Question

A cylinder steel plug is inserted into a circular hole of diameter $2.60cm$ in a brass plate. When the plug and the plates are at a temperature of ${20}^{\circ }C$, the diameter of the plug is $0.010cm$ smaller than that of the hole. The temperature at which the plug will just fit in it is:
[Given, ${\alpha }_{\text{steel}}=\frac{11\times {10}^{-6}}{{}^{\circ }C}$ and ${\alpha }_{\text{brass}}=\frac{19\times {10}^{-6}}{{}^{\circ }C}$]

• A. $-{48}^{\circ }C$
• B. $-{20}^{\circ }C$
• C. $-{10}^{\circ }C$
• D. $-{458}^{\circ }C$

Answer 1

The correct option is C $-{10}^{\circ }C$

$2+2\times \frac{19}{{10}^6}T=1.99+1.99\times \frac{11}{{10}^6}T$

$\Rightarrow 1000=(1.99\times 11-2\times 19)T$

$\Rightarrow T\approx -{10}^{o}C$