A dumbbell is placed in water of density rho. | vedclass

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(B) For equilibrium,the total downward force (weight) must equal the total upward force (buoyant force).Let $V$ be the volume of each sphere. The buoy

Vedclass · Accepted Answer

$\frac{d(\rho V - 2M)}{2(\rho V - 3M)}$

Vedclass · Answer

(B) For equilibrium,the total downward force (weight) must equal the total upward force (buoyant force).Let $V$ be the volume of each sphere. The buoyant force on each half-submerged sphere is $F_B = ho (V/2) g$.Total weight = $Mg + 2Mg + mg = (3M + m)g$.Total buoyant force = $2 	imes (ho V/2) g = ho V g$.Equating forces: $(3M + m)g = ho V g \implies m = ho V - 3M$.Taking torque about the position of mass $m$ (at distance $l$ from the sphere of mass $2M$ and distance $(d-l)$ from the sphere of mass $M$):Torque due to sphere $M$: $	au_1 = (ho V/2 g - Mg)(d - l)$.Torque due to sphere $2M$: $	au_2 = (ho V/2 g - 2Mg)l$.For rotational equilibrium,$	au_1 = 	au_2$:$(ho V/2 - M)g(d - l) = (ho V/2 - 2M)gl$.$(ho V - 2M)(d - l) = (ho V - 4M)l$.$(ho V - 2M)d - (ho V - 2M)l = (ho V - 4M)l$.$(ho V - 2M)d = (ho V - 4M + ho V - 2M)l = (2ho V - 6M)l$.$l = \frac{d(ho V - 2M)}{2(ho V - 3M)}$.

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