Advertisements
Advertisements
प्रश्न
A body of volume 100 cm3 weighs 1 kgf in air. Find:
- Its weight in water and
- Its relative density.
Advertisements
उत्तर
Volume of body = 100 cm3
Weight in air, W1 = 1 kgf = 1000 gf
Mass of body = 1 kg = 1000 g
R.D. of solid = 10
R.D. of water = 1
(i) Let W2 be the weight of the body in water.
R.D. of body = `W_1/(W_1 - W_2) xx "R.D. of water"`
or , 10 = `1000/((1000 - W_2)) xx 1`
or, 10(1000 - W2) = 1000
or, 1000 - W2 = 100
or, W2 = 900 gf = 0.9 kgf
(ii) R.D. of body = Density in C.G.S. (without unit)
or, R.D. = `"Mass"/"Volume" = 1000/100 = 10`
APPEARS IN
संबंधित प्रश्न
What is the unit of relative density?
The unit of relative density is :
The density of iron is 7.8 x 103 kg m-3. What is its relative density?
Calculate the mass of a body whose volume is 2 m3 and relative density is 0.52.
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?
