Question

An iron tyre 100 cm in diameter is to be fitted on to a wooden wheel of 100.4 cm in diameter if $\alpha$ of Iron is $1\times {10}^{-5}{/}^{\circ }C$, the temperature increase required for this purpose is :

• A. ${40}^{\circ }C$
• B. ${100}^{\circ }C$
• C. ${400}^{\circ }C$
• D. ${1000}^{\circ }C$

Answer 1

Answer: (C)

${400}^{\circ }C$

In this case, $l_1=100cm=1m$
$l_2=100.4cm=1.004m$
So, $l_2-l_1=0.004m$
$\alpha =1\times {10}^{-5}$ per degree centigrade
$\alpha =(l_2-l_1)/\left(l_{1}dt\right)$
$dt=(l_2-l_1)/\left(l_1\alpha \right)$
$dt=\left(0.004\right)/(1\times 1\times {10}^{-5})$
$dt=400$
Putting the values, we get, change in temperature $=$ 400 degree centigrade.