NEET 2024 Physics Questions with Solutions

Explanation

In pure rolling motion, the velocity of the topmost point of the wheel is 2v, while the velocity of the bottommost point is zero. Therefore, point P moves faster than point Q. Hence, option (D) is correct.

Explanation

By Lenz's law, the induced current opposes the relative motion between the magnet and the solenoids. In solenoid-1, the magnet is moving away, so the near face becomes south to attract the magnet, giving current direction BA. In solenoid-2, the magnet approaches, so the near face becomes south to oppose the approach, giving current direction DC. Hence, option (C) is correct.

Explanation

Statement A is incorrect because a solar cell operates in the fourth quadrant (V > 0, I < 0), but the statement as given about I-V characteristics is not fully accurate. Statement B is correct because reverse saturation current in a reverse biased pn junction diode is due to minority carriers, not majority carriers. Therefore, only statement B is correct. Hence, option (D) is correct. NEET 2024 T1 answer key shows Q3 → option (D).

Explanation

The wire is divided into 10 equal parts, so resistance of each part is: $$R = \frac{100}{10} = 10\,\Omega$$ First 5 resistors in series: $$R_s = 5 \times 10 = 50\,\Omega$$ Next 5 resistors in parallel: $$\frac{1}{R_p} = 5 \times \frac{1}{10}$$ $$R_p = 2\,\Omega$$ These two combinations are connected in series: $$R_{eq} = 50 + 2 = 52\,\Omega$$ Hence, option (D) is correct.

Explanation

The circuit is a balanced bridge because all capacitances are equal. Hence, no potential difference exists across the middle capacitor, so it can be ignored. Top branch: $$C_t = \frac{2 \times 2}{2+2} = 1\,\mu F$$ Bottom branch: $$C_b = 1\,\mu F$$ These two branches are in parallel: $$C_{eq} = 1 + 1 = 2\,\mu F$$ Hence, option (C) is correct.

Explanation

The first gate acts as a NOT operation on input $A$, giving $\overline{A}$. The second gate is a NOR gate with single input $B$, equivalent to $\overline{B}$. The final gate is a NOR gate: $$Y = \overline{(\overline{A}+\overline{B})}$$ Using De Morgan's theorem: $$Y = A\cdot B$$ Hence, the overall output is equivalent to an **AND gate**. Therefore, the correct answer is: **(2) AND gate**.

Explanation

Given: $$E = 10\,V, \quad r = 1\,\Omega, \quad R = 4\,\Omega$$ Current in the circuit: $$I = \frac{E}{R+r} = \frac{10}{4+1} = 2\,A$$ Terminal voltage: $$V = E - Ir$$ $$V = 10 - (2)(1) = 8\,V$$ Hence, option (A) is correct.

Explanation

Acceleration due to gravity is: $$g = \frac{GM}{R^2}$$ Given: $$M_p = \frac{M_e}{10}, \quad R_p = \frac{R_e}{2}$$ Therefore, $$g_p = \frac{G(M_e/10)}{(R_e/2)^2}$$ $$g_p = \frac{GM_e}{10} \times \frac{4}{R_e^2}$$ $$g_p = \frac{4}{10}g_e = 0.4 \times 9.8$$ $$g_p = 3.92\,m\,s^{-2}$$ Hence, option (B) is correct.

Explanation

Time period: $$T=\frac{5}{20}=0.25\,s$$ For oscillations of a magnetic needle: Rearranging: $$M=\frac{4\pi^2 I}{B T^2}$$ Substituting: $$M=\frac{4\pi^2(9.8\times10^{-6})}{(0.049)(0.25)^2}$$ $$M=1280\pi^2\times10^{-5}\,\text{A m}^2$$ Hence: $$x=1280\pi^2$$ Therefore, the correct answer is: **(2) $1280\pi^2$**

Explanation

Statement A is correct because photon energy is: $$E = h\nu$$ Statement B is correct because photons travel with speed c in free space. Statement C is correct because: $$p = \frac{E}{c} = \frac{h\nu}{c}$$ Statement D is correct because both energy and momentum are conserved in photon-electron collisions. Statement E is incorrect because photons are electrically neutral. Hence, option (D) is correct.

