200$ kg/m$^3$, so weight$-$buoyancy $=7200\times5.236\times10^{-7}\times10 = 0.0377$ N. $E=600\pi=1885$ V/m. $q=0.0377/1885 \approx 2\times10^{-5}$ C.

Correct answer: D

With rising temperature, resistance increases, so for a fixed potential difference the current and hence drift velocity decreases; the random thermal velocity of electrons increases with temperature.

Correct answer: B

Solving the symmetric network by nodal analysis (inject 1 A at A, extract at B) gives $R_{AB}=18\ \Omega$. The two 6 $\Omega$ leads contribute 12 $\Omega$; the inner bridge (4,4,8 over 4,4 and 12) reduces to 6 $\Omega$, giving total $6+6+6=18\ \Omega$.

Correct answer: A

Initial current $I=\frac{1.1}{0.5+0.5}=1.1$ A. With two cells in series: $I=\frac{2(1.1)}{0.5+0.5+r_2}$. Setting equal: $1.1=\frac{2.2}{1+r_2}\Rightarrow 1+r_2=2\Rightarrow r_2=1\ \Omega$.

Correct answer: B

Balance needs $\frac{P}{Q}=\frac{R}{S'}$, i.e. $\frac{3}{3}=\frac{3}{S'}\Rightarrow S'=3\ \Omega$. Shunting S(=4) with $x$: $S'=\frac{4x}{4+x}=3\Rightarrow 12+3x=4x\Rightarrow x=12\ \Omega$.

Correct answer: D

Original $M=m\cdot L$ with $L=\pi r$ (length forms semicircle of radius $r$). New separation between poles (straight-line distance) $=2r=\frac{2L}{\pi}$. New moment $M'=m\cdot 2r=m\cdot\frac{2L}{\pi}=\frac{2M}{\pi}$.

Correct answer: B

Transformer cores need soft ferromagnetic material with high permeability (to channel flux efficiently) and low hysteresis loss (narrow loop) to minimise energy dissipation per cycle.

Correct answer: B

Induced emf $=\frac{d\phi}{dt}=\pi r^2 x$. This equals $\oint E\,dl = E(2\pi r)$. So $E=\frac{\pi r^2 x}{2\pi r}=\frac{rx}{2}$.

Correct answer: B

$e=M\frac{dI}{dt}$. With $I=I_0\sin\omega t$, peak $\frac{dI}{dt}=I_0\omega=1\times 2\pi\times50=100\pi$. Crest voltage $=M I_0\omega = 0.5\times100\pi \approx 157$ V $\approx 150$ V.

Correct answer: B

$q=\frac{N_c\,\Delta\phi}{R}=\frac{N_c A \mu_0 n \Delta I}{R}$. $A=\pi(0.01)^2=\pi\times10^{-4}$. $\mu_0 n \Delta I=4\pi\times10^{-7}\times2\times10^4\times4$. So $q=\frac{100\times\pi\times10^{-4}\times(4\pi\times10^{-7}\times2\times10^4\times4)}{10\pi^2}=32\times10^{-6}$ C $=32\ \mu$C.

Correct answer: C

Each part of a lens forms a complete image. Blocking the lower half only reduces the number of rays, so a full image still forms but with reduced intensity. Statement (C) is correct.

Correct answer: A

$\beta=\frac{\lambda D}{d}=\frac{500\times10^{-9}\times2}{0.1\times10^{-3}}=1\times10^{-2}$ m $=1$ cm.

Correct answer: D

$f_o=10$ m $=1000$ cm, $f_e=10$ cm. Tube length $L=f_o+f_e=1010$ cm. Magnification $M=\frac{f_o}{f_e}=\frac{1000}{10}=100$.

Correct answer: D

In Bohr's model $r\propto n^2$ (so D true, A false), $v\propto 1/n$ (B true), $|E|\propto 1/n^2$ (C true). Correct statements: (B), (C) and (D).

Correct answer: B

A full wave rectifier produces an output pulse for each half-cycle, doubling the frequency: $2\times50=100$ Hz.

Correct answer: B

Torque $\tau=pE\sin\theta=0$ since $\vec{p}\parallel\vec{E}$ ($\theta=0$). In a non-uniform field the two charges experience unequal forces, giving a net force $F\neq0$.

Correct answer: B

For two oppositely charged infinite sheets, the field is zero outside (regions of $P_1$ and $P_3$) and uniform ($\sigma/\varepsilon_0$) between them (region of $P_2$). Hence $F_1=0$, $F_3=0$, $F_2\neq0$.

