Question

An iron ball of diameter $6cm$ and is $0.01mm$ too large to pass through a hole in a brass plate when the ball and plate are at a temperature of ${20}^{\circ }C$. The temperature at which (both for ball and plate) the ball just pass through the hole is:
$({\alpha }_{iron}=12\times {10}^{-6}/1^{\circ }C;{\alpha }_{brass}=18\times {10}^{-6}/1^{\circ }C)$

• A. ${68}^{\circ }C$
• B. ${48}^{\circ }C$
• C. ${28}^{\circ }C$
• D. ${40}^{\circ }C$

Answer 1

Answer: (A)

${68}^{\circ }C$

$\begin{array}{cc}\text{given, diameter of iron ball}& =30mm\\ \text{size of brass hole}=& 30-0.01\\ & =29.99mm\end{array}$

$\begin{array}{c}\text{For the ball to pass through hole}\\ \text{both must have same size.}\\ 30(1+12\times {10}^{-6}\Delta \theta )=29.99(1+18\times {10}^{-6}\Delta \theta )\\ \frac{30}{29.99}=\frac{1+18\times {10}^{-6}\Delta \theta }{1+12\times {10}^{-6}\Delta \theta }\end{array}$

$\begin{array}{c}\text{Multiplying}RHS\text{by}{10}^6\text{in numerator}\\ \text{and denominator both}\\ \begin{array}{cc}1.0003=& \frac{{10}^6+18\Delta \theta }{{10}^6+12\Delta \theta }\\ 1000300+12.0036\Delta \theta ={10}^6+18\Delta \theta & \end{array}\\ \begin{array}{cc}300& =5.9964\Delta \theta \\ \Rightarrow & △\theta ={48}^{\circ }C\\ \Rightarrow T-20& =48\\ \Rightarrow & T={68}^{\circ }C\end{array}\end{array}$