A student measures the value of g with the help of a simple pendulum using the formula ` g = (4pi^(2)L)/(T^(2))`. He measures length L with a meter scale having least count 1 mm and finds it 98.0 cm . The time period is measured with the help of a watch of least count 0.1s . The time of 20 oscillations is found to be 40.4 s. The error `Deltag` in the measurment of g is `(" in" m//s^(2))`.

Options

A. `9.68[(0.1)/(98) +0.1]`
B. `9.86 [(1)/(98) + 0.1]`
C. `9.68 [(0.1)/(98) +(0.1)/(20)]`
D. `9.68 [(1)/(98) +(1)/(20)]`

Answer

a