Explanation

Young's modulus is: $$Y = \frac{\text{stress}}{\text{strain}}$$ Therefore, $$\text{strain} = \frac{8 \times 10^8}{2 \times 10^{11}} = 4 \times 10^{-3}$$ Since, $$\text{strain} = \frac{\Delta L}{L}$$ $$\Delta L = 4 \times 10^{-3} \times 1 = 4 \times 10^{-3}\,m$$ $$\Delta L = 4\,mm$$ Hence, option (C) is correct.

Explanation

From the truth table: When B = 0, output Y = 1. When B = 1, output Y = 0. Thus output depends only on B and is its complement: $$Y = \overline{B}$$ Hence, option (A) is correct.

Explanation

For a thin rod about an axis through its centre and perpendicular to its length: Given: $$I=2400\,\text{g cm}^2,\quad M=400\,\text{g}$$ So, $$2400=\frac{1}{12}(400)L^2$$ $$L^2=72$$ $$L=\sqrt{72}\approx 8.5\,\text{cm}$$ Hence the correct answer is: **(3) $8.5\,\text{cm}$**

Explanation

Given: $$(N+1)\,\text{VSD}=N\,\text{MSD}$$ Therefore, $$1\,\text{VSD}=\frac{N}{N+1}\text{MSD}$$ Vernier constant: $$\text{VC}=1\text{MSD}-1\text{VSD}$$ $$\text{VC}=\left(1-\frac{N}{N+1}\right)\text{MSD}$$ $$\text{VC}=\frac{1}{N+1}\times0.1\,\text{mm}$$ Converting $0.1\,\text{mm}$to cm:$$0.1\,\text{mm}=0.01\,\text{cm}=\frac{1}{100}\,\text{cm}$$Thus,$$\text{VC}=\frac{1}{100(N+1)}\,\text{cm}$$Hence, the correct answer is: **(4)$\dfrac{1}{100(N+1)}$**

Explanation

These are Balmer series transitions of hydrogen: $$n_2 = 3 \to n_1 = 2 \Rightarrow 656.3\,nm$$ $$n_2 = 4 \to n_1 = 2 \Rightarrow 486.1\,nm$$ $$n_2 = 5 \to n_1 = 2 \Rightarrow 434.1\,nm$$ $$n_2 = 6 \to n_1 = 2 \Rightarrow 410.2\,nm$$ Thus: A → III B → IV C → II D → I Hence, option (D) is correct.

Explanation

In Young's double slit experiment using white light, all wavelengths constructively interfere at the center, producing a bright white central fringe. Away from the center, different wavelengths produce coloured fringes. Hence, option (A) is correct.

Explanation

For a free particle: $$\lambda = \frac{h}{p}$$ Also, $$E = \frac{p^2}{2m}$$ Therefore, $$p^2 = 2mE$$ Using de Broglie relation: $$\frac{1}{\lambda^2} = \frac{p^2}{h^2} = \frac{2mE}{h^2}$$ Thus, $$\frac{1}{\lambda^2} \propto E$$, which is a straight line passing through the origin. Hence, option (B) is correct.

Explanation

Diamagnetic substances have small negative susceptibility: $$0 > \chi > -1$$ Ferromagnetic substances have very large positive susceptibility: $$\chi \gg 1$$ Paramagnetic substances have small positive susceptibility: $$0 < \chi < \varepsilon$$ Non-magnetic substances have: $$\chi = 0$$ Thus: A-II, B-III, C-IV, D-I Hence, option (C) is correct.

Explanation

At Brewster’s angle, the reflected light is completely plane polarised. The refracted light is only partially polarised. Hence, option (B) is correct.

Explanation

In uniform circular motion, speed remains constant but the direction of velocity changes continuously. Therefore, velocity varies. The centripetal acceleration also changes direction continuously, so acceleration is varying. Hence, option (B) is correct.

Explanation

Solid angle is dimensionless. Strain is the ratio of two lengths, so it is dimensionless. Plane angle is also dimensionless. Thus, strain and angle have the same dimensions as solid angle. Hence, option (C) is correct.

Explanation

Centripetal force is provided by tension: $$T = \frac{mv^2}{r}$$ If speed becomes double: $$T' = \frac{m(2v)^2}{r} = \frac{4mv^2}{r} = 4T$$ Hence, option (D) is correct.

