NEET 2021 Physics Questions with Solutions

Explanation

In SHM, displacement x = A sin(ωt + φ). Potential energy U = (1/2)kx² = (1/2)kA² sin²(ωt + φ). Since sin²θ has frequency twice that of sinθ, frequency of potential energy = 2n.

Explanation

In polar molecules, the centres of positive and negative charges do not coincide. Therefore they possess a permanent electric dipole moment.

Explanation

When two conducting spheres are connected by a wire, their potentials become equal: $$V_1 = V_2$$ $$\frac{kQ_1}{R_1} = \frac{kQ_2}{R_2}$$ $$\Rightarrow \frac{Q_1}{Q_2} = \frac{R_1}{R_2}$$ Surface charge density of a sphere is: $$\sigma = \frac{Q}{4\pi R^2}$$ Therefore, $$\frac{\sigma_1}{\sigma_2} = \frac{Q_1}{Q_2} \cdot \frac{R_2^2}{R_1^2}$$ Substituting: $$\frac{\sigma_1}{\sigma_2} = \frac{R_1}{R_2} \cdot \frac{R_2^2}{R_1^2}$$ $$\frac{\sigma_1}{\sigma_2} = \frac{R_2}{R_1}$$ Hence, option (C) is correct.

Explanation

Energy stored in a capacitor is given by: $U = \dfrac{1}{2}CV^2$ For a parallel plate capacitor: $C = \dfrac{\varepsilon_0 A}{d}$ Also, electric field between plates: $E = \dfrac{V}{d} \Rightarrow V = Ed$ Substituting: $U = \dfrac{1}{2} \times \dfrac{\varepsilon_0 A}{d} \times (Ed)^2$ $U = \dfrac{1}{2} \varepsilon_0 E^2 Ad$ Hence, option (D) is correct.

Explanation

Energy: $$[E] = [ML^2T^{-2}]$$ Gravitational constant: From Newton’s law: $$F = \frac{Gm_1m_2}{r^2}$$ $$[G] = \frac{[F][r^2]}{[M]^2} = \frac{[MLT^{-2}] \cdot [L^2]}{[M]^2} = [M^{-1}L^3T^{-2}]$$ Therefore, $$\left[\frac{E}{G}\right] = \frac{[ML^2T^{-2}]}{[M^{-1}L^3T^{-2}]}$$ $$= [M^{2}L^{-1}T^{0}]$$ Hence, the correct answer is **Option (B)**.

Explanation

A large aperture objective lens improves: 1. Light gathering power → helps observe faint celestial objects. 2. Resolving power → helps distinguish closely spaced objects. 3. Image brightness and visibility. Thus all statements are correct, so option D is the correct answer.

Explanation

Root mean square speed: vrms = √(3RT/M) → (Q) Pressure of ideal gas: P = (1/3)nmv² → (P) Average kinetic energy per molecule: K = (3/2)kBT → (S) For 1 mole of diatomic gas: U = (f/2)RT For diatomic gas, f = 5 U = (5/2)RT → (R) Hence correct matching is: (A)-(Q), (B)-(P), (C)-(S), (D)-(R)

Explanation

A zener diode is always operated in reverse bias breakdown region when used as a voltage regulator. Statement (A) is correct. The potential barrier for a silicon p-n junction is approximately 0.7 V (not 0.1–0.3 V), so statement (B) is incorrect.

Explanation

Potential energy of a dipole in an electric field is U = -P·E = -PEcosθ. Here θ = 180°, so U = +PE. The electric field decreases towards the right, hence the dipole moves right to lower its potential energy.

Explanation

The convex lens converges parallel rays to its focus at 20 cm. For rays to emerge parallel after passing through the concave lens, this image must coincide with the focal point of the concave lens (5 cm). Therefore, d = 20 - 5 = 15 cm.

Explanation

Escape velocity is given by: $$v_e = \sqrt{\frac{2GM}{R}}$$ For a planet of uniform density: $$M = \frac{4}{3}\pi R^3 \rho$$ Substituting: $$v_e = \sqrt{\frac{2G\left(\frac{4}{3}\pi R^3\rho\right)}{R}}$$ $$v_e = \sqrt{\frac{8\pi G\rho R^2}{3}}$$ Since density remains the same: $$v_e \propto R$$ New planet radius = 4R $$\frac{v_1}{v} = \frac{4R}{R} = 4$$ $$v_1 = 4v$$ Hence, option (A) is correct.

