The time period of a simple pendulum is T. When the length is increased by 10 cm, its period is `T_(1)`. When the length is decreased by 10 cm, its period is `T_(2)`. Then, the relation between T, `T_(1)` and `T_(2)` is

Options

A. `(2)/(T^(2)) = (1)/(T_(1)^(2)) + (1)/(T_(2)^(2))`
B. `(2)/(T^(2)) = (1)/(T_(1)^(2)) - (1)/(T_(2)^(2))`
C. `2T^(2) = T_(1)^(2) + T_(2)^(2)`
D. `2T^(2) = T_(1)^(2) - T_(2)^(2)`

Answer

C

Explanation

`T alpha sqrt(l)`
`T_(1) alpha sqrt(l + 10) , T_(2) alpha sqrt(l - 10)`
`(T_(1)^(2))/(T^(2)) = (l + 10)/(l), (T_(2)^(2))/(T^(2)) = (l - 10)/(l)`
`T_(1)^(2) + T_(2)^(3) = 2T^(2)`