1.

A lift is ascending with acceleration (g)/(3). Find the time period of simple pendulum of length L kept in lift.

Answer»

`2PI sqrt((3L)/(g))`
`pi sqrt((3L)/(g))`
`2pi sqrt((3L)/(2g))`
`2pi sqrt((2L)/(3g))`

Solution :`implies g_(eff)= g-(g)/(3)= (2g)/(3)`
`therefore T= 2pi sqrt((L)/(g_(eff)))= 2pi sqrt((L)/(2g"/"3))`
`=2pi sqrt((3L)/(2g))`.


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