A wooden block of density 0.5 g/cc is tied t | vedclass

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(B) At equilibrium, the forces acting on the block are the buoyant force $(F_B)$ acting upwards, the weight of the block $(W)$ acting downwards, and t

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$2$

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(B) At equilibrium, the forces acting on the block are the buoyant force $(F_B)$ acting upwards, the weight of the block $(W)$ acting downwards, and the tension in the string $(T)$ acting downwards.For equilibrium: $F_B = W + T$.Thus, $T = F_B - W$.Given: Density of block $ho_s = 0.5 \ g/cc = 500 \ kg/m^3$, density of liquid $ho_l = 1 \ g/cc = 1000 \ kg/m^3$, $T = 20 \ N$, and $g = 10 \ m/s^2$.$T = ho_l V g - ho_s V g = V g (ho_l - ho_s)$.Substituting the values: $20 = V 	imes 10 	imes (1000 - 500)$.$20 = V 	imes 10 	imes 500$.$20 = 5000 V$.$V = \frac{20}{5000} = \frac{1}{250} \ m^3$.The mass of the block is $m = ho_s V$.$m = 500 	imes \frac{1}{250} = 2 \ kg$.

$A$ wooden block of density $0.5 \ g/cc$ is tied to a string. The other end of the string is fixed to the bottom of a tank. The tank is filled with a liquid of density $1 \ g/cc$. If the tension of the string is $20 \ N$, then the mass of the block is (Take $g = 10 \ m/s^2$) (in $kg$)

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