Question

If the length of the simple pendulum is increased by 44%, then what is the change in time period of pendulum?

• A. 22%
• B. 20%
• C. 33%
• D. 44%

Answer 1

The correct option is B 20%

Times period of a simple pendulum is directly proportional to the square root of the length of the pendulum.
The relation for time period of a simple pendulum id
$T=2\pi \sqrt{\frac{l}{g}}$ .... (i)
or $T\propto \sqrt{l}$ ...(ii)
Hence, $T^{\prime }\propto \sqrt{l^{\prime }}$ ...(iii)
(Given, $l^{\prime }=l+44\%$ of $l=l+\frac{44}{100}l=\frac{144}{100}l$)
Putting given value of $l^{\prime }$ in Eq. (iii), we get
$T^{\prime }=\sqrt{\frac{144}{100}}l$ ...(iv)
From Eqs. (ii) and (iv), we get
$\frac{T^{\prime }}{T}=\sqrt{\frac{144}{100}}=\frac{12}{10}$
Hence, $T^{\prime }=\frac{12}{10}T$
Now percentage change in time period is
$(\frac{T^{\prime }-T}{T}\times 100)=\left(\frac{\frac{12}{10}T-T}{T}\right)\times 100$
$\left(\frac{T^{\prime }-T}{T}\right)\times 100=\frac{2}{10}\times 100=20\%$