Relative Density
1.

A flat–bottomed test tube with some lead shots floats upright on the surface of water with 20 cm of its length immersed. The tube and lead shots were found to weigh 20 gf when weighed in a beam balance. The same test tube when floated in kerosene was found to float with 24 cm immersed.

(i) What mass of water does it displace when floating in water ?
(ii) What is the area of cross–section of the tube?
(iii) Calculate the R.D. of kerosene.

Sol:
(i) When floating, weight of water displaced = weight of floating test tube
= 20 gf
∴ Mass of water displaced = 20 g
(ii) Mass of water displaced = length immersed × area of cross–section of tube × density of water.
20 = 20 × a × 1(since density of water = 1 g cm–3)
Area of cross–section of tube a = 1.0 cm2
(iii) R.D. of kerosene = (Length immersed in water) / (Length immersed in kerosene)
= 20/24
= 5/6
= 0.833
Alternatively Weight of test tube = Weight of water displaced
= Weight of kerosene displaced
20 × a × dw × g = 24 × a × dk × g
∴ R.D. of kerosene = dk/dw
= 20/24
= 5/6
= 0.833.
2.

A test tube is loaded with shots so that it floats in alcohol immersed to a mark on the tube. The tube and the shots weigh 15.3 gf. The tube is then placed in water and shots are added to sink the tube to the same mark. Now the tube and the shots weigh 18 gf. Calculate the relative density of alcohol.

Sol:
When the test tube, loaded with shots, floats in alcohol,
weight of the displaced alcohol = weight of the tube and shots = 15.3 gf.
When the tube floats in water,
weight of the water displaced for the same volume = weight of the tube and shots in second case = 18 gf
R.D. of alcohol = (Weight of alcohol) / (Weight of water of same volume)
= 15.3/18
= 0.85
3.

A solid weighs 25 gf in air and 21 gf when completely immersed in a liquid of relative density 0.8. Find:

(i) The volume of the solid
(ii) The relative density of the solid

Sol:
(i) Let V be the volume of the solid.
Weight of liquid displaced = volume of liquid displaced × density of liquid × g
= V × 0.8 gf
