Question

A steel scale measures the length of a copper wire as $80.0\, cm$, when both are at $20^{\circ} C$, the calibration temperature for the scale. What would the scale read for the length of the rod when both are at $40^{\circ} C ?$ Given: $\alpha$ for steel $=11 \times 10^{-6}$ per ${ }^{\circ} C$ and $\alpha$ for $Cu =11 \times 10^{-6} \text{per} ^{\circ} C$

• A. 80.0096 cm
• B. 80.0272 cm
• C. 1 cm
• D. 25.2 cm

Answer 1

Answer: (A)

80.0096 cm

Answer 2

Using the relation $l_{t}=l_{0}(1+\alpha t)$ $=1 \times\left[1+11 \times 10^{-6} \times\left(40^{\circ}-20^{\circ}\right)\right]$ $=1.00022 \,cm$ Now length of copper rod at $40^{\circ} C$ $R_{t}'=l_{0}'(1+\alpha' t)$ $=80\left[1+17 \times 10^{-6}\left(40^{\circ}-20^{\circ}\right)\right]$ $=80.0272 \,cm$ Now number of $cms$ observed on the scale $=\frac{80.0272}{1.00022}=80.0096$