The coefficient of apparent expansion of mercury in a glass vessel is {153 × 10^{-6}}{^{}{C}}and in a steel vessel is {144 × 10^{-6}}{^{}{C}}. If α for steel is {12 × 10^{-6}}{^{}{C}} , then that of glass is

Correct Answer is : 9×10−6∘C{9 × 10^{-6}}{^{}{C}}∘C9×10−6

Correct Answer is : 9×10−6∘C{9 × 10^{-6}}{^{}{C}}∘C9×10−6

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VITEEE 2021 Paper

Section: Physics

Question : 10 of 80

Marks: +1, -0

\frac{153 \times 10^{-6}}{^{\circ}\mathrm{C}}

\frac{144 \times 10^{-6}}{^{\circ}\mathrm{C}}

\frac{12 \times 10^{-6}}{^{\circ}\mathrm{C}}

, then that of glass is

Solution:

\gamma_{\text{real}} = \gamma_{\text{app.}} + \gamma_{\text{vessel}}

(\gamma_{\text{app.}} + \gamma_{\text{vessel}})_{\text{glass}} = (\gamma_{\text{app}} + \gamma_{\text{vessel}})_{\text{steel}}

\Rightarrow 153 \times 10^{-6} + (\gamma_{\text{vessel}})_{\text{glass}}

= 144 \times 10^{-6} + (\gamma_{\text{vessel}})_{\text{steel}}

Further,

(\gamma_{\text{vessel}})_{\text{steel}} = 3 \alpha = 3 \times (12 \times 10^{-6})

= \frac{36 \times 10^{-6}}{^{\circ}\mathrm{C}}

\Rightarrow 153 \times 10^{-6} + (\gamma_{\text{vessel}})_{\text{glass}}

= 144 \times 10^{-6} + 36 \times 10^{-6}

\Rightarrow (\gamma_{\text{vessel}})_{\text{glass}} = 3 \alpha = \frac{27 \times 10^{-6}}{^{\circ}\mathrm{C}}

\Rightarrow \alpha = \frac{9 \times 10^{-6}}{^{\circ}\mathrm{C}}

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