Question

A steel scale is correct at $0^\circ C$, the length of a brass tube measured by it at $40^\circ C$ is 5m. The correct length of the tube at $0^\circ C$is (Coefficients of linear expansion of steel and brass are $11 \times {10^{ - 6}}/^\circ C$ and $19 \times {10^{ - 6}}/^\circ C$respectively).
(a) 4.001m
(b) 5.001m
(c) 4.999m
(d) 4.501m

Answer 1

Steel scale and brass tube both are expanded when the temperature is $40^\circ C$. At $0^\circ C$, both are at their correct lengths.

Formula used:
1. Change in length due to thermal expansion: $\Delta l = l\alpha \Delta T$ ……(1)
Where,
$l$ is the length of the conductor at the initial temperature.
$\alpha$ is the coefficient of thermal expansion
$\Delta T$ is the difference between initial and final temperatures.
2. Change in temperature: $\Delta T = {T_f} - {T_i}$ ……(2)
Where,
${T_i}$ is the initial temperature.
${T_f}$ is the final temperature.

Complete step by step answer:
Given:
1. The length of a brass tube measured by a steel scale at $40^\circ C$: $l' = 5m$
2. Coefficient of linear expansion of steel ${\alpha _S} = 11 \times {10^{ - 6}}/^\circ C$
3. Coefficient of linear expansion of brass ${\alpha _B} = 19 \times {10^{ - 6}}/^\circ C$

To find: The correct length of the tube at $0^\circ C(l)$

Step 1 of 3:
Find $\Delta T$ using eq (2):
$\Delta T = 40 - 0$
$\Delta T = 40^\circ C$
Let the lengths of the brass tube and steel scale be $l$.
Use eq (1) to find the thermal expansion of brass:
$\Delta {l_B} = l(19 \times {10^{ - 6}}/^\circ C)(40^\circ C)$
Use eq (1) to find the thermal expansion of steel:
$\Delta {l_S} = l(11 \times {10^{ - 6}}/^\circ C)(40^\circ C)$

Step 2 of 3:
Change in measured tube length will be the difference between the thermal expansion of brass and thermal expansion of steel.
$\Delta l = \Delta {l_B} - \Delta {l_S}$
$\Delta l = l(19 \times {10^{ - 6}}/^\circ C - 11 \times {10^{ - 6}}/^\circ C)(40^\circ C)$
$\Delta l = l(320 \times {10^{ - 6}})$

Step 3 of 3:
New length measured by scale would be:
$l{'} = l + \Delta l$
$5 = l + l(320 \times {10^{ - 6}})$
Rearranging to find $l$ :
$l = \dfrac{5}{{1 + (320 \times {{10}^{ - 6}})}}$
$l \approx 4.999m$

The correct length of the tube at $0^\circ C$ is 4.999m. Hence, option (c) is correct.

Note:
The metal expands on heating. Steel and brass are metal alloys and they will have a greater length at a higher temperature. In the case of a steel scale, the length measured by it during summer will always have an error due to its expansion by heating.