Advertisements
Advertisements
प्रश्न
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
उत्तर
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
संबंधित प्रश्न
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].
What is the unit of relative density?
Differentiate between density and relative density of a substance.
The relative density of mercury is 13.6. State its density in
(i) C.G.S. unit and
(ii) S.I. unit.
The density of iron is 7.8 x 103 kg m-3. What is its relative density?
A stone of density 3000 kgm3 is lying submerged in water of density 1000 kgm3. If the mass of stone in air is 150 kg, calculate the force required to lift the stone. [g = 10 ms2]
A solid of density 7600 kgm3 is found to weigh 0.950 kgf in air. If 4/5 volume of solid is completely immersed in a solution of density 900 kgm3, find the apparent weight of solid in a liquid.
A cube of the lead of side 8 cm and R.D. 10.6 is suspended from the hook of a spring balance. Find the reading of spring balance. The cube is now completely immersed in sugar solution of R.D. 1.4. Calculate the new reading of spring balance.
An iceberg floats in sea water of density 1.17 g cm 3, such that 2/9 of its volume is above sea water. Find the density of iceberg.
