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

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3.316×10−5/∘C

2.316×10−5/∘C

4.316×10−5/∘C

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=V2×1.2ρl×g ………..(ii)Now dividing (i) by (ii), we get 1615.5=V1V2×1.241.2But V1V2=1+3αt2−t1=15.5×1.2416×1.2=1.001042⇒ 3α42∘−27∘=0.001042⇒α=2.316×10−5/∘C

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