If a cylinder of diameter 1.0 , cm at 30^{cir | vedclass

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(A) The diameter of the hole $D_0 = 0.9997 \, cm$ must increase to $D = 1.0 \, cm$ to allow the cylinder to fit.Using the formula for linear expansion

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(A) The diameter of the hole $D_0 = 0.9997 \, cm$ must increase to $D = 1.0 \, cm$ to allow the cylinder to fit.Using the formula for linear expansion: $\Delta D = D_0 \alpha \Delta T$.Here,$\Delta D = D - D_0 = 1.0 - 0.9997 = 0.0003 \, cm$.Substituting the values: $0.0003 = 0.9997 	imes (12 	imes 10^{-6}) 	imes \Delta T$.$\Delta T = \frac{0.0003}{0.9997 	imes 12 	imes 10^{-6}} \approx \frac{0.0003}{0.0000119964} \approx 25.007^{\circ}C$.Rounding to the nearest integer,the minimum required rise in temperature is $25^{\circ}C$.

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