Question

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

Answer 1

$\large{\boxed{Hello\:Brainlics\:!!}}$

Given,

=$T _{1} = 27°C$

=$L_{T1} = 5.231 \: m$

=$L_{T2} = 5.243 \: {}m$

So,

$L_{T2} = L_{T1}(1 + \alpha _{1}( T_{2} - T_{1}))$

= 5.243m = 5.231 m [1+1.20×10‐⁵ K-¹$( T _{2} - 27°C)$ ]

Or,

$T_{2} = 218°C$