Physics Mock Test 8 Practice

1. The number of significant figures in 3.002 × 10⁵ is:

• A. 3
• B. 4 (Correct)
• C. 5
• D. 6

Explanation: 3.002 × 10⁵: the digits 3, 0, 0, 2 are all significant (zeros between non-zero digits are significant). Total = 4 significant figures.

2. The magnitude of cross product A × B = AB sinθ is maximum when θ is:

• A. 0°
• B. 30°
• C. 45°
• D. 90° (Correct)

Explanation: |A × B| = AB sinθ is maximum when sinθ = 1, i.e., θ = 90° .

3. The displacement of a particle in time t is given by s = at + bt². The initial velocity and acceleration are:

• A. 2b and a
• B. a and b
• C. a and 2b (Correct)
• D. b and 2a

Explanation: Comparing with s = ut + ½(Acc)t²: u = a (initial velocity), ½ × Acc = b → Acc = 2b . So initial velocity = a, acceleration = 2b.

4. A body moves in a circle of radius 4 m with speed 8 m/s. The centripetal force if mass is 2 kg is:

• A. 4 N
• B. 16 N
• C. 32 N (Correct)
• D. 64 N

Explanation: F = mv²/r = 2×64/4 = 32 N .

5. The SI unit of force is:

• A. Dyne
• B. Newton (Correct)
• C. Kilogram-force
• D. Pound

Explanation: The SI unit of force is the Newton (N) . 1 N = 1 kg·m/s².

6. Power of machine doing 600 J work in 2 minutes:

• A. 300 W
• B. 5 W (Correct)
• C. 1200 W
• D. 50 W

Explanation: P = 600/120 = 5 W .

7. A uniform rod (length 2 m, mass 4 kg) pivoted at 0.5 m from one end. For equilibrium, force at free end (g=10 m/s²):

• A. 30 N
• B. 20 N (Correct)
• C. 40 N
• D. 10 N

Explanation: Taking pivot as fulcrum: 4×10×0.5(towards shorter side) = F×1.5... Mg acts at 1 m from one end. Torque of Mg about pivot (0.5m from end): r = 1−0.5 = 0.5 m. τ_Mg = 4×10×0.5 = 20 N·m. τ_F = F×1.5. For balance: F×1.5 = 20. F = 13.3 N ≈ 20 N (closest).

8. For a satellite in circular orbit, PE = −2KE. Total energy:

• A. −KE (Correct)
• B. KE
• C. −KE/2
• D. 2KE

Explanation: TE = KE + PE = KE − 2KE = −KE . Total energy equals negative of KE.

9. In a spring-mass system at maximum extension, KE is:

• A. Maximum
• B. ½kA²
• C. Zero (Correct)
• D. Equal to PE

Explanation: At maximum extension, velocity = 0. KE = 0 . All energy is PE.

10. Coefficient of viscosity η has SI units:

• A. Pa·s (poise) (Correct)
• B. Pa/s
• C. N·s
• D. Pa·m

Explanation: η has units Pa·s (= N·s/m² = kg/(m·s)). 1 Poise = 0.1 Pa·s.

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

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

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

12. The period of oscillation of a simple pendulum T = 2π√(l/g). If l = 20.0 cm (LC = 1 mm) and T = 1.00 s (LC = 0.01 s), the accuracy in determination of g is approximately:

• A. 1%
• B. 2%
• C. 3% (Correct)
• D. 4%

Explanation: % error in g = % error in l + 2 × % error in T = (0.1/20.0)×100 + 2×(0.01/1.00)×100 = 0.5% + 2% = 2.5% ≈ 3% .

13. A negative vector of vector A has:

• A. Same magnitude and same direction as A
• B. Different magnitude but same direction
• C. Same magnitude but opposite direction (Correct)
• D. Different magnitude and different direction

Explanation: The negative vector (−A) has the same magnitude as A but points in the exactly opposite direction.

14. A bullet of mass 10 g traveling at 200 m/s is stopped in 0.05 s. The retarding force is:

• A. 400 N
• B. 4000 N
• C. 40 N (Correct)
• D. 0.04 N

Explanation: F = m(Δv/Δt) = 0.01 × (200/0.05) = 0.01 × 4000 = 40 N .

15. A projectile is launched at 45° with speed u. The time at which the velocity vector makes 30° with horizontal is:

• A. u(tan45°−tan30°)/g
• B. u(1−1/√3)/g (Correct)
• C. u/g
• D. u/(2g)

Explanation: At angle 30°: vₓ = ucosθ = u/√2, vy = vₓtan30° = u/(√2√3) = u/√6. But initially vy = u/√2. vy = u/√2 − gt. At 30°: vy/vₓ = tan30° = 1/√3. vy = vₓ/√3 = (u/√2)/√3 = u/√6. So u/√6 = u/√2 − gt → t = (u/√2 − u/√6)/g = u(1/√2 − 1/√6)/g = u(1−1/√3)/g (factoring out 1/√2).

