KnowledgeBoat Logo

Physics

A body of mass 70 kg, when completely immersed in water, displaces 20,000 cm3 of water. Find (i) the weight of body in water and (ii) the relative density of material of body.

Fluids Upthrust

72 Likes

Answer

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 = 110,00,000\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:

Density=MassVolume=700.02=3500 Kg m3\text{Density} = \dfrac{\text{Mass}}{\text{Volume}} \\[0.5em] = \dfrac{70}{0.02} \\[0.5em] = 3500 \text { Kg m}^{-3}

R.D. of body=Density of body in kg m31000 kg cm3=35001000=3.5\text{R.D. of body} = \dfrac{\text {Density of body in kg m}^{-3}}{\text{1000 kg cm}^{-3}} \\[0.5em] = \dfrac{3500}{1000} \\[0.5em] = 3.5

Therefore,

Relative density of body = 3.5

Answered By

30 Likes


Related Questions