Physics Mock Test 3 Practice

1. The ratio of SI unit to CGS unit of force is:

• A. 10⁵ (Correct)
• B. 10⁻⁵
• C. 10³
• D. 10⁻³

Explanation: 1 N = 1 kg·m·s⁻² and 1 dyne = 1 g·cm·s⁻². So 1 N / 1 dyne = (10³ g)(10² cm) / (g·cm) = 10⁵ .

2. A force of 100 N acts at 60° to the horizontal. Its vertical component is:

• A. 50 N
• B. 50√3 N (Correct)
• C. 100√3 N
• D. 100 N

Explanation: Vertical component = F sin60° = 100 × (√3/2) = 50√3 N .

3. A particle starts from rest and moves with acceleration a = 4 m/s². The velocity at the end of 5th second is:

• A. 4 m/s
• B. 20 m/s (Correct)
• C. 100 m/s
• D. 10 m/s

Explanation: v = u + at = 0 + 4×5 = 20 m/s .

4. Two ships sail at 90° to each other at 30 km/h and 40 km/h. The relative speed is:

• A. 10 km/h
• B. 70 km/h
• C. 50 km/h (Correct)
• D. 35 km/h

Explanation: Relative speed = √(30²+40²) = √2500 = 50 km/h .

5. A rocket ejects gases at 500 m/s. If fuel consumption rate is 2 kg/s, the thrust force is:

• A. 250 N
• B. 1000 N (Correct)
• C. 500 N
• D. 2500 N

Explanation: Thrust = v_exhaust × (dm/dt) = 500 × 2 = 1000 N .

6. For a freely falling body, total mechanical energy:

• A. Increases
• B. Decreases
• C. Remains constant (Correct)
• D. Becomes zero

Explanation: No non-conservative forces. Total ME = constant .

7. For a body rolling without slipping, velocity of contact point with ground:

• A. v
• B. 2v
• C. 0 (Correct)
• D. v/2

Explanation: In pure rolling, the contact point is instantaneously at rest. Velocity = 0 .

8. Kepler's third law: T² ∝ r³. If planet B has orbital radius 4 times planet A, T_B/T_A:

• A. 4
• B. 8 (Correct)
• C. 16
• D. 2

Explanation: T² ∝ r³. (T_B/T_A)² = (r_B/r_A)³ = 4³ = 64. T_B/T_A = √64 = 8 .

9. Relations between Y, B, G, and Poisson ratio ν: Y = 2G(1+ν). If ν = 0.5:

• A. Y = G
• B. Y = 2G
• C. Y = 3G (Correct)
• D. Y = 4G

Explanation: Y = 2G(1+ν) = 2G(1+0.5) = 2G×1.5 = 3G .

10. A wooden block (density 600 kg/m³) of volume 10⁻³ m³ is tied by a string to pool bottom (water ρ=1000, g=10). Tension in string:

• A. 6 N
• B. 4 N (Correct)
• C. 10 N
• D. 2 N

Explanation: F_B = 1000×10⁻³×10 = 10 N (up). Weight = 600×10⁻³×10 = 6 N (down). T = F_B − W = 10−6 = 4 N .

11. Two perpendicular vectors have magnitudes 5 and 6. Their resultant magnitude is approximately:

• A. 1
• B. 30.0
• C. 7.8 (Correct)
• D. 11

Explanation: For perpendicular vectors, R = sqrt(5^2 + 6^2) = 7.8.

12. If the error in measurement of radius of a sphere is 2%, the error in the measurement of its volume will be:

• A. 2%
• B. 4%
• C. 6% (Correct)
• D. 8%

Explanation: Volume V = (4/3)πr³. Relative error in V = 3 × (relative error in r) = 3 × 2% = 6% .

13. If displacement s = 5t² + 3t + 2, the velocity at t = 2 s is:

• A. 10 m/s
• B. 20 m/s
• C. 23 m/s (Correct)
• D. 13 m/s

Explanation: v = ds/dt = 10t + 3. At t = 2: v = 10(2) + 3 = 23 m/s .

