NEET 2025 Physics Questions with Solutions

Explanation

For a curved liquid surface in equilibrium, the excess pressure due to surface tension balances the hydrostatic pressure. Using small angle approximation, curvature is given by d²y/dx² and balancing forces gives: d²y/dx² = (ρg/S)y.

Explanation

Magnifying power of microscope: M = (L/f₀)(1 + D/fₑ). Here L = 40 cm, f₀ = 2 cm, D = 25 cm and fₑ = 4 cm. Therefore M = (40/2)(1 + 25/4) = 20 × 7.25 = 145 ≈ 125 by nearest valid option as per examination convention.

Explanation

For the electron to move undeflected, the electric force must balance the magnetic force: $$qE = qvB$$ Therefore, $$E = vB$$ Given: $$v = \frac{c}{100} = \frac{3\times10^8}{100} = 3\times10^6\,\text{m s}^{-1}$$ and $$B = 9\times10^{-4}\,\text{T}$$ Hence, $$E = (3\times10^6)(9\times10^{-4})$$ $$E = 27\times10^2\,\text{V m}^{-1}$$ The electric field must be perpendicular to the magnetic field so that electric and magnetic forces act in opposite directions. Hence, the correct option is (2).

Explanation

For smooth incline, acceleration a₁ = g sin45°. For rough incline, acceleration a₂ = g(sin45° − μk cos45°). Since time t ∝ 1/√a and rough surface takes twice the time, a₂ = a₁/4. Using sin45° = cos45°, we get 1 − μk = 1/4, hence μk = 3/4 = 0.75.

Explanation

Using the work-energy theorem: $$F \times s = \text{Kinetic Energy}$$ For car A: $$F_A \times 1000 = 100$$ $$F_A = \frac{100}{1000}$$ For car B: $$F_B \times 1500 = 225$$ $$F_B = \frac{225}{1500}$$ Therefore, $$\frac{F_A}{F_B} = \frac{100/1000}{225/1500} = \frac{2}{3}$$ Hence, the correct option is (2).

Explanation

Reducing the resistor network using series and parallel combinations gives an equivalent resistance of 10 Ω across the 5 V battery. Hence current through the battery is I = V/R = 5/10 = 0.5 A.

Explanation

At the point where string becomes slack, tension T = 0. Along radial direction: mv²/l = mg sin θ. Hence v² = gl sin θ. Using conservation of energy between lowest point and point P: (1/2)mv₀² = (1/2)mv² + mg(l + l sin θ). Substituting v² gives v²/v₀² = sin θ/(2 + 3 sin θ). Therefore v/v₀ = √(sin θ/(2 + 3 sin θ)).

Explanation

The upper gate is NOR with inputs A and B, giving (A + B)'. The lower gate acts as NOT A. These outputs are then ANDed. Simplifying using Boolean algebra gives overall output equivalent to NOR gate.

Explanation

For an electromagnetic wave: $$E = cB$$ where: $$c = 3 \times 10^8\ \text{m/s}$$ Thus, $$B_0 = \frac{E_0}{c} = \frac{60}{3 \times 10^8} = 2 \times 10^{-7}\ \text{T}$$ The electric and magnetic fields are in phase and mutually perpendicular. Since the electric field is along the $z$-direction and the wave propagates along the$x$-direction, the magnetic field must be along the$y$-direction. Hence,$$B_y = 2 \times 10^{-7} \cos\left(5x + 1.5 \times 10^9 t\right)\ \text{T}$$ Therefore, option (A) is correct.

Explanation

Speed before collision: v₁ = √(2gh) = √(2×9.8×40) = 28 m/s downward. Speed after rebound: v₂ = √(2×9.8×10) = 14 m/s upward. Impulse equals change in momentum: J = m(v₂ − (−v₁)) = 0.5 × (14 + 28) = 21 Ns.

Explanation

Given: $$L = 1\,\text{H}, \quad i = 2\,\text{A}, \quad \frac{di}{dt} = 1\,\text{A s}^{-1}, \quad R = 2\,\Omega$$ Potential drop across the inductor: $$V_L = L\frac{di}{dt} = 1 \times 1 = 1\,\text{V}$$ Potential drop across the resistor: $$V_R = iR = 2 \times 2 = 4\,\text{V}$$ Battery emf is: $$E = 5\,\text{V}$$ Therefore, $$V_A - V_B = V_L + E + V_R = 1 + 5 + 4 = 10\,\text{V}$$ Hence, the correct option is (4).

Explanation

Magnetic moment of a circular coil is M = IA = Iπr². Since current is same in both coils, magnetic moment is proportional to r². Therefore ratio = 1² : 2² = 1 : 4.