Explanation

Step 1: $$^{A}_{Z}X \rightarrow \ ^{A}_{Z-1}B$$ Atomic number decreases by 1 while mass number remains unchanged. This corresponds to **$\beta^+$decay (positron emission)**.$$p \rightarrow n + e^+ + \nu$$Step 2:$$^{A}_{Z-1}B \rightarrow \ ^{A-4}_{Z-3}C$$Mass number decreases by 4 and atomic number decreases by 2. This corresponds to **$\alpha$decay**.$$^{4}_{2}He$$is emitted. Step 3:$$^{A-4}_{Z-3}C \rightarrow \ ^{A-4}_{Z-2}D$$Atomic number increases by 1 while mass number remains unchanged. This corresponds to **$\beta^-$decay**.$$n \rightarrow p + e^- + \bar{\nu}$$Therefore, the correct sequence is:$$\beta^+ ,\ \alpha ,\ \beta^-$$ Hence, **Option D** is correct.

Explanation

Pitch of screw gauge = 1 mm Number of circular divisions = 100 Least count: LC = 1/100 mm = 0.01 mm = 0.001 cm Diameter = MSR + (CSR × LC) = 0 + (52 × 0.001) = 0.052 cm Hence option (A) is correct.

Explanation

Given: VL = 40 V VC = 10 V VR = 40 V For a series LCR circuit: Vrms = √[VR² + (VL - VC)²] Vrms = √[(40)² + (40 - 10)²] Vrms = √(1600 + 900) Vrms = √2500 = 50 V Current amplitude: I0 = 10√2 A Irms = I0/√2 = 10 A Using: Z = Vrms / Irms Z = 50 / 10 = 5 Ω Therefore, the correct answer is option (A).

Explanation

Let the particle be at height $h$ above Earth's surface when the condition is satisfied. Initial potential energy at height $S$: $PE_i = mgS$ At height $h$: $PE = mgh$ Loss in potential energy = Gain in kinetic energy: $KE = mg(S-h)$ Given: $KE = 3PE$ $mg(S-h) = 3mgh$ $S-h = 3h$ $S = 4h$ $h = \dfrac{S}{4}$ Now, $KE = mg\left(S-\dfrac{S}{4}\right)$ $KE = mg\left(\dfrac{3S}{4}\right)$ $\dfrac{1}{2}mv^2 = \dfrac{3mgS}{4}$ $v^2 = \dfrac{3gS}{2}$ $v = \sqrt{\dfrac{3gS}{2}}$ Hence, option (A) is correct.

Explanation

For motion down a smooth incline, acceleration a = gsinθ. Distance travelled in nth second: Sₙ = u + (a/2)(2n−1) Since u = 0, Sₙ = (gsinθ/2)(2n−1) Similarly, Sₙ₊₁ = (gsinθ/2)(2n+1) Therefore, Sₙ/Sₙ₊₁ = (2n−1)/(2n+1).

Explanation

Radioactive decay law: A = A₀(1/2)^(t/T₁/₂) Given: T₁/₂ = 100 hours t = 150 hours A/A₀ = (1/2)^(150/100) = (1/2)^(3/2) = 1/(2√2)

Explanation

The middle capacitor is short-circuited because both its terminals are at the same potential. Hence it does not contribute. The remaining two capacitors C and C are connected in parallel. Equivalent capacitance: Ceq = C + C = 2C

Explanation

Initial total binding energy of the nucleus: $$BE_i = 240 \times 7.6 = 1824\,\text{MeV}$$ Final total binding energy of the two fragments: $$BE_f = 2 \times (120 \times 8.5) = 2040\,\text{MeV}$$ Gain in binding energy: $$\Delta BE = BE_f - BE_i$$ $$\Delta BE = 2040 - 1824 = 216\,\text{MeV}$$ Hence, the correct answer is **Option A**.

