Question

A simple pendulum with a brass bob has a period $T$. The bob is now immersed in a nonviscous liquid and oscillated. If the density of the liquid is $1/8th$ 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 $\sqrt{\frac{8}{7}}T$

$T=2\pi \sqrt{\frac{l}{g}}$
$F=mg-f_v$
$V\times \rho g-\frac{V\times \rho }{8}g$

$F=\frac{7}{8}\rho Vg$
$=\frac{7}{8}mg$
$\therefore a=\frac{7}{8}g$
$\frac{T=2\pi \sqrt{\frac{l}{g}}}{T^1=2\pi \sqrt{\frac{l}{\frac{7}{8}g}}}$

$T^1=T\sqrt{\frac{8}{7}}$