NEET 2023 Physics Questions with Solutions

Explanation

Longitudinal stress is defined as force per unit area. Hence stress = W/A.

Explanation

For a solid sphere, k = sqrt(2/5)R. For a hollow sphere, k = sqrt(2/3)R. Therefore the ratio is sqrt(2/5) : sqrt(2/3) = sqrt(3/5) = 3 : 5 approximately, but based on the official answer key, option (B) is treated as correct.

Explanation

The two 3 μF capacitors inside the box are in parallel, giving 6 μF. This 6 μF is in series with the left 3 μF capacitor. Hence equivalent capacitance = (3×6)/(3+6) = 2 μF.

Explanation

Force acts in the direction of change in momentum. The velocity changes from southward to eastward, so the change in velocity is toward north-east.

Explanation

Zero net electric flux means the number of electric field lines entering the surface equals the number leaving the surface.

Explanation

Potential energy stored in a spring is proportional to x². Stretching changes from 2 cm to 8 cm, i.e. 4 times. Hence energy becomes 4² = 16 times.

Explanation

For zero deflection in the galvanometer, the potential difference across it must be zero. Applying circuit balance gives R = 100 Ω.

Explanation

For an ideal transformer, input power equals output power. Therefore primary current I = P/V = 60/220 ≈ 0.27 A.

Explanation

In a full wave rectifier, the capacitor acts as a filter circuit. It smoothens the pulsating dc output and removes the ac ripple component from the rectified output.

Explanation

Speed of light in air: v₁ = x/t₁ Speed of light in denser medium: v₂ = 10x/t₂ Critical angle C satisfies: sin C = v₂/v₁ = (10x/t₂)/(x/t₁) = 10t₁/t₂ Therefore, the critical angle is: C = sin⁻¹(10t₁/t₂)

Explanation

22000 Ω = 22 × 10³ Ω. For resistor colour coding: - First band = 2 → Red - Second band = 2 → Red - Third band = multiplier 10³ → Orange Hence, the third band colour is Orange.

Explanation

Photovoltaic devices such as solar cells convert light energy into electrical energy. Zener diodes are specially designed to operate in reverse bias under breakdown conditions for voltage regulation. Therefore, both statements are correct.

Explanation

Energy stored in an inductor is: U = (1/2)LI² Given: L = 4 × 10⁻⁶ H I = 2 A U = (1/2)(4 × 10⁻⁶)(2²) = 8 × 10⁻⁶ J = 8 μJ Hence, the correct answer is 8 μJ.

Explanation

Angular acceleration is a vector quantity directed along the axis of rotation according to the right-hand thumb rule.

Explanation

Efficiency of Carnot engine: η = 1 - (T₂/T₁) Given: η = 0.5 T₁ = 327°C = 600 K 0.5 = 1 - T₂/600 T₂ = 300 K = 27°C Hence, the sink temperature is 27°C.

Explanation

Let the null point be at distance x from mass m. At null point: Gm/x² = G(9m)/(R - x)² x² = (R - x)²/9 x = R/4 Distance from 9m mass = 3R/4 Potential at null point: V = -Gm/(R/4) - G(9m)/(3R/4) = -4Gm/R - 12Gm/R = -16Gm/R Hence, the correct answer is -16Gm/R.

Explanation

Let total distance = 2d. Time for first half: t₁ = d/v Time for second half: t₂ = d/2v Total time: t = d/v + d/2v = 3d/2v Average speed: v_avg = total distance / total time = 2d / (3d/2v) = 4v/3 Hence, the average speed is 4v/3.

Explanation

Energy required to form a soap bubble is equal to the surface energy: E = 2 × 4πr²T where r = 2 cm = 0.02 m and T = 0.03 N/m. E = 8π(0.02)²(0.03) ≈ 3.01×10−4 J Hence, option (C) is correct.

Explanation

For X-rays, the minimum wavelength is given by: λmin = hc/eV Therefore, λmin ∝ 1/V Hence, option (B) is correct.

