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(C) Let $V$ be the volume of the body,$\rho_b$ be the density of the body,$\rho_w$ be the density of water $(1000 \,kg/m^3)$,and $\rho_l$ be the densi

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$2000 \,kg/m^3$

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(C) Let $V$ be the volume of the body,$ho_b$ be the density of the body,$ho_w$ be the density of water $(1000 \,kg/m^3)$,and $ho_l$ be the density of the liquid.In air: $T_1 = V ho_b g = 40.2 \,N$.In water: $T_2 = V(ho_b - ho_w)g = 28.4 \,N$.Buoyant force in water: $F_{Bw} = T_1 - T_2 = 40.2 - 28.4 = 11.8 \,N$.Since $F_{Bw} = V ho_w g$,we have $V g = 11.8 / 1000 = 0.0118 \,m^3 \cdot kg/m^3 \cdot m/s^2$.In liquid: $T_3 = V(ho_b - ho_l)g = 16.6 \,N$.Buoyant force in liquid: $F_{Bl} = T_1 - T_3 = 40.2 - 16.6 = 23.6 \,N$.Since $F_{Bl} = V ho_l g$,we have $V ho_l g = 23.6 \,N$.Dividing the two buoyant force equations: $\frac{V ho_l g}{V ho_w g} = \frac{23.6}{11.8} = 2$.Therefore,$ho_l = 2 ho_w = 2 	imes 1000 \,kg/m^3 = 2000 \,kg/m^3$.

$A$ body is suspended by a light string. The tensions in the string when the body is in air,when the body is totally immersed in water,and when the body is totally immersed in a liquid are respectively $40.2 \,N$,$28.4 \,N$,and $16.6 \,N$. The density of the liquid is

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