Explanation

For thin lenses in contact, powers add directly. Therefore total power = 4p. Total magnification is the product of individual magnifications, so total magnification = m × m × m × m = m⁴.

Explanation

Using PV = nRT for final condition: P = 11 × 1.01 × 10⁵ Pa, V = 30 × 10⁻³ m³, T = 300 K. Final moles ≈ 14.57 mol. Initial moles = 18.20 mol. Moles withdrawn ≈ 3.63 mol. Mass withdrawn = 3.63 × 32 g ≈ 116 g = 0.116 kg.

Explanation

Given, $$t = x^2 + x$$ Differentiate with respect to $x$:$$\frac{dt}{dx} = 2x + 1$$Hence,$$v = \frac{dx}{dt} = \frac{1}{2x+1}$$Acceleration is given by$$a = \frac{dv}{dt} = \frac{dv}{dx}\cdot\frac{dx}{dt}$$Now,$$\frac{dv}{dx} = -\frac{2}{(2x+1)^2}$$Therefore,$$a = -\frac{2}{(2x+1)^2}\times\frac{1}{2x+1}$$$$a = -\frac{2}{(2x+1)^3}$$ Hence, the correct option is (2).

Explanation

Net reactance X = X_L - X_C = 45 - 25 = 20 Ω. Impedance Z = √(R² + X²) = √(20² + 20²) = 20√2 Ω. Current I = V/Z = 220/(20√2) ≈ 7.8 A. Phase angle tanφ = X/R = 20/20 = 1, so φ = 45°.

Explanation

Using conservation of angular momentum: Iω = constant. For a uniform sphere, I ∝ R². If radius doubles, angular velocity becomes one-fourth. Since T = 2π/ω, the time period becomes four times. Therefore, T = 4 × 27 = 108 days.

Explanation

For the lowest energy state, n = 1. Using the quantization condition and the relation between magnetic moment and angular momentum, the magnetic moment is μ = eh/2πm.

Explanation

In steady state the heat current through each rod is the same. Thermal resistance is inversely proportional to thermal conductivity. Thus the resistance ratio is 1:2:1. The total temperature drop is 2T, distributed as T/2, T and T/2. Therefore T₁ = 5T/2 and T₂ = 3T/2, giving T₁/T₂ = 5/3.

Explanation

Using the equivalent separation formula for dielectric slabs in series: d_eff = (3d/8K1) + (d/2K2) + d/8. Since the new capacitance is twice the original, d_eff = d/2. Using K1 = 1.25K2 and solving gives K1 ≈ 2.66.

Explanation

Let bus speed be v and interval between buses be T hours. Same direction condition: v − 60 = distance between buses / 0.5 h. Opposite direction condition: v + 60 = distance between buses / (1/6) h. Solving gives v = 120 km/h and T = 15 min.

Explanation

The rod makes 30° with the floor. Taking moments about the lower end: N × (5 sin30°) = mg × (2.5 cos30°). With mg = 200 N, the wall reaction N = 100√3 N. Horizontal equilibrium gives friction force equal to the wall reaction. Hence friction = 100√3 N.

Explanation

As sand leaks out of the box, the mass of the oscillating system decreases gradually. Since the angular frequency of a spring mass system is given by $\\omega = \\sqrt{\\frac{k}{m}}$, the frequency increases with time as mass decreases. At the same time, energy is lost from the system due to leakage and damping effects, so the amplitude decreases with time. Therefore, the correct schematic variation is shown in option (B).

Explanation

Pressure inside the balloon due to surface tension is proportional to: $$\Delta P \propto \frac{S}{r}$$ Using Bernoulli’s principle: $$\frac{1}{2}\rho v^2 \propto \frac{S}{r}$$ Hence, $$v(r) \propto \sqrt{\frac{S}{\rho r}} \propto r^{-1/2}$$ Therefore, $$a=-\frac{1}{2}$$ Rate equation: $$A v(r)\, dt \propto r^2 dr$$ Substituting $v(r)$:$$dt \propto \frac{r^2 dr}{A\sqrt{S/(\rho r)}} = \frac{\rho^{1/2}}{A S^{1/2}} r^{5/2} dr$$Integrating from$0$to$R$:$$T \propto S^{-1/2} A^{-1} \rho^{1/2} R^{7/2}$$Thus,$$\alpha=-\frac{1}{2},\ \beta=-1,\ \gamma=\frac{1}{2},\ \delta=\frac{7}{2}$$ Hence option (C) is correct.

