Question

A solid weighs 32 g in air and 28.8 g in water. Find:
(i) The volume of solid
(ii) R.D. of the solid
(iii) The weight of the solid in a liquid of density $ 0.9g/c{m^3} $

Answer 1

Hint
A solid has a definite mass and shape. If we ignore the air resistance, the mass of solid in air is the actual weight of the solid. But for a solid in some liquid, buoyancy forces also act.
Formula used: $ V = \dfrac{M}{\rho } $, where V is the volume of substance, M is the mass, and $ \rho $ is the density of the substance.

Complete step by step answer
In this question, we are provided with the weight of a solid in both air and water, and are asked to calculate its volume. The data given to us includes:
Weight of solid in air $ {W_1} = 32 $ g
Weight of solid in water $ {W_2} = 28.8 $ g
We know that the volume of a solid can be calculated as:
 $\Rightarrow V = \dfrac{M}{\rho } $
But we are not aware of the density. Hence, we calculate the relative density of this solid first as:
$\Rightarrow RD = \dfrac{{{\text{Weight of solid in air}}}}{{{\text{Loss in weight of solid in water}}}} \times {\text{RD of water}} $
This gives us the relative density RD as:
 $\Rightarrow RD = \dfrac{{{W_1}}}{{{W_1} - {W_2}}} \times 1 = \dfrac{{32}}{{32 - 28.8}} $ [Relative density of water is 1]
 $\Rightarrow RD = \dfrac{{32}}{{3.2}} = 10 $
This relative density is the same as the density of the solid because the relative density of water is 1.
Hence, the volume of the solid will be:
 $\Rightarrow V = \dfrac{{{W_1}}}{{10}} = \dfrac{{32}}{{10}} = 3.2g/c{m^3} $
Moving on to the third part, we are required to find the weight of the solid in a different liquid. Let us assume the weight in this liquid to be $ W $ . We know:
 $\Rightarrow RD = \dfrac{{{\text{Weight of solid in air}}}}{{{\text{Loss in weight of solid in liquid}}}} \times {\text{RD of liquid}} $
Putting the known values in this equation:
 $\Rightarrow RD = \dfrac{{{W_1}}}{{{W_1} - W}} \times 0.9 = \dfrac{{32}}{{32 - W}} \times 0.9 = 10 $
Cross multiplying give us:
 $\Rightarrow \dfrac{{32}}{{10}} \times 0.9 = 32 - W $
Solving for W gives us:
 $\Rightarrow W = 32 - \dfrac{{32}}{{10}} \times 0.9 = 32 - 2.88 = 29.12 $ g
To summarise:
1. Volume of the solid $ = 3.2g/c{m^3} $
2. RD of the solid $ = 10 $
3. Weight of solid in liquid $ = 29.12 $ g

Note
Even though a solid has a well-defined mass and shape, the weight we perceive of it changes depending on the medium it is put in. When we measure the weight of a quantity in air, no external forces act on it and the true weight is shown. Whereas, if we measure the weight of the same entity in water, the upward thrust of water tends to reduce the perceived weight of the matter.