Question

A steel scale is correct at $0^{o}C$, the length of a brass tube measured by it at ${40}^{o}C$ is $5m$. The correct length of the tube at $0^{o}C$ is (Coefficients of linear expansion of steel and brass are $11\times {10}^{-6{/}^{o}C}$ and $19\times {10}^{-6{/}^{o}C}$ respectively).

Answer 1

$l_s^{\prime }=l_b^{\prime }=5$ m

$T_f={40}^{0}c$

Actual length of brass tube, $l_b=?$

$T_i=0$

For Length of scale at $0^0=l_s,$ $l_s^{\prime }=l_s(1+{\alpha }_s\Delta T)$

$5=l_s(1+40{\alpha }_s)$

$l_s=\frac{5}{1+40{\alpha }_s}$

Actual length of tube at ${40}^0$

$l_b^{\prime }=l_s^{\prime }+\Delta l_s$

$=l_s^{\prime }+l_s^{\prime }-l_s$

$=2l_s^{\prime }-l_s$

$l_b^{\prime }=(2\times 5-\frac{5}{1+40{\alpha }_s})$

Actual length of tube at $0^{0}c=l_b$

$l_b^{\prime }=l_b(1+{\alpha }_{0}4T)$

$l_b=\frac{l_b^{\prime }}{1+{\alpha }_{b}4T}=(10-\frac{5}{1+40{\alpha }_s})\times \frac{1}{1+40{\alpha }_b}$

$=(10-\frac{5}{1+40{\alpha }_s})\times \frac{1}{1+40{\alpha }_b}$

$=(10-\frac{5}{1+40\times 11\times {10}^{-6}})\times \left(\frac{1}{1+40\times 19\times {10}^{-6}}\right)$

$=(10-\frac{5}{1+440\times {10}^{-6}})\times \left(\frac{1}{1+760\times {10}^{-6}}\right)$

$=(10-\frac{5}{1.000440})\times \left(\frac{1}{1+00076}\right)$

$=4.998\cong 4.999m$

$\text{or}$

Let the length of tube and scale at $0^o$C be l. Then,

Change in measured tube length

$\Delta l=$Thermal expansion of brass$-$ Thermal expansion of steel

$\Delta l=l{\alpha }_B\Delta T-l\Delta {\alpha }_B\Delta T$

$\Delta l=l\Delta T(\alpha B-\alpha S)$

$\Delta l=l(40-0)(19\times {10}^{-6}-11\times {10}^{-6})$

So, new length measured by scale$=l+\Delta l$

$\Rightarrow 5=l+l\left(40\right)(7\times {10}^{-6})$

$\Rightarrow \frac{5}{1+40\times 7\times {10}^{-6}}=l$

$\Rightarrow l=4.9986\approx 4.999$m

Option C is correct.