Physics Mock Test 5 Practice

1. The dimensions of surface tension are:

• A. [MT⁻²] (Correct)
• B. [MLT⁻²]
• C. [ML⁰T⁻²]
• D. [ML⁻¹T⁻²]

Explanation: Surface tension = Force/Length = [MLT⁻²]/[L] = [MT⁻²] .

2. The resultant of two forces P and Q is R. If Q is doubled, the new resultant is perpendicular to P. Then:

• A. P = Q
• B. P = R
• C. Q = R (Correct)
• D. P = 2Q

Explanation: If new resultant (with 2Q) is perpendicular to P, using the triangle law and the condition of perpendicularity leads to Q = R .

3. A particle at rest starts moving with acceleration a = (4 − t) m/s². At what time does the velocity become maximum?

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

Explanation: Velocity is maximum when a = 0: 4 − t = 0 → t = 4 s . After this, a becomes negative and velocity decreases.

4. The ratio of angular velocities of the hour hand and minute hand of a clock is:

• A. 1:12 (Correct)
• B. 12:1
• C. 1:24
• D. 24:1

Explanation: ω_h/ω_m = (2π/12 h)/(2π/1 h) = 1:12 .

5. A 0.5 kg ball changes velocity from 20 m/s to −30 m/s. Contact time = 0.01 s. Average force is:

• A. 500 N
• B. 1000 N
• C. 2500 N (Correct)
• D. 100 N

Explanation: Impulse = 0.5×50 = 25 N·s. F = 25/0.01 = 2500 N .

6. Ball thrown at 30° with KE = 10 J. KE at maximum height:

• A. 10 J
• B. 7.5 J (Correct)
• C. 2.5 J
• D. 5 J

Explanation: At max height, KE = KE₀cos²30° = 10×(3/4) = 7.5 J .

7. MI of thin circular ring (M, R) about axis through centre perpendicular to plane:

• A. MR²/2
• B. MR² (Correct)
• C. 2MR²
• D. MR²/4

Explanation: I_ring (through centre, perpendicular) = MR² .

8. A body is weightless in a satellite because:

• A. Gravity is zero
• B. It is in free fall (both satellite and body have same acceleration) (Correct)
• C. Satellite is in vacuum
• D. Far from Earth

Explanation: Weightlessness in satellite is due to free fall — both body and satellite accelerate at g toward Earth.

9. The region between elastic limit and fracture point is called:

• A. Elastic region
• B. Plastic region (Correct)
• C. Linear region
• D. Safe region

Explanation: Beyond elastic limit, permanent deformation occurs — this is the plastic region .

10. Venturimeter works on:

• A. Pascal's law
• B. Archimedes' principle
• C. Bernoulli's equation (Correct)
• D. Newton's 2nd law

Explanation: Venturimeter uses the pressure difference created by velocity change — based on Bernoulli's equation .

11. A body moves in a circle of radius 3 m with speed 4 m/s. Centripetal acceleration is:

• A. 5.333333333333333 m/s^2 (Correct)
• B. 7.333333333333333 m/s^2
• C. 3.333333333333333 m/s^2
• D. 12 m/s^2

Explanation: Centripetal acceleration = v^2/r = 16/3 = 5.333333333333333 m/s^2.

12. The resistance R = V/I. If V = (100 ± 5) V and I = (10 ± 0.2) A, the maximum percentage error in R is:

• A. 5%
• B. 7% (Correct)
• C. 2%
• D. 9%

Explanation: % error in R = % error in V + % error in I = (5/100)×100% + (0.2/10)×100% = 5% + 2% = 7% .

13. The displacement of a particle is given by x = A sin(ωt). The maximum velocity is:

• A. Aω²
• B. Aω (Correct)
• C. A/ω
• D. A²ω

Explanation: v = dx/dt = Aω cos(ωt). Maximum value of cos is 1, so v_max = Aω .

14. A particle's x-t graph is an upward-opening parabola. This means the particle has:

• A. Constant positive velocity
• B. Increasing positive acceleration
• C. Constant positive acceleration (Correct)
• D. Decreasing velocity

Explanation: x = ut + ½at² is a parabola. An upward-opening parabola indicates constant positive acceleration .

15. A gun fires a bullet at 45° and hits a target 1 km away (same level). The muzzle speed is (g = 10 m/s²):

• A. 10 m/s
• B. 100 m/s (Correct)
• C. 1000 m/s
• D. 316 m/s

Explanation: R = u²sin90°/g = u²/g. 1000 = u²/10 → u² = 10000 → u = 100 m/s .

