A block of steel of size 5 times 5 times 5 , | vedclass

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(A) The apparent weight of an object submerged in a fluid is given by the formula: $W_{app} = W_{actual} - F_B$,where $F_B$ is the buoyant force.$W_{a

Vedclass · Accepted Answer

$6 	imes 5 	imes 5 	imes 5$

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(A) The apparent weight of an object submerged in a fluid is given by the formula: $W_{app} = W_{actual} - F_B$,where $F_B$ is the buoyant force.$W_{app} = V_b ho_s g - V_b ho_w g = V_b g (ho_s - ho_w)$.Given:Volume of the block $V_b = 5 	imes 5 	imes 5 \, cm^3 = 125 \, cm^3$.Relative density of steel $	ext{RD}_s = \frac{ho_s}{ho_w} = 7$,so $ho_s = 7 \, g/cm^3$ and $ho_w = 1 \, g/cm^3$.Substituting the values:$W_{app} = (5 	imes 5 	imes 5) 	imes (7 - 1) \, g wt$$W_{app} = 6 	imes 5 	imes 5 	imes 5 \, g wt$.Thus,the correct option is $A$.

$A$ block of steel of size $5 \times 5 \times 5 \, cm^3$ is weighed in water. If the relative density of steel is $7$,its apparent weight is ........... $g wt$.

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