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 $



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