In an experiment of simple pendulum, the errors in the measurement of length of the pendulum `(L)` and time period `(T)` are `3%` and `2%` respectively. The maximum percentage error in the value of `L//T^(2)` is

Options

A. `5%`
B. `7%`
C. `8%`
D. `1%`

Answer

B

Explanation

The time period of simple pendulum is `T= 2pisqrt((L)/(g))`
Squaring both sides, we get `T^(2)= 4pi(L)/(g)` or `g= 4pi^(2)(L)/(T^(2))` …..(i)
the maximum percentage error in g is
`(Deltag)/(g) xx 100= (DeltaL)/(L) xx 100+ 2((DeltaT)/(T)) xx 100= 3%+2xx 2%= 7%`
From (i) we get `(L)/(T^(2))= (g)/(4pi^(2))`
The maximum percentage error in `(L)/(T^(2))` is
`(Delta((L)/(T^(2))))/((L)/(T^(2)))xx 100= (Deltag)/(g)xx100= 7%`