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प्रश्न
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.
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उत्तर
Density of solid = 5000 kg m-3
Weight of solid = 0.5 kgf
Density of water = 1000 kg m-3
Here , Upthrust = Volume of the solid × density of water × g
= `(0.5/"g")/5000 xx 1000 xx "g"`
= `0.5/(5000 xx "g") xx 1000 xx "g" = 0.1` kgf
Apparent weight = True weight - Upthrsut
= 0.5 - 0.1 = 0.4 kgf
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:
(i) The weight of the body, (ii) The upthrust on the body,
(iii) The apparent weight of the body in liquid, (iv) The loss in weight of the body.
A body weighs W1gf in air and when immersed in a liquid it weighs W2gf, while it weights W3gf on immersing it in water. Find:
- volume of the body
- upthrust due to liquid
- relative density of the solid
- relative density of the liquid
The unit of relative density is :
Calculate the mass of a body whose volume is 2 m3 and relative density is 0.52.
A piece of stone of mass 15.1 g is first immersed in a liquid and it weighs 10.9 gf. Then on immersing the piece of stone in water, it weighs 9.7 gf. Calculate:
- The weight of the piece of stone in air,
- The volume of the piece of stone,
- The relative density of stone,
- The relative density of the liquid.
A solid weighs 120 gf in air and and 105 gf when it is completely immersed in water. Calculate the relative density of solid.
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 piece of wax floats in brine. What fraction of its volume will be immersed?
R.D. of wax = 0.95, R.D. of brine = 1.1.
