Physics Mock Test 4 Practice

1. A screw gauge has 50 divisions on its circular scale and pitch 0.5 mm. Its least count is:

• A. 0.01 mm (Correct)
• B. 0.001 mm
• C. 0.1 mm
• D. 0.005 mm

Explanation: Least count = Pitch / Number of divisions on circular scale = 0.5 mm / 50 = 0.01 mm .

2. If A and B are parallel vectors, then A × B equals:

• A. AB
• B. Zero vector (Correct)
• C. A unit vector
• D. AB/2

Explanation: When θ = 0° (parallel), |A × B| = AB sin0° = 0. The cross product of parallel vectors is the zero vector .

3. For uniform motion, the position-time (x-t) graph is:

• A. A parabola
• B. A horizontal line
• C. A straight line with nonzero slope (Correct)
• D. A curved line

Explanation: For uniform motion (constant velocity), x = x₀ + vt is linear. The x-t graph is a straight line with nonzero slope .

4. For a projectile fired at angle θ, the ratio of range to maximum height is:

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

Explanation: R = 2u²sinθcosθ/g; H = u²sin²θ/2g. R/H = 4cosθ/sinθ = 4/tanθ .

5. Two masses M and m (M > m) on a pulley; system acceleration = g/3. The ratio M/m is:

• A. 2 (Correct)
• B. 3
• C. 4
• D. 5

Explanation: a = (M−m)g/(M+m) = g/3. 3(M−m) = M+m → 2M = 4m → M/m = 2 .

6. Spring compressed by x has PE = E. Compressed by 2x, PE becomes:

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

Explanation: PE ∝ x². x→2x: PE → 4E .

7. A disc (I=0.1 kg·m²) rotating at 5 rad/s. Another disc (I=0.1 kg·m²) placed coaxially at rest. New ω:

• A. 5 rad/s
• B. 2.5 rad/s (Correct)
• C. 10 rad/s
• D. 1.25 rad/s

Explanation: L = 0.1×5 = 0.5 kg·m²/s. I_total = 0.2. ω = 0.5/0.2 = 2.5 rad/s .

8. A satellite in orbit has total energy:

• A. Positive
• B. Zero
• C. Negative (Correct)
• D. Infinite

Explanation: Total E = −GMm/2r negative .

9. For a perfectly rigid body, Y is:

• A. Zero
• B. Finite
• C. Infinite (∞) (Correct)
• D. Equal to G

Explanation: Perfectly rigid body: no strain for any stress → Y = stress/0 = ∞ .

10. Plimsoll line on ships indicates:

• A. Speed limit
• B. Maximum safe loading level (related to buoyancy) (Correct)
• C. Depth of ocean
• D. Engine location

Explanation: Plimsoll line shows the maximum safe loading level to ensure adequate buoyancy.

11. 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.

12. While measuring the thickness of a glass plate with a screw gauge, the main scale reads 2 mm and the circular scale reads 30 divisions (LC = 0.01 mm). The thickness is:

• A. 2.30 mm (Correct)
• B. 2.03 mm
• C. 23.0 mm
• D. 0.230 mm

Explanation: Thickness = Main scale reading + (CSR × LC) = 2 mm + (30 × 0.01 mm) = 2 + 0.30 = 2.30 mm .

13. The slope of a position-time graph at any instant gives:

• A. Distance
• B. Acceleration
• C. Velocity (Correct)
• D. Displacement

Explanation: v = ds/dt. The derivative of position with respect to time is velocity , which equals the slope of the position-time graph.

14. A bullet fired from a gun has muzzle velocity 500 m/s and decelerates at 100 m/s² in a sand bag. Distance penetrated before stopping:

• A. 500 m
• B. 2500 m
• C. 1250 m (Correct)
• D. 250 m

Explanation: v² = u² − 2as → 0 = 250000 − 200s → s = 1250 m .

15. The angular momentum of a particle of mass m in a circle of radius r with angular velocity ω is:

• A. mr²ω (Correct)
• B. mrω
• C. mr²ω²
• D. mω/r

Explanation: L = mvr = m(rω)r = mr²ω . Also L = Iω where I = mr².

16. A coin on a rotating disc stays due to:

• A. Centrifugal force
• B. Centripetal force provided by friction (Correct)
• C. Normal force
• D. Gravity

Explanation: The coin needs centripetal force (toward center) for circular motion. This is provided by static friction .

17. 2 kg block slides 5 m on rough floor (μ = 0.3, g = 10 m/s²). Energy lost to friction:

• A. 15 J
• B. 30 J (Correct)
• C. 6 J
• D. 3 J

Explanation: f = μmg = 6 N. E_lost = 6×5 = 30 J .

