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A body is allowed to slide from the top ...

A body is allowed to slide from the top along a smooth inclined plane of length `5m` at an angle of inclination `30^(@)`.If `g=10ms^(-2)`, time taken by the body to reach the bottom of the plane is

A

`(sqrt3)/(2)s`

B

1.414s

C

`(1)/(sqrt2)s`

D

2s

Text Solution

Verified by Experts

The correct Answer is:
B
`t=sqrt((2l)/(gsintheta))`
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Knowledge Check

  • An inclined plane of height h and length l have the angle of inclination theta . The time taken by a body to come from the top of the bottom of this inclined plane will be

    A
    `sin theta sqrt((2h)/(g))`
    B
    `(1)/(sin theta) sqrt((2h)/(g))`
    C
    `sqrt((2h)/(g))`
    D
    `sqrt((2l)/(g))`

  • A body is slipping from an inclined plane of height h and length l . If the angle of inclination is theta , the time taken by the body to come from the top to the bottom of this inclined plane is

    A
    `sqrt((2h)/(g))`
    B
    `sqrt((2l)/(g))`
    C
    `(1)/(sin theta) sqrt((2h)/(g))`
    D
    `sin theta sqrt((2h)/(g))`

  • A block released from rest from the tope of a smooth inclined plane of angle of inclination theta_(1) = 30^(@) reaches the bottom in time t_(1) . The same block, released from rest from the top of another smooth inclined plane of angle of inclination theta_(2) = 45^(@) reaches the bottom in time t_(2) . Time t_(1) and t_(2) are related as ( Assume that the initial heights of blocks in the two cases are equal ).

    A
    `( t_(2))/( t_(1)) = ( 1)/( sqrt( 2))`
    B
    `( t_(2))/( t_(1)) = ( 1)/(2)`
    C
    `(t_(2))/( t_(1)) = 1`
    D
    `(t_(1))/(t_(2)) = ( 1)/(2)`