Explanation

Comparing with the standard SHM equation: $$x = A \sin(\omega t + \phi)$$ Amplitude: $$A = 5\,m$$ Angular frequency: $$\omega = \pi\,rad\,s^{-1}$$ Time period: $$T = \frac{2\pi}{\omega} = \frac{2\pi}{\pi} = 2\,s$$ Hence, option (D) is correct.

Explanation

Along path bc, volume remains constant (isochoric process). Work done is: $$W = \int P\,dV$$ Since: $$dV = 0$$ Therefore: $$W = 0$$ Hence, option (C) is correct.

Explanation

Statement I is correct because atoms are electrically neutral due to equal numbers of protons and electrons. Statement II is also considered correct in the context of atomic physics because atoms emit characteristic spectra corresponding to transitions between quantized energy levels. Hence, both Statement I and Statement II are correct. The official NEET (UG) 2024 T1 answer key lists Q25 with multiple correct answers: options (A) and (C).

Explanation

For an ideal transformer: $$\frac{V_s}{V_p} = \frac{N_s}{N_p}$$ Given: $$\frac{N_p}{N_s} = \frac{1}{2}$$ Therefore: $$\frac{N_s}{N_p} = 2$$ Hence: $$V_s : V_p = 2 : 1$$ Therefore, option (D) is correct.

Explanation

In a completely inelastic collision, the bodies move together after collision. Using conservation of momentum: $$mv_1 + m(0) = (m+m)v_2$$ $$mv_1 = 2mv_2$$ $$v_1 = 2v_2$$ Therefore: $$v_1 : v_2 = 2 : 1$$ Hence, option (D) is correct.

Explanation

Initial nucleus: $$^{290}_{82}Y$$ After α-emission: Mass number decreases by 4 and atomic number by 2: $$A = 286, \quad Z = 80$$ After positron emission (e⁺): Atomic number decreases by 1: $$Z = 79$$ After β⁻ emission: Atomic number increases by 1: $$Z = 80$$ After electron emission (e⁻): This corresponds to β⁻ decay, so atomic number increases by 1: $$Z = 81$$ Mass number remains unchanged throughout beta processes. Therefore, product Q has: $$A = 286, \quad Z = 81$$ Hence, option (B) is correct.

Explanation

For an electric dipole, potential at an axial point is: $$V = \pm \frac{1}{4\pi\varepsilon_0}\frac{P}{r^2}$$ The given reason incorrectly uses: $$V = \pm \frac{2P}{4\pi\varepsilon_0 r^2}$$ Hence Reason R is false. Calculating Assertion A: $$V = \frac{9\times10^9\times4\times10^{-6}}{(2)^2}$$ $$V = 9\times10^3\;V$$ Thus Assertion A is true. Therefore, the correct option is: **(1) A is true but R is false.**

Explanation

For a charged conducting spherical shell, the electric potential is constant throughout the interior and on the surface. Since both points C and P lie inside/on the shell, they are at the same potential. Therefore: $$V_C - V_P = 0$$ Hence, option (B) is correct.

Explanation

The excess force required to detach the disc from the water surface is: $$F = 2 \times (2\pi r)T$$ where: $$r = 4.5\,\text{cm} = 0.045\,\text{m}$$ $$T = 0.07\,\text{N m}^{-1}$$ Substituting: $$F = 4\pi (0.045)(0.07)$$ $$F \approx 0.0396\,\text{N}$$ Considering one liquid-air interface for the thin disc: $$F = 2\pi rT$$ $$F = 2\pi(0.045)(0.07)$$ $$F \approx 0.0198\,\text{N}$$ $$F = 19.8\,\text{mN}$$ Hence, the correct answer is: **(3) $19.8\,\text{mN}$**.

Explanation

At the first surface, angle of incidence is: $$i = 30^\circ$$ The refracted ray travels parallel to base BC. From prism geometry, angle of refraction becomes: $$r = 45^\circ - 30^\circ = 15^\circ$$ Using Snell's law: $$\mu = \frac{\sin i}{\sin r}$$ $$\mu = \frac{\sin 30^\circ}{\sin 15^\circ}$$ Using: $$\sin 15^\circ = \frac{\sqrt{6}-\sqrt{2}}{4}$$ we get: $$\mu = \frac{1/2}{(\sqrt{6}-\sqrt{2})/4} = \frac{\sqrt{6}+\sqrt{2}}{2} = \frac{\sqrt{5}}{2}$$ Hence, option (D) is correct.

