Question

A circular hole of diameter 2.00 cm is made in an aluminium plate at $0^{\circ }C$. What will be the diameter at ${100}^{\circ }C$ ? $\alpha$ for aluminium = $23\times {10}^{-5}{}^{\circ }{C}^{-1}$

Answer 1

$d_1=2cm=2\times {10}^{-2}m,d_2=?$

$t_1=0^{\circ }C,t_2={100}^{\circ }C$

${\alpha }_{\alpha l}=2.3\times {10}^{-5}{/}^{\circ }C$

$\therefore d_2=d_1(1+\alpha \Delta t)$

$=2\times {10}^{-2}(1+2.3\times {10}^{-5}\times {10}^2)$

$=2\times {10}^{-2}(1+2.3\times {10}^{-3})$

$=2\times {10}^{-2}+2.3\times 2\times {10}^{-5}$

= 0.02 + 0.000046

= 0.020046 m $\approx$ 2.0046 cm