Upthrust in Fluids, Archimedes' Principle and Floatation — Question 9

Given,

Mass of body = 70 kg

Volume of water displaced = 20,000 cm³

Converting cm³ to m³, we get,

100 cm = 1m

100 cm x 100 cm x 100 cm = 1 m³

Therefore, 20,000 cm³ = $\dfrac{1}{10,00,000}$ x 20,000 = 0.02 m³

Hence, volume of water displaced = 0.02 m³

(i) Mass of body immersed in water = mass of the water displaced = volume of water displaced x density of water
= 0.02 x 1000
= 20 kg

Weight of the body = mg = 70 x g = 70 kgf

Weight of water displaced = mg = 20 x g = 20 kgf

Weight of body in water = weight of body in air - upthrust due to liquid
= 70 kgf - 20 kgf    [∵ upthrust is equal to weight of water displaced]
= 50 kgf

(ii) Formula for density is:

$\text{Density} = \dfrac{\text{Mass}}{\text{Volume}} \\[1 em] = \dfrac{70}{0.02} \\[1 em] = 3500 \text { Kg m}^{-3}$

$\text{R.D. of body} = \dfrac{\text {Density of body in kg m}^{-3}}{\text{1000 kg cm}^{-3}} \\[1 em] = \dfrac{3500}{1000} \\[1 em] = 3.5$

Therefore,

Relative density of body = 3.5