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

Options

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

Answer

b

Explanation

If h be the height then length of inclined plane in two
cases is`l_(1)=(h)/(sin theta_(1))`and `l_(2) = (h)/(sin  theta_(2)`and acceleration `a_(1) = g sin theta-(1)`, and `a_(2) = g sin theta_(2)`
`l=ut+(1)/(2)at^(2)`
`l_(1)=0+(1)/(2)a_(1)t_(1)^(2)...(i)`
and`l_(2)=0+(1)/(2)a_(2)t_(2)^(2)...(ii)`
Dividing (ii) by (i)
`(l_(2))/(l_(1))=(a_(2))/(a_(1))xx(t_(2)^(2))/(t_(2)^(2))`
`:.(t_(2)^(2))/(t_(1)^(2))=(l_(2))/(l_(1))xx(a_(20))/(a_(1))=(sin theta_(1))/(sin theta_(2))xx(g sin theta_(1))/(g sintheta_(2))`
`:.(t_(2))/(t_(1))=((sin theta_(1))/(sin theta_(2)))`
Hence correct choice is (b)