The length of a simple pendulum executing simple harmonic motion is increased by `21%`. The percentage increase in the time period of the pendulum of increased length is.

Options

A. `10%`
B. `21%`
C. `30%`
D. `50%`

Answer

A

Explanation

If initial length `l_(1) = 100` then `t_(2) = 121`
By using `T = 2pi sqrt((1)/(g)) rrArr (T_(1))/(T_(2)) = sqrt((I_(1))/(I_(2)))`
Hence `(T_(1))/(T_(2)) = sqrt((100)/(121)) rArr T_(2) = 1.1T_(1)`
`%` increase `= (T_(2) - T_(1))/(T_(1)) xx 100= 10%`
Alternative Time period of simple pendulum
`T = 2pi sqrt((1)/(g))`
`:. (Delta T)/(T) = (1)/(2) (DeltaI)/(I)`
since `(DeltaI)/(I) = 21%`
`:.Delta T)/(T) = (1)/(2) xx 21% = 10%`