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A piece of metal weight $46\, gm$ in air, when it is immersed in the liquid of specific gravity $1.24$ at $27°C$ it weighs $30\, gm.$ When the temperature of liquid is raised to $42°C$ the metal piece weight $30.5\, gm,$ specific gravity of the liquid at $42°C$ is $1.20,$ then the linear expansion of the metal will be
${3.316 × 10-5/ }{°C^{-1}}$
${2.316 × 10-5 }{°C^{-1}}$
${4.316 × 10-5 }{°C^{-1}}$
None of these
Solution
(b)Loss of weight at $27°C$ is
= $46 -30 = 16 = V_1 × 1.24 \rho_l × g$…$(i)$
Loss of weight at $42°C$ is
=$ 46 -30.5 = 15.5 = V_2 × 1.2 \rho_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 $\frac{{{V_2}}}{{{V_1}}} = 1 + 3\alpha (t_2 -t_1)$ = $\frac{{15.5 \times 1.24}}{{16 \times 1.2}}$ $= 1.001042$
==> $ 3\alpha (42º -27º) = 0.001042 $
==> $\alpha = 2.316 × 10^{-5}{°C^{-1}}$