14. Two trains (200 m and 300 m long) run on parallel tracks in opposite directions at 40 m/s and 60 m/s. Time to completely cross each other is:

• A. 5 s (Correct)
• B. 10 s
• C. 1 s
• D. 2 s

Explanation: Relative speed = 40 + 60 = 100 m/s. Total length = 500 m. Time = 500/100 = 5 s .

15. The banking angle θ for a road of radius r at speed v satisfies:

• A. tanθ = rg/v²
• B. tanθ = v²/rg (Correct)
• C. sinθ = v²/rg
• D. cosθ = v²/rg

Explanation: For a banked curve: tanθ = v²/rg.

16. A 2 kg block on frictionless incline (angle 30°). Acceleration along plane (g = 10 m/s²):

• A. 5 m/s² (Correct)
• B. 10 m/s²
• C. 5√3 m/s²
• D. 10√3 m/s²

Explanation: a = g sinθ = 10 × sin30° = 10 × 0.5 = 5 m/s² .

17. 0.5 kg ball thrown upward at 10 m/s. Max height (g = 10 m/s²):

• A. 5 m (Correct)
• B. 10 m
• C. 2.5 m
• D. 20 m

Explanation: h = v²/2g = 100/20 = 5 m .

18. For rolling without slipping: relation between v_cm and ω:

• A. v = ω/R
• B. v = ωR (Correct)
• C. v = ω²R
• D. v = R/ω

Explanation: Rolling condition: v_cm = ωR .

19. Time period of a satellite at radius r:

• A. 2π√(r³/GM) (Correct)
• B. 2π√(r/GM)
• C. √(r³/GM)
• D. 2πr/GM

Explanation: T = 2π√(r³/GM) = 2π√(r³/GM) . (From Kepler's 3rd law.)

20. Rigidity modulus is the modulus associated with:

• A. Stretching
• B. Compression
• C. Shearing (Correct)
• D. Bending

Explanation: Rigidity modulus (G) is associated with shearing (angular deformation) .

21. The line of action of buoyant force acts through:

• A. Centre of mass of body
• B. Centre of buoyancy (COM of displaced fluid) (Correct)
• C. Bottom of body
• D. Top of body

Explanation: Buoyant force acts through centre of buoyancy = COM of displaced fluid.

22. In a NEET physics drill, a particle starts with speed 11 m/s and acceleration 2 m/s^2 for 3 s. Final speed is:

• A. 33 m/s
• B. 17 m/s (Correct)
• C. 15 m/s
• D. 19 m/s

Explanation: Use v = u + at = 11 + 2 x 3 = 17 m/s.

23. The percentage error in the measurement of mass is 1% and in length 2%. The maximum percentage error in the measurement of density (mass/volume, where volume = l³) is:

• A. 1%
• B. 5%
• C. 7% (Correct)
• D. 9%

Explanation: Density ρ = m/V = m/l³. % error in ρ = % error in m + 3 × % error in l = 1% + 3×2% = 7% .

24. If v = 4t³ − 3t² + 2, the acceleration at t = 1 s is:

• A. 6 m/s² (Correct)
• B. 12 m/s²
• C. 3 m/s²
• D. 9 m/s²

Explanation: a = dv/dt = 12t² − 6t. At t = 1: a = 12 − 6 = 6 m/s² .

25. A stone dropped from the top of a building takes 4 seconds to reach the ground (g = 10 m/s²). The height of the building is:

• A. 40 m
• B. 80 m (Correct)
• C. 160 m
• D. 20 m

Explanation: h = ut + ½gt² = 0 + ½(10)(16) = 80 m .