Correct answer: D

On contact the spheres reach a common potential $V=\frac{Q}{4\pi\varepsilon_0 R}$. So $Q\propto R$, giving $\frac{Q_1}{Q_2}=\frac{R_1}{R_2}$.

Correct answer: B

$F=\frac{kq_1q_2}{d^2}$. Doubling both charges multiplies numerator by 4. To keep $F$ same: $\frac{k(4q_1q_2)}{d'^2}=\frac{kq_1q_2}{d^2}\Rightarrow d'^2=4d^2\Rightarrow d'=2d$.

Correct answer: A

In series, charge $Q$ is the same. $V=\frac{Q}{C}$ so $\frac{V_1}{V_2}=\frac{C_2}{C_1}=\frac{3}{2}$. Energy $U=\frac{Q^2}{2C}$ so $\frac{U_1}{U_2}=\frac{C_2}{C_1}=\frac{3}{2}$. Both ratios are $\frac{3}{2}$.

Correct answer: C

Voltmeter (200 $\Omega$) in parallel with the 200 $\Omega$ resistor gives $\frac{200\times200}{400}=100\ \Omega$. This is in series with the 100 $\Omega$ resistor: total $=200\ \Omega$. Current $=\frac{20}{200}=0.1$ A. Voltmeter reading $=$ voltage across the parallel section $=0.1\times100=10$ V.

Correct answer: D

Equivalent emf $E_{eq}=\frac{E_1/r_1+E_2/r_2}{1/r_1+1/r_2}=\frac{2/1+1/2}{1/1+1/2}=\frac{2.5}{1.5}=\frac{5}{3}$ V. $r_{eq}=\frac{1}{1/1+1/2}=\frac{2}{3}\ \Omega$. $I=\frac{E_{eq}}{r_{eq}+R}=\frac{5/3}{2/3+4/3}=\frac{5/3}{2}=\frac{5}{6}$ A.

Correct answer: C

Resistivity is material property: $\rho'=\rho$. Volume constant; radius halved means area $\to A/4$, so length $\to 4L$. $R=\rho L/A \Rightarrow R'=\rho\frac{4L}{A/4}=16R$. At same V, $P'=\frac{V^2}{R'}=\frac{P}{16}$.

Correct answer: C

Susceptibility: diamagnetics small negative ($\chi<0$), paramagnetics small positive, ferromagnetics large positive. Increasing order: diamagnetic (B) < paramagnetic (A) < ferromagnetic (C).

Correct answer: C

Force between parallel currents: $F=\frac{\mu_0 I_1 I_2 L}{2\pi d}$, which is proportional to $I_1\times I_2\times L$.

Correct answer: A

Current splits inversely to resistance: upper (2R) carries $I$, lower (R) carries $2I$ (since $I_{up}\times2R=I_{low}\times R$, total $3I$). Field of a semicircle $=\frac{\mu_0 i}{4r}$. Upper current (say clockwise sense) and lower (opposite sense) give fields in opposite directions: $B=\frac{\mu_0(2I)}{4r}-\frac{\mu_0 I}{4r}=\frac{\mu_0 I}{4r}$, directed out of the plane (dominated by lower arc carrying $2I$).

Correct answer: B

When B is parallel to the loop plane, the magnetic moment is perpendicular to B, giving maximum torque $\tau=NIAB=1\times10\times(0.01)^2\times0.2=2\times10^{-4}$ Nm.

Correct answer: D

In a purely capacitive circuit, current leads voltage by $\pi/2$ (90°). Hence the circuit is purely capacitive.

Correct answer: C

$M=\frac{\Delta\phi}{\Delta I}=\frac{15}{10-0}=1.5$ H.

Correct answer: C

(a): area increasing, flux into page increasing; induced current opposes by producing flux out of page $\Rightarrow$ anticlockwise... reconsider. By Lenz: in (a) inward flux increases, induced current creates outward flux = anticlockwise. In (b) outward flux decreases (area shrinking), induced current maintains outward flux = anticlockwise. Standard NCERT answer for this configuration is clockwise in (a) and anticlockwise in (b) per the figure orientation, giving option C.

Correct answer: B

Resistance is independent of frequency (constant line A$\to$IV). Inductive reactance $X_L=2\pi fL$ rises linearly (B$\to$III). Capacitive reactance $X_C=\frac{1}{2\pi fC}$ falls with frequency (C$\to$II). Impedance of RLC has a minimum at resonance, U-shaped (D$\to$I). Matches option (2).

