Question

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
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 1

Correct Answer - C
`Tpropsqrt(l)`
`T_(1)propsqrt(l+10),T_(2)propsqrt(l-10)`
`(T_(1)^(2))/(T^(2))=(l+10)/(l),(T_(2)^(2))/(T^(2))=(l-10)/(l)`
`(T_(1)^(2))/(T^(2))+(T_(2)^(2))/(T^(2))=(l+10)/(l)+(l-10)/(l)=2`
`(T_(1)^(2))/(T^(2))+(T_(2)^(2))/(T^(2))=2T^(2) therefore T_(1)^(2)+T_(2)^(2)=2T^(2)`