Question

A brass disc at $0^{\circ }C$ has a diameter of 50 cm and a hole of diameter 10 cm. For Brass $\alpha =18\times {10}^{-6}{/}^{\circ }6$, when the disc is heated to ${100}^{\circ }C$ the diameter of the hole becomes.

• A. $10.018cm$
• B. $10.180cm$
• C. $10.8cm$
• D. $10.81cm.$

Answer 1

Answer: (A)

$10.018cm$

Let $d_2=$ diameter of the hole finally
$d_2=d_1(1+\alpha \Delta T)$
$d_2=10(1+18\times {10}^{-6}\times 100)$
$=10(1+18\times {10}^{-4})$
$=10(1+0.0018)$
$=\left(1.0018\right)=10.018)=10.018cm$