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A block is released from point A as show...

A block is released from point `A` as shown in figure .All surfaces are smooth and there is no loss of mechnical energy anywhere. Find the time period of oscillation of block.

Text Solution

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The correct Answer is:
A, B, C
As there is no loss of mechanical energy, the block rises upto the same height 'h' on other side also.
`AB = (h)/(sinalpha)' BC = (h)/(sinbeta)`
`a_(1) = g sin a`, `a_(2) = g sin beta`
`u_(A) = u_(C) = 0`
Using `t = sqrt((2s)/(a))`
(with `u = 0`)
:. Time period of oscillation,
`T = 2t_(AB) + 2t_(CB) = 2(t_(AB) + t_(CB))`
`= 2 [sqrt((2xx(h)/(sin alpha))/(gsin alpha)) + sqrt((2xx(h)/(sin beta))/(g sin beta))]`
`= 2 sqrt((2h)/(g))[cosec alpha + cosec beta]`
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