Question

A solid cube floats in water half immersed and h small vertical oscillations of time period pi/5s. Find the mass (in kg) (Take g=10m/s^(2).

Answer 1

Mass of the cube = 4 kg

Explanation:

Let the side of the cube be a

The volume of the half of the cube will be $a\times a\times \frac{a}{2}=\frac{a^3}{2}$

The cube will displace water equal to this volume

If the density of water is $\rho_w$ then the weight of displaced water

$=\rho_w\times \frac{a^3}{2}\times g$

This will be equal to the buoyant force on the block

If the mass of the block is M and density $\rho$ then

$M=a^3\times \rho$

Since the block floats half immersed

Therefore,

$\frac{a^3}{2}\times\rho_w\times g=a^3\times \rho\times g$

$\implies \rho_w=2\rho$

If block is given small displacement x downwards then

Net force on the block

$F=\rho_w\times a^2(\frac{a}{2}+x)\times g-\rho\times a^3 \times g$

$\implies F=\rho_w\times \frac{a^3}{2}\times g+\rho_w\times a^2xg-\frac{\rho_w}{2}\times a^3\times g$

$\implies F=\rho_w a^2gx$

Comparing with $F=kx$

We get

$k=\rho_w\times a^2g$

Time period of oscillations is given by

$\boxed{T=2\pi\sqrt{m/k}}$

$\implies \frac{\pi}{5}=2\pi\sqrt{\frac{m}{k}}$

$\implies \frac{m}{k}=\frac{1}{25\times 4}$

$\implies \frac{a^3\times \rho}{\rho_w\times a^2g}=\frac{1}{100}$

$\implies \frac{a\times\rho}{2\rho\times g}=\frac{1}{100}$

$\implies \frac{a}{20}=\frac{1}{100}$

$\implies a=0.2$ m

Thus, the mass of the cube

$m=a^3\times \rho$

$\implies m=0.2^3\times\frac{\rho_w}{2}$

$\implies m=8\times 10^{-3}\times\frac{1000}{2}$

$\implies m=4$ kg

Hope this answer is helpful.

Know More:

Q: Solid sphere of radius r is floating in a liquid of density p with half of its volume submerged. If the sphere is slightly pushed an released, it starts performing s.H.M. Frequency of these oscillations is:

Click Here: https://brainly.in/question/7206141

Q: A cubical block of wood edge 'a' and density 'p' floats in water of density 2P.The Lower surface of the cube just touches the free end of massless spring of force constant K fixed at the bottom of the vessel.The weight W put over the block so that it is completely immersed in water with out wetting the weight is:

Click Here: https://brainly.in/question/6925920