1.

The period of oscillation of a simple pendulum is T in a stationary lift. It thelift moves upward with acceleration of 8g the time period will :

Answer»

becomes T/2
becomes T/3
remains same
none of these.

Solution :In stationary LIFT `T=2pi sqrt((L)/(g))`
When lift ASCENDS then apparent ACCELERATION is `g.=g+a=g+8g=9g`
`:.""T.=2pi sqrt((l)/(g.))=2pi sqrt((l)/(9g))impliesT.=(T)/(3)`
Correct CHOICE is (b).


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