A uniform solid cylinder of density 0.8 g/cm | vedclass

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(B) For the cylinder to be in equilibrium,the total buoyant force must equal the weight of the cylinder.Let $A$ be the cross-sectional area of the cyl

Vedclass · Accepted Answer

$0.25$

Vedclass · Answer

(B) For the cylinder to be in equilibrium,the total buoyant force must equal the weight of the cylinder.Let $A$ be the cross-sectional area of the cylinder,$L$ be the total length of the cylinder,and $h$ be the length of the cylinder in air.The total length of the cylinder is $L = h_A + h_B + h$.The weight of the cylinder is $W = (A \cdot L) \cdot ho_C \cdot g = A(h_A + h_B + h) ho_C g$.The buoyant force provided by liquid $A$ is $F_A = A \cdot h_A \cdot ho_A \cdot g$.The buoyant force provided by liquid $B$ is $F_B = A \cdot h_B \cdot ho_B \cdot g$.Equating the forces: $F_A + F_B = W$.$A \cdot h_A \cdot ho_A \cdot g + A \cdot h_B \cdot ho_B \cdot g = A(h_A + h_B + h) ho_C g$.Dividing by $A \cdot g$: $h_A ho_A + h_B ho_B = (h_A + h_B + h) ho_C$.Substituting the given values: $(1.2 	imes 0.7) + (0.8 	imes 1.2) = (1.2 + 0.8 + h) 	imes 0.8$.$0.84 + 0.96 = (2.0 + h) 	imes 0.8$.$1.8 = 1.6 + 0.8h$.$0.2 = 0.8h$.$h = 0.2 / 0.8 = 0.25 \ cm$.

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