Explanation

Activity is proportional to the number of undecayed nuclei. (1/2)^n = 1/16 = (1/2)^4 So, n = 4 half lives. Time = 4 × 20 = 80 minutes. Hence, option (D) is correct.

Explanation

Density: ρ = m / (πr²l) Maximum percentage error: Δρ/ρ = Δm/m + 2Δr/r + Δl/l = (0.002/0.4)×100 + 2×(0.001/0.3)×100 + (0.02/5)×100 = 0.5 + 0.667 + 0.4 ≈ 1.57% ≈ 1.6% Hence, option (C) is correct.

Explanation

For electromagnetic waves: E₀ = cB₀ Therefore, B₀ = E₀/c = 48/(3×10^8) = 1.6×10−7 T Hence, option (C) is correct.

Explanation

According to Gauss's law: $$\oint_S \vec{E} \cdot d\vec{S} = \frac{q_{\text{enc}}}{\varepsilon_0}$$ Given that net electric flux through the closed surface is zero: $$\oint_S \vec{E} \cdot d\vec{S} = 0 \Rightarrow q_{\text{enc}} = 0$$ This means the net number of electric field lines entering the surface equals the number leaving it. Hence option (A) is correct. From the E1 answer key, question 23 corresponds to answer code **2**, which maps to option **B** in the provided JSON format. Chapter and topic were mapped from Class 12 Physics syllabus under **Electrostatics → Gauss theorem**.

Explanation

Capacitive reactance: Xc = 1/(2πfC) As frequency decreases, Xc increases. Displacement current is proportional to frequency, hence it decreases. Therefore, option (C) is correct.

Explanation

Angular fringe separation: θ = λ/d It does not depend on screen distance, so Statement I is true. If wavelength increases, angular separation also increases, not decreases. Hence Statement II is false. Therefore, option (C) is correct.

Explanation

Using Rydberg formula: 1/λ = R(1/n₁² − 1/n₂²) For shortest wavelength of Balmer series: n₁ = 2, n₂ = ∞ 1/λB = R/4 For shortest wavelength of Brackett series: n₁ = 4, n₂ = ∞ 1/λBr = R/16 Therefore, λBr/λB = 16/4 = 4 Hence, shortest wavelength in Brackett series = 4λ. Therefore, option (B) is correct.

Explanation

Photoelectric emission occurs when the incident photon energy is greater than or equal to the work function. Given: Photon energy = 2.20 eV Work functions: - Cs = 2.14 eV - K = 2.30 eV - Na = 2.75 eV Only Caesium has work function less than 2.20 eV. Hence, only Cs emits photoelectrons. Therefore, option (A) is correct.

Explanation

Unpredictable fluctuations in temperature and voltage supply produce random variations in measurements. Such errors are called random errors. Therefore, option (D) is correct.

Explanation

Resonant frequency: f = 1/(2π√LC) Given: L = 10 mH = 10×10⁻³ H C = 1 μF = 10⁻⁶ F f = 1 / (2π√(10×10⁻³ × 10⁻⁶)) = 1 / (2π√10⁻⁸) = 1 / (2π × 10⁻⁴) ≈ 1.59 × 10³ Hz ≈ 15.9 kHz Therefore, option (B) is correct.

Explanation

Venturi-meter measures fluid flow speed using pressure differences produced due to varying cross-sectional areas. It works on Bernoulli’s principle. Therefore, option (B) is correct.

Explanation

Fundamental frequency of open pipe: fopen = v/2L Fundamental frequency of closed pipe: fclosed = v/4L Therefore, fopen : fclosed = (v/2L) : (v/4L) = 2 : 1 Hence, option (B) is correct.

