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.



Mass of stone = 113 g

The rise in the level of water is equivalent to the volume occupied by the stone

Rise in water level = 40 ml – 30 ml = 10 ml

∴ The volume occupied by the stone = 10 cm3

Density = Mass\over Volume

= 113\over 10 = 11.3 g cm-3

Therefore, density of stone = 11.3 g cm-3

RD of stone = 11.3\over 1 = 11.3

Relative density of stone = 11.3

Was this answer helpful?

Didn't liked the above answer ?

≫ Some Related Questions

Upthrust in Fluids

A weather forecasting plastic balloon of volume 15 m3 contains hydrogen of density 0.09 kg m-3. The volume of an equipment carried by the balloon is negligible compared to it’s own volume. The mass of the empty balloon alone is 7.15 kg. The balloon is floating in air of density 1.3 kg m-3. Calculate : (i) the mass of hydrogen in the balloon, (ii) the mass of hydrogen and balloon, (iii) the total mass of hydrogen, balloon and equipment if the mass of equipment is x kg, (iv) the mass of air displaced by balloon and (v) the mass of equipment using the law of floatation.

Open Answer »