Advertisements
Advertisements
Question
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]
Advertisements
Solution
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
APPEARS IN
RELATED QUESTIONS
A solid of density 5000 kg m-3 weighs 0.5 kgf in air. It is completely immersed in water of density 1000 kg m-3. Calculate the apparent weight of the solid in water.
Complete the following sentence.
Density in kg m-3 = ............ × density in g cm-3
The relative density of mercury is 13.6. State its density in
(i) C.G.S. unit and
(ii) S.I. unit.
A piece of stone of mass 113 g sinks to the bottom in water contained in a measuring cylinder and water level in cylinder rises from 30 ml to 40 ml. Calculate R.D. of stone.
A body of volume 100 cm3 weighs 1 kgf in air. Find:
- Its weight in water and
- Its relative density.
A solid of density 7600 kgm3 is found to weigh 0.950 kgf in air. If 4/5 volume of solid is completely immersed in a solution of density 900 kgm3, find the apparent weight of solid in a liquid.
An iceberg floats in sea water of density 1.17 g cm 3, such that 2/9 of its volume is above sea water. Find the density of iceberg.
A piece of metal weighs 44.5 gf in air, 39.5 gf in water. What is the R.D. of the metal?
A body of volume 100 cm3 weighs 1 kgf in air. Calculate its weight in water. What is its relative density?
