• # question_answer A simple pendulum with a solid metal bob has a period T. The metal bob is now immersed in a liquid having density one-tenth that of the metal of the bob. The liquid is non-viscous. Now the period of the same pendulum with its bob remaining all the time in the liquid will be A) $\frac{9}{10}\,T$                     B) $T\sqrt{\frac{10}{9}}$C) unchanged                    D) $T\sqrt{\frac{9}{10}}$

$T=2\pi \sqrt{\frac{\ell }{g}}$ In a liquid $T'=2\pi \sqrt{\frac{\ell }{g\left\{ 1-\frac{1}{\sigma } \right\}}}$ $\sigma =\frac{density\,\,of\,\,bob}{density\,\,of\,\,liquid}$ $\sigma =10$ $T'=2\pi \sqrt{\frac{\ell }{g\left\{ 1-\frac{1}{\sigma } \right\}}}\,\,=\,\,2\pi \sqrt{\frac{\ell }{g}}\,\sqrt{\frac{10}{9}}$ $T'=T\sqrt{\frac{10}{9}}$