A) \[T\]
B) \[\sqrt{\frac{9}{10}}T\]
C) \[\sqrt{\frac{10}{9}T}\]
D) \[\frac{T}{10}\]
Correct Answer: C
Solution :
Net weight of pendulum equals \[w=w-{{w}_{th}}\] \[\Rightarrow \] \[w=V\rho g-\frac{V\rho g}{10}\] \[\Rightarrow \] \[w=\frac{9}{10}w\] \[\Rightarrow \] \[w=\frac{9}{10}mg\] \[(\because w=mg)\] \[\Rightarrow \] \[g=\frac{9}{10}g\] \[\therefore \] \[T=2\pi \sqrt{\frac{l}{g}}\Rightarrow T\sqrt{\frac{10}{9}}T\]
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