Question

Example 12.2 A blacksmith fixes iron ring
on the rim of the wooden wheel of a bullock
cart. The diameter of the rim and the Iron
ring are 5.243 m and 5.231 m respectively
at 27 °C. To what temperature should the
ring be heated so as to fit the rim of the
wheel?​

Answer 1

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Answer 2

Given :-

Initial temperature = 27° C

Initial length = 5.231 m

Final length = 5.243 m

To Find :-

Temperature the ring should be heated so as to fit the rim of the  wheel.

Solution :-

We know that,

• t = Temperature
• l = Length

According to the question,

$\underline{\boxed{\sf \alpha =\dfrac{\Delta l}{l_1 \Delta T}= \dfrac{l_1-l_2}{l_1 \Delta T} }}$

Given that,

Initial temperature (T) = 27° C

Initial length (l₁) = 5.231 m

Final length (l₂) = 5.243 m

Substituting their values,

$\sf l_2=l_1[1+ \alpha_1 (T_2=T_1)]$

$\sf 5.243=5.231[1+1.20 \times 10^{-5}(T_2-27)]$

$\sf T_2=\dfrac{5.243-5.231}{5.231 \times 1.2 \times 10^5} +27$

$\sf T_2=191.1+27$

$\sf T_2=218.1 \approx 218^o \ C$

Therefore, the temperature required for the ring to be heated so as to fit the rim of the  wheel is 218° C.