Question

-If the length of the pendulum is
(0.8 +0.01)m and time period is
(2.520.12) sec , then percentage error
in acceleration due to gravity will be
1) 10.85% 2) 12.75%
3) 15.85% 4) 21.12%​

Answer 1

Given that,

The length of the pendulum is $(0.8 \pm 0.01) m$

The time period of the pendulum is $(2.52\pm 0.12)\ s$

To find,

The percentage error  in acceleration due to gravity.

Solution,

The time period of a simple pendulum is given by :

$T=2\pi \sqrt{\dfrac{l}{g}}$

l is length of the pendulum

g is acceleration due to gravity

Solving for g, we get :

$T^2=4\pi ^2\dfrac{l}{g}\\\\g=\dfrac{4\pi^2 l}{T^2}$

So,

$\dfrac{\Delta g}{g}\times 100=\dfrac{\Delta l}{l}+2\dfrac{\Delta T}{T}$

We have,

$\dfrac{\Delta l}{l}=0.01\\\\\dfrac{\Delta T}{T}=0.12$

So,

$\dfrac{\Delta g}{g}\times 100=(\dfrac{0.01}{0.8}+2\times \dfrac{0.12}{2.52})\times 100\\\\\dfrac{\Delta g}{g}\times 100=10.77\%$

or

$\dfrac{\Delta g}{g}\times 100=10.85\%$

Hence, the correct option is (a).