Question

A cuboidal piece of wood has dimensions $a$, $b$ and $c$. Its relative density is $d$. It is floating in a large body of water such that the side $a$ is vertical. It is pushed down a bit and released. The time period of SHM executed by it is
A. $2\pi \sqrt {\dfrac{{abc}}{g}}$
B. $2\pi \sqrt {\dfrac{g}{{da}}}$
C. $2\pi \sqrt {\dfrac{{bc}}{{dg}}}$
D. $2\pi \sqrt {\dfrac{{da}}{g}}$

Answer 1

Use the law of floatation of an object floating on a liquid. From this law determine the length of the wooden cube immersed in the water. Use this length as the length of the oscillation in the formula for the time period of the simple pendulum and determine the time period of the SHM executed by the cube.

Formulae used:
The time period $T$ of simple pendulum is
$T = 2\pi \sqrt {\dfrac{L}{g}}$ …… (1)
Here, $L$ is the length of the simple pendulum and $g$ is the acceleration due to gravity.
The density $\rho$ of an object is
$\rho = \dfrac{M}{V}$ …… (2)
Here, $M$ is the mass of the object and $V$ is the volume of the object.
The upward thrust $F$ acting on an object floating on water is
$F = \rho Vg$ …… (3)
Here, $\rho$ is density of water, $V$ is the volume of water displaced by the floating object and $g$ is acceleration due to gravity.

Complete step by step answer:
We have given that the dimensions of the wooden cube are $a$, $b$ and $c$. The relative density of the wooden cube is $d$. The length of the wooden block immersed in the water is the length of the cube for oscillations.
Let $M$ be the mass of the wooden cube.
The weight of the wooden cube is
$W = Mg$
According to equation (2), the above equation becomes
$W = dVg$
Substitute $abc$ for $V$ volume of the cube in the above equation.
$W = dabcg$
The buoyant force acting on the wooden cube is given by equation (3).
$F = {\rho _w}\left( {ybc} \right)g$
Here, ${\rho _w}$ is density of water and $y$ is the length of the wooden cube immersed in water from the length $a$.
Substitute $1\,{\text{g/cc}}$ for ${\rho _w}$ in the above equation.
$F = \left( {1\,{\text{g/cc}}} \right)\left( {ybc} \right)g$
$\Rightarrow F = ybcg$
According to the law of floatation, the weight of the wooden block immersed in water is equal to the buoyant force acting on the wooden cube immersed in the water.
$W = F$
Substitute $abcdg$ for $W$ and $ybcg$ for $F$ in the above equation.
$abcdg = ybcg$
$\Rightarrow y = da$
Therefore, the length of oscillation of the cube in the water is $da$.
We can determine the time period of oscillation of the cube using equation (1).
Rewrite equation (1) for the time period of oscillation of the cube.
$T = 2\pi \sqrt {\dfrac{y}{g}}$
Substitute $da$ for $y$ in the above equation.
$\therefore T = 2\pi \sqrt {\dfrac{{da}}{g}}$
Therefore, the time period for the SHM of the cube is $2\pi \sqrt {\dfrac{{da}}{g}}$.

Hence, the correct option is D.

Note: The students should be careful while using floatation law. The value of the volume of water displaced by the wooden cube in the formula for upward thrust acting on the wooden cube is equal to the volume of the cube immersed in the water and not the total volume of the wooden cube.