16. Rolling friction is generally ________ than sliding friction.

• A. Greater
• B. Equal
• C. Lesser (Correct)
• D. Unpredictable

Explanation: Rolling friction is much less than sliding friction. This is why wheels reduce friction in transport.

17. Work done by non-conservative force in a closed path is:

• A. Zero
• B. Positive
• C. Negative
• D. Non-zero (Correct)

Explanation: Non-conservative forces dissipate energy; work in closed path is non-zero .

18. For rolling without slipping, kinetic friction between body and surface is:

• A. Kinetic friction acts
• B. Static friction acts (no slipping) (Correct)
• C. No friction needed
• D. Rolling friction only

Explanation: In pure rolling, contact point has zero relative velocity — static friction acts (not kinetic).

19. Binding energy of satellite in orbit (radius r, mass m, planet mass M):

• A. GMm/r
• B. GMm/2r (Correct)
• C. −GMm/2r
• D. GMm/2r²

Explanation: Binding energy = −(Total E) = −(−GMm/2r) = GMm/2r . Energy needed to free the satellite.

20. Ultimate tensile strength (UTS) represents:

• A. Stress at which elasticity ends
• B. Maximum stress the material can withstand (Correct)
• C. Stress at fracture
• D. Elastic limit

Explanation: UTS = maximum stress before the material begins to neck/fail.

21. Water (ρ=1000 kg/m³) flows in pipe. At section 1: A=10 cm², v=2 m/s, P=2×10⁵ Pa. At section 2: A=5 cm², same height. v₂:

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

Explanation: Continuity: A₁v₁ = A₂v₂. v₂ = 10×2/5 = 4 m/s .

22. Two perpendicular vectors have magnitudes 4 and 6. Their resultant magnitude is approximately:

• A. 10
• B. 2
• C. 24.0
• D. 7.2 (Correct)

Explanation: For perpendicular vectors, R = sqrt(4^2 + 6^2) = 7.2.

23. If the units of length and force are doubled, the unit of energy will change by a factor of:

• A. 2
• B. 4 (Correct)
• C. 8
• D. 1

Explanation: Energy = Force × Length. If both force and length are doubled, new unit of energy = 2F × 2L = 4FL = 4 times the old unit.

24. A particle's velocity is v = 3t² − 12t + 9. When is the acceleration zero?

• A. t = 1 s
• B. t = 2 s (Correct)
• C. t = 3 s
• D. t = 4 s

Explanation: a = dv/dt = 6t − 12. Setting a = 0: 6t = 12 → t = 2 s .

25. A ball is dropped from a height H. After falling through H/2, its speed is:

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

Explanation: v² = 2g(H/2) = gH → v = √(gH) .

26. For circular motion, the work done by centripetal force in one revolution is:

• A. 2πmv²/r
• B. mv²/r
• C. Zero (Correct)
• D. mv²r

Explanation: Centripetal force is always perpendicular to velocity. Work = F·d·cos90° = zero .

27. A rocket of initial mass M burns fuel at rate r kg/s and ejects gas at speed v. The thrust is:

• A. Mv
• B. rv (Correct)
• C. Mr
• D. Mrv

Explanation: Thrust = v_exhaust × (mass ejection rate) = v × r = rv .

28. Potential energy is a property of:

• A. Single body alone
• B. System of interacting bodies (Correct)
• C. Only massive bodies
• D. Only charged bodies

Explanation: PE belongs to a system of interacting bodies .

29. Torque on a spinning top is zero. Its angular momentum vector:

• A. Changes direction
• B. Increases in magnitude
• C. Remains constant (direction and magnitude) (Correct)
• D. Decreases

Explanation: dL/dt = τ = 0 → L = constant in both direction and magnitude.

30. KE of satellite in circular orbit (mass m, radius r):

• A. GMm/r
• B. GMm/2r (Correct)
• C. GMm/2r²
• D. 2GMm/r

Explanation: mv²/r = GMm/r². v² = GM/r. KE = ½mv² = GMm/2r .

31. For a ductile material, fracture occurs:

• A. Immediately after UTS
• B. Well below elastic limit
• C. After significant plastic deformation (Correct)
• D. Without necking

Explanation: Ductile materials show significant plastic deformation (necking) before fracture.

