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Physics Question on thermal properties of matter

A piece of metal weighs 45 g45\, g in air and 25 g25\, g in a liquid of density 1.5×103 kg−m−31.5 \times 10^{3} \,kg - m ^{-3} kept at 30∘C30^{\circ} C. When the temperature of the liquid is raised to 40∘C40^{\circ} C, the metal piece weighs 27 g27\, g. The density of liquid at 40∘C40^{\circ} C is 1.25×103 kg−m−31.25 \times 10^{3} \,kg - m ^{-3}. The coefficient of linear expansion of metal is

A

1.3×10−3/∘C1.3 \times 10^{-3} /^{\circ} C

B

5.2×10−3/∘C5.2 \times 10^{-3} /^{\circ} C

C

2.6×10−3/∘C2.6 \times 10^{-3} /^{\circ} C

D

0.26×10−3/∘C0.26 \times 10^{-3} /^{\circ} C

Answer

2.6×10−3/∘C2.6 \times 10^{-3} /^{\circ} C

Explanation

Solution

Volume of the metal at 30∘C30^{\circ} C V30=loss of weightspecific gravity×gV_{30} = \frac{\text{loss} \, \text{of} \, \text{weight} }{\text{specific} \, \text{gravity} \times g} =(45−25)g1.5×g=13.33 cm3= \frac{\left(45 -25\right)g}{1.5 \times g} = 13.33 \,cm^{3} Similarly, volume of metal at 40∘C40^{\circ}C V40=(45−27)g1.25×g=14.40 cm3 V_{40}= \frac{\left(45 -27\right)g}{1.25 \times g} = 14.40\, cm^{3} Now , V40=V30[1+γ(t2−t1)]V_{40} = V_{30} \left[1+ \gamma\left(t_{2}-t_{1}\right)\right] or γ=V40−V30V30(t2−t1)\gamma = \frac{V_{40} - V_{30}}{V_{30} \left(t_{2} -t_{1}\right)} =14.40−13.3313.33(40−30)= \frac{14.40 - 13.33}{13.33\left(40-30\right)} =8.03×10−3/∘C= 8.03 \times10^{-3} /^{\circ}C ∴\therefore Coefficient of linear expansion of the metal α=γ3=8.03×10−33\alpha = \frac{\gamma}{3} = \frac{8.03 \times10^{-3}}{3} ≈2.6×10−3/∘C\approx 2.6 \times10^{-3}/^{\circ}C