A) \[\sqrt{2}\]
B) 2
C) \[2\sqrt{2}\]
D) None of these
Correct Answer: B
Solution :
The effective acceleration of a bob in water \[=g=g\left[ 1-\frac{\sigma }{\rho } \right]\] where a and \[\rho \] are the densities of water and the bob respectively. Since, the periods of oscillation of the bob in air and water are given as \[T=2\pi \sqrt{\frac{l}{g}}\] and \[T=2\pi \sqrt{\frac{l}{g}}\] \[\therefore \] \[\frac{T}{T}=\sqrt{\frac{g}{g}}=\sqrt{\frac{g\left( 1-\frac{\sigma }{\rho } \right)}{g}}\] \[=\sqrt{1-\frac{\sigma }{\rho }}=\sqrt{1-\frac{1}{\rho }}\] Putting \[\frac{T}{T}=\frac{1}{\sqrt{2}}\] We obtain, \[\frac{1}{2}=1-\frac{1}{\rho }\] \[\Rightarrow \] \[\rho =2\]
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