Question

A block of wood floats in a liquid with four fifth with its volume submerged. If the relative density is 0.8. What is the density of the liquid in ${\text{kg}}{{\text{m}}^{ - 3}}$?
(a) 750 (b) 1000
(c ) 1250 (d) 1500

Answer 1

Hint: The above problem is based on the concept of the density. The density is the mass of the liquid in per unit volume. The relative density is the ratio of density of liquid with density of the standard liquid. The standard liquid for all the liquid is water.

Complete step by step answer
Given: The submerged volume of the block in water is ${v_s} = \dfrac{4}{5}v$, the relative density of the block is ${\rho _r} = 0.8$.
The formula to find the density of the liquid is given as:
$\rho = \dfrac{{{\rho _w}}}{{{\rho _r}}}$
Here, ${\rho _w}$ is the density of the water and its value in SI units is $1000\;{\text{kg}} \cdot {{\text{m}}^{ - 3}}$.
Substitute $1000\;{\text{kg}} \cdot {{\text{m}}^{ - 3}}$for ${\rho _w}$ and 0.8 for ${\rho _r}$in the above expression to find the density of the liquid.
$\rho = \dfrac{{1000\;{\text{kg}} \cdot {{\text{m}}^{ - 3}}}}{{0.8}}$
$\rho = 1250\;{\text{kg}} \cdot {{\text{m}}^{ - 3}}$

Thus, the density of the liquid is $1250\;{\text{kg}} \cdot {{\text{m}}^{ - 3}}$and the option (c) is the correct answer for the above problem.

Additional Information: The standard fluid for the gasses is air. The density of the liquid decreases with increase in the temperature. The floating objects follow the law of floatation. This law states that the weight of the liquid displaced by the object is equal to the weight of the object.

Note: Substitute the value of the density of liquid in SI units to find the correct answer. Always use the law of floatation to solve the problem related to the submerged object or floating object in the liquid.
The density is also the same as the mass of the object in the one unit volume.