HERE SHEET IS HEATED HENCE RADIUS OF THE HOLE INCREASES

THEN INCREASE IN RADIUS \(\displaystyle\triangle{R}={R}\alpha\triangle{T}\)

\(\displaystyle\triangle{T}\) = INCREASE IN TEMPERATURE =80K

\(\displaystyle\alpha=\) The coefficient of linear expansion of copper is \(\displaystyle{17}\times{10}^{{-{6}}}\ {K}^{{-{1}}}\)

R = INITIAL IN TEMPERATURE =80K

plug the values in equation we get \(\displaystyle\triangle{R}={3}{m}\cdot{17}\times{10}^{{-{6}}}\ {K}^{{-{1}}}\cdot{80}{K}={408}\cdot{10}^{{-{5}}}{m}={4.1}{m}{m}\)

radius of the hole increases by 4.1mm

THEN INCREASE IN RADIUS \(\displaystyle\triangle{R}={R}\alpha\triangle{T}\)

\(\displaystyle\triangle{T}\) = INCREASE IN TEMPERATURE =80K

\(\displaystyle\alpha=\) The coefficient of linear expansion of copper is \(\displaystyle{17}\times{10}^{{-{6}}}\ {K}^{{-{1}}}\)

R = INITIAL IN TEMPERATURE =80K

plug the values in equation we get \(\displaystyle\triangle{R}={3}{m}\cdot{17}\times{10}^{{-{6}}}\ {K}^{{-{1}}}\cdot{80}{K}={408}\cdot{10}^{{-{5}}}{m}={4.1}{m}{m}\)

radius of the hole increases by 4.1mm