The period of a simple pendulum will increase by what factor of its inextensible length increased by a factor of four ____________
A. \(2 \pi\)
B. 4
C. 2
D. \(1 / 4\)
Correct Answer: C
Explanation
\(T=2 \pi \sqrt{\frac{l}{g}}\)
i.e. \(T \alpha \sqrt{l}\)
i.e. \(T \alpha \sqrt{l} ; \frac{T_{1}}{T_{2}}=\sqrt{\frac{l_{1}}{l_{2}}}\)
\(\frac{T_{1}^{2}}{T_{2}^{2}}=\frac{I_{1}}{l_{2}} ; \frac{T_{2}^{2}}{T_{1}^{2}}=\frac{l_{2}}{l_{1}}\) now if \(l_{2}=4 l_{1}\)
\(T_{2}^{2}=\frac{4 l_{1}}{I_{1}} \times T_{1}^{2} ; T_{2}=\sqrt{4 T_{1}^{2}}=2 T_{1}\)
\(T_{2}\) will increase by 2