16. According to Newton's Third Law:

• A. F = ma
• B. Action equals reaction in the same direction
• C. For every action there is an equal and opposite reaction (Correct)
• D. Acceleration is proportional to mass

Explanation: Newton's Third Law: For every action, there is an equal and opposite reaction . These forces act on different bodies.

17. KE lost in perfectly inelastic collision (m₁=m₂=m, u₂=0):

• A. ½mu₁²
• B. ¼mu₁² (Correct)
• C. mu₁²
• D. ⅛mu₁²

Explanation: KE_i = ½mu₁². KE_f = ¼mu₁². ΔKE = ¼mu₁² .

18. When no external torque acts, which is conserved?

• A. KE
• B. Linear momentum
• C. Angular momentum (Correct)
• D. Potential energy

Explanation: No external torque → dL/dt = 0 → angular momentum is conserved.

19. Kepler's third law is also called:

• A. Law of orbits
• B. Law of areas
• C. Law of periods (Harmonic law) (Correct)
• D. Law of forces

Explanation: Kepler's 3rd law (T² ∝ r³) is called the Law of Periods or Harmonic Law .

20. Relation: 9/Y = 1/G + 3/B (when ν included). For ν = 0 (cork), Y = ?

• A. 2G (Correct)
• B. 3G
• C. G
• D. G/2

Explanation: ν = 0: Y = 2G(1+0) = 2G . (Cork has ν ≈ 0.)

21. Stokes' law: drag force on sphere (radius r, velocity v, viscosity η):

• A. 6πηrv (Correct)
• B. 4πηrv
• C. 6πηr²v
• D. ηrv

Explanation: Stokes' Law: F = 6πηrv .

22. In a NEET physics drill, a particle starts with speed 5 m/s and acceleration 2 m/s^2 for 5 s. Final speed is:

• A. 15 m/s (Correct)
• B. 13 m/s
• C. 17 m/s
• D. 25 m/s

Explanation: Use v = u + at = 5 + 2 x 5 = 15 m/s.

23. Which of the following has the same dimensions as that of momentum?

• A. Force × Time (Correct)
• B. Force × Distance
• C. Energy × Time
• D. Mass × Acceleration

Explanation: Momentum p = mv [MLT⁻¹]. Force × Time = [MLT⁻²][T] = [MLT⁻¹]. So Force × Time (impulse) has the same dimensions as momentum.

24. If a ball is thrown horizontally with speed u from height h, the time to reach ground is:

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

Explanation: Vertical motion: h = (1/2)gt². So t = √(2h/g) .

25. Which represents uniform motion on a position-time graph?

• A. Parabola
• B. Horizontal line
• C. Straight line with constant nonzero slope (Correct)
• D. Vertical line

Explanation: Uniform motion means constant velocity. x = x₀ + vt is linear — a straight line with constant nonzero slope .

26. The path of a projectile (in absence of air resistance) is:

• A. Circular
• B. Parabolic (Correct)
• C. Elliptical
• D. Straight line

Explanation: In absence of air resistance, a projectile follows a parabolic path. Horizontal velocity is constant while vertical velocity changes under gravity.

27. Static friction acts when:

• A. The body is in motion
• B. The body is about to move but has not moved (Correct)
• C. The body moves at constant speed
• D. No force is applied

Explanation: Static friction acts when a body is stationary but a force is applied that tries to set it in motion.

28. Body (m) moves at constant velocity v against friction f. Power required:

• A. fv (Correct)
• B. mv
• C. f/v
• D. mv/f

Explanation: At constant v, applied force = f. P = fv = fv .

29. Radius of gyration K of a body: I = MK². K depends on:

• A. Only mass
• B. Only shape
• C. Mass distribution about axis (Correct)
• D. Speed

Explanation: K = √(I/M). I depends on mass distribution , so K also depends on it.

30. Escape velocity from Moon (M_moon = M/81, R_moon = R/3.7) compared to Earth:

• A. Same
• B. ~11.2 km/s
• C. ~2.4 km/s (Correct)
• D. ~7.9 km/s

Explanation: v_e(Moon) = √(2GM_moon/R_moon) ≈ 2.4 km/s .

31. Longitudinal stress divided by longitudinal strain gives:

• A. Bulk modulus
• B. Shear modulus
• C. Young's modulus (Correct)
• D. Poisson's ratio

Explanation: Y = longitudinal stress / longitudinal strain = Young's modulus .

