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Q.

A piece of metal weight 46 gm in air, when it is immersed in the liquid of specific gravity 1.24 at 27°C it weighs 30 gm. When the temperature of liquid is raised to 42°C the metal piece weight 30.5 gm, specific gravity of the liquid at 42°C is 1.20, then the linear expansion of the metal will be

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a

3.316×10-5/0C

b

2.316×10-5/0C

c

4.316×10-5/0C

d

None of these

answer is B.

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Detailed Solution

Loss of weight at 27°C is= 46 - 30 = 16 = V1×1.24ρl×g-----(i)Loss of weight at 42°C is= 46 - 30.5 = 15.5 = V2×1.2ρ1×g-----(ii)Now dividing (i) by (ii), we get 1615.5 = V1V2×1.241.2 But V2V1 = 1+3α(t2-t1) = 15.5×1.2416×1.2 = 1.001042⇒3α(420-270) = 0.001042 ⇒α = 2.316×10-5/0C

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