A brass disc has a hole of diameter 2.5 cm at `27^(@)C`. Find the change in the diameter of the hole of the disc when heated to `327^(@)C`. Given coefficient of linear expansion of brass = ` 1.9 xx 10^(-5) .^(@)C^(-1)`

Explanation

Here, `D_(27) = 2.5 cm` ,
`Delta T = 327 - 27 = 300^(@)C`
`alpha = 1.9 xx 10^(-5) .^(@)C^(-1) , D_(327)-D_(27) = ?`
`D_(327) = D_(27)[1+alpha Delta T] D_(27) + D_(27) alpha Delta T`
Change in diameter = `D_(327)-D_(27) = D_(27) alpha Delta T`
`=2.5 xx 91.9 xx 10^(-5) xx 300`
` = 0.014 cm`.