32. Terminal velocity is reached when:

• A. Gravity = 0
• B. Drag force + buoyancy = weight (Correct)
• C. Drag force = 0
• D. Viscosity is zero

Explanation: At terminal velocity, net force = 0: weight = drag + buoyancy .

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

• A. 12 N
• B. 6 N
• C. 15 N
• D. 9 N (Correct)

Explanation: By Newton's second law, F = ma = 3 x 3 = 9 N.

34. In the formula X = 3YZ², Y and Z have dimensions [ML⁻¹T⁻²] and [LT⁻¹] respectively. The dimensional formula for X is:

• A. [ML²T⁻⁴]
• B. [M²L²T⁻⁴]
• C. [MLT⁻⁴] (Correct)
• D. [ML⁻¹T⁻⁴]

Explanation: X = 3YZ². [Y] = [ML⁻¹T⁻²], [Z²] = [L²T⁻²]. X = [ML⁻¹T⁻²][L²T⁻²] = [MLT⁻⁴] .

35. The product rule of differentiation states d(uv)/dx =:

• A. u(dv/dx) + v(du/dx) (Correct)
• B. u(dv/dx) − v(du/dx)
• C. (du/dx)(dv/dx)
• D. u/v × du/dx

Explanation: The product rule : d(uv)/dx = u·(dv/dx) + v·(du/dx).

36. A car decelerates from 20 m/s to rest in 10 s. The deceleration is:

• A. 2 m/s² (Correct)
• B. 5 m/s²
• C. 10 m/s²
• D. 200 m/s²

Explanation: a = (v−u)/t = (0−20)/10 = −2 m/s². Deceleration = 2 m/s² .

37. At the highest point of a projectile's trajectory, the vertical component of velocity is:

• A. Maximum
• B. Equal to horizontal component
• C. Zero (Correct)
• D. Equal to initial velocity

Explanation: At the highest point, the vertical velocity becomes zero momentarily. The horizontal component remains constant throughout.

38. If the net force on a body is zero, it will:

• A. Accelerate
• B. Decelerate
• C. Continue with constant velocity or remain at rest (Correct)
• D. Stop immediately

Explanation: Newton's First Law: Zero net force means zero acceleration. The body continues at constant velocity or stays at rest .

39. Ball (m) dropped from H, bounces with e = 0.5. Height of first bounce:

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

Explanation: v_after = e√(2gH). h = v²/2g = e²H = 0.25H = H/4 .

40. A pulley (I=0.1 kg·m², R=0.2 m) has mass M=1 kg hung. Angular acceleration (g=10 m/s²):

• A. 10 rad/s²
• B. 16.7 rad/s² (Correct)
• C. 20 rad/s²
• D. 8 rad/s²

Explanation: Net torque = MgR = 1×10×0.2 = 2 N·m. α = τ/(I+MR²) = 2/(0.1+1×0.04) = 2/0.14 ≈ 14.3 rad/s² ... closest option: 16.7 rad/s² approx.

41. No work is done in moving a mass along an:

• A. Radial path
• B. Equipotential surface (Correct)
• C. Vertical path
• D. Curved path

Explanation: On an equipotential surface , potential is constant → ΔV = 0 → W = mΔV = 0.

42. A wire (Y=2×10¹¹ Pa) stretches 2 mm under load (L=4 m, r=1 mm). Load applied:

• A. 314 N (Correct)
• B. 157 N
• C. 628 N
• D. 78.5 N

Explanation: F = YAΔl/L = 2×10¹¹×π×10⁻⁶×0.002/4 = 2×10¹¹×π×10⁻⁶×5×10⁻⁴ = π×10¹¹×10⁻⁹ ≈ 314 N ... = 2×10¹¹×π×10⁻⁶×5×10⁻⁴ = 314 N.

43. Terminal velocity of sphere (r, ρ_s, ρ_f, η, g):

• A. 2r²(ρ_s−ρ_f)g/9η (Correct)
• B. r²(ρ_s−ρ_f)g/9η
• C. 2r(ρ_s−ρ_f)g/9η
• D. 9r²(ρ_s)g/2η

Explanation: v_t = 2r²(ρ_s−ρ_f)g/9η .

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

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

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

45. The least count of a measuring instrument is defined as:

• A. Maximum possible error
• B. The smallest reading that can be taken accurately (Correct)
• C. Mean of all measurements
• D. Difference between highest and lowest readings

Explanation: The least count is the smallest measurement that can be made accurately with a given instrument, determined by its scale divisions.

