Question

# The time period of oscillations of a simple pendulum is $$1$$ minute. If its length is increased by $$44\%$$ then its new time period of oscillation will be

A
96 s
B
58 s
C
12 s
D
72 s

Solution

## The correct option is C $$12\ s$$We know $$T= 2\pi \sqrt{\dfrac{l}{g}}$$$$T \propto \sqrt{l_{1}}$$Given data $$T_{1} = 1\, sec$$$$T_{2}=2$$$$l_{1}=l$$$$l_{2} = 44\% = 100+44 = 144l$$$$\therefore \dfrac{T_{1}}{T_{2}}=\sqrt{\dfrac{l_{1}}{l_{2}}}\Rightarrow \dfrac{1}{T_{2}}=\sqrt{\dfrac{1}{144}}$$$$T_{2}= 12\, sec$$Physics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More