A stone of density 3000 kg/m³ is lying submerged in water of density 1000 kg/m³, if the mass of stone in air is 150 Kg, calculate the force required to lift the stone. Take g = 10 m/s²

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Solution

Step 1. Given data
Density of stone, $ρ_{s} = 3000 k g / m^{3}$
Density of water, $ρ_{w} = 1000 k g / m^{3}$
Mass of stone in the air, $m_{s} = 150 k g$
Acceleration due to gravity, $g = 10 m / s^{2}$
Force required to lift the stone, $f = ?$
Step 2. Calculation of the force required to lift the stone
Volume of stone, $v_{s} = \frac{m_{s}}{ρ_{s}} = \frac{150 k g}{3000 k g / m^{3}} = 0.05 m^{3}$
Buoyant force acting on the stone, $F_{b} = ρ_{w_{}} \times v_{s} \times g$ …………………(a)
Substituting the given values in equation (a), we get
$F_{b} = 1000 k g / m^{3} \times 0.05 m^{3} \times 10 m / s^{2} F_{b} = 500 N$
Weight of the stone, $w_{s} = m_{s} \times g$
$w_{s} = 150 k g \times 10 m / s^{2} = 1500 N$
Apparent weight of the stone in water, $w_{a} = w_{s} - F_{b} = 1500 N - 500 N = 1000 N$
To lift the stone in the water, the force applied must be equal to the apparent weight of the stone in the water.
$f = W_{a} = 1000 N$
Hence, the force required to lift the stone is equal to 1000 N.

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