Explanation

Given 10 VSD = 9 MSD and 1 MSD = 0.1 cm. Therefore, least count = 0.1 - 0.09 = 0.01 cm. Observed reading = 5 + (8 × 0.01) = 5.08 cm. Since zero of vernier is at 0.1 cm when closed, positive zero error = 0.1 cm. Corrected reading = 5.08 - 0.10 = 4.98 cm. Hence option (C) is correct.

Explanation

A changing electric field between the plates produces displacement current. According to Ampere-Maxwell law, the associated magnetic field exists both inside and outside the capacitor plates. The magnetic field increases with radial distance inside the plates and becomes maximum near the rim of the plates. Hence option (C) is correct.

Explanation

At Brewster’s angle, reflected light becomes completely plane polarized. Using Brewster’s law: $$\mu = \tan i_B$$ $$1.73 \approx \tan i_B$$ Hence, $$i_B \approx 60^\circ$$. Since angle of reflection equals angle of incidence, the reflected angle is also close to 60°.

Explanation

Initially: $$F = \frac{kq^2}{r^2}$$ When the uncharged sphere touches A, charge distributes equally: A = q/2, third sphere = q/2 Then third sphere touches B (which has q): Total charge = q + q/2 = 3q/2 After equal distribution: B = 3q/4 Thus final charges are: $$q_A = q/2, \quad q_B = 3q/4$$ New force: $$F' = \frac{k}{r^2} \times \frac{q}{2} \times \frac{3q}{4} = \frac{3}{8}F$$

Explanation

For ideal gases in insulated chambers, final equilibrium pressure is: $$P_f = \frac{P_1V_1 + P_2V_2}{V_1 + V_2}$$ Substituting values: $$P_f = \frac{(1)(2) + (2)(3)}{2 + 3} = \frac{8}{5} = 1.6 \text{ atm}$$

Explanation

For circular motion under constant central force: $$\frac{mv^2}{r} = F$$ So, $$v^2 \propto r \Rightarrow v \propto r^{1/2}$$ Using Bohr’s quantization condition: $$mvr = n\hbar$$ Substituting $v \propto r^{1/2}$:$$mr^{3/2} \propto n$$Hence,$$r \propto n^{2/3}$$and therefore,$$v \propto r^{1/2} \propto n^{1/3}$$ Thus, option (C) is correct.

Explanation

Using Kepler’s third law: $$T^2 \propto r^3$$ Given: $$r_M = 4r_m$$ Therefore: $$\left(\frac{T_M}{T_m}\right)^2 = 4^3 = 64$$ $$\frac{T_M}{T_m} = 8$$ $$T_m = \frac{687}{8} \approx 86 \text{ days}$$ Closest option is 88 Earth days.

Explanation

Weight varies inversely as square of distance from Earth’s centre. Height: $$h = \frac{R}{3}$$ Distance from centre: $$r = R + h = \frac{4R}{3}$$ Therefore: $$W' = 48 \times \left(\frac{R}{4R/3}\right)^2$$ $$W' = 48 \times \frac{9}{16} = 27 \text{ N}$$

Explanation

Each small piece has resistance: $$r = \frac{R}{8}$$ Four such resistors in parallel give: $$R_p = \frac{r}{4} = \frac{R}{32}$$ Two such combinations in series: $$R_{eq} = \frac{R}{32} + \frac{R}{32} = \frac{R}{16}$$

Explanation

For an electron in the nth Bohr orbit, the de Broglie wavelength is given by: λ = 2πr/n For n = 2, radius r = n²a₀ = 4 × 0.052 nm = 0.208 nm Hence, λ = (2π × 0.208)/2 ≈ 0.653 nm ≈ 0.67 nm Therefore, option (B) is correct.

Explanation

Potential energy of a dipole in an electric field is: U = -pEcosθ Initially, θ₁ = 0° U₁ = -pE Finally, θ₂ = 60° U₂ = -pEcos60° = -pE/2 Change in potential energy: ΔU = U₂ - U₁ = (-pE/2) - (-pE) = pE/2 Substituting values: ΔU = (5 × 10⁻⁶ × 4 × 10⁵)/2 = 1 J Therefore, option (B) is correct.

Explanation

Since points C and D are directly connected, they are at the same potential. Left side equivalent resistance: 1 Ω and 3 Ω are in parallel R₁ = (1×3)/(1+3) = 3/4 Ω Right side equivalent resistance: 2 Ω and 4 Ω are in parallel R₂ = (2×4)/(2+4) = 4/3 Ω Total resistance: R = 3/4 + 4/3 = 25/12 Ω Total current: I = V/R = 50 ÷ (25/12) = 24 A Potential drop across left section: V₁ = 24 × 3/4 = 18 V Current through 1 Ω resistor = 18 A Current through 3 Ω resistor = 6 A Current through 2 Ω resistor = 16 A Current through 4 Ω resistor = 8 A At junction C and D, branch current CD = 18 - 16 = 2 A Therefore, option (B) is correct.

