Question

The length of a simple pendulum executing simple harmonic motion is increased by 21%. The percentage increase in the time period of the pendulum of increased lengths is :

• A. 11%
• B. 21%
• C. 42%
• D. 10%

Answer 1

The correct option is D 10%

$\begin{array}{cc}& \text{original equation:}\\ & \begin{array}{c}T=2=\sqrt{\frac{l}{g}}...........\left(i\right)\\ \text{after increasing}l:\end{array}\\ & T^{\prime }=2\pi \sqrt{\frac{l^{\prime }}{g}}\\ & \text{given}\to \frac{△l}{d}\times 100=21\%\text{.}\end{array}$

$\begin{array}{c}\Rightarrow \Delta l=\frac{21}{100}\\ \text{also,}{l}^{\prime }=l+\Delta l\\ \Rightarrow l^{\prime }=\frac{121}{100}d\\ \text{dividing}\left(i\right)\text{by}\left(ii\right)\end{array}$

$\begin{array}{c}\frac{T}{T^{\prime }}=\sqrt{\frac{l}{l^{\prime }}}\\ \frac{T}{T^{\prime }}=\sqrt{\frac{100}{121}}\\ T^{\prime }=\sqrt{1.21}T\\ \Delta T=T^{\prime }-T\\ =(\sqrt{1.21}-1)T\end{array}$

$\begin{array}{c}\frac{\Delta T}{T}=0.1\\ \text{\% change}=0.1\times 100\%=10\%\end{array}$