Question

A pendulum is constructed as a thin walled sphere of radius R filled up with water and suspended from a point O. Centre of sphere from O is l apart. How many times will the time period of small oscillation change when the water freezes?

• A. $\frac{2}{5}{\left(\frac{R}{l}\right)}^2$
• B. $1+\frac{2}{5}{\left(\frac{R}{l}\right)}^2$
• C. $\sqrt{1+\frac{2}{5}{\left(\frac{R}{l}\right)}^2}$
• D. $\sqrt{\frac{1}{1+\frac{2}{5}{\left(\frac{R}{l}\right)}^2}}$

Answer 1

Answer: (C)

$\sqrt{1+\frac{2}{5}{\left(\frac{R}{l}\right)}^2}$

Case (i): when water is in liquid state and its viscosity is zero the pendulum behaves like a normal pendulum so its time period will be
$T=2\pi \sqrt{\frac{l}{g}}$
Case (ii): when the water freezes the sphere will become solid and it will be like a physical pendulum
$T_1=2\pi \sqrt{\frac{I_{support}}{mg{l}_{cm}}}=2\pi \sqrt{\frac{m{l}^2+\frac{2}{5}m{R}^2}{mgl}}{T}_1=2\pi \sqrt{\frac{l}{g}}\sqrt{1+\frac{2}{5}{\left(\frac{R}{l}\right)}^2}{T}_1=T\sqrt{1+\frac{2}{5}{\left(\frac{R}{l}\right)}^2}$
So the correct answer is 'C'