NEET 2026 Physics Questions with Solutions

Explanation

Speed of light in the new unit system is taken as 1. Time taken = 6 min 40 s = 6×60 + 40 = 400 s Distance = speed × time Since v = 1, d = 1 × 400 = 400 Hence the correct answer is option (D).

Explanation

Young’s modulus = Stress/Strain = FL/[A(ΔL)] → II Compressibility = 1/(Bulk modulus) = -(1/ΔP)(ΔV/V) → III Bulk modulus = -P(V/ΔV) → IV Poisson’s ratio = Lateral strain/Longitudinal strain = Δd/ΔL × (L/d) → I Hence the correct matching is A-II, B-III, C-IV, D-I (Option D).

Explanation

For ideal diodes: forward resistance = 0 and reverse resistance = $\infty$. Only the branches containing $4\Omega$ and $2\Omega$ conduct. Hence the equivalent circuit has two parallel resistors across $10V$. Current: $$I = \frac{10}{4} + \frac{10}{2} = \frac{5}{2} + 5 = \frac{15}{2}\,A$$ Therefore, the correct answer is $\boxed{\frac{15}{2}\,A}$.

Explanation

Convert angular speeds into SI units: $$\omega_1 = 600\times\frac{2\pi}{60}=20\pi\,\text{rad/s}$$ $$\omega_2 = 1200\times\frac{2\pi}{60}=40\pi\,\text{rad/s}$$ Angular acceleration: $$\alpha=\frac{\omega_2-\omega_1}{t} =\frac{40\pi-20\pi}{10} =2\pi\,\text{rad/s}^2$$ Using rotational equation of motion: $$\omega_2^2=\omega_1^2+2\alpha\theta$$ Substitute values: $$(40\pi)^2=(20\pi)^2+2(2\pi)\theta$$ $$1600\pi^2=400\pi^2+4\pi\theta$$ $$1200\pi^2=4\pi\theta$$ $$\theta=300\pi\,\text{rad}$$ Number of revolutions: $$N=\frac{\theta}{2\pi} =\frac{300\pi}{2\pi} =150$$ Therefore, the correct answer is: $$\boxed{150}$$

Explanation

For SHM, velocity is: $$v=A\omega\cos(\omega t+\phi)$$ Hence kinetic energy: $$K=\frac12 mv^2$$ $$K=\frac12 mA^2\omega^2\cos^2(\omega t+\phi)$$ Thus, $$K\propto \cos^2(\omega t+\phi)$$ Since $\cos^2\theta$ is always non-negative and has period $T/2$, kinetic energy becomes zero twice in one oscillation and reaches maximum twice. Therefore the correct variation is **Option (C)**.

Explanation

Terminal voltage of a battery is: V = E - ir Substituting values: V = 12 - (0.6 × 2) V = 12 - 1.2 = 10.8 V Hence, option (D) is correct.

Explanation

For gases at the same temperature: Thus, $v_{rms} \propto \dfrac{1}{\sqrt{M}}$ Therefore, Substituting: Hence, $\dfrac{v_{rms}^{Ar}}{v_{rms}^{Cl}}=\dfrac{\sqrt7}{2}$ Therefore, option (A) is correct.

Explanation

Since the prism is equilateral, prism angle A = 60°. The refracted ray inside the prism is parallel to the base, hence the path is symmetric and: i = e = 50° Using prism relation: Substituting: Therefore: Hence, option (C) is correct.

Explanation

["A. E = hν → Energy of photon → IV","B. Diffraction and interference confirm wave nature of light → III","C. λ = h/p represents de Broglie wavelength → I","D. Compton effect confirms particle nature of light → II","","Correct matching:"]

Explanation

For hydrogen atom: Total energy, E = -k e² / (2r) Magnitude: |E| = k e² / (2r) Substitute: 3.4 eV = (9×10⁹ × (1.6×10⁻¹⁹)²)/(2r) Converting eV to joule: 3.4 × 1.6×10⁻¹⁹ = (9×10⁹ × (1.6×10⁻¹⁹)²)/(2r) r = (9×10⁹ × 1.6×10⁻¹⁹)/(2×3.4) r = 2.12 × 10⁻¹⁰ m ≈ 2.1 × 10⁻¹⁰ m Hence option (C) is correct.

Explanation

For the box to remain at rest relative to the trolley, static friction must provide the required acceleration. Maximum static friction: fₘₐₓ = μₛN Since: N = mg Maximum acceleration occurs when: ma = μₛmg Therefore: a = μₛg Substitute: μₛ = 0.12, g = 10 m/s² ⇒ a = 0.12 × 10 ⇒ a = 1.2 m/s² Hence option (D) is correct.

Explanation

Capacitors C₁, C₂, C₃ and C₄ (each 10 μF) are connected in series. Equivalent capacitance of series combination: 1/C = 1/10 + 1/10 + 1/10 + 1/10 ⇒ C = 2.5 μF This 2.5 μF combination is in parallel with C₅ = 2.5 μF. Therefore: Ceq = 2.5 + 2.5 = 5 μF Charge on equivalent branch: Q = CV = 2.5 × 50 Q = 125 μC For capacitors in series, charge remains same: q₁ = q₂ = q₃ = q₄ = 125 μC For capacitor C₅: q₅ = 2.5 × 50 = 125 μC Hence all capacitors carry 125 μC and equivalent capacitance is 5 μF. Therefore option (B) is correct.

