Questions
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
detailed solution
Correct option is B
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/∘CSimilar Questions
A glass vessel is partially filled with mercury and when both are heated together, the volume of the unfilled part of vessel remains constant at all temperature. Find the initial volume (in cm3) of mercury if the empty part measures 34 cm3
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