Question

A metal sphere 10.01 cm in diameter is placed on a brass ring of internal diameter 10 cm and at the same temperature of 12° C. The temperature up to which they should be heated together so that the metal sphere just passes through the ring is ${\displaystyle [{\alpha }_{\text{metal}}=12\times {10}^{-6}/°\text{C}\text{and}{\alpha }_{\text{brass}}=18\times {10}^{-6}/°\text{C}]}$ ____________.

पर्याय

• 167° C

• 179° C

• 191° C

• 197° C

Answer 1

A metal sphere 10.01 cm in diameter is placed on a brass ring of internal diameter 10 cm and at the same temperature of 12° C. The temperature up to which they should be heated together so that the metal sphere just passes through the ring is 179° C.

Explanation:

Diameter of sphere = Diameter of ring

${\displaystyle 10.01[1+12\times {10}^{-6}({\text{t}}_2-12)]}$

${\displaystyle =10[1+18\times {10}^{-6}({\text{t}}_2-12)]}$

${\displaystyle 1.001+1.001\times 12\times {10}^{-6}({\text{t}}_2-12)=1+18\times {10}^{-6}({\text{t}}_2-12)}$

${\displaystyle 1.001-1=18\times {10}^{-6}({\text{t}}_2-12)-1.001\times 12\times {10}^{-6}({\text{t}}_2-12)}$

${\displaystyle {10}^{-3}=({\text{t}}_2-12)\times {10}^{-6}[18-12.012]}$

${\displaystyle \therefore {\text{t}}_2-12=\frac{{10}^{-3}}{5.988}\times {10}^{-6}}$

${\displaystyle \therefore {\text{t}}_2\approx 179°\text{C}}$