Question

A simple pendulum has a time period T in vacuum. Its time period when it is completely immersed in a liquid of density one-eighth of the density of material of the bob is

• A. $\sqrt{\frac{7}{8}}T$
• B. $\sqrt{\frac{5}{8}}T$
• C. $\sqrt{\frac{3}{8}}T$
• D. $\sqrt{\frac{8}{7}}T$

Answer 1

Answer: (D)

$\sqrt{\frac{8}{7}}T$

Answer 2

Let l be the length of simple pendulum, V be the volume and $\rho$ be the density of the material of bob.
$\therefore \, \, \, \, \, \,$ Density of the liquid $= \frac{\rho}{8}$
Time period of simple pendulum in vacuum
$\, \, \, \, \, \, \, \, \, \, \, T=2\pi \sqrt{\frac{l}{g}}$
When the simple pendulum is completely immersed in a liquid then forces acting on bob are :
(i) V$\rho$g, in downward direction
(ii) V $\frac{\rho}{8}g$, in upward direction
Therefore, the resultant force acting on bob in downward direction = $V\rho g -V\frac{\rho}{8}g$
$\, \, \, \, \, \, \, \, \, \, \, =\frac{7V\rho g}{8}$
Therefore, effective value of g in liquid is$\frac{7}{8}g.$
$\therefore \, \, \, \,$ Time penirods, T'=$2\pi \sqrt{\frac{l}{(7g/8)}}$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\sqrt{\frac{8}{7}}T$