When a ball is thrown up vertically with velocity vo, it reaches a maximum height of h. If one wishes triple the maximum height then the ball should be thrown with velocity
vo
3vo
9vo
3/2 vo
A person is standing in an elevator. In which situation he finds his weight less?
When the elevator moves upward with constant acceleration
When the elevator moves downward with constant acceleration
When the elevator moves upward with uniform velocity
When the elevator moves downward with uniform velocity
Using mass (M), length (L), time (T) and current (A) as fundamental quantities, the dimension of permittivity is
[ M L-2 T2 A ]
[ M-1 L-3 T4 A2 ]
[ M L T-2 A ]
[ M L2 T-1 A2 ]
A body starting from rest moves along a straight line with a constant acceleration. The variation of speed (v) with distance ( s ) is represented by the graph
A particle starts from rest and has an acceleration of 2 m/s2 for 10 sec. After that, it travels for 30 sec with constant speed and then undergoes a retardation of 4 m/s2 and comes back to rest. The total distance covered by the particle is
650 m
750 m
700 m
800 m
B.
750 m
Initial velocity ( u ) = 0
Acceleration ( ) = 2 m/s2
time during acceleration (t1) = 10 sec
Time during constant velocity (t2 ) = 30 sec
and retardation ( ) = 4 m/s2
( negative sign due to retardation )
Distance covered by the particle during acceleration
s1 = ut1 +
= ( 0 × 10 ) + × 2 × (10)2
s1 = 100 m ......(i)
And velocity of the particle a the end of acceleration
v = u +
= 0 + ( 2 × 10 )
v = 20 m/s
Therefore distance covered by the particle during constant velocity
s2 = v × t2
= 20 × 30⇒
s2 = 600 m .....(ii)
Relation for the distance covered by the particle during retardation ( s3 ) is
v2 = u2 + 2s3
⇒ ( 0 )2 = ( 20 )2 + 2 × ( 4 ) × s3
⇒ 400 = 8 s3
⇒ s3 =
⇒ s3 = 50 m
Therefore total distance covered by the particle
s = s1 + s2 + s3
= 100 +600 + 50
s = 750 m
A particle moving with velocity (1/10)th of light will cross a nucleus in about
10-8 s
10-12 s
10-47 s
10-21 s
A coin is dropped in a lift. It takes time t1 to reach the floor when the lift is stationary. It takes times t2 when the lift is moving up with constant acceleration. Then:
t1 > t2
t2 > t1
t1 = t2
t1 >> t2