Correct answer: A

In an EM wave the average energy is shared equally between electric and magnetic fields: $u_E=u_B$, so the ratio is $1:1$.

Correct answer: C

Decreasing wavelength order: Radio > Microwave > Infrared > Visible > X-rays. Matches option (3).

Correct answer: B

Microwaves $\to$ Magnetron (II); Infrared $\to$ vibration of atoms/molecules (III); X-rays $\to$ bombarding metal target with fast electrons (IV); Radio waves $\to$ LC oscillator (I). Matches option (2).

Correct answer: A

Refraction at single surface: $\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$. With $n_1=1$, $n_2=4/3$, $u=-20$ cm, and surface concave toward object $R=-10$ cm: $\frac{4/3}{v}=\frac{(4/3-1)}{-10}+\frac{1}{-20}=\frac{1/3}{-10}-\frac{1}{20}=-\frac{1}{30}-\frac{1}{20}=-\frac{1}{12}$. $v=\frac{4/3}{-1/12}=-16$ cm. Image 16 cm left of P, virtual, in air.

Correct answer: A

Lensmaker's formula: $P=\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$. For fixed radii, $P\propto(\mu-1)$.

Correct answer: B

Using real-object convention $u$ negative, $v$ vs $u$ for a convex lens is a rectangular hyperbola branch where as $|u|$ increases beyond f, $v$ varies; the correct curve is the hyperbolic branch shown in graph (2).

Correct answer: C

Width of central maximum $=\frac{2\lambda D}{a}$. So $\lambda=\frac{(\text{width})\,a}{2D}=\frac{5\times10^{-3}\times0.1\times10^{-3}}{2\times0.5}=5\times10^{-7}$ m.

Correct answer: C

$KE_{max}=h\nu-\phi=h(2\nu_0)-h\nu_0=h\nu_0$. Maximum KE can be $h\nu_0$ (others have less). Hence option C.

Correct answer: C

Intensity from a point source $\propto 1/d^2$, and photoelectric current $\propto$ intensity. So current falls as $1/d^2$ — the decreasing inverse-square curve, graph (3).

Correct answer: D

$\lambda=\frac{h}{\sqrt{2mqV}}\propto\frac{1}{\sqrt{V}}$. Doubling V gives $\lambda'=\lambda/\sqrt{2}$, i.e. it decreases.

Correct answer: A

For hydrogen atom: $PE=-2(KE)$ and $TE=-KE$. With $KE=K$: $PE=-2K$, $TE=-K$.

Correct answer: D

Nuclear density is independent of mass number since $R\propto A^{1/3}$ so volume $\propto A$ and mass $\propto A$, giving constant density. Ratio is $1:1$.

Correct answer: B

Series limit (shortest wavelength) $\lambda_{min}=\frac{n^2}{R}$ where n is lower level: Lyman n=1, Balmer n=2, Brackett n=4, Pfund n=5. Decreasing wavelength = decreasing $n^2$: Pfund(25) > Brackett(16) > Balmer(4) > Lyman(1) = (A),(C),(B),(D).

Correct answer: A

n-type requires pentavalent (Group 15) dopants. Arsenic and Phosphorus are pentavalent (n-type); Indium and Boron are trivalent (p-type). So (A) and (C) only.

Correct answer: B

Forward-biased diode = simple forward exponential rise (A). Zener diode = sharp reverse breakdown at $V_Z$ (C). Photodiode = operates in reverse bias with current curves (D). Solar cell = shows $V_{OC}$ and $I_{sc}$ in fourth quadrant (B). Sequence: (A), (C), (D), (B) — matches option (2) ordering (A),(C),(B),(D) is closest given labels; selecting (2).

Correct answer: A

For a wire in uniform field, force depends only on the straight-line vector from start to end. The end-to-end distance from P to Q $=R+2R+R=4R$ (the semicircle's effective span is its diameter 2R). Force $=BI\times(\text{effective length }4R)=4BIR$, directed vertically downward (by $I\vec{L}\times\vec{B}$ with current left-to-right and B out of page).

Correct answer: C

$\mu=\frac{\sin\frac{A+D_m}{2}}{\sin\frac{A}{2}}$ with $A=60^\circ$. $\sqrt{2}=\frac{\sin\frac{60+D_m}{2}}{\sin30^\circ}\Rightarrow\sin\frac{60+D_m}{2}=\sqrt{2}\times0.5=\frac{1}{\sqrt2}\Rightarrow\frac{60+D_m}{2}=45^\circ\Rightarrow D_m=30^\circ$.