An iron tyre is to be fitted on to a wooden wheel 1m in diameter. The diameter of tyre is 6 mm smaller than that of wheel. The tyre should be heated so that its temperature increases by a minimum of (the coefficient of cubical expansion of iron is 3.6 × 10 –5 /ºC)
Text Solution
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Initial diameter of tyre = (1000 – 6) mm = 994 mm , so initial radius of tyre $R = \frac{994}{2} = 497 \text{ mm}$
and change in diameter Δ Δ D = 6 mm so $\Delta R = \frac{6}{2} = 3\, mm$
After increasing temperature by Δ Δ θ θ tyre will fit onto wheel
Increment in the length (circumference) of the iron tyre
Δ Δ L = L × × α α × × Δ Δ θ θ $= L \times \frac{\gamma}{3} \times \Delta \theta$ [As $\alpha = -\left|\frac{\gamma}{3}\right|$
$2 \pi \Delta R = 2 \pi R \left(\frac{\gamma}{3}\right) \Delta \theta$ ⇒ ⇒ $\Delta \theta = \frac{3}{\gamma} \frac{\Delta R}{R} = \frac{3 \times 3}{3.6 \times 10^{-5} \times 497}$
⇒ ⇒ $\Delta \theta = 500^\circ \mathrm{C}$
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