Question

The bob of simple pendulum having density $\frac{4000}{3}{\text{kg/m}}^3$ executes $\text{SHM}$ in water with a period ${}^{\prime }{t}^{\prime }\text{s}$. If the period of oscillation of bob in air is $10\text{s}$, find the value of ${}^{\prime }{t}^{\prime }$.
(Neglect frictional force of water, take density of water as $1000{\text{kg/m}}^3$)

• A. $10$
• B. $20$
• C. $30$
• D. $40$

Answer 1

The correct option is B $20$

Given,
Period of oscillation of bob in air $\left(T\right)=10\text{s}$
Density of bob $\left(\rho \right)=\frac{4000}{3}{\text{kg/m}}^3$
Density of water $\left(\sigma \right)=1000{\text{kg/m}}^3$
We know that,
Time period of oscillation of bob in water
$\left(t\right)=\frac{T}{\sqrt{1-\frac{\sigma }{\rho }}}$
$t=\frac{10}{\sqrt{1-\frac{3000}{4000}}}$
$t=20\text{s}$