The coefficient of linear expansion of copper is 17 x 10-6 K-1. A sheet of copper has a round hole with a radius of 3.0 m cut out of it. If the sheet is heated and undergoes a change in temperature of 80 K, what is the change in the radius of the hole? It decreases by 4.1 mm. It increases by 4.1 mm. It decreases by 8.2 mm. It increases by 8.2 mm. It does not change.

The coefficient of linear expansion of copper is 17 x 10-6 K-1. A sheet of copper has a round hole with a radius of 3.0 m cut out of it. If the sheet is heated and undergoes a change in temperature of 80 K, what is the change in the radius of the hole? It decreases by 4.1 mm. It increases by 4.1 mm. It decreases by 8.2 mm. It increases by 8.2 mm. It does not change.

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asked 2021-05-06
The coefficient of linear expansion of copper is 17 x 10-6 K-1. A sheet of copper has a round hole with a radius of 3.0 m cut out of it. If the sheet is heated and undergoes a change in temperature of 80 K, what is the change in the radius of the hole? It decreases by 4.1 mm. It increases by 4.1 mm. It decreases by 8.2 mm. It increases by 8.2 mm. It does not change.

Answers (1)

2021-05-08
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
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