KnowledgeBoat Logo

Physics

A body of volume 100 cm3 weighs 1 kgf in air. Find (i) it's weight in water and (ii) it's relative density.

Fluids Upthrust

57 Likes

Answer

Given,

Volume of body = 100 cm3

Weight of the body in air (W1) = 1 kgf = 1000 gf

Let weight of the body in water be W2

Density=MassVolume=1000100=10 g cm3\text{Density} = \dfrac{\text{Mass}}{\text{Volume}} \\[0.5em] = \dfrac{1000}{100} \\[0.5em] = 10 \text { g cm}^{-3}

Hence,

Density of body = 10 g cm-3

R.D. of body=Density of body in g cm31.0 g cm3=101=10\text{R.D. of body} = \dfrac{\text {Density of body in g cm}^{-3}}{\text{1.0 g cm}^{-3}} \\[0.5em] = \dfrac{10}{1} \\[0.5em] = 10

Therefore,

Relative Density of body = 10

R.D. of body=W1W1W210=10001000W210×(1000W2)=10001000010W2=100010W2=9000W2=900010W2=900 gf\text{R.D. of body} = \dfrac{W{1}}{W{1} − W{2}} \\[0.5em] \Rightarrow 10 = \dfrac{1000}{1000 − W{2}} \\[0.5em] \Rightarrow 10 \times (1000 − W{2}) = 1000 \\[0.5em] \Rightarrow 10000 - 10W{2} = 1000 \\[0.5em] \Rightarrow 10W{2} = 9000 \\[0.5em] \Rightarrow W{2} = \dfrac{9000}{10} \\[0.5em] \Rightarrow W_{2} = 900 \text { gf}

Hence,

(i) Weight in water = 900 gf

(ii) Relative density of body = 10

Answered By

34 Likes


Related Questions