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

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(C) The apparent weight loss is equal to the weight of the displaced liquid (Archimedes' principle). Let $V_{30}$ and $V_{40}$ be the volumes of the m

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). Let $V_{30}$ and $V_{40}$ be the volumes of the metal at $30^{\circ} C$ and $40^{\circ} C$ respectively. At $30^{\circ} C$: Loss in weight $= 45 - 25 = 20 \ g$. $V_{30} = \frac{	ext{Loss in weight}}{ho_{30}} = \frac{20 \ g}{1.5 \ g/cm^3} = 13.33 \ cm^3$. At $40^{\circ} C$: Loss in weight $= 45 - 27 = 18 \ g$. $V_{40} = \frac{	ext{Loss in weight}}{ho_{40}} = \frac{18 \ g}{1.25 \ g/cm^3} = 14.40 \ cm^3$. Using the volume expansion formula: $V_{40} = V_{30}(1 + \gamma \Delta T)$,where $\Delta T = 10^{\circ} C$. $\gamma = \frac{V_{40} - V_{30}}{V_{30} \Delta T} = \frac{14.40 - 13.33}{13.33 	imes 10} = \frac{1.07}{133.3} \approx 8.027 	imes 10^{-3} /^{\circ} C$. The coefficient of linear expansion $\alpha = \frac{\gamma}{3} = \frac{8.027 	imes 10^{-3}}{3} \approx 2.67 	imes 10^{-3} /^{\circ} C$. Rounding to the nearest option,we get $2.6 	imes 10^{-3} /^{\circ} C$.

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