Solveeit Logo
Login

Question

Physics Question on Thermal Expansion

The volume of a metal sphere increases by 0.24%0.24\% when its temperature is raised by 40∘C40^{\circ}C. The coefficient of linear expansion of the metal is

A

2×10−5 ∘C−12 \times 10^{-5} \,^{\circ}C^{-1}

B

6×10−5 ∘C−16 \times 10^{-5} \,^{\circ}C^{-1}

C

18×10−5 ∘C−118 \times 10^{-5} \,^{\circ}C^{-1}

D

1.2×10−5 ∘C−11.2 \times 10^{-5} \,^{\circ}C^{-1}

Answer

2×10−5 ∘C−12 \times 10^{-5} \,^{\circ}C^{-1}

Explanation

Solution

From, VT=V0(1+γΔT)V_{T}=V_{0}\left(1+\gamma\Delta T\right) ⇒VT−V0V0=γΔT\Rightarrow \frac{V_{T}-V_{0}}{V_{0}}=\gamma\Delta T 0.24100=γ 40 ∘C\frac{0.24}{100}=\gamma\,40\,^{\circ}C γ=0.24100×40=6×10−5 ∘C−1\gamma=\frac{0.24}{100 \times 40}=6 \times 10^{-5}\,^{\circ}C^{-1} Coefficient of linear expansion, α=γ3=6×10−53\alpha=\frac{\gamma}{3}=\frac{6\times10^{-5}}{3} =2×10−5 ∘C−1=2 \times 10^{-5}\,^{\circ}C^{-1}