Question

A wooden block floating in a bucket of water has 4/5 of its volume submerged. When certain amount of an oil is poured into the bucket, it is found that the block is just under the oil surface with half of its volume under water and half in oil. The density of oil relative to that of water is:
1. 0.5
2. 0.8
3. 0.6
4. 0.7

Answer 1

Correct Answer - Option 3 : 0.6

Concept:

Archimedes' principle:

Archimedes' principle is the statement that the buoyant force on an object is equal to the weight of the fluid displaced by the object and it is denoted by the following formula,

Buoyant force = Vρg

where,

V = volume of fluid displaced,

ρ = density of fluid

Or, Buoyant force = weight of fluid displaced.

Calculation:

By applying Archimedes principal, the weight of the block will be equal to the weight of the liquid displaced.

For case 1:

mg = f₃

mg = m' g

\({P_{block}} = \frac{4}{5}{P_w}\)

\(\frac{4}{5}V{\rho _w}g = mg\)

For case 2:

mg = F_bw + F_bo

\(mg = \frac{V}{2}{\rho _w}g + \frac{V}{2}p'g\)

When block floats fully in water and oil, then

\(\frac{V}{2}{\rho _w}g + \frac{V}{2}p'g = mg\)

\(\Rightarrow \;\frac{V}{2}{\rho _w}g + \frac{V}{2}p'g = \frac{4}{5}V{\rho _w}g\)

\(\Rightarrow \frac{1}{2}{\rho _w} + \frac{{p'}}{2} = \frac{4}{5}{\rho _w}\)

\(\frac{{p'}}{2} = \left[ {\frac{4}{5} - \frac{1}{2}} \right]{\rho _w}\)

\(\frac{{p'}}{2} = \frac{3}{{10}}{\rho _w}\)

\(\rho ' = \frac{3}{5}{\rho _w} = 0.6{\rho _w}\)

Therefore, the density of oil relative to that of water is 0.6.