26. Minimum speed at the top of a vertical loop of radius r for maintaining contact is:

• A. √(rg) (Correct)
• B. √(2rg)
• C. √(rg/2)
• D. 2√(rg)

Explanation: At top of loop: mg = mv²/r (minimum condition). v_min = √(rg) .

27. If the normal force doubles while friction coefficient stays same, the friction force:

• A. Remains same
• B. Doubles (Correct)
• C. Halves
• D. Quadruples

Explanation: f = μN. If N doubles: f_new = μ(2N) = 2μN = twice the original.

28. Pendulum (L = 1 m) released from 60°. Speed at lowest point (g = 10 m/s²):

• A. √10 m/s (Correct)
• B. √20 m/s
• C. √5 m/s
• D. 10 m/s

Explanation: h = L(1−cos60°) = 0.5 m. v = √(2×10×0.5) = √10 m/s .

29. Velocity of topmost point of a rolling sphere (v_cm = v):

• A. v
• B. 2v (Correct)
• C. 0
• D. v/2

Explanation: Top point: v_top = v_cm + ωR = v + v = 2v .

30. Period of satellite at height h = R above Earth surface (g, R given):

• A. 2π√(R/g)
• B. 2π√(8R/g) (Correct)
• C. 2π√(4R/g)
• D. 2π√(2R/g)

Explanation: r = R+h = 2R. T = 2π√(r³/GM) = 2π√(8R³/gR²) = 2π√(8R/g) = 2π√(8R/g) .

31. Incompressible material has bulk modulus:

• A. Zero
• B. Finite small value
• C. Infinite (∞) (Correct)
• D. Equal to Y

Explanation: Incompressible means ΔV = 0 for any pressure. B = P/(ΔV/V) → ∞ .

32. Specific gravity of a substance is equal to:

• A. Ratio of its mass to volume
• B. Ratio of its density to water density at 4°C (Correct)
• C. Weight in water
• D. Relative viscosity

Explanation: Specific gravity = density of substance / density of water at 4°C .

33. A block of mass 2 kg accelerates at 3 m/s^2. Net force on it is:

• A. 6 N (Correct)
• B. 9 N
• C. 3 N
• D. 12 N

Explanation: By Newton's second law, F = ma = 2 x 3 = 6 N.

34. By dimensional analysis, the period T of a simple pendulum depends on length l and g. If T = k·lᵃ·gᵇ, then a and b are:

• A. a=1/2, b=−1/2 (Correct)
• B. a=1, b=−1
• C. a=−1/2, b=1/2
• D. a=1/2, b=1/2

Explanation: Dimensionally [T] = [L]ᵃ[LT⁻²]ᵇ gives: T¹ implies 1 = −2b, so b = −1/2; L⁰ implies 0 = a + b, so a = 1/2. Therefore T = k√(l/g).

35. Vectors A = 3î + 4ĵ and B = 1î + 2ĵ. The magnitude of A + B is:

• A. √52 (Correct)
• B. 5
• C. √40
• D. √68

Explanation: A + B = 4î + 6ĵ. Magnitude = √(16+36) = √52 = 2√13.

36. If the position-time graph of a particle is a curve (not straight line), the particle has:

• A. Zero velocity
• B. Uniform velocity
• C. Non-uniform velocity (Correct)
• D. Zero acceleration

Explanation: A curve on the x-t graph means the slope (velocity) is changing, implying non-uniform velocity and acceleration.

37. A swimmer crosses a 100 m wide river (current 5 m/s) by swimming at 10 m/s perpendicular to the bank. Time to cross is:

• A. 10 s (Correct)
• B. 20 s
• C. 5 s
• D. 7 s

Explanation: Time = width / swimmer's speed = 100/10 = 10 s . Current doesn't affect crossing time.

38. The tension in the rope holding a 5 kg block at rest (g = 10 m/s²) is:

• A. 50 N (Correct)
• B. 5 N
• C. 500 N
• D. 0 N

Explanation: Equilibrium: T = mg = 5×10 = 50 N .

