Question

A simple pendulum with a brass bob has a period T. The bob is now immersed in a non-viscous liquid and oscillated. If the density of the liquid is $\frac{1}{8}th$ of brass, the time period of the same pendulum will be

• A. $\sqrt{\frac{8}{7}}T$
• B. $\frac{8}{7}T$
• C. $\frac{64}{49}T$
• D. T

Answer 1

The correct option is A $\sqrt{\frac{8}{7}}T$

$m{g}^{\prime }=(m-M_0)g=m(1-\frac{m_0}{m})g{g}^{\prime }(1-\frac{{\rho }_0}{\rho })g=(1-\frac{1}{8})g=\frac{7}{8}g;[\because {\rho }_g=\frac{\rho }{8}]T^{\prime }=2\pi \sqrt{\frac{l}{g^{\prime }}}=2\pi \sqrt{\frac{l}{g}\times \frac{8}{7}}=\sqrt{\frac{8}{7}}T[\therefore T=2\pi \frac{\sqrt{l}}{g}]$