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ICSE Class 9 Physics Question 2 of 16

Upthrust in Fluids, Archimedes' Principle and Floatation — Question 3

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Question 3

A body weighs W1 gf in air and when immersed in a liquid, it weighs W2 gf, while it weighs W3 gf on immersing it in water. Find: (i) volume of the body (ii) upthrust due to liquid (iii) relative density of the solid and (iv) relative density of the liquid.

Answer

Given,

Weight of the body in air = W1 gf

Weight of the body in liquid = W2 gf

Weight of the body in water = W3 gf

(i) Let V be the volume of the body.

Upthrust due to water = loss in weight when immersed in water = (W1 - W3) gf    ...[Eq 1]

Weight of water displaced = Volume of water displaced x density of water x g

Volume of water displaced = Volume of the body = V

Density of water = 1 g cm-3

∴ Weight of water displaced = V x 1 x g = V gf     ...[Eq 2]

But weight of water displaced is equal to upthrust due to water

∴ From eqns 1 & 2,

V = (W1 - W3) cm3

∴ Volume of the body = V = (W1 - W3) cm3

(ii) Upthrust due to liquid = loss in weight when immersed in liquid = (W1 - W2) gf

(iii) Relative density of the solid

R.D. of solid=(Wt of body)<em>air(Wt of body)</em>air−(Wt of body)water=W1W1−W3\text{R.D. of solid} \\[0.5em] = \dfrac{(\text{Wt of body})<em>\text{air}}{(\text{Wt of body})</em>\text{air}-(\text{Wt of body})_\text{water}} \\[0.5em] = \dfrac{\text W_1}{\text W_1 - \text W_3}

(iv) Relative density of the liquid

R.D. of liquid=(Wt of body)<em>air−(Wt of body)</em>liquid(Wt of body)<em>air−(Wt of body)</em>water=W1−W2W1−W3\text{R.D. of liquid} = \dfrac{(\text{Wt of body})<em>\text{air}-(\text{Wt of body})</em>\text{liquid}}{(\text{Wt of body})<em>\text{air}-(\text{Wt of body})</em>\text{water}} \\[0.5em] = \dfrac{\text W_1 - \text W_2}{\text W_1 - \text W_3}