A block released from rest from the top of a smooth inclined plane of angle `theta_1` reaches the bottom in time `t_1`. The same block released 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 the two inclined planes are
`l_1 = (h)/( sin theta_1 ) and l_2 =(h )/( sin theta_2)`
acceleration 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 = 1/2 a_2 t_(2)^(2)`
` therefore (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_1) = ( g sin theta_1) /( g sin theta_2) xx ( sin theta_1)/( sin theta_2) = ( sin^2 theta_1)/( sin^2 theta_2)`
p or ` (t_2)/(t_1)= ( sin theta_1)/( sin theta_2)`