Q. A piece of wood has dimensions $a,\ b\ and\ c.$ Its relative density is d. It is floating in water such that the side c is vertical. It is now pushed down gently and released. The time period is
A) $T = 2\pi\sqrt{\left( \frac{abc}{g} \right)}$
B) $T = 2\pi\sqrt{\left( \frac{b\ a}{dg} \right)}$
C) $T = 2\pi\sqrt{\left( \frac{g}{dc} \right)}$
D) $T = 2\pi\sqrt{\left( \frac{a\ c}{g} \right)}$
Explanation
(d) Let the distance of vertical disc $c$ of block be pushed in liquid, when block is floating, then Buoyancy force
$= abxx_{\omega},\ g = abxg$
The mass of piece of wood$= abcd$
So acceleration$= - ab\frac{xg}{abcd} = - \left( \frac{g}{cd} \right)x$
Hence, time period, $T = 2\pi\sqrt{\frac{dc}{g}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.