Advertisements
Advertisements
प्रश्न
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.
उत्तर
Relative density of wax = 0.95
Relative density of brine = 1.1
(Density of wax/ Density of brine) = fraction submerged
0.95/1.1 = fraction of volume submerged
Fraction of volume submerged = 0.86
APPEARS IN
संबंधित 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.
The relative density of mercury is 13.6. State its density in
(i) C.G.S. unit and
(ii) S.I. unit.
Calculate the mass of a body whose volume is 2 m3 and relative density is 0.52.
A body weighs 20 gf in air and 18.0 gf in water. Calculate the relative density of the material of the body.
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 solid of area of cross-section 0.004 m2 and length 0.60 m is completely immersed in water of density 1000 kgm3. Calculate:
- Wt of solid in SI system
- Upthrust acting on the solid in SI system.
- Apparent weight of solid in water.
- Apparent weight of solid in brine solution of density 1050 kgm3.
[Take g = 10 N/kg; Density of solid = 7200 kgm3]
A solid weighs 105 kgf in air. When completely immersed in water, it displaces 30,000 cm3 of water, calculate relative density of solid.
A body of volume 100 cm3 weighs 1 kgf in air. Calculate its weight in water. What is its relative density?
