Question

A simple pendulum is used to determine the acceleration due to gravity at the surface of a planet. The pendulum has a length of 2 m and its period is measured to be 2 s. The value of g obtained in this investigation is most nearly __________.A. 1 m/s²B. 2 m/s²C. 5 m/s²D. 10 m/s²E. 20 m/s²

Answer 1

Acceleration due to gravity is 20
`m/sec^2`

So option (E) will be correct answer

Step-by-step explanation:

We have given length of the pendulum l = 2 m

Time period of the pendulum T = 2 sec

We have to find acceleration due to gravity g

We know that time period of pendulum is given by

`T=2\pi \sqrt{(l)/(g)}`

`2=2* 3.14 \sqrt{(2)/(g)}`

`0.3184= \sqrt{(2)/(g)}`

Squaring both side

`0.1014= {(2)/(g)}`

`g=19.71=20m/sec^2`

So acceleration due to gravity is 20
`m/sec^2`

So option (E) will be correct answer.