Courses
Courses for Kids
Free study material
Offline Centres
More
Store

A cylindrical block of wood of mass ‘m’ and cross section ‘A’ is floating in a liquid of density$\sigma$ with its axis vertical. It is depressed a little and then released. Then its frequency is:A. $f = \dfrac{1}{\pi }\sqrt {\dfrac{{A\sigma g}}{m}}$B. $f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{A\sigma g}}{m}}$C. $f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{m}{{A\sigma g}}}$D. $f = \dfrac{1}{\pi }\sqrt {\dfrac{m}{{A\sigma g}}}$

Last updated date: 20th Jun 2024
Total views: 404.4k
Views today: 7.04k
Verified
404.4k+ views
Hint:- Density is defined as the ratio of mass and volume of the body. Frequency is the number of oscillations in one cycle by the body. Whenever anybody is submerged inside any liquid then the upward force that acts on the body is buoyancy force.

Formula used: The formula of buoyancy force that acts on the cylinder which is submerged in a liquid of density$\sigma$ and submerged height h is given by,
${F_B} = \pi {r^2}h\sigma g$
Where h is the height upto which the body is submerged in the liquid. The density is $\sigma$ of the liquid and r is the radius of the cylinder.

Complete step-by-step solution
It is given that a cylindrical block of wood of mass ‘m’ and cross section ‘A’ which is floating with a density $\sigma$ with its axis vertical and it is depressed a little and then released then the frequency of the oscillations has to be calculated.
The cylinder is dipped in the liquid at a height of h and an extra x height the block is depressed and then released then the cylindrical block starts to oscillate. Let us first calculate the value of the weight that the cylindrical block possesses.
Volume of the cylindrical block is given by,
$W = \pi {r^2}h\sigma g$………eq. (1)
Where h is the height of the block inside the liquid $\sigma$ is the density of the liquid and r is the radius of the cylindrical block.
The buoyancy force that acts on the cylindrical block when the block is depressed x meters into the liquid is given by,
${F_B} = \pi {r^2}\sigma g \cdot \left( {h + x} \right)$………eq. (2)
Where h is the height upto which the body is submerged in the liquid. The density is $\sigma$ of the liquid and r is the radius of the cylinder.
Now when the block is x meter more depressed and then released then the net force that acts on the block is given by,
$F = W - {F_B}$
Replace the value of $W$ and ${F_B}$ in the above relation from the equation (1) and equation (2).
$\Rightarrow F = W - {F_B}$
$\Rightarrow F = \left( {\pi {r^2}h\sigma g} \right) - \pi {r^2}\sigma g \cdot \left( {h + x} \right)$
$\Rightarrow F = \pi {r^2}h\sigma g - \pi {r^2}\sigma gh - \pi {r^2}\sigma gx$
$\Rightarrow F = - \pi {r^2}\sigma gx$………eq. (3)
Now the force that acts in the case of simple harmonic motion is given by,
$F = - kx$………eq. (4)
Where k is constant and x is the displacement from the mean position.
Equating the equation (3) and equation (4) we get.
$\Rightarrow - kx = - \pi {r^2}\sigma gx$
$\Rightarrow k = \pi {r^2}\sigma g$………eq. (5)
The frequency of the body in simple harmonic motion is given by,
$f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}}$
Where f is the frequency and k is constant m is mass and f is the frequency. replace the value of k into the above relation.
$\Rightarrow f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}}$
$\Rightarrow f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{\pi {r^2}\sigma g}}{m}}$
Where the area of the cross section is equal to$A = \pi {r^2}$.
$\Rightarrow f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{A\sigma g}}{m}}$

The frequency of the cylindrical block is$f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{A\sigma g}}{m}}$. The correct answer for this problem is option B.

Note:-
The buoyant force that acts on the cylindrical was initially equal to the weight of the block but when the equilibrium was distributed then the oscillations took place. The extra height that the cylindrical block was depressed was the main reason that caused the cylindrical block to oscillate.