A block released from rest from the top of a smooth inclined plane of angle ` theta_(1)` reaches the bottom in `t_(1)` .
The same block relased from rest from the top of another smooth inclined plane of angle ` theta_(2)` , reaches the bottom in time ` t_(2)` . If the two inclined planes have the same height, the relation between ` t_(1) and t_(2) ` is

Options

A. ` t_(2)/ t_(1) = (sin theta_(1))/(sin theta_(2))^(1/2)`
B. `t_(2)/t_(1) = 1 `
C. `t_(2)/t_(1) = (sin theta_(1))/(sin theta_(2))`
D. `t_(2)/t_(1)= (sin^(2) theta_(1))/(sin^(2)theta_(2))`

Answer

C

Explanation

Lengths of two inclined planes are
`l_(1) = h/(sin theta_(1)) and l_(2) = h/(sin theta_(2))`
Accelerations of the block down the two planes are ` a_(1) =g sin theta_(1) and a_(2) = g sin theta_(2)`
As ` l_(1) = 1/2 a_(1)t_(1)^(2) and l_(2) a_(2)t_(2)^(2)`
` l_(1)/l_(2) = (a_(1)t_(1)^(2))/(a_(2)t_(2)^(2)) or t_(2)^(2)/(t_(1)^(2)) = (a_(1)l^(2))/(a_(2)l^(2)) = ( sin theta_(1)/( g sin theta_(2)) xx ( sin theta_(1))/(sin theta_(2))`
` t_(2)/t_(1) = (sin theta_(1))/( sin theta_(2))`