Properties Of Solids And Liquids Question 563
Question: A piece of metal weight 46 gm in air, when it is immersed in the liquid of specific gravity 1.24 at $ 27{}^\circ C $ it weighs 30 gm. When the temperature of liquid is raised to $ 42{}^\circ C $ , the metal piece weighs 30.5 gm, specific gravity of the liquid at $ 42{}^\circ C $ is 1.20, then the linear expansion of the metal will be
Options:
A) $ 3.316\times {{10}^{-5}}/{}^\circ C $
B) $ 2.316\times {{10}^{-5}}/{}^\circ C $
C) $ 4.316\times {{10}^{-5}}/{}^\circ C $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Loss of weight at $ 27{}^\circ C $ is
$ =46-30=16=V _{1}\times 1.24{{\rho} _{1}}\times g $
Loss of weight at $ 42{}^\circ C $ is
$ =46-30.5=15.5=V _{2}\times 1.24{{\rho} _{1}}\times 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+3\alpha ( t _{2}-t _{1} )=\frac{15.5\times 1.24}{16\times 1.2}=1.001042 $
$ \Rightarrow 3\alpha ( 42{}^\circ -27{}^\circ )=0.001042 $
$ \Rightarrow \alpha =2.316\times {{10}^{-5}}/{}^\circ C $
BETA