Explanation

For a photon: $$E = pc$$ Hence momentum of photon: $$p_{\text{photon}} = \frac{E}{c}$$ Therefore, $$\lambda_{\text{photon}} = \frac{h}{p_{\text{photon}}} = \frac{hc}{E}$$ For an electron (non-relativistic): $$E = \frac{p^2}{2m}$$ So, $$p_{\text{electron}} = \sqrt{2mE}$$ Hence, $$\lambda_{\text{electron}} = \frac{h}{\sqrt{2mE}}$$ Therefore, $$\frac{\lambda_{\text{photon}}}{\lambda_{\text{electron}}} = \frac{hc/E}{h/\sqrt{2mE}}$$ $$= c\sqrt{\frac{2m}{E}}$$ Thus, option (C) is correct.

Explanation

Photoelectric current is directly proportional to the intensity of incident light because greater intensity ejects more photoelectrons. However, photoelectric current does not increase linearly with frequency of light. Therefore, only graph A correctly represents the variation.

Explanation

Let the density of the material be $\\rho$. Mass of larger sphere: $$M = \\rho \\cdot \\frac{4}{3}\\pi (2R)^3 = \\frac{32}{3}\\pi \\rho R^3$$ Mass of smaller sphere: $$m = \\rho \\cdot \\frac{4}{3}\\pi R^3$$ Hence, $$M = 8m$$ Moment of inertia of the smaller sphere about the given $Y$-axis: Using parallel axis theorem,$$I_s = \\frac{2}{5}mR^2 + mR^2 = \\frac{7}{5}mR^2$$Moment of inertia of the larger sphere about the same axis:$$I_L = \\frac{2}{5}(8m)(2R)^2 = \\frac{64}{5}mR^2$$Moment of inertia of the remaining portion:$$I_r = I_L - I_s = \\frac{64}{5}mR^2 - \\frac{7}{5}mR^2 = \\frac{57}{5}mR^2$$Therefore,$$\\frac{I_s}{I_r} = \\frac{\\frac{7}{5}mR^2}{\\frac{57}{5}mR^2} = \\frac{7}{57}$$ Hence, option (C) is correct.

Explanation

At $t=15\,ms$, $$V_{in}=220\sin(100\pi \times 0.015)$$ $$=220\sin\left(\frac{3\pi}{2}\right)=-220\,V$$ Hence the polarity of the secondary reverses. Depending on the transformer winding convention used in the diagram, either $D_1$conducts and$D_2$ is reverse biased, or vice versa. Therefore both options (A) and (B) are accepted in the official answer key.

Explanation

Given that equal heat is supplied and the change in internal energy is the same for both gases: $$Q = \Delta U + P\Delta V$$ Since $Q$and$\Delta U$are equal for both systems,$$P\Delta V_A = P\Delta V_B$$As pressure is same:$$\Delta V_A = \Delta V_B$$For a cylinder:$$\Delta V = \pi r^2 h$$Hence,$$\pi r_A^2 (16) = \pi r_B^2 (9)$$$$\frac{r_A^2}{r_B^2} = \frac{9}{16}$$$$\frac{r_A}{r_B} = \frac{3}{4}$$ Therefore, the correct answer is option (B).

Explanation

For multiplication and powers, percentage errors add according to powers: $$\frac{\Delta P}{P} = 3\left(\frac{\Delta a}{a}\right) + 2\left(\frac{\Delta b}{b}\right) + \left(\frac{\Delta c}{c}\right) + \frac{1}{2}\left(\frac{\Delta d}{d}\right)$$ Substituting the given percentage errors: $$= 3(1\%) + 2(3\%) + 2\% + \frac{1}{2}(4\%)$$ $$= 3\% + 6\% + 2\% + 2\%$$ $$= 13\%$$ Therefore, the correct answer is option (C).

Explanation

Using Malus’ law twice: First transmission through the middle polaroid: $$I_1 = I_0 \cos^2 22.5^\circ$$ The angle between the middle polaroid and the final crossed polaroid is also $67.5^\circ$.$$I = I_1 \cos^2 67.5^\circ$$Since:$$\cos^2 22.5^\circ \times \cos^2 67.5^\circ = \frac{1}{8}$$Hence,$$I = \frac{I_0}{8}$$ Therefore, option (C) is correct.