Explanation

Work done equals change in gravitational potential energy: W = U₂ − U₁ Initial position: at Earth's surface (r = R) U₁ = −GMm/R Final position: height = R above surface Thus final distance from Earth's center: r = 2R U₂ = −GMm/(2R) Therefore: W = (−GMm/2R) − (−GMm/R) W = GMm/2R Using: GM/R² = g ⇒ GM = gR² Substitute: W = (gR²m)/(2R) W = mgR/2 Hence option (D) is correct.

Explanation

Density: ρ = Mass / Volume Given: Mass = 5.580 kg Side = 9.0 cm = 9.0 × 10⁻² m Volume: V = (9.0 × 10⁻²)³ V = 729 × 10⁻⁶ m³ Density: ρ = 5.580 / (729 × 10⁻⁶) ρ = 7.654 × 10³ kg m⁻³ Applying significant figures: Mass has 4 significant figures, Side measurement (9.0 cm) has 2 significant figures. Therefore final answer must have 2 significant figures. Thus: ρ ≈ 7.7 × 10³ kg m⁻³ Hence X = 7.7 and option (D) is correct.

Explanation

For a ball thrown vertically upward and returning under gravity, acceleration remains constant and downward throughout the motion: Acceleration = −g Therefore the slope of the velocity-time graph must remain constant and negative during the entire journey. Hence velocity decreases linearly with time: - Positive during upward motion - Zero at highest point - Negative during downward motion Thus the correct graph is a straight line with constant negative slope crossing the time axis. Only graph C satisfies this condition. Hence option (A) is correct.

Explanation

Total mechanical energy of the pendulum remains constant: E = KE + PE = 0.02 J At equilibrium position: Potential energy = 0 Therefore: Total energy = Kinetic energy ½mv² = 0.02 Given: m = 20 g = 0.02 kg Substitute: ½ × 0.02 × v² = 0.02 0.01v² = 0.02 v² = 2 v = √2 v ≈ 1.41 m/s Hence option (B) is correct.

Explanation

Intensity in YDSE is given by: I = I₀ cos²(φ/2) where phase difference: φ = (2π/λ)Δx For path difference Δx = λ: φ = 2π Therefore: I = I₀ cos²(π) I = I₀ Given: K = I₀ Now for path difference: Δx = λ/3 Phase difference: φ = (2π/λ)(λ/3) φ = 2π/3 Hence: I = I₀ cos²(π/3) I = I₀(1/2)² I = I₀/4 Since K = I₀: I = K/4 Hence option (A) is correct.

Explanation

The diode becomes reverse biased during the positive half cycle of the input AC signal. When reverse biased: - Current does not flow through the circuit. - Entire input voltage appears across the diode. During the negative half cycle: - Diode becomes forward biased. - Voltage drop across an ideal diode becomes approximately zero. Therefore, voltage across the diode exists only during one half cycle and becomes zero during the other half cycle. Hence the waveform across the diode is a positive half-wave followed by zero. Thus option (D) is correct.

Explanation

For a series LCR circuit, resonance frequency is: Given: L = 1 mH = 10⁻³ H C = 0.1 μF = 10⁻⁷ F Substitute: f₀ = 1 / [2π√(10⁻³ × 10⁻⁷)] = 1 / [2π × 10⁻⁵] ≈ 1.59 × 10⁴ Hz = 15.9 kHz Hence option (D) is correct.

Explanation

In interference and diffraction, energy is redistributed and conserved, so statement A is true. Interference and diffraction occur in all wave phenomena including sound waves, hence statement B is false. Therefore, option (C) is correct.

Explanation

Given y(x,t)=2.0 cos 2π(10t−0.0080x+0.35). Wave number k=2π×0.008. For Δx=0.5 m=50 cm, phase difference Δϕ=kΔx=2π×0.008×50=0.8π rad. Hence option (B) is correct.

Explanation

Resultant force: F = √(8²+6²)=10 N. Acceleration: a=F/m=10/5=2 m s⁻². Direction with respect to 8 N force: tanθ=6/8=3/4, therefore θ=tan⁻¹(3/4). Hence option (D) is correct.

Explanation

Energy lost = (1/2)[C1C2/(C1+C2)]V². For C1=C2=200 pF and V=100 V: Energy lost = (1/2)(200×200/400)×10⁻¹²×100² = 0.5×10⁻⁶ J. Hence option (A) is correct.

Explanation

Power = Work/Time = mgh/t = (1000×9.8×20)/10 = 19600 W = 19.6 kW. Hence option (C) is correct.

Explanation

20 VSD = 16 MSD ⇒ 1 VSD = 16/20 MSD. Least count = 1 MSD − 1 VSD = 1 − 16/20 = 4/20 mm = 0.2 mm = 0.02 cm. Hence option (C) is correct.

