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Question

Physics Question on thermal properties of matter

A blacksmith fixes iron ring on the rim of the wooden wheel of bullock cart . the diameter of the rim and the iron ring are 5.243 m and 5.231 m respectively at 27∘C27^{\circ} C . The temperature should the ring be heated so as to fit the rim of the wheel?

A

218∘ C218^{\circ} \,C

B

512∘ C512^{\circ} \,C

C

312∘ C312^{\circ} \,C

D

281∘ C281^{\circ} \,C

Answer

218∘ C218^{\circ} \,C

Explanation

Solution

Given, Initial Temperature, T1=270CT _{1}=27^{0} C Initial length, l1=5.231ml _{1}=5.231 m Final length, l2=5.243ml _{2}=5.243 m Now, α1=Δll1ΔT=l2−l1l1ΔT\alpha_{1}=\frac{\Delta l }{ l _{1} \Delta T }=\frac{ l _{2}- l _{1}}{ l _{1} \Delta T } ⟹l2=l1[1+α1(T2−T1)]\Longrightarrow l _{2}= l _{1}\left[1+\alpha_{1}\left( T _{2}- T _{1}\right)\right] That is, 5.243m=5.231m[1+1.20×10−5K−1(T2−270C)]5.243 m =5.231 m \left[1+1.20 \times 10^{-5} K ^{-1}\left( T _{2}-27^{0} C \right)\right] This gives, T2=218∘CT _{2}=218^{\circ} C