Properties Of Solids And Liquids Question 104
Question: A piece 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 [BHU 1995]
Options:
A) 3.316 × 10-5/ºC
B) 2.316 × 10-5/ºC
C) 4.316 × 10-5/ºC
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Loss of weight at 27ºC is = 46 - 30 = 16 = V1 × 1.24 rl × g -(i)
Loss of weight at 42ºC is = 46 - 30.5 = 15.5 = V2 × 1.2 r l × g -(ii)
Now dividing (i) by (ii), we get $ \frac{16}{15.5} $ = $ \frac{V _{1}}{V _{2}}\times \frac{1.24}{1.2} $
But $ \frac{V _{2}}{V _{1}} $ = 1 + 3a (t2 - t1) = $ \frac{15.5\times 1.24}{16\times 1.2} $ = 1.001042
therefore 3a (42º - 27º) = 0.001042 therefore a = 2.316 × 10-5/ºC.
