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A block at rest slides down a smooth inc...

A block at rest slides down a smooth inclined plane which makes an angle `60^(@)` with the vertical and it reaches the ground in `t_(1)` seconds. Another block is dropped vertically from the same point and reaches the ground in `t_(2)` seconds. Then the ratio of `t_(1) : t_(2)` is

A

`1 :2`

B

`2 :1`

C

`1 : 3`

D

`1 : sqrt2`

Text Solution

Verified by Experts

The correct Answer is:
B
Let L be the length and H be height of the inclined plane respectively

Acceleration of the block slide down the smooth incline plane is
`a = g cos 60^(@)`
`therefore L = (1)/(2) g cos 60^(@) t_(1)^(2) " " [ because u = 0] .... (i)`
Acceleration of another block dropped vertically down from the same inclined plane is
a = g
`therefore H = (1)/(2) at_(2)^(2) = (1)/(2) "gt"_(2)^(2) " " [ because u = 0]`
From figure ,
`cos 60^(@) = (H)/(L) implies H = L cos 60^(@)`
`therefore L cos 60^(@) = (1)/(2) "gt"_(2)^(2) " " .... (ii)`
Divide (i) by (ii) we get
`(t_(1)^(2) cos 60^(@))/(t_(2)^(2)) = (1)/(cos 60^(@))`
`implies (t_(1)^(2))/(t_(2)^(2)) = (1)/(cos^(2) 60^(@)) = (4)/(1)`
`implies (t_(1))/(t_(2)) = (2)/(1)`
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