Advertisements
Advertisements
प्रश्न
A solid weighs 0.08 kgf in air and 0.065 kgf in water. Find
(1) R.D. of solid
(2) Density of solid in SI system. [Density of water = 1000 kgm3]
Advertisements
उत्तर
Weight of solid in air = 0.08 kgf
Weight of solid in water = 0.065 kgf
Density of water = 1000 kgm3
(1) Relative density (RD) of solid
= Weight of solid in air
wt. of solid in air-wt. of solid in water
`= "Weight of solid in air"/("wt. of solid in air" - "wt. of solid in water")`
R.D. = `0.08/(0.08 - 0.065)`
R.D. = `0.08/0.015 = 5.3333`
(2) R.D. of solid = `"Density of solid"/"Density of water"`
`5.3333 = "Density of solid"/1000`
Density of solid = 5.3333 × 1000 = 5333.3 kgm-3
APPEARS IN
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.
How are the (i) Mass, (ii) Volume and (iii) Density of a metallic piece affected, if at all, with an increase in temperature?
Differentiate between density and relative density of a substance.
The relative density of silver is 10.8. Find its density.
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 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.
A piece of metal weighs 44.5 gf in air, 39.5 gf in water. What is the R.D. of the metal?
