Unit 2: Kinematics — Online MCQ Test
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Practice Unit 2: Kinematics with a free chapter-wise online MCQ test.
This chapter covers: The frame of reference motion in a straight line Position- time graph speed and velocity; Uniform and non-uniform motion average speed and instantaneous velocity uniformly accelera....
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Unit 2: Kinematics — Important Questions & Answers
A ball is thrown vertically downward with a velocity of 20 m/s from the top of a tower. It hits the ground after some time with a velocity of 80 m/s. The height of the tower is: (g= 10 m/s²)
- A. 340 m
- B. 320 m
- C. 300 m
- D. 360 m
Answer: C. 300 m
Using v² = u² + 2gh, where v=80, u=20, g=10. 80² = 20² + 2(10)h => 6400 = 400 + 20h => 6000 = 20h => h = 300 m.
Using v² = u² + 2gh, where v=80, u=20, g=10. 80² = 20² + 2(10)h => 6400 = 400 + 20h => 6000 = 20h => h = 300 m.
There are two inclined surfaces of equal length (L) and same angle of inclination 45º with the horizontal. One of them is rough and the other is perfectly smooth. A given body takes 2 times as much time to slide down on rough surface than on the smooth surface. The coefficient of kinetic friction (μk) between the object and the rough surface is close to
- A. 0.25
- B. 0.40
- C. 0.5
- D. 0.75
Answer: D. 0.75
Time t = sqrt(2L/a). t_rough = 2 * t_smooth. So acceleration a_smooth = 4 * a_rough. a_smooth = g sin 45, a_rough = g(sin 45 - μ cos 45). Solving g/sqrt(2) = 4 * g(1 - μ)/sqrt(2) leads to μ = 0.75.
Time t = sqrt(2L/a). t_rough = 2 * t_smooth. So acceleration a_smooth = 4 * a_rough. a_smooth = g sin 45, a_rough = g(sin 45 - μ cos 45). Solving g/sqrt(2) = 4 * g(1 - μ)/sqrt(2) leads to μ = 0.75.
In some appropriate units, time (t) and position (x) relation of a moving particle is given by t = x^2 + x. The acceleration of the particle is
- A. -3/2(x+2)^-3
- B. -3/2(2x+1)^-3
- C. 3/2(x+1)^-3
- D. 2/21+x
Answer: B. -3/2(2x+1)^-3
dt/dx = 2x + 1. v = dx/dt = 1/(2x+1). a = dv/dt = (dv/dx)*(dx/dt) = d/dx(2x+1)^-1 * (2x+1)^-1 = -1(2x+1)^-2 * 2 * (2x+1)^-1 = -2/(2x+1)^3.
dt/dx = 2x + 1. v = dx/dt = 1/(2x+1). a = dv/dt = (dv/dx)*(dx/dt) = d/dx(2x+1)^-1 * (2x+1)^-1 = -1(2x+1)^-2 * 2 * (2x+1)^-1 = -2/(2x+1)^3.
Two cities X and Y are connected by a regular bus service with a bus leaving in either direction every T min. A girl driving at 60 km/h from X to Y notices a bus passing her every 30 minutes in her direction and every 10 minutes in the opposite. Find T and bus speed v.
- A. 9 min, 40 km/h
- B. 25 min, 100 km/h
- C. 10 min, 90 km/h
- D. 15 min, 120 km/h
Answer: D. 15 min, 120 km/h
Relative speed in same direction: v-60 = d/30. Relative speed in opposite: v+60 = d/10. Solving for v and T (where T=d/v) yields 15 min and 120 km/h.
Relative speed in same direction: v-60 = d/30. Relative speed in opposite: v+60 = d/10. Solving for v and T (where T=d/v) yields 15 min and 120 km/h.
A ball is thrown vertically downward with a velocity of 20 m/s from the top of a tower. It hits the ground after some time with a velocity of 80 m/s. The height of the tower is: (g = 10 m/s²)
- A. 320 m
- B. 300 m
- C. 360 m
- D. 340 m
Answer: B. 300 m
Using v² = u² + 2gh: (80)² = (20)² + 2×10×h → 6400 = 400 + 20h → 20h = 6000 → h = 300 m.
Using v² = u² + 2gh: (80)² = (20)² + 2×10×h → 6400 = 400 + 20h → 20h = 6000 → h = 300 m.