A cube of edge length 10 , cm is just balance | vedclass

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(C) For the cube to be in equilibrium,the weight of the cube must be equal to the total buoyant force exerted by the two liquids.Weight of the cube $W

Vedclass · Accepted Answer

$480$

Vedclass · Answer

(C) For the cube to be in equilibrium,the weight of the cube must be equal to the total buoyant force exerted by the two liquids.Weight of the cube $W = m g$,where $m$ is the mass of the cube.Buoyant force $F_B = F_{B,A} + F_{B,B} = V_A ho_A g + V_B ho_B g$.Here,the edge length of the cube is $L = 10 \, cm = 0.1 \, m$. The cross-sectional area of the cube is $A_{cube} = L^2 = 100 \, cm^2 = 0.01 \, m^2$.The volume of the cube submerged in liquid $A$ is $V_A = A_{cube} 	imes h_A = 100 \, cm^2 	imes 4 \, cm = 400 \, cm^3 = 400 	imes 10^{-6} \, m^3$.The volume of the cube submerged in liquid $B$ is $V_B = A_{cube} 	imes h_B = 100 \, cm^2 	imes 6 \, cm = 600 \, cm^3 = 600 	imes 10^{-6} \, m^3$.The densities of the liquids are $ho_A = 0.6 	imes 1000 \, kg/m^3 = 600 \, kg/m^3$ and $ho_B = 0.4 	imes 1000 \, kg/m^3 = 400 \, kg/m^3$.Equating weight and buoyant force: $m g = (V_A ho_A + V_B ho_B) g$.$m = V_A ho_A + V_B ho_B = (400 	imes 10^{-6} \, m^3 	imes 600 \, kg/m^3) + (600 	imes 10^{-6} \, m^3 	imes 400 \, kg/m^3)$.$m = (240,000 	imes 10^{-6}) + (240,000 	imes 10^{-6}) \, kg = 0.24 \, kg + 0.24 \, kg = 0.48 \, kg$.Converting to grams,$m = 0.48 	imes 1000 \, g = 480 \, g$.

$A$ cube of edge length $10 \, cm$ is just balanced at the interface of two liquids $A$ and $B$ as shown in the figure. If liquids $A$ and $B$ have specific gravities $0.6$ and $0.4$ respectively,then the mass of the cube is ................ $g$.

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