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प्रश्न
A stone of density 3000 kgm3 is lying submerged in water of density 1000 kgm3. If the mass of stone in air is 150 kg, calculate the force required to lift the stone. [g = 10 ms2]
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उत्तर
Density of stone =ρ = 3000 kgm3
Density of water = ρ’ = 1000 kgm3
Mass of stone = m = 150 kg
Acceleration due to gravity = g = 10 ms-2
Volume of stone = V = `"m"/ρ`
V = `150/3000 = 1/20` = 0.05 m3
Actual weight of stone = mg = 150 x 10 = 1500 N
Volume of water displaced = Volume of stone
V = 0.05 m3
Mass of water displaced = m’ = V x p’
m’ = 0.05 x 1000 = 50 kg
Upthrust = m’g = 50 x 10 = 500 N
Force required to lift the stone
= Actual weight of stone – upthrust
= 1500 -500 = 1000 N
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संबंधित प्रश्न
A body of volume V and density ρ is kept completely immersed in a liquid of density ρL. If g is the acceleration due to gravity, then write expressions for the following:
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