First slide

Thermal expansion

Question

Apiece of metal weight 46 gm in air, when it is immersed in the liquid of specific gravity 1.24 at 27^oC it weighs 30 gm. When the temperature of liquid is raised to 42^oC the metal piece weight 30.5 gm, specific gravity of the liquid at 42^oC is 1.20, then the linear expansion of the metal will be

Moderate

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

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
$= 46 - 30 .5 = 15 .5 = V_{2} \times 1 .2 ρ_{l} \times g$ ………..(ii)
Now dividing (i) by (ii), we get $\frac{16}{15 .5} = \frac{V_{1}}{V_{2}} \times \frac{1 .24}{1 .2}$
But $\frac{V_{1}}{V_{2}} = 1 + 3 α (t_{2} - t_{1}) = \frac{15 .5 \times 1 .24}{16 \times 1 .2} = 1 .001042$
$\Rightarrow 3 α (42^{\circ} - 27^{\circ}) = 0 .001042 \Rightarrow α = 2 .316 \times 10^{- 5} /^{\circ} C$

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