A piece of metal weighs 46 g in air and 30 g in lipuid of density `1.24 xx 10^3 kg m^(-3)` kept at `27^0C`. When the temperature of the liquid is raised to `42^0C` the metal piece weights 30.5 g . The density of the liqued at `42^0 C` is `1.20 xx 10^3 kg m^(-3)`. Calculate the coefficient of linear expandsion of the metal.

Options

A. `3.316 xx 10^(-5)//^(@)C`
B. `2.316 xx 10^(-5)//^(@)C`
C. `4.316 xx 10^(-5)//^(@)C`
D. None of these

Answer

B

Explanation

Loss of weight at `27^(@)C` is
`=46-30 = 16 = V_(1) xx 1.24 rho_(l) xx g` ...(i)
Loss of weight at `42^(@)C` is
`=46 - 30.5 = 15.5 = V_(2) xx 1.2 rho_(l) xxg` ..(ii)
Now dividing (i) by (ii) , we get , `(16)/(15.5) = (V_1)/(V_2) xx (1.24)/(1.2)`
But `(V_2)/(V_1) = 1+ 3 alpha (t_(2)-t_(1))=(15.5xx1.24)/(16xx1.2) = 1.001042`
`rArr 3 alpha(42^(@)-27^(@)) = 0.001042 rArr alpha = 2.316 xx 10^(-5)//^(@)C`.