Explanation

Current in a semiconductor depends on charge carrier concentration and mobility: I ∝ nqμ Given: - Electron concentration in n-type = Hole concentration in p-type - Applied electric field is same across both Since electron mobility is greater than hole mobility in semiconductors: μₑ > μₕ Therefore, current in n-type semiconductor is greater than current in p-type semiconductor. Hence, option (D) is correct.

Explanation

Energy of one photon is given by: $$E = \frac{hc}{\lambda}$$ Substituting: $$E = \frac{(6.6 \times 10^{-34})(3 \times 10^8)}{600 \times 10^{-9}}$$ $$E = 3.3 \times 10^{-19}\ \text{J}$$ Given power of source: $$P = 3.3 \times 10^{-3}\ \text{W} = 3.3 \times 10^{-3}\ \text{J/s}$$ Number of photons emitted per second: $$n = \frac{P}{E}$$ $$n = \frac{3.3 \times 10^{-3}}{3.3 \times 10^{-19}} = 10^{16}$$ Hence, option (D) is correct.

Explanation

At terminal velocity, net force = 0. Mg = Fb + Fv Buoyant force: Fb = density of liquid × volume × g Given density of glycerin = d/2 Since M = dV Fb = (d/2)Vg = Mg/2 Therefore, Fv = Mg - Mg/2 = Mg/2 Hence option (B) is correct.

Explanation

From the prism geometry, angle of incidence at second face = 30°. Using Snell’s law: sin i / sin e = 1/√3 sin 30° / sin e = 1/√3 1/2sin e = 1/√3 sin e = √3/2 Therefore, e = 60° Hence option (B) is correct.

Explanation

For a capacitor: $$q = CV$$ Given: $$V = V_0 \sin \omega t$$ Therefore, $$q = C V_0 \sin \omega t$$ Displacement current is: $$I_d = \frac{dq}{dt}$$ Differentiating: $$I_d = \frac{d}{dt}(C V_0 \sin \omega t)$$ $$I_d = C V_0 \omega \cos \omega t$$ Thus, $$I_d = V_0 \omega C \cos \omega t$$ Hence, option **(B)** is correct.

Explanation

Energy = Force × Distance. Using A = LT⁻² → L = AT². Therefore Energy = F × AT² = [F][A][T²].

Explanation

Using Hooke’s law: k = F/x = 10/(0.05) = 200 N/m. Time period T = 2π√(m/k) = 2π√(2/200) = 0.628 s.

Explanation

Direction of propagation of an EM wave is given by E × B. For propagation along x-axis, only option C gives a cross product along x-direction.

Explanation

Using Newton's law of cooling: $$\frac{dT}{dt}=-k(T-T_s)$$ where $T_s=20^\circ C$is the surrounding temperature. For cooling from$90^\circ C$to$80^\circ C$:$$t=\frac{1}{k}\ln\left(\frac{90-20}{80-20}\right)=\frac{1}{k}\ln\left(\frac{70}{60}\right)=\frac{1}{k}\ln\left(\frac{7}{6}\right)$$For cooling from$80^\circ C$to$60^\circ C$:$$t'=\frac{1}{k}\ln\left(\frac{80-20}{60-20}\right)=\frac{1}{k}\ln\left(\frac{60}{40}\right)=\frac{1}{k}\ln\left(\frac{3}{2}\right)$$Thus,$$\frac{t'}{t}=\frac{\ln(3/2)}{\ln(7/6)}\approx \frac{13}{5}$$Therefore,$$t'=\frac{13}{5}t$$ Hence, option (C) is correct.

Explanation

For negligible work function, the entire energy of the incident photon is converted into kinetic energy of the emitted electron. Photon energy: $$E = \frac{hc}{\lambda}$$ Kinetic energy of emitted electron: $$K = \frac{p^2}{2m}$$ Using de-Broglie relation: $$p = \frac{h}{\lambda_d}$$ Therefore, $$K = \frac{1}{2m}\left(\frac{h}{\lambda_d}\right)^2 = \frac{h^2}{2m\lambda_d^2}$$ Since photon energy = kinetic energy: $$\frac{hc}{\lambda} = \frac{h^2}{2m\lambda_d^2}$$ Rearranging: $$\lambda = \left(\frac{2mc}{h}\right)\lambda_d^2$$ Hence, option (D) is correct.

