Q. A wheel is 80.3 cm in circumference. An iron tyre measures 80.0 cm around its inner face. If the coefficient of linear expansion for iron is $12 \times 10^{- 6}℃^{- 1}$, the temperature of the tyre must be raised by

A) 105$℃$
B) 417$℃$
C) 312$℃$
D) 223$℃$
Explanation

(c) Here, $\mathrm{\Delta}l = 80.3 - 80.0 = 0.3\ $cm
$l = 80\ cm,\ \alpha = 12\ \times 10^{- 6}℃^{- 1}$
Rise in temperature $\mathrm{\Delta}T = \frac{\mathrm{\Delta}l}{l\alpha}$
$\mathrm{\Delta}T = \ \frac{0.3}{80 \times 12 \times 10^{- 6}} = 312.5℃$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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