Question

An iron tyre is to be fitted on to a wooden wheel of diameter 1 m. 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 epansion of iron is $3.6\times {10}^{-5}{\left(°C\right)}^{-1}$ and consider iron tyre to be isotropic.)

• A. $167°C$
• B. $334°C$
• C. $500°C$
• D. $200°C$

Answer 1

The correct option is C $500°C$

Initial diameter of tyre = 1000-6=994 mm.
$\Rightarrow R=\frac{994}{2}=497$ mm.
$\Delta R=\frac{\Delta D}{2}=\frac{6}{2}=3$ mm.
After increasing temperature by $\Delta T$, tyre will fit by increment in its length (circumference)
$\Rightarrow \Delta L=L\alpha \Delta T=2\pi \Delta R$
$\Rightarrow \Delta T=\frac{2\pi \Delta R}{L\alpha }$
Since, $L=2\pi R$
$\Rightarrow \Delta T=\frac{2\pi \Delta R}{\left(2\mathrm{\pi R}\right)\alpha }=\frac{\Delta R}{R\alpha }$
As, $\alpha =\frac{\gamma }{3}=\frac{3.6\times {10}^{-5}}{3}=1.2\times {10}^{-5}$ (since iron is isotropic)
$\Rightarrow \Delta T=\frac{3}{497\times \left(1.2\times {10}^{-5}\right)}=500°C$ (approx)