1.

A block released fromrestfromthe top of a smooth inclined planeof angletheta_(1)reaches the bottom int_(1) . The same block relased from rest from the top of another smooth inclined planeof angletheta_(2) , reachesthe bottom in timet_(2). If thetwo inclined planes havethe same height, the relation betweent_(1) and t_(2)is

Answer»

` t_(2)/ t_(1) =(sin theta_(1))/(sin theta_(2))^(1/2)`
`t_(2)/t_(1) = 1 `
`t_(2)/t_(1) = (sin theta_(1))/(sin theta_(2))`
`t_(2)/t_(1)= (sin^(2) theta_(1))/(sin^(2)theta_(2))`

SOLUTION :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 planesare` 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))`


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