First slide

Thermal expansion

Question

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

Moderate

A

$3.316 \times 10^{- 5} /^{0} C$
B

$2.316 \times 10^{- 5} /^{0} C$
C

$4.316 \times 10^{- 5} /^{0} C$
D

None of these

Solution

Loss of weight at 27°C is
= 46 - 30 = 16 = $V_{1} \times 1.24 ρ_{l} \times g - - - - - (i)$
Loss of weight at 42°C is
= 46 - 30.5 = 15.5 = $V_{2} \times 1.2 ρ_{1} \times g - - - - - (ii)$
Now dividing (i) by (ii), we get $\frac{16}{15.5} = \frac{V_{1}}{V_{2}} \times \frac{1.24}{1.2}$

But $\frac{V_{2}}{V_{1}} = 1 + 3 α (t_{2} - t_{1}) = \frac{15.5 \times 1.24}{16 \times 1.2} = 1.001042$
$\Rightarrow 3 α (42^{0} - 27^{0}) = 0.001042 \Rightarrow α = 2.316 \times 10^{- 5} /^{0} C$

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