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Question: A piece of metal weighs 46 g in air. When immersed in a liquid of specific gravity 1.24 at 27 \[{}^\...

A piece of metal weighs 46 g in air. When immersed in a liquid of specific gravity 1.24 at 27 ∘C{}^\circ C it weighs 30 g. When the temperature of the liquid is raised to 42 ∘C{}^\circ C in a liquid of specific gravity 1.20, it weighs 30.5g. The coefficient of linear expansion of metal is
(A) 2.23×10−5∘C 2.23\times {{10}^{-5}}{}^\circ C\\\
(B) 6.7×10−5∘C 6.7\times {{10}^{-5}}{}^\circ C\\\
(C) 4.46×10−5∘C 4.46\times {{10}^{-5}}{}^\circ C\\\
(D) none of the above

Explanation

Solution

In this problem, a body is weighed in air and then it is weighed in some liquid. From Archimedes principle, it is clear that there will be a loss in weight which is also mentioned in the question. We are given specific gravities at two different temperatures.

Complete Step by Step Solution:
Weight in air= 46g
Weight in liquid at temperature 27 ∘C{}^\circ C= 30g
Loss in weight= 46g-30g= 16g
We know specific gravity is defined as the ratio of the density of the body to the density of the air.
The volume of metal at 27 ∘C{}^\circ C= 161.24=12.90322cm3 \dfrac{16}{1.24}=12.90322c{{m}^{3}}\\\
The volume of metal at 42∘C{}^\circ C=46−30.51.20=12.91666cm3 \dfrac{46-30.5}{1.20}=12.91666c{{m}^{3}}\\\
Volume expansion coefficient is represented by the Greek symbol γ \gamma \\\
So, γ\gamma = V2−V1V1(T2−T1) \dfrac{{{V}_{2}}-{{V}_{1}}}{{{V}_{1}}({{T}_{2}}-{{T}_{1}})}\\\
Putting the values,

γ=12.91666−12.9032212.90322(42−27)=6.944×10−5/∘C \gamma = \dfrac{12.91666-12.90322}{12.90322(42-27)}=6.944 \times {{10}^{-5}}/{}^\circ C\\\
So, the value of the coefficient of volume expansion of metal is 6.944×10−5/∘C 6.944\times {{10}^{-5}}/{}^\circ C\\\
Now we need to find the value of the coefficient of linear expansion of the metal α \alpha \\\
There exists a relationship between the coefficient of volume expansion and coefficient of linear expansion which is given by α=γ3 \alpha =\dfrac{\gamma }{3}\\\
Putting the value, we get α=γ3=2.314×10−5/∘C \alpha =\dfrac{\gamma }{3}=2.314\times {{10}^{-5}}/{}^\circ C\\\

So, the correct option is approx (A) 2.23×10−5∘C 2.23\times {{10}^{-5}}{}^\circ C\\\

Note:
In this problem in the options the answers were given in per degree Celsius. If the units were not mentioned we would have taken all the temperature values in Kelvin scale. Also, the loss of weight is attributed to the amount of the fluid displaced.