18. MI of solid cylinder (M, R) about its axis:

• A. MR²
• B. MR²/2 (Correct)
• C. 2MR²/5
• D. MR²/3

Explanation: I_solid cylinder (axis) = MR²/2 .

19. Energy required to move satellite from orbit r₁ to r₂ (r₂ > r₁):

• A. GMm(1/r₁−1/r₂)/2 (Correct)
• B. GMm(1/r₁−1/r₂)
• C. GMm(1/r₂−1/r₁)/2
• D. 0

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

20. Shear modulus is related to the ability of a material to:

• A. Resist compression
• B. Resist volume change
• C. Resist twisting/shearing (Correct)
• D. Conduct heat

Explanation: G (shear/rigidity modulus) measures resistance to twisting and shearing .

21. Object (W_air=100 N, W_water=80 N, W_liquid=85 N). Relative density of liquid:

• A. 0.75 (Correct)
• B. 0.5
• C. 1.25
• D. 0.25

Explanation: RD = (W_air−W_liquid)/(W_air−W_water) = (100−85)/(100−80) = 15/20 = 0.75 .

22. 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.

23. The velocity of sound in a medium depends on pressure P and density ρ. By dimensional analysis, v is proportional to:

• A. √(P/ρ) (Correct)
• B. √(Pρ)
• C. P/ρ
• D. P²/ρ

Explanation: [v] = LT⁻¹, [P] = ML⁻¹T⁻², [ρ] = ML⁻³. If v = kPᵃρᵇ: LT⁻¹ = (ML⁻¹T⁻²)ᵃ(ML⁻³)ᵇ gives a = 1/2, b = −1/2, so v ∝ √(P/ρ) .

24. ∫sin x dx equals:

• A. cos x + C
• B. −cos x + C (Correct)
• C. sin x + C
• D. tan x + C

Explanation: ∫sin x dx = −cos x + C . Verified: d(−cos x)/dx = sin x.

25. Two cars A and B start from the same point. A goes north at 60 km/h, B goes east at 80 km/h. After 1 hour the distance between them is:

• A. 20 km
• B. 100 km (Correct)
• C. 140 km
• D. 50 km

Explanation: After 1 h: A is 60 km north, B is 80 km east. Distance = √(60²+80²) = 100 km .

26. Two projectiles thrown at same speed u at angles (45° + α) and (45° − α). Their ranges are:

• A. Equal (Correct)
• B. Differ by 2α
• C. In ratio sinα:cosα
• D. Unequal

Explanation: R₁ = u²sin(90°+2α)/g = u²cos2α/g. R₂ = u²sin(90°−2α)/g = u²cos2α/g. Both ranges are equal .

27. Two bodies (4 kg at 3 m/s and 6 kg at 2 m/s in opposite directions) collide and stick. Velocity after collision:

• A. 0 m/s (Correct)
• B. 0.5 m/s (dir of 4 kg)
• C. 0.5 m/s (dir of 6 kg)
• D. 1 m/s

Explanation: Taking 4 kg direction positive: p = 4×3 − 6×2 = 12 − 12 = 0. v = 0/(10) = 0 m/s .

28. Work done by conservative force depends on:

• A. Path taken
• B. Time taken
• C. Initial and final positions only (Correct)
• D. Speed

Explanation: Conservative force work is path-independent — depends only on initial and final positions .

29. Torque due to gravity on a uniform rod (M, L) pivoted at one end, making angle θ with vertical:

• A. MgL sinθ
• B. MgL cosθ
• C. MgL sinθ/2 (Correct)
• D. MgL cosθ/2

Explanation: Gravity acts at L/2 from pivot. Perpendicular distance = (L/2)sinθ. τ = Mg×(L/2)sinθ = MgL sinθ/2 .

30. When satellite orbit radius increases, its speed:

• A. Increases
• B. Decreases (Correct)
• C. Stays same
• D. Becomes zero

Explanation: v₀ = √(GM/r). As r increases, v₀ decreases .

31. When pressure increases on a gas, its bulk modulus:

• A. Decreases
• B. Stays same (ideal gas)
• C. Increases (for isothermal ideal gas) (Correct)
• D. Becomes zero

Explanation: Isothermal ideal gas: B = P. As pressure increases, B increases .

32. Equation of continuity for fluid flow states:

• A. P + ½ρv² = constant
• B. A₁v₁ = A₂v₂ (Correct)
• C. F = ma
• D. P = ρgh

Explanation: Continuity equation: A₁v₁ = A₂v₂ (conservation of mass for incompressible fluid).

