A particle starts from rest and has an acceleration  of 2 m/

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 Multiple Choice QuestionsMultiple Choice Questions

161. 'Parsec' is the unit of'
  • time

  • distance

  • frequency

  • angular acceleration


162.

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

  • 3 vo

  • 3vo

  • 9vo

  • 3/2 vo


163.

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


164.

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 ]


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165.

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


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166.

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 (α1 ) = 2 m/s2 

time during acceleration (t1) = 10 sec

Time during constant velocity (t2 ) = 30 sec

and retardation ( α2 ) = - 4 m/s2 

( negative sign due to retardation )

Distance covered by the particle during acceleration

    s1 = ut112 α1t1 2

         = ( 0 × 10 ) + 12 × 2 × (10)2

   s1 = 100 m                 ......(i)

And velocity of the particle a the end of acceleration 

  v = u  + 12 α1t1 2   

      = 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 ( s) is

   v2 = u2 + 2α2s3

⇒   ( 0 )2 = ( 20 )2 + 2 × ( - 4 ) × s3

⇒    400 = 8 s3

⇒   s34008

⇒   s3 = 50 m

Therefore total distance covered by the particle 

  s = s1  + s2 + s3

    = 100 +600 + 50

 s = 750 m


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167.

Speed in kilometre per hour in S.I unit is represented as

  • KMPH

  • Kmhr-1

  • Kmh-1

  • kilometre/hour


168.

S.I unit of velocity is 

  • ms

  • m sec-1

  • m hr-1

  • m/hr


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169.

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


170.

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


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