Two cubes of size 1.0 m sides,one of relativ | vedclass

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(B) Given two cubes of side $1 \ m$ each with relative densities $\rho_A = 0.60$ and $\rho_B = 1.15$ respectively.Since $\rho_A  1$,

Vedclass · Accepted Answer

$25$

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(B) Given two cubes of side $1 \ m$ each with relative densities $ho_A = 0.60$ and $ho_B = 1.15$ respectively.Since $ho_A  1$,the lighter cube $(A)$ tends to float while the heavier cube $(B)$ tends to sink. In equilibrium,the system is connected by a wire,so both cubes are in the water.Let $x$ be the height of the lighter cube $A$ above the water surface. The volume of $A$ submerged is $(1-x) \ m^3$.The total downward force (weight) is $W = W_A + W_B = (ho_A V + ho_B V) ho_w g = (0.60 + 1.15) 	imes 1^3 	imes ho_w g = 1.75 ho_w g$.The total upward force (buoyancy) is $F_B = B_A + B_B = ho_w g V_{sub,A} + ho_w g V_{sub,B} = ho_w g (1-x) + ho_w g (1) = ho_w g (2-x)$.Equating upward and downward forces: $ho_w g (2-x) = 1.75 ho_w g$.$2 - x = 1.75$.$x = 0.25 \ m = 25 \ cm$.Thus,the lighter cube projects $25 \ cm$ above the water surface.

Two cubes of size $1.0 \ m$ sides,one of relative density $0.60$ and another of relative density $1.15$ are connected by a weightless wire and placed in a large tank of water. Under equilibrium,the lighter cube will project above the water surface to a height of ........ $cm$.

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