Question

A hole of radius $2.12cm$ is drilled in a metal sheet maintained at ${30}^{\circ }C$. The coefficient of linear expansion of the metal is $1.7\times {10}^{-5}{K}^{-1}$. The sheet is now heated to ${230}^{\circ }C$. Then

• A. Area of the hole decreases to $9.31\times {10}^{-2}c{m}^2$
• B. The change in the radius of the hole is $7.00\times {10}^{-3}cm$
• C. Area of the hole increases by $6.65\times {10}^{-2}c{m}^2$
• D. The change in the radius of the hole is $5.00\times {10}^{-3}cm$

Answer 1

The correct option is B The change in the radius of the hole is $7.00\times {10}^{-3}cm$

Radius of hole $=2.12cmat{30}^{o}C$

$\alpha$ (of metal) $=1.7\times {10}^{-5}{K}^{-1}$

For Aerial expansion

$A=A_o(1+\beta △T)$

$\beta =$ Aerial coefficient of expansion

So, $\beta =1.7\times {10}^{-5}{K}^{-1}$

Now, $△T=(230-30)+273=473K$

Now,

$A_o=\pi {R_o}^2\left(at30℃\right)A=\pi R^2\left(at230℃\right)\pi R^2=\pi {R_o}^2(1+1.7\times {10}^{-5}\times 473)R^2={R_o}^2(1+0.008041)R=R_o\sqrt{1.008041}R=2.128506$

Change in radius $=R-R_o\simeq 7.00\times {10}^{-3}cm$