Explanation

Input power: P = mgh/t = ṁgh = 15 × 10 × 60 = 9000 W Losses = 10% Generated power = 90% of 9000 = 8100 W = 8.1 kW Hence option (C) is correct.

Explanation

Potential of a drop: V = kq/r For 27 identical drops: Total charge Q = 27q Volume conservation: R³ = 27r³ R = 3r Potential of bigger drop: V' = kQ/R = k(27q)/(3r) = 9(kq/r) = 9 × 220 = 1980 V Hence option (A) is correct.

Explanation

Using conservation of mechanical energy and Vₑ = √(2GM/R), we get h = Rk²/(1-k²).

Explanation

Magnetic field due to larger loop at center is B = μ₀I/2R₁. Flux through smaller loop = B·πR₂². Hence M ∝ R₂²/R₁.

Explanation

For a complete ring, the moment of inertia about an axis through its centre and perpendicular to its plane is: $$I_{\text{full}} = MR^2$$ The removed arc corresponds to $90^\circ$out of$360^\circ$, so the removed mass is:$$M_{\text{removed}} = \frac{90}{360}M = \frac{M}{4}$$Since every particle of the ring is at the same distance$R$from the axis, the moment of inertia of the removed arc is:$$I_{\text{removed}} = \frac{M}{4}R^2$$Thus, moment of inertia of the remaining part:$$I_{\text{remaining}} = MR^2 - \frac{M}{4}R^2$$$$I_{\text{remaining}} = \frac{3}{4}MR^2$$Comparing with:$$I = KMR^2$$we get:$$K = \frac{3}{4}$$ Hence, the correct answer is **Option B**.

Explanation

Velocity of car at t=4s: v = u + at = 0 + 5×4 = 20 m/s. Ball retains horizontal velocity = 20 m/s. After 2 s, vertical velocity = gt = 20 m/s. Resultant velocity = √(20²+20²)=20√2 m/s. Acceleration = g = 10 m/s².

Explanation

For ideal transformer: Input power = Output power. 220 × Ip = 44 ⇒ Ip = 0.2 A.

Explanation

Velocity before impact = √(2gh)=10√2 m/s downward. After rebound same magnitude upward. Impulse = change in momentum = 2mv = 2×0.15×10√2 ≈ 4.2 kg m/s.

Explanation

Resonant frequency ω₀ = 1/√LC = 50 rad/s. Half power frequencies: ω = ω₀ ± R/2L = 50 ± 40/10 = 50 ± 4. Hence frequencies are 46 rad/s and 54 rad/s.

Explanation

Taking moments about 40 cm mark: Torque due to 2 kg mass = 2×10×0.2 = 4 Nm. Torque due to rod's weight = 0.5×10×0.6 = 3 Nm. Torque due to mass m at 160 cm = m×10×1.2 = 12m Nm. Equilibrium: 4 = 3 + 12m ⇒ m = 1/12 kg.

Explanation

For circular motion, speed of the particle is: $$v = \frac{2\pi R}{T}$$ For projectile motion, maximum height: $$H = \frac{v^2 \sin^2\theta}{2g}$$ Given: $$H = 4R$$ Substituting: $$\frac{v^2 \sin^2\theta}{2g} = 4R$$ $$\frac{\left(\frac{2\pi R}{T}\right)^2 \sin^2\theta}{2g} = 4R$$ $$\frac{4\pi^2 R^2 \sin^2\theta}{2gT^2} = 4R$$ $$\frac{2\pi^2 R \sin^2\theta}{gT^2} = 4$$ $$\sin^2\theta = \frac{2gT^2}{\pi^2 R}$$ $$\theta = \sin^{-1}\left(\sqrt{\frac{2gT^2}{\pi^2 R}}\right)$$ Hence, option (A) is correct.

Explanation

For one mole of an ideal gas, Mayer's relation gives: where: - $C_p$ = molar specific heat at constant pressure - $C_v$ = molar specific heat at constant volume - $R$ = universal gas constant Hence, the correct option is **(3)**.

Explanation

Using λ = c/f: λ = (3 × 10^8)/(1368 × 10^3) = 219.3 m