Explanation

Distance covered in free fall is proportional to t². Hence larger reaction time gives larger distance. Ordering reaction times: 0.22 > 0.21 > 0.20 > 0.19 > 0.18 ⇒ B > E > A > D > C. Therefore option (C) is correct.

Explanation

Total wire resistance = 4 Ω, so each side has 1 Ω. The network forms a balanced Wheatstone bridge, hence no current flows through the 2 Ω diagonal resistor. Equivalent resistance across A and C becomes 1 Ω. Thus current I = V/R = 2/1 = 2 A. Therefore option (A) is correct.

Explanation

For a heater of constant resistance: P ∝ V². Therefore, P=(200/220)²×400 ≈ 331 W. Hence option (C) is correct.

Explanation

Magnetic field at centre of coil: B=μ₀Ni/2R. Therefore i=(2RB)/(μ₀N)=2.5 A. Magnetic moment M=NiA=100×2.5×π×(0.05)²≈2 A m². Hence option (D) is correct.

Explanation

Induced emf ε = Bℓv. Here B=0.3 T, ℓ=3×10⁻² m (shorter side), v=2×10⁻² m s⁻¹. Therefore ε=0.3×3×10⁻²×2×10⁻²=1.8×10⁻⁴ V. Hence option (D) is correct.

Explanation

Nuclear radius r=r₀A^(1/3). Hence volume V∝r³∝A, so statement B is true and A is false. Mass defect is the difference between the mass of a nucleus and the sum of masses of its constituent nucleons, so D is true and C is false. Hence option (D) is correct.

Explanation

Using R = R₀A^(1/3) and volume=(4/3)πR³=M/ρ, solving gives A≈12. Hence option (A) is correct.

Explanation

Time period T=60/30=2 s. Using T=2π√(l/g), l=gT²/(4π²)=1 m. Hence option (D) is correct.

Explanation

Using first law of thermodynamics: Q=ΔU+W ⇒ 100=ΔU+75 ⇒ ΔU=25 W. Hence option (D) is correct.

Explanation

Mass of ring M=mL. Radius r=L/(2π). Using parallel axis theorem: Iyy′=ICM+Mr²=(Mr²/2)+Mr²=(3/2)Mr²=(3/2)(mL)(L/2π)²=3mL³/(8π²). Hence option (B) is correct.

Explanation

Galvanometer current Ig=1 mA=0.001 A, resistance Rg=100 Ω. Shunt current Is≈10 A. Since IgRg=IsRs, Rs=(0.001×100)/10=0.01 Ω. Hence option (A) is correct.

Explanation

Interchanging the cell and galvanometer does not affect the balance condition of the metre bridge. At balance point, galvanometer shows no deflection; otherwise, deflection can be either side. Hence option (D) is correct.

Explanation

For AC: i=I₀sin(ωt), where ω=2πf=120π rad/s. Peak occurs at sin(ωt)=1 ⇒ ωt=π/2. Therefore t=(π/2)/(120π)=1/240 s. Hence option (D) is correct.

Explanation

Inside the conductor (r<a), magnetic field B=(μ₀Ir)/(2πa²), therefore B∝r. Outside the conductor (r>a), B=μ₀I/(2πr), therefore B∝1/r. Hence magnetic field increases linearly inside and decreases inversely outside, matching graph (1).

Explanation

Above threshold voltage in forward bias, diode current rises sharply, so statement A is true. This current is forward current, not reverse saturation current. Hence statement B is false. Therefore option (B) is correct.

Explanation

A is true: electric field inside a conductor is zero. B is false: E=σ/ε₀ depends on surface charge density. C is true: excess charge resides only on surface. D is true: electric field is normal to surface. E is false: potential inside conductor is constant, not necessarily zero. Hence option (C) is correct.

Explanation

Threshold wavelength λ₀=hc/W₀=(6.6×10⁻³⁴×3×10⁸)/(6.6×1.6×10⁻¹⁹)≈187.5 nm. Photoelectric effect occurs only for λ<187.5 nm. Therefore 200 nm does not produce photoelectric emission. Hence option (C) is correct.

Explanation

For a concave lens, a ray incident parallel to the principal axis diverges after refraction and appears to come from the first principal focus. Hence option (B) is correct.

Explanation

Absolute pressure: P=P₀+ρgh. Therefore 100×10⁵=10⁵+1000×10×h ⇒ h=990 m. Hence option (A) is correct.

Explanation

Microwaves → klystron/magnetron (IV); Visible light → electron transitions (I); Gamma rays → radioactive decay (II); Infrared → molecular vibrations (III). Hence option (C) is correct.

Explanation

Energy converted to light per second = 150×0.08=12 J. Number of photons = 12/(4.42×10⁻¹⁹)=2.71×10¹⁹ photons/s. Hence option (A) is correct.

Explanation

Using first law: ΔU=q+w. Here q=+500 J and w=−200 J (work done by system). Therefore ΔU=500−200=300 J. Hence option (B) is correct.