A piece of metal has a weight of 49 gm in ai | vedclass

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(A) The apparent weight of an object in a liquid is given by $W_{app} = W_{air} - F_B$, where $F_B = V \rho g$ is the buoyant force.At $T_1 = 32^{\cir

Vedclass · Accepted Answer

$\frac{8}{3} 	imes 10^{-3} /^{\circ}C$

Vedclass · Answer

(A) The apparent weight of an object in a liquid is given by $W_{app} = W_{air} - F_B$, where $F_B = V ho g$ is the buoyant force.At $T_1 = 32^{\circ}C$: $W_{app1} = 39 \ gm$, $W_{air} = 49 \ gm$, $ho_1 = 1.2 	imes 10^3 \ kg/m^3$.$V_1 = \frac{W_{air} - W_{app1}}{ho_1} = \frac{(49 - 39) 	imes 10^{-3} \ kg}{1.2 	imes 10^3 \ kg/m^3} = \frac{10 	imes 10^{-3}}{1.2 	imes 10^3} = \frac{1}{12} 	imes 10^{-5} \ m^3 \approx 8.33 	imes 10^{-6} \ m^3$.At $T_2 = 42^{\circ}C$: $W_{app2} = 40 \ gm$, $W_{air} = 49 \ gm$, $ho_2 = 1.0 	imes 10^3 \ kg/m^3$.$V_2 = \frac{W_{air} - W_{app2}}{ho_2} = \frac{(49 - 40) 	imes 10^{-3} \ kg}{1.0 	imes 10^3 \ kg/m^3} = 9 	imes 10^{-6} \ m^3$.The change in volume is $\Delta V = V_2 - V_1 = V_1 (3\alpha \Delta T)$.$9 	imes 10^{-6} - 8.33 	imes 10^{-6} = 8.33 	imes 10^{-6} 	imes 3 \alpha 	imes (42 - 32)$.$0.67 	imes 10^{-6} = 8.33 	imes 10^{-6} 	imes 3 \alpha 	imes 10$.$\alpha = \frac{0.67}{8.33 	imes 30} \approx 2.67 	imes 10^{-3} /^{\circ}C = \frac{8}{3} 	imes 10^{-3} /^{\circ}C$.

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