A) \[T\times \frac{\sqrt{3}}{2}\,s\]
B) \[T\times \frac{2}{\sqrt{3}}\,s\]
C) \[\frac{T}{2}s\]
D) \[\sqrt{T}\,s\]
Correct Answer: B
Solution :
Effective acceleration due to gravity \[g=g-\frac{g}{4}=\frac{3g}{4}\] Time period \[T=2\pi \sqrt{\left( \frac{l}{g} \right)}\] .? (i) \[T=2\pi \sqrt{\left( \frac{4l}{3g} \right)}=2\pi \times \frac{2}{\sqrt{3}}\sqrt{\left( \frac{l}{g} \right)}\] ... (ii) Now dividing Eq. (ii) by Eq. (i), we get \[\frac{T}{T}=\frac{2}{\sqrt{3}}\] or \[T=T\times \frac{2}{\sqrt{3}}\]
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