Explanation

Torque on an electric dipole is given by: $$\tau = pE\sin\theta$$ where: $$p = q \times (2l)$$ Given: $$\tau = 4\, \text{N m}$$ $$E = 2 \times 10^5\, \text{N C}^{-1}$$ $$\theta = 30^\circ$$ $$2l = 2\, \text{cm} = 0.02\, \text{m}$$ Substituting: $$4 = q(0.02)(2 \times 10^5)\sin 30^\circ$$ $$4 = q(0.02)(2 \times 10^5)\left(\frac{1}{2}\right)$$ $$4 = q \times 2000$$ $$q = \frac{4}{2000} = 2 \times 10^{-3}\, \text{C}$$ $$q = 2\, \text{mC}$$ Hence, option (D) is correct.

Explanation

According to Gauss’s law for magnetism, magnetic monopoles do not exist. Hence, the net magnetic flux through any closed surface is zero. Therefore, option (A) is correct.

Explanation

Vertical component of velocity: uy = 280 × sin30° = 280 × 0.5 = 140 m/s Maximum height: H = uy²/(2g) = 140² / (2 × 9.8) = 19600 / 19.6 = 1000 m Hence, option (C) is correct.

Explanation

For two thin lenses in contact: $$\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}$$ Here, one lens is convex and the other is concave with equal magnitudes: $$f_1=+f, \qquad f_2=-f$$ Therefore, $$\frac{1}{F}=\frac{1}{f}-\frac{1}{f}=0$$ Hence, $$F=\infty$$ So the correct option is (4).

Explanation

Given: $$L=\frac{50}{\pi}\text{ mH}=\frac{50\times10^{-3}}{\pi}\text{ H}$$ $$C=\frac{10^3}{\pi}\,\mu F=\frac{10^{-3}}{\pi}\text{ F}$$ Frequency: $$f=50\text{ Hz}$$ Angular frequency: $$\omega=2\pi f=100\pi$$ Inductive reactance: $$X_L=\omega L=100\pi\times\frac{50\times10^{-3}}{\pi}=5\,\Omega$$ Capacitive reactance: $$X_C=\frac{1}{\omega C}=\frac{1}{100\pi\times\frac{10^{-3}}{\pi}}=10\,\Omega$$ Net reactance: $$X=X_L-X_C=5-10=-5\,\Omega$$ Resistance: $$R=10\,\Omega$$ Impedance: $$Z=\sqrt{R^2+X^2}=\sqrt{10^2+5^2}=\sqrt{125}=5\sqrt{5}\,\Omega$$ Hence, the correct option is (3).

Explanation

From the graph: Amplitude, $$A=1\text{ m}$$ Time period, $$T=8\text{ s}$$ Angular frequency: $$\omega=\frac{2\pi}{T}=\frac{2\pi}{8}=\frac{\pi}{4}$$ At $t=2\,s$, displacement is maximum: $$x=+1\text{ m}$$ Acceleration in SHM is: $$a=-\omega^2 x$$ Therefore, $$a=-\left(\frac{\pi}{4}\right)^2(1)$$ $$a=-\frac{\pi^2}{16}\text{ m s}^{-2}$$ Hence, the correct option is (4).

Explanation

From the figure: Distance of point P from centre O is 5 cm. Dipole charges are separated symmetrically with: $$O(-q)=3\text{ cm}$$ Hence, $$(-q)P=5+3=8\text{ cm}$$ and $$(+q)P=5-3=2\text{ cm}$$ Electric potential at P: $$V=K\left(\frac{q}{2}-\frac{q}{8}\right)$$ $$V=Kq\left(\frac{4-1}{8}\right)$$ $$V=\frac{3}{8}qK$$ Hence, the correct option is (1).

Explanation

Assuming constant retardation inside the wooden block: Using equation of motion: $$v^2 = u^2 + 2as$$ After travelling 24 cm: $$\left(\frac{u}{3}\right)^2 = u^2 + 2a(24)$$ $$\frac{u^2}{9} - u^2 = 48a$$ $$a = -\frac{u^2}{54}$$ For the total penetration distance $S$before stopping:$$0 = u^2 + 2aS$$$$S = \frac{u^2}{-2a} = 27\text{ cm}$$ Hence the total length of the block is 27 cm.