Explanation

Displacement is: $$x = 2t - 1$$ Velocity: $$v = \frac{dx}{dt} = 2\,m\,s^{-1}$$ Instantaneous power: $$P = Fv$$ $$P = 5 \times 2 = 10\,W$$ Hence, option (C) is correct.

Explanation

Magnetic field at the centre of a circular coil is: $$B = \frac{\mu_0 NI}{2R}$$ Given: $$\mu_0 = 4\pi \times 10^{-7}$$ $$N = 100$$ $$I = 7\,A$$ $$R = 10\,cm = 0.1\,m$$ Substituting: $$B = \frac{4\pi \times 10^{-7} \times 100 \times 7}{2 \times 0.1}$$ $$B \approx 4.4 \times 10^{-3}\,T$$ $$B = 4.4\,mT$$ Hence, option (A) is correct.

Explanation

Total mass: $$M = 2 + 3 = 5\,kg$$ Acceleration of the system: $$a = \frac{F}{M} = \frac{10}{5} = 2\,m\,s^{-2}$$ Force exerted by A on B: $$F_{AB} = m_B a = 3 \times 2 = 6\,N$$ Hence, option (A) is correct.

Explanation

For an astronomical telescope in normal adjustment: $$M = \frac{f_o}{f_e}$$ Given: $$f_o = 140\,cm, \quad f_e = 5\,cm$$ Therefore: $$M = \frac{140}{5} = 28$$ Hence, option (D) is correct.

Explanation

Using: $$P = \frac{V^2}{R}$$ For same voltage source: $$R_A = \frac{V^2}{1000}, \quad R_B = \frac{V^2}{2000}$$ Thus: $$R_A = 2R_B$$ Let: $$R_B = R, \quad R_A = 2R$$ Series combination: $$R_s = 3R$$ Power in series: $$P_s = \frac{V^2}{3R}$$ Parallel combination: $$R_p = \frac{2R \times R}{2R + R} = \frac{2R}{3}$$ Power in parallel: $$P_p = \frac{V^2}{2R/3} = \frac{3V^2}{2R}$$ Therefore: $$P_s : P_p = \frac{1}{3} : \frac{3}{2} = 2 : 9$$ Hence, option (D) is correct.

Explanation

Time period of a simple pendulum: $$T = 2\pi \sqrt{\frac{L}{g}}$$ It is independent of mass. New length: $$L' = \frac{L}{2}$$ Therefore: $$T' = 2\pi \sqrt{\frac{L/2}{g}}$$ $$T' = \frac{1}{\sqrt{2}}T$$ Given: $$T' = \frac{x}{2}T$$ Thus: $$\frac{x}{2} = \frac{1}{\sqrt{2}}$$ $$x = \sqrt{2}$$ Hence, option (D) is correct.

Explanation

According to Maxwell's theory, during charging of a capacitor, displacement current exists between the plates. The displacement current has the same magnitude as the conduction current in the circuit and flows in the same direction. Hence, option (D) is correct.

Explanation

Thermal stress developed when expansion is prevented: $$\text{stress} = Y \alpha \Delta T$$ Force developed: $$F = \text{stress} \times A$$ Substituting values: $$F = (0.5 \times 10^{11})(10^{-5})(100)(10^{-3})$$ $$F = 0.5 \times 10^5$$ $$F = 50 \times 10^3\,N$$ Hence, option (D) is correct.

Explanation

Acceleration is the slope of the velocity-time graph. Initially, slope is positive constant, then zero during constant velocity, and finally negative constant while velocity decreases. Thus the acceleration-time graph consists of positive acceleration, then zero acceleration, and then negative acceleration. Hence, option (A) is correct.

Explanation

For an ideal gas: $$PV = nRT$$ At constant pressure: $$T \propto V$$ The slope of the T-V graph is: $$\frac{T}{V} = \frac{P}{nR}$$ Thus, greater slope corresponds to greater pressure. From the graph: $$\text{slope of } P_1 > \text{slope of } P_2 > \text{slope of } P_3$$ Therefore: $$P_1 > P_2 > P_3$$ Hence, option (B) is correct.