32. In the pipe problem above, P₂ at section 2 (same height):

• A. 2×10⁵ Pa
• B. 1.94×10⁵ Pa (Correct)
• C. 2.06×10⁵ Pa
• D. 1.9×10⁵ Pa

Explanation: Bernoulli: P₁+½ρv₁² = P₂+½ρv₂². P₂ = 2×10⁵+½×1000×(4−16) = 2×10⁵−6000 = 1.94×10⁵ Pa .

33. In a NEET physics drill, a particle starts with speed 5 m/s and acceleration 2 m/s^2 for 3 s. Final speed is:

• A. 11 m/s (Correct)
• B. 9 m/s
• C. 13 m/s
• D. 15 m/s

Explanation: Use v = u + at = 5 + 2 x 3 = 11 m/s.

34. The expression (1/2)εE² where ε is permittivity and E is electric field has dimensions of:

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

Explanation: (1/2)εE² represents energy density (energy per unit volume) = [ML²T⁻²]/[L³] = [ML⁻¹T⁻²] .

35. The scalar triple product A·(B × C) represents:

• A. Area of parallelogram
• B. Volume of parallelepiped (Correct)
• C. Length of diagonal
• D. Surface area of cube

Explanation: The scalar triple product A·(B × C) gives the volume of the parallelepiped formed by vectors A, B, and C as edges.

36. In the nth second, a body starting from rest covers a distance of (2n − 1) units. The acceleration is:

• A. 1 unit/s²
• B. 2 unit/s² (Correct)
• C. 3 unit/s²
• D. 0.5 unit/s²

Explanation: Distance in nth second: sₙ = u + a(2n−1)/2 = 0 + a(2n−1)/2. Given sₙ = 2n−1: a/2 = 1 → a = 2 unit/s² .

37. A projectile is fired at 60° with speed 40 m/s (g=10 m/s²). Time to reach maximum height is:

• A. 2 s
• B. 4 s
• C. 2√3 s (Correct)
• D. 4√3 s

Explanation: t = usinθ/g = 40×sin60°/10 = 40×(√3/2)/10 = 2√3 s .

38. A person stands on a scale in an elevator. The scale reads zero when:

• A. Elevator is at rest
• B. Elevator moves up at constant speed
• C. Elevator is in free fall (Correct)
• D. Elevator accelerates upward

Explanation: In free fall, apparent weight = m(g−g) = 0. The scale reads zero — this is weightlessness.

39. 5 kg body slides frictionless incline (h = 4 m, g = 10 m/s²). Speed at bottom:

• A. √80 m/s (Correct)
• B. √40 m/s
• C. √20 m/s
• D. 4√5 m/s

Explanation: v = √(2gh) = √80 m/s ≈ 8.94 m/s.

40. A body has both translational (KE_t) and rotational (KE_r) KE. Total KE:

• A. KE_t − KE_r
• B. KE_t × KE_r
• C. KE_t + KE_r (Correct)
• D. KE_t / KE_r

Explanation: Total KE = KE_translational + KE_rotational = KE_t + KE_r .

41. Gravitational potential at distance r from mass M:

• A. GM/r
• B. −GM/r (Correct)
• C. GM/r²
• D. −GM/r²

Explanation: V = −GM/r. Gravitational potential is always negative .

42. Elastic potential energy per unit volume = ½ × Stress × Strain = ½ × Stress²/Y. This equals:

• A. ½ × Y × strain² (Correct)
• B. Y × strain
• C. Y × stress
• D. ½ × Y/strain²

Explanation: u = ½ × stress × strain = ½ × Y × strain². Both forms are equivalent. = ½ × Y × strain² .

43. Tank (H=2 m full) with hole at depth h=1 m from top. Horizontal range on ground (floor at base of tank):

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

Explanation: v = √(2gh) = √(2×10×1) = √20. Height to fall = H−h = 1 m. t = √(2×1/10) = √0.2. Range = v×t = √20×√0.2 = √4 = 2 m .

44. 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.

45. In an experiment, the refractive index n = sin((A+D)/2) / sin(A/2) where A = 60° ± 1° and D = 30° ± 1°. The error in n arises from:

• A. Error in A only
• B. Error in D only
• C. Error in both A and D (Correct)
• D. Cannot be determined

Explanation: Both A and D appear in the formula. The total error in n is a combined effect of errors in both prism angle A and angle of minimum deviation D. Both contribute to the uncertainty in n.

