1.

A simple pendulum has a time period T_(1)when on the earth's surface and T_(2) when taken to a height R above the earth's surface where R is the radius of the earth. The value of T_(2)//T_(1) is

Answer»

1
`sqrt(2)`
4
2

Solution :`T_(1) = 2PI sqrt(L/g), T_(2) = 2pi sqrt(l/(g//4)) = 2 xx 2pi sqrt(l/g) = 2T_(2)`
`RARR T_(2) = 2T_(1) rArr T_(2)/T_(1) =2`


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