A) \[\frac{5}{2}a\]
B) \[\frac{3}{2}a\]
C) \[\frac{7}{2}a\]
D) \[\frac{1}{2}a\]
Correct Answer: B
Solution :
We know that period of a physical pendulum \[T=2\pi \sqrt{\frac{{{l}_{0}}}{mgd}}=2\pi \sqrt{\frac{\frac{1}{2}m{{a}^{2}}+m{{a}^{2}}}{mga}}=2\pi \sqrt{\frac{3a}{2g}}\] T for simple pendulum \[=2\pi \sqrt{\frac{l}{g}}\] ?(ii) From Eqs. (i) and (ii), we get \[l=\frac{3}{2}a\]
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