A cube having a side of 10 cm with unknown m | vedclass

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(A) Given: Side of the cube $a = 10 \ cm = 0.1 \ m$. Volume of the cube $V = a^3 = (0.1)^3 = 10^{-3} \ m^3$.Let the mass of the cube be $m$ and its de

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$3$

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(A) Given: Side of the cube $a = 10 \ cm = 0.1 \ m$. Volume of the cube $V = a^3 = (0.1)^3 = 10^{-3} \ m^3$.Let the mass of the cube be $m$ and its density be $ho$. Thus,$m = ho V = ho 	imes 10^{-3} \ kg$.The distance of the unknown mass from the wedge is $2 \ cm = 0.02 \ m$,and the distance of the $200 \ g$ $(0.2 \ kg)$ mass from the wedge is $25 \ cm = 0.25 \ m$.When half the volume of the cube is submerged,the buoyant force $F_B$ acting on it is given by $F_B = ho_{water} 	imes V_{submerged} 	imes g = 1000 \ kg/m^3 	imes (0.5 	imes 10^{-3} \ m^3) 	imes 10 \ m/s^2 = 5 \ N$.For the rod to be in rotational equilibrium about the wedge point $O$,the net torque must be zero:$	au_{net} = (mg - F_B) 	imes 0.02 \ m - (0.2 \ kg 	imes 10 \ m/s^2) 	imes 0.25 \ m = 0$$(m 	imes 10 - 5) 	imes 0.02 = 2 	imes 0.25$$(10m - 5) 	imes 0.02 = 0.5$$10m - 5 = 0.5 / 0.02 = 25$$10m = 30$$m = 3 \ kg$.

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