Advertisements
Advertisements
प्रश्न
A body weighs 20 gf in air and 18.0 gf in water. Calculate the relative density of the material of the body.
Advertisements
उत्तर
Weight of body in air , W1 = 20 gf
Weight of body when completely immersed in water W2 = 18 gf
R.D. of body = `W_1/(W_1 - W_2) xx "R.D. of water"`
R.D. of body = `20/(20 - 18) xx 1`
R.D. of body = 10
APPEARS IN
संबंधित प्रश्न
What do you understand by the term relative density of a substance?
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 1.5 kgf in air and 0.9 kgf in a liquid of density 1.2 × 103 kg m-3. Calculate R. D. of solid.
A solid weighs 32 gf in air and 28.8 gf in water. Find: (i) The volume of solid, (ii) R.D. of solid and (iii) The weight of solid in a liquid of density 0.9 g cm-3.
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 glass cylinder of length 12 x 10-2 m and area of crosssection 5 x 10-4 m2 has a density of 2500 kgm-3. It is immersed in a liquid of density 1500 kgm-3, such that 3/8. of its length is above the liquid. Find the apparent weight of glass cylinder in newtons.
A solid of R.D. = 2.5 is found to weigh 0.120 kgf in water. Find the wt. of solid in air.
A piece of metal weighs 44.5 gf in air, 39.5 gf in water. What is the R.D. of the metal?