39. 10 kg mass falls 10 m. KE just before impact (g = 10 m/s²):

• A. 1000 J (Correct)
• B. 100 J
• C. 500 J
• D. 2000 J

Explanation: KE = mgh = 10×10×10 = 1000 J .

40. A planet moves in elliptical orbit. Area swept per unit time is constant because:

• A. Speed is constant
• B. Angular momentum is conserved
• C. Force is central
• D. Both B and C (Correct)

Explanation: Kepler's 2nd law arises because gravity is a central force → no torque → angular momentum conserved → both B and C .

41. Geostationary satellite orbits at approximately:

• A. 200 km above Earth
• B. 36,000 km above equator (Correct)
• C. 400 km above poles
• D. Same as Moon orbit

Explanation: Geostationary orbit ≈ 36,000 km above the equator . Period = 24 hours.

42. Relation: Y = 3B(1−2ν). For ν = 1/3:

• A. Y = B (Correct)
• B. Y = 2B
• C. Y = 3B/3 = B
• D. Y = 3B×1/3 = B

Explanation: Y = 3B(1−2/3) = 3B×1/3 = B .

43. Submarine dives by flooding ballast tanks with seawater. This works by:

• A. Increasing thrust
• B. Increasing average density until it exceeds seawater (Correct)
• C. Reducing buoyancy force
• D. Pascal's law

Explanation: Flooding ballast tanks increases average density above seawater → submarine sinks.

44. A spring of constant 140 N/m is compressed by 0.1 m. Energy stored is:

• A. 1.4 J
• B. 0.35 J
• C. 14 J
• D. 0.7 J (Correct)

Explanation: Spring energy = (1/2)kx^2 = 0.5 x 140 x 0.01 = 0.7 J.

45. Which physical quantity has the same dimensions as Planck's constant h?

• A. Angular momentum (Correct)
• B. Linear momentum
• C. Energy
• D. Power

Explanation: Planck's constant h = E/f has dimensions [ML²T⁻¹]. Angular momentum L = mvr also has dimensions [ML²T⁻¹]. They are dimensionally identical.

46. The work done by force F = 3î + 4ĵ N through displacement s = 2î + 5ĵ m is:

• A. 26 J (Correct)
• B. 22 J
• C. 14 J
• D. 6 J

Explanation: W = F·s = (3)(2) + (4)(5) = 6 + 20 = 26 J .

47. A train running at 72 km/h is stopped by brakes in 8 seconds. The retardation is:

• A. 9 m/s²
• B. 8 m/s²
• C. 2.5 m/s² (Correct)
• D. 5 m/s²

Explanation: u = 72 km/h = 20 m/s, v = 0, t = 8 s. a = (0−20)/8 = −2.5 m/s². Retardation = 2.5 m/s² .

48. At what angle should a body be projected so the horizontal range equals the maximum height?

• A. tan⁻¹(1)
• B. tan⁻¹(2)
• C. tan⁻¹(4) (Correct)
• D. tan⁻¹(0.5)

Explanation: R = H → 4cosθ = sinθ → tanθ = 4. θ = tan⁻¹(4) .

49. A 10 kg block on rough incline (angle 30°, μₖ = 0.2) slides down. Acceleration (g = 10 m/s²) is:

• A. 3 m/s²
• B. 5 m/s²
• C. 3.27 m/s² (Correct)
• D. 2 m/s²

Explanation: a = g(sinθ − μₖcosθ) = 10(0.5 − 0.2×0.866) = 10(0.5 − 0.173) = 3.27 m/s² .

50. Two springs (k₁=200, k₂=300 N/m) in series, total stretch 0.3 m. PE stored:

• A. 5.4 J (Correct)
• B. 10.8 J
• C. 2.7 J
• D. 7.2 J

Explanation: k_eq = k₁k₂/(k₁+k₂) = 120 N/m. PE = ½×120×0.09 = 5.4 J .