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Q.1. A ball is projected with a velocity, 10 ms^{–1} , at an angle of 60° with the vertical direction. Its speed at the highest point of its trajectory will be [2022]
(1) 5√3 ms^{–1}
(2) 5 ms–1
(3) 10 ms–1
(4) Zero
Answer (1)
Sol. At highest point vertical component of velocity become zero.
At highest point speed of object = 10cos30°
= 5√3 ms^{–1}
Q. 2. The angular speed of a fly wheel moving with uniform angular acceleration changes from 1200 rpm to 3120 rpm in 16 seconds. The angular acceleration in rad/s^{2} is [2022]
(1) 4𝝅
(2) 12𝝅
(3) 104𝝅
(4) 2𝝅
Ans: (1)
Sol:
= 4𝝅
Q 3. A car starts from rest and accelerates at 5 m/s2. At t = 4 s, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at t = 6 s ? [2021]
A: 20√2m/s, 0 m/s^{2}
B: 20√2m/s, 10 m/s^{2}
C: 20 m/s,5 m/s^{2}
D: 20 m/s, 0
Ans: B
Solutions:
velocity of car at t = 4 sec is
v = u + at v = 0 + 5(4)
= 20 m/s
At t = 6 sec
acceleration is due to gravity
∴ a = g = 10 m/s
vx = 20 m/s (due to car)
vy = u + at
= 0 + g(2) (downward)
= 20 m/s (downward)
= 20√2 m/s.
Q 4. A particle moving in a circle of radius R with a uniform speed takes a time T to complete one revolution. If this particle were projected with the same speed at an angle 'θ'to the horizontal, the maximum height attained by it equals 4R. The angle of projection,θ, is then given by: [2021]
A:
B:
C:
D:
Ans: B
Solutions:
Q 5. The speed of a swimmer in still water is 20 m/s. The speed of river water is 10 m/s and is flowing due east. If he is standing on the south bank and wishes to cross the river along the shortest path, the angle at which he should make his strokes w.r.t. north is given by: [2019]
A: 30° west
B: 0° west
C: 60° west
D: 45° west
Ans: A
Solution:
V_{SR} = 20m/s
V_{RG} = 10m/s
sin θ = 10/20
sin θ = 1/2
θ = 30^{0} west
Q 6. A particle moving with velocity is acted by three forces shown by the vector triangle PQR. The velocity of the particle will: [2019]
A: Increase
B: Decrease
C: Remain constant
D: Change according to the smallest force
Ans: C
Solution
As forces are forming closed loop in same order
Q 7. The x and y coordinates of the particle at any time are x = 5t – 2t^{2} and y = 10t respectively, where x and y are in meters and t in seconds. The acceleration of the particle at t = 2s is: [2017]
A: 5 m/s^{2}
B: – 4 m/s^{2}
C: – 8 m/s^{2}
D: 0
Ans: B
Solution:
Q 8. One end of string of length l is connected to a particle of mass 'm' and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in circle with speed 'v' the net force on the particle (directed towards centre) will be (T represents the tension in the string): [2017]
A:
B:
C: Zero
D: T
Ans: D
Solution: Net force on the particle in uniform circular motion is centripetal force, which is provided by the tension in string.
Q 9. If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the angle between these vectors is: [2016]
A: 180 deg
B: 0 deg
C: 90 deg
D: 45 deg
Ans: C
Solution:
Q 10. A ship A is moving Westwards with a speed of 10 km h^{1} and a ship B 100 km South of A, is moving Northwards with a speed of 10 km h^{1}. The time after which the distance between them becomes shortest, is: [2015]
A: 10√2h
B: 0 h
C: 5h
D: 5√2h
Ans: C
Solution:
Q 11. A particle is moving such that its position coordinates (x, y) are (2m, 3m) at time t = 0, (6m, 7m) at time t = 2s and (13m, 14m) at time t= 5s. Average velocity vector from t = 0 to t = 5s is: [2014]
A:
B:
C:
D:
Ans: B
Solution:
Average velocity Vector,
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