33. 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.

34. In the formula T = 2π√(l/g), if l is measured with 1% error and T with 2% error, the error in g is:

• A. 5% (Correct)
• B. 3%
• C. 1%
• D. 6%

Explanation: g = 4π²l/T². % error in g = % error in l + 2×% error in T = 1% + 2×2% = 5% .

35. The unit vector along the direction of A = 3î + 4ĵ is:

• A. (3î + 4ĵ)/5 (Correct)
• B. (3î + 4ĵ)/7
• C. 3î + 4ĵ
• D. (3î − 4ĵ)/5

Explanation: |A| = √(9+16) = 5. Unit vector â = A/|A| = (3î + 4ĵ)/5 .

36. A particle moves along x-axis with acceleration a = 2t m/s². If v = 0 at t = 0, the velocity at t = 3 s is:

• A. 6 m/s
• B. 9 m/s (Correct)
• C. 18 m/s
• D. 3 m/s

Explanation: v = ∫a dt = ∫2t dt = t² + C. At t=0, v=0 → C=0. At t=3: v = 9 = 9 m/s .

37. A particle moves in a circle of radius r. In half a revolution, the displacement is:

• A. 0
• B. πr
• C. 2r (Correct)
• D. r

Explanation: In half revolution, the particle goes from one end to the other of a diameter. Displacement = 2r .

38. A block of mass m pressed against wall by force F at angle θ to wall. Minimum F to prevent sliding (friction μ):

• A. mg/μ
• B. mg/(μcosθ + sinθ) (Correct)
• C. mg/sinθ
• D. mg/(μsinθ + cosθ)

Explanation: Normal force N = Fcosθ. Friction = μFcosθ. Equilibrium: Fsinθ + μFcosθ = mg → F = mg/(sinθ + μcosθ) .

39. Which is a non-conservative force?

• A. Gravity
• B. Spring force
• C. Friction (Correct)
• D. Coulomb force

Explanation: Friction dissipates energy — it is non-conservative.

40. Acceleration of a sphere rolling down incline (angle θ, no slipping, g):

• A. g sinθ
• B. 5g sinθ/7 (Correct)
• C. 2g sinθ/3
• D. g sinθ/2

Explanation: For solid sphere: a = g sinθ/(1+I/MR²) = g sinθ/(1+2/5) = 5g sinθ/7 = 5g sinθ/7 .

41. Kepler's third law is derived from:

• A. Conservation of energy
• B. Newton's law of gravitation (Correct)
• C. Conservation of angular momentum
• D. Newton's 3rd law

Explanation: Kepler's 3rd law (T² ∝ r³) follows directly from Newton's law of gravitation .

42. Poisson's ratio for an incompressible material (rubber-like):

• A. 0
• B. 0.3
• C. 0.5 (Correct)
• D. 1

Explanation: Incompressible material: no volume change. ν → 0.5 (theoretical limit).

43. Pipe narrows from 4 cm² to 1 cm² area. Fluid enters at 2 m/s. Exit velocity:

• A. 0.5 m/s
• B. 8 m/s (Correct)
• C. 2 m/s
• D. 4 m/s

Explanation: A₁v₁ = A₂v₂. 4×2 = 1×v₂. v₂ = 8 m/s .

44. Two perpendicular vectors have magnitudes 3 and 4. Their resultant magnitude is approximately:

• A. 5.0 (Correct)
• B. 7
• C. 1
• D. 12.0

Explanation: For perpendicular vectors, R = sqrt(3^2 + 4^2) = 5.0.

45. The length and width of a rectangle are 5.7 cm and 3.4 cm respectively. Its area, with correct significant figures, is:

• A. 19.38 cm²
• B. 19.4 cm²
• C. 19 cm² (Correct)
• D. 20 cm²

Explanation: Area = 5.7 × 3.4 = 19.38 cm². Since both measurements have 2 significant figures, the result should be reported with 2 significant figures: 19 cm² .

46. A projectile is launched at 45°. Its horizontal and vertical components of initial velocity u are:

• A. u and 0
• B. u/√2 and u/√2 (Correct)
• C. u/2 and u/2
• D. u and u

Explanation: uₓ = u cos45° = u/√2, uᵧ = u sin45° = u/√2. At 45°, both components are equal: u/√2 each .