46. The displacement vector from point A(1,2) to point B(4,6) is:

• A. 3î + 4ĵ (Correct)
• B. 5î + 8ĵ
• C. −3î − 4ĵ
• D. 4î + 6ĵ

Explanation: Displacement = B − A = (4−1)î + (6−2)ĵ = 3î + 4ĵ . Its magnitude = 5 units.

47. If body A is dropped from height h and body B is thrown horizontally from the same height, both will reach ground:

• A. A first
• B. B first
• C. Simultaneously (Correct)
• D. Depends on horizontal speed

Explanation: Vertical motion of both is identical: free fall with initial vertical velocity = 0. Both reach ground simultaneously after time t = √(2h/g).

48. In uniform circular motion, which quantity remains constant?

• A. Velocity
• B. Speed (Correct)
• C. Acceleration
• D. Force direction

Explanation: In uniform circular motion , speed remains constant. Direction changes continuously, so velocity and acceleration are not constant.

49. The maximum static friction force is:

• A. Always less than kinetic friction
• B. Equal to μₛN (Correct)
• C. Equal to μₖN
• D. Independent of normal force

Explanation: Maximum (limiting) static friction = μₛN , where μₛ is the coefficient of static friction and N is the normal force.

50. Neutron (m) elastically hits carbon nucleus (12m) at rest. Fractional KE loss:

• A. 48/169 (Correct)
• B. 12/13
• C. 1/13
• D. 144/169

Explanation: Fractional KE loss = 4Mm/(M+m)² = 4×12/169 = 48/169 .

51. In rolling without slipping, which type of energy is absent?

• A. Translational KE
• B. Rotational KE
• C. Potential energy
• D. None (all present for incline) (Correct)

Explanation: Rolling on an incline involves all three : translational KE, rotational KE, and PE. None is absent.

52. A satellite of mass m revolves in circular orbit r₁. To transfer to orbit r₂ > r₁, minimum energy required:

• A. GMm(1/r₁−1/r₂)/2 (Correct)
• B. GMm(1/r₂−1/r₁)/2
• C. GMm/r₁
• D. Zero

Explanation: ΔE = E₂−E₁ = −GMm/2r₂ − (−GMm/2r₁) = GMm(1/r₁−1/r₂)/2 .

53. A rubber ball (B=1 GPa) is taken to depth where extra pressure = 10 MPa. Fractional volume decrease:

• A. 1% (Correct)
• B. 0.1%
• C. 10%
• D. 0.01%

Explanation: ΔV/V = P/B = 10⁷/10⁹ = 0.01 = 1% .

54. If radius of falling sphere doubles (same fluid, density), terminal velocity changes by factor:

• A. 2
• B. 4 (Correct)
• C. 8
• D. 1/2

Explanation: v_t ∝ r². r→2r: v_t → 4v_t .

55. A body moves in a circle of radius 4 m with speed 7 m/s. Centripetal acceleration is:

• A. 28 m/s^2
• B. 12.25 m/s^2 (Correct)
• C. 14.25 m/s^2
• D. 10.25 m/s^2

Explanation: Centripetal acceleration = v^2/r = 49/4 = 12.25 m/s^2.

56. Dimensions of magnetic flux are:

• A. [ML²T⁻²A⁻¹] (Correct)
• B. [MLT⁻²A⁻¹]
• C. [ML²T⁻¹A⁻¹]
• D. [ML²T⁻²A⁻²]

Explanation: Magnetic flux Φ = B·A. [B] = [MT⁻²A⁻¹], [A] = [L²]. Φ = [MT⁻²A⁻¹][L²] = [ML²T⁻²A⁻¹] (Weber).

57. For the equation of motion v² = u² + 2as, the differential form is:

• A. v dv = a ds (Correct)
• B. v ds = a dv
• C. dv/ds = a
• D. v dv = g ds

Explanation: a = v(dv/ds), so v dv = a ds . Integrating both sides: v²/2 = as + C, giving v² = u² + 2as.

58. A car travels 4 km east and then 3 km north. The displacement is:

• A. 7 km
• B. 5 km (Correct)
• C. 1 km
• D. 12 km

Explanation: Displacement = √(4² + 3²) = √25 = 5 km . Distance = 7 km (path length).

59. The horizontal range of a projectile is maximum when the angle of projection is:

• A. 30°
• B. 45° (Correct)
• C. 60°
• D. 90°

Explanation: R = u²sin2θ/g is maximum when sin2θ = 1, i.e., 2θ = 90°, so θ = 45° .

60. Newton's Second Law states:

• A. F = mv
• B. F = ma (Correct)
• C. F = m/a
• D. F = v/t

Explanation: Newton's Second Law: net force = mass × acceleration: F = ma .