Question

A bob of simple pendulum is suspended by a thread of length $l=20\text{cm}$ , if it is placed in a liquid whose density is $\eta$ times less than that of the bob. If $\eta =3$, the time period of small oscillations is nearly equal to
Neglect the viscosity of liquid. Take $g=10{\text{m/s}}^2$

• A. $1.1\text{s}$
• B. $2\text{s}$
• C. $3.1\text{s}$
• D. $4.2\text{s}$

Answer 1

The correct option is A $1.1\text{s}$

Given,
Length of simple pendulum $l=20\text{cm}$
$\eta =3$
Let $V=$ volume of pendulum bob,
$\rho =$ density of material of the bob.
$g^{\prime }=$ effective value of acceleration due to gravity as effective weight of bob will change

$\text{density of liquid}=\frac{1}{\eta }\times \text{density of material of bob}$
$\Rightarrow \sigma =\frac{\rho }{\eta }$
$\text{Effective weight}=\text{weight of bob - upthrust of liquid}$
$\Rightarrow m{g}^{\prime }=mg-F_B$
$V\rho g^{\prime }=V\rho g-V\sigma g$
$V\rho g^{\prime }=V\rho g-\frac{V\rho g}{\eta }$
$\Rightarrow g^{\prime }=(1-\frac{1}{\eta })g$
Applying the formula for time period of pendulum placed in liquid:
$T^{\prime }=2\pi \sqrt{\frac{l}{g^{\prime }}}$
substituting value of $g^{\prime }$,
$T^{\prime }=2\pi \sqrt{\frac{\eta l}{(\eta -1)g}}$
$T^{\prime }=2\times 3.14\times \sqrt{\frac{3\times 20\times {10}^{-2}}{(3-1)\times 10}}$
$\therefore T^{\prime }\simeq 1.1\text{s}$