47. A particle's position is x = 3t³ − 2t² + t. The acceleration at t = 2 s is:

• A. 28 m/s²
• B. 30 m/s²
• C. 32 m/s² (Correct)
• D. 34 m/s²

Explanation: v = dx/dt = 9t² − 4t + 1. a = dv/dt = 18t − 4. At t=2: a = 36−4 = 32 m/s² .

48. Condition for a vehicle not to skid on curved road of radius r with friction coefficient μ is:

• A. v ≤ √(μrg) (Correct)
• B. v ≥ √(μrg)
• C. v ≤ μrg
• D. v ≤ √(rg/μ)

Explanation: For circular motion: μmg ≥ mv²/r → v² ≤ μrg → v ≤ √(μrg) .

49. The minimum force needed to move mass m on surface with μ, applied at optimal angle, is:

• A. μmg
• B. μmg/√(1+μ²) (Correct)
• C. mg/√(1+μ²)
• D. μmg√(1+μ²)

Explanation: Minimum force at optimal angle θ = tan⁻¹(μ): F_min = μmg/√(1+μ²) .

50. At midpoint of freely falling body from height H, its KE equals:

• A. mgh
• B. mgH/2 (Correct)
• C. 2mgH
• D. mgH/4

Explanation: At h = H/2: KE = Total − PE = mgH − mgH/2 = mgH/2 .

51. Angular momentum of a particle moving in a straight line through origin:

• A. Maximum
• B. Zero (Correct)
• C. Constant
• D. Infinite

Explanation: L = r × p. If particle passes through origin, r = 0 → L = 0 . Actually if line passes through origin, r and p are parallel → L = rp sin0° = 0.

52. Moon has no atmosphere because:

• A. No water
• B. Escape velocity is low (≈2.4 km/s) (Correct)
• C. Too far from Sun
• D. No magnetic field

Explanation: Moon's escape velocity ≈ 2.4 km/s — less than typical gas molecule speeds. Atmosphere escapes. Low escape velocity .

53. For liquids, Young's modulus Y =:

• A. Very high
• B. Zero (Correct)
• C. Infinite
• D. Same as B

Explanation: Liquids cannot support tensile stress — they flow. Y = 0 for liquids.

54. Bernoulli's equation is a statement of:

• A. Newton's 3rd law
• B. Conservation of mass
• C. Conservation of energy for ideal fluid (Correct)
• D. Conservation of momentum

Explanation: Bernoulli's equation = conservation of energy applied to ideal (non-viscous, incompressible) fluid flow.

55. In a NEET physics drill, a particle starts with speed 7 m/s and acceleration 2 m/s^2 for 4 s. Final speed is:

• A. 13 m/s
• B. 17 m/s
• C. 28 m/s
• D. 15 m/s (Correct)

Explanation: Use v = u + at = 7 + 2 x 4 = 15 m/s.

56. The SI unit of pressure is Pascal. 1 atmosphere = 1.013 × 10⁵ Pa. The pressure of 2 atm in Pa is:

• A. 2.026 × 10⁵ Pa (Correct)
• B. 1.013 × 10⁵ Pa
• C. 0.506 × 10⁵ Pa
• D. 4.052 × 10⁵ Pa

Explanation: 2 atm = 2 × 1.013 × 10⁵ Pa = 2.026 × 10⁵ Pa .

57. If acceleration a = 6t, and v = 0 at t = 0, then v at t = 3 s is:

• A. 18 m/s
• B. 27 m/s (Correct)
• C. 9 m/s
• D. 54 m/s

Explanation: v = ∫a dt = ∫6t dt = 3t² + C. At t=0, v=0 → C=0. At t=3: v = 3(9) = 27 m/s .

58. The velocity-time graph of a particle is a straight line making 45° with the time-axis. If the particle starts from rest, the displacement in 4 s is:

• A. 8 m (Correct)
• B. 16 m
• C. 4 m
• D. 32 m

Explanation: Acceleration = tan45° = 1 m/s². s = ½at² = ½(1)(16) = 8 m .

59. In a vertical circle of radius R, the minimum speed at the bottom to complete the loop is:

• A. √(gR)
• B. √(2gR)
• C. √(5gR) (Correct)
• D. √(3gR)

Explanation: Minimum speed at top = √(gR). Energy conservation: v_bottom² = gR + 4gR = 5gR → v_bottom = √(5gR) .

60. An elevator descends with acceleration g/4. Apparent weight of 60 kg person (g = 10 m/s²):

• A. 600 N
• B. 450 N (Correct)
• C. 750 N
• D. 150 N

Explanation: Apparent weight = m(g−a) = 60(10 − 2.5) = 60×7.5 = 450 N .