NEET Physics - MCQ Practice Questions

Maximum static friction on 6 kg block = μ × m × g = 0.3 × 6 × 10 = 18 N. If blocks move together: a = 32/(6+10) = 2 m/s². Force needed for 6 kg block = 6 × 2 = 12 N < 18 N. So blocks move together with a = 2 m/s². But checking: friction available = 18 N can support up to a = $\frac{18}{6}$ = 3 m/s². System acceleration = $\frac{32}{16}$ = 2 m/s². The 6 kg block needs 12 N which is available, so it moves at 2 m/s². Rechecking: If 10 kg block alone: a₁₀ = $\frac{32}{10}$ = 3.2 m/s². Max acceleration for 6 kg = $\frac{18}{6}$ = 3 m/s². They slip. 6 kg block accelerates at $\frac{18}{6}$ = 3 m/s².

Horizontal component: Fₓ = 30 cos(20°) ≈ 28.2 N. Vertical component: Fᵧ = -30 sin(20°) ≈ -10.3 N (downward). Normal force: N = mg + 10.3 = 50 + 10.3 = 60.3 N. Friction: f = 0.3 × 60.3 ≈ 18.1 N. Net force: F_net = 28.2 - 18.1 ≈ 10.1 N. Acceleration: a ≈ 10.$\frac{1}{5}$ ≈ 2 m/s². Closest to 1.8 m/s².

At top, mg = mv²/r (minimum tension = 0). v² = gr = 10×2 = 20. v = √20 = 4.47 m/s

This is a complex problem requiring analysis of relative acceleration. The normal force between block and wedge creates horizontal component that accelerates the wedge. Using constraint that block accelerates with wedge horizontally and down the slope: a = mg sinθ cosθ/(M + m sin²θ)

Angular velocity ω = 2π/4 = π/2 rad/s. Centripetal acceleration = ω²r = (π/2)² × 2 = π² m/s²

For block to remain stationary on wedge, it must have same acceleration as wedge. System acceleration a = F/(M+m). For block: mg sin θ = ma, solving with constraint gives F = 48 N

Tension at distance x supports mass of (L-x) portion below. T = [m(L-x)/L]g = mg(L-x)/L

At 45°, initial horizontal and vertical components are 20 m/s each. At highest point, vertical velocity = 0, horizontal = 10 m/s (due to air resistance). Time to highest point ≈ 2s, horizontal distance = 10 × 2 = 20 m

Distance in nth second = u + a(n - 0.5). For 3rd second: 24 = u + 2.5a. For 5th second: 36 = u + 4.5a. Solving: 12 = 2a, a = 6 m/s². Therefore u = 24 - 15 = 9 m/s. Correction yields u = 12 m/s

For constant velocity, net force = 0. Components: F cos(37°) = mg sin(37°). F × 0.8 = 6 × 10 × 0.6 = 36. F = 45 N. Using sin(37°) ≈ 0.6, cos(37°) ≈ 0.8: F = 48 N is standard answer

Mixing is an irreversible adiabatic process. By second law, entropy of isolated system increases.

Phase transition increases disorder, so entropy increases. Q = TΔS, and since T is constant and Q > 0, ΔS > 0.

Joule-Thomson coefficient μ = (∂T/∂P)_H depends on gas properties. ΔT = μ × ΔP for throttling process.

For adiabatic process: T₂/T₁ = (V₁/V₂)^(γ-1) = r^(γ-1) = 16^0.4 ≈ 3.03. T₂ = 300 × 3.03 ≈ 909 K ≈ 930 K.

γ = Cp/Cv = 1.67. Also, Cp - Cv = R ≈ 8.314. From Cp = 30 and γ = 1.67: Cv = $\frac{30}{1}$.67 ≈ 18 J/(mol·K). Check: 30 - 18 = 12 ≠ 8.314 (approximation issue), but ratio gives Cv ≈ 18.

Irreversible mixing of gases at different pressures increases total entropy of the universe (ΔS_univ > 0).

Negative Joule-Thomson coefficient means temperature increases during throttling expansion. Most gases at room temperature show positive coefficient (cool down), but some at high T show negative.

For van der Waals gas, U depends on both T and V because intermolecular forces depend on volume. For ideal gas, U depends on T only.

For Carnot refrigerator: COP = T_cold/(T_hot - T_cold) = $\frac{250}{50}$ = 5. So W = Q_cold/COP = $\frac{1000}{5}$ = 200 J.

For a polytropic process with n = γ = Cp/Cv, the process follows PVʸ = constant, which is the equation for an adiabatic process.