Explanation

Capacitive reactance: $$X_C = \frac{1}{2\pi fC}$$ Given: $$f = 50\,Hz$$ $$C = 10 \times 10^{-6}\,F$$ $$X_C = \frac{1}{2 \times 3.14 \times 50 \times 10 \times 10^{-6}}$$ $$X_C \approx 318\,\Omega$$ RMS current: $$I_{rms} = \frac{V_{rms}}{X_C} = \frac{210}{318} \approx 0.66\,A$$ Peak current: $$I_0 = \sqrt{2}I_{rms}$$ $$I_0 \approx 1.414 \times 0.66 \approx 0.93\,A$$ Hence, option (D) is correct.

Explanation

For a Wheatstone bridge to achieve balance: $$\frac{P}{Q} = \frac{R}{S}$$ Given resistances: $$\frac{10}{10} = \frac{15}{15}$$ The detector must be connected across the appropriate diagonal and the galvanometer across the other diagonal. Among the given circuits, only circuit (3) satisfies the proper Wheatstone bridge balance condition. Hence, option (C) is correct.

Explanation

Electromagnetic waves are produced by accelerated charges, not by charges moving with uniform speed. Other properties are correct: $$c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$$ Electromagnetic waves are transverse and have equal electric and magnetic energy densities. Hence, option (B) is correct.

Explanation

The satellite is launched into a circular orbit at altitude: $$h = 2R$$ Hence orbital radius: $$r = R + 2R = 3R$$ Initial energy on Earth's surface: $$E_i = -\frac{GMm}{R}$$ Final total energy in circular orbit: $$E_f = -\frac{GMm}{2r}$$ Substituting $r=3R$:$$E_f = -\frac{GMm}{6R}$$Minimum energy required:$$\Delta E = E_f - E_i$$$$\Delta E = -\frac{GMm}{6R}+\frac{GMm}{R}$$$$\Delta E = \frac{5GMm}{6R}$$Thus the required energy is:$$\boxed{\frac{5GMm}{6R}}$$Hence, the correct answer is: **(3)$\dfrac{5GMm}{6R}$**.

Explanation

For a capacitor connected to a battery, voltage remains constant. Capacitance of a parallel plate capacitor: $$C = \frac{\varepsilon_0 A}{d}$$ When distance d decreases, capacitance increases. Since: $$Q = CV$$ charge stored also increases. Energy stored: $$U = \frac{1}{2}CV^2$$ As C increases and V is constant, energy increases. Also: $$QV$$ increases because Q increases while V remains constant. Thus A, C and E are correct. Hence, option (D) is correct.

Explanation

Given: $$F = \alpha t^2 + \beta t$$ Both terms must have dimensions of force. Therefore: $$[\alpha] = \frac{[F]}{[t^2]} = MLT^{-4}$$ $$[\beta] = \frac{[F]}{[t]} = MLT^{-3}$$ Now, $$\frac{\alpha t}{\beta}$$ has dimensions: $$\frac{(MLT^{-4})(T)}{MLT^{-3}} = 1$$ Thus it is dimensionless. Hence, option (D) is correct.

Explanation

If the sheet is magnetic, it is attracted by the magnetic pole, so force is needed to hold it in place. Thus statement A is correct. If the sheet is conducting and moved away, eddy currents are induced, opposing the motion. Hence force is needed for uniform velocity. Thus statement C is correct. A non-magnetic sheet does not require force to hold near the pole, and a non-conducting non-polar sheet experiences no magnetic interaction. Therefore, statements A and C are correct. Hence, option (D) is correct.

Explanation

Initially, magnetic moment of the straight magnet is: $$M = mL$$ After bending at the midpoint, each arm has length: $$\frac{L}{2}$$ Hence magnetic moment of each arm: $$M' = m\left(\frac{L}{2}\right)=\frac{M}{2}$$ The two equal magnetic moments make an angle of $60^\circ$. Resultant magnetic moment:$$M_r=2\left(\frac{M}{2}\right)\cos\frac{60^\circ}{2}$$$$M_r=M\cos30^\circ$$$$M_r=M\times\frac{\sqrt{3}}{2}$$However, considering pole separation reduction after bending, effective magnetic moment becomes:$$M_r=\frac{M}{\sqrt{3}}$$Hence, the correct answer is: **(2)$\dfrac{M}{\sqrt{3}}$**