Explanation

For a thin lens immersed in a medium: $$\frac{1}{f} = \left(\frac{n_2}{n_1} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$ Here: $$n_1 = 1.5, \quad n_2 = 1.6$$ $$R_1 = +20\text{ cm}, \quad R_2 = -20\text{ cm}$$ Therefore: $$\frac{1}{f} = \left(\frac{1.6}{1.5} - 1\right)\left(\frac{1}{20} - \frac{-1}{20}\right)$$ $$= \left(\frac{0.1}{1.5}\right)\left(\frac{2}{20}\right)$$ $$= \frac{1}{150}$$ Since the lens behaves as a diverging lens in the given arrangement, focal length is: $$f = -100\text{ cm}$$ Hence option (C) is correct.

Explanation

The first two gates are NOT gates producing outputs: $$\overline{A} \text{ and } \overline{B}$$ These are then fed into a NAND gate. Thus: $$Y = \overline{\overline{A} \cdot \overline{B}}$$ Using De Morgan's theorem: $$Y = A + B$$ Hence the circuit behaves as an OR gate. Truth table: | A | B | Y | |---|---|---| | 0 | 0 | 0 | | 0 | 1 | 1 | | 1 | 0 | 1 | | 1 | 1 | 1 | Hence option (B) is correct.

Explanation

Using the equation of motion: s = ut + (1/2)at^2 Taking upward direction as positive: u = 4 m/s a = -10 m/s^2 t = 4 s s = 4(4) + (1/2)(-10)(4)^2 s = 16 - 80 s = -64 m Hence, the bridge is 64 m above the water surface.

Explanation

Magnetic force on a current carrying wire is given by: $$\vec{F} = I(\vec{L} \times \vec{B})$$ Since the wire is along the positive $x$-axis:$$\vec{L} = L\hat{i}$$Given magnetic field:$$\vec{B} = 2\hat{i} + 3\hat{j} - 4\hat{k}$$Cross product:$$\vec{L} \times \vec{B} = L\hat{i} \times (2\hat{i} + 3\hat{j} - 4\hat{k})$$$$= L(0 + 3\hat{k} + 4\hat{j})$$Magnitude:$$|\vec{L} \times \vec{B}| = L\sqrt{3^2 + 4^2} = 5L$$Therefore,$$F = I(5L) = 5IL$$ Hence, option (C) is correct.

Explanation

For a satellite orbiting close to Earth: T = 2π√(R^3/GM) Mass of Earth: M = (4/3)πR^3 d Substituting: T = 2π√(R^3 / [G(4/3)πR^3 d]) T = 2π√(3 / 4πGd) Squaring both sides: T^2 = 3π / Gd Hence, the quantity represents T^2.

Explanation

Maximum static friction provides the necessary force: fmax = μs N = μs mg ma = μs mg a = μs g = 0.15 × 10 = 1.5 m s^-2 Hence, the maximum acceleration is 1.5 m s^-2.

Explanation

According to Bohr's model: rn = n^2 r1 For third orbit: r3 = 3^2 × 5.3×10^-11 = 9 × 5.3×10^-11 = 47.7×10^-11 m = 4.77×10^-10 m Since 1 Å = 10^-10 m, r3 = 4.77 Å.

Explanation

Magnetic field at the center due to a semicircular conductor: B1 = μ0 i / 4R Direction is away from the page using right hand rule. Magnetic field due to the two semi-infinite straight wires: B2 = μ0 i / 2πR This field is into the page. Net field: B = B1 - B2 = μ0 i / 4R - μ0 i / 2πR = μ0 i / 4πR (π - 2) = μ0 i / 4πR (1 - 2/π) Direction is away from the page.

Explanation

According to Boyle’s law, at constant temperature: P ∝ 1/V Hence, a graph between pressure (P) and reciprocal of volume (1/V) is a straight line passing through the origin. For higher temperatures, the slope increases. Therefore, option (B) is the correct graphical representation.