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(C) Let $V_s$ be the volume of the solid metal and $V_c$ be the volume of the cavities. The total volume of the ball is $V = V_s + V_c$.The mass of th

Vedclass · Accepted Answer

$16$

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(C) Let $V_s$ be the volume of the solid metal and $V_c$ be the volume of the cavities. The total volume of the ball is $V = V_s + V_c$.The mass of the ball is $M = 9.8 \ kg$. Since the density of the metal is $ho_m = 7800 \ kg/m^3$, the volume of the solid part is $V_s = M / ho_m = 9.8 / 7800 \ m^3$.When immersed in water, the buoyant force equals the weight of the displaced water, which is $1.5 \ kg 	imes g$. Thus, the total volume $V = V_s + V_c = 1.5 / ho_w$, where $ho_w = 1000 \ kg/m^3$.So, $V = 1.5 / 1000 = 0.0015 \ m^3$.The volume of the solid part is $V_s = 9.8 / 7800 \approx 0.001256 \ m^3$.The volume of the cavities is $V_c = V - V_s = 0.0015 - 0.001256 = 0.000244 \ m^3$.The fraction of the volume of the cavities is $(V_c / V) 	imes 100 = (0.000244 / 0.0015) 	imes 100 \approx 16.26 \%$.Therefore, the fraction is approximately $16 \%$.

$A$ metal ball of density $7800 \ kg/m^3$ is suspected to have a large number of cavities. It weighs $9.8 \ kg$ when weighed in air and $1.5 \ kg$ less when immersed in water. The fraction by volume of the cavities in the metal ball is approximately ....... $\%$

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