A piece of metal weighs 45 g in air and 25 | vedclass

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Question

(C) The apparent weight loss is equal to the weight of the displaced liquid (Archimedes' Principle).At $30^{\circ} C$: Loss in weight = $45 - 25 = 20

Vedclass · Accepted Answer

$2.6 	imes 10^{-3} /^{\circ} C$

Vedclass · Answer

(C) The apparent weight loss is equal to the weight of the displaced liquid (Archimedes' Principle).At $30^{\circ} C$: Loss in weight = $45 - 25 = 20 \ g$.Volume of metal $V_{30} = \frac{	ext{Loss}}{ho_{30}} = \frac{20 \ g}{1.5 \ g/cm^3} = 13.33 \ cm^3$.At $40^{\circ} C$: Loss in weight = $45 - 27 = 18 \ g$.Volume of metal $V_{40} = \frac{	ext{Loss}}{ho_{40}} = \frac{18 \ g}{1.25 \ g/cm^3} = 14.40 \ cm^3$.The coefficient of volume expansion $\gamma$ is given by $V_{40} = V_{30}(1 + \gamma \Delta T)$.$\gamma = \frac{V_{40} - V_{30}}{V_{30} \Delta T} = \frac{14.40 - 13.33}{13.33 	imes (40 - 30)} = \frac{1.07}{133.3} \approx 8.03 	imes 10^{-3} /^{\circ} C$.The coefficient of linear expansion $\alpha = \frac{\gamma}{3} = \frac{8.03 	imes 10^{-3}}{3} \approx 2.67 	imes 10^{-3} /^{\circ} C$.

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