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If T(1) and T(2) are the time-periods of...

If `T_(1)` and `T_(2)` are the time-periods of oscillation of a simple pendulum on the surface of earth (of radius R) and at a depth d, the d is equal to

A

`(1-(T_(1)^(2))/(T_(2)^(2)))R`

B

`(1-(T_(2)^(2))/(T_(1)^(2)))R`

C

`(1-(T_(1)/(T_(2)))R`

D

`(1-(T_(2))/(T_(1)))R`

Text Solution

Verified by Experts

The correct Answer is:
A
(a) We know that the time period of a simple pendulum
`T=2pisqrt((l)/(g))`
`impliesT prop (l)/(sqrt(g))`
Here in given condition,
`T_(1)=(k)/(sqrt(g))" "…(i)`
and `T_(2)=(K)/(sqrt(g(1-(d)/(R))))" "...(ii)`
From Eqs. (i) and (ii) , we get
`(T_(1))/(T_(2))=(K//sqrt(3))/(K//sqrt(g(1-(d)/(R))))`
or, `(T_(1))/(T_(2))=sqrt((1-(d)/(R)))`
or, `(T_(1)^(2))/(T_(2)^(2))=1-(d)/(R)`
`implies d=[1-(T_(1)^(2))/(T_(2)^(2))]R`
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