Advertisements
Advertisements
Question
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.
Advertisements
Solution
Mass of stone = 113 g
Rise in water level = (40 - 30) ml = 10 ml
This rise is equal to the space occupied (volume) by the stone .
∴ volume of stone = 10 cm-3
Density of stone in C.G.S. = `"Mass"/"Volume" = 113/10 = 11.3` gcm-3
R.D. = 11.3
APPEARS IN
RELATED QUESTIONS
A sphere of iron and another sphere of wood of the same radius are held under water. Compare the upthrust on the two spheres.
[Hint: Both have equal volume inside the water].
Differentiate between density and relative density of a substance.
Relative density of a substance is expressed by comparing the density of that substance with the density of :
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 body of volume 100 cm3 weighs 1 kgf in air. Find:
- Its weight in water and
- Its relative density.
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. 4.2 is found to weigh 0.200 kgf in air. Find its apparent weight in water.
An aluminium cube of side 5 cm and RD. 2.7 is suspended by a thread in alcohol of relative density 0.80. Find the tension in thread.
A solid weighs 105 kgf in air. When completely immersed in water, it displaces 30,000 cm3 of water, calculate relative density of solid.