46. î × ĵ equals:

• A. 1
• B. −k̂
• C. k̂ (Correct)
• D. 0

Explanation: î × ĵ = k̂ (right-hand rule). The cyclic order is î → ĵ → k̂ → î.

47. The velocity of a particle is v = √(4 + 4x), where x is in metres. The acceleration is:

• A. 2 m/s² (Correct)
• B. 4 m/s²
• C. 1 m/s²
• D. 8 m/s²

Explanation: a = v(dv/dx). v² = 4 + 4x → 2v(dv/dx) = 4 → v(dv/dx) = 2. So a = 2 m/s² (constant).

48. When a stone whirled in a circle breaks free from the string, it:

• A. Falls vertically
• B. Moves in a circular path
• C. Flies off tangentially (Correct)
• D. Moves toward center

Explanation: When centripetal force disappears, no horizontal force acts. The stone flies in the direction of its instantaneous velocity: tangentially .

49. A body moves up an incline (angle θ) with deceleration a. The coefficient of kinetic friction μₖ is:

• A. (a−g sinθ)/(g cosθ) (Correct)
• B. (a+g sinθ)/(g cosθ)
• C. (g sinθ − a)/(g cosθ)
• D. g cosθ/(a+g sinθ)

Explanation: Going up: net deceleration a = g sinθ + μₖg cosθ. So μₖ = (a−g sinθ)/(g cosθ) .

50. Spring (k=1000 N/m, x=0.1 m) fires 10 g ball. Ball's speed:

• A. 5 m/s
• B. √1000 m/s (Correct)
• C. 10 m/s
• D. √500 m/s

Explanation: ½kx² = ½mv². v² = kx²/m = 1000×0.01/0.01 = 1000. v = √1000 m/s ≈ 31.6 m/s.

51. Condition for rotational equilibrium:

• A. Σv = 0
• B. Σa = 0
• C. Στ = 0 (Correct)
• D. ΣL = 0

Explanation: For rotational equilibrium, net torque = Στ = 0 .

52. Relation between gravitational field (g) and gravitational potential (V):

• A. g = V/r
• B. g = −dV/dr (Correct)
• C. V = −g/r
• D. g = dV/dr

Explanation: g = −dV/dr . Gravitational field is negative gradient of potential.

53. Elastic PE stored in wire (Y=2×10¹¹ Pa, V=1 cm³, strain=10⁻³):

• A. 100 J
• B. 0.1 J (Correct)
• C. 10 J
• D. 0.01 J

Explanation: u = ½Yε² = ½×2×10¹¹×10⁻⁶ = 10⁵ J/m³. PE = u×V = 10⁵×10⁻⁶ = 0.1 J .

54. Aeroplane wings are designed so that air moves faster over top. This creates:

• A. Upward lift (lower pressure on top) (Correct)
• B. Downward force
• C. No pressure difference
• D. Increased drag

Explanation: Faster air on top → lower pressure on top. Pressure difference → upward lift (Bernoulli effect).

55. 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.

56. The ratio h/e (Planck's constant / electronic charge) has dimensions of:

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

Explanation: [h] = [ML²T⁻¹], [e] = [AT]. So [h/e] = [ML²T⁻¹]/[AT] = [ML²T⁻²A⁻¹] .

57. The distance covered by a particle with v = 2t from t = 0 to t = 4 s is:

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

Explanation: s = ∫₀⁴ 2t dt = [t²]₀⁴ = 16 − 0 = 16 m .

58. The displacement of a particle is s = 6 + 12t − 2t². The particle comes to rest at t =

• A. 1 s
• B. 2 s
• C. 3 s (Correct)
• D. 6 s

Explanation: v = ds/dt = 12 − 4t. Setting v = 0: 12 − 4t = 0 → t = 3 s .

59. At what angle to horizontal should a cliff projectile (height H) be launched so it lands at greatest horizontal distance?

• A. 30°
• B. 45°
• C. Less than 45° (Correct)
• D. Greater than 45°

Explanation: For projectiles from height, the optimal angle for maximum range is less than 45° . The exact angle depends on the height H and initial speed u.

60. The property of a body to resist change in its state is called:

• A. Force
• B. Momentum
• C. Inertia (Correct)
• D. Friction

Explanation: Inertia is the property of a body to resist any change in its state of rest or motion. It is measured by mass.