Question

A barometer reads 75.0 cm on a steel scale. The room temperature is ${30}^{\circ }C$. The scale is correctly graduated for $0^{\circ }C$. The coefficient of linear expansion of steel is $\alpha$ = $1.2\times {10}^{-5}{/}^{\circ }C$ and the coefficient of volume expansion of mercury is $\gamma$ = $1.8\times {10}^{-4}{/}^{\circ }C$. Find the correct atmospheric pressure.

• A. 75 cm
• B. 74.89 cm
• C. 74.35 cm
• D. 75.35 cm

Answer 1

The correct option is B

74.89 cm

The 75 cm length of steel at $0^{\circ }C$ will become $l_{\theta }$ at ${30}^{\circ }C$ where,

$I_{\theta }$ = $\left(75cm\right)[1+\alpha ({30}^{\circ }C\left)\right]$. ................(1)

The length of mercury column at ${30}^{\circ }C$ is $l_{\theta }$. Suppose the length of the mercury column, if it were at $0^{\circ }C$, is $l_0$. Then,

$I_{\theta }$ = $I_0[1+\frac{\gamma }{3}({30}^{\circ }C\left)\right]$. ................(2)

By (1) and (2),

$I_0[1+\frac{\gamma }{3}({30}^{\circ }C\left)\right]$ = $75cm[1+({30}^{\circ }C\left)\right]$

Or, $I_0$ = $75cm\frac{[1+\alpha ({30}^{\circ }C\left)\right]}{[1+\frac{\gamma }{3}({30}^{\circ }C\left)\right]}$

= $75cm$ $[1+(\alpha -\frac{\gamma }{3}\left)\right({30}^{\circ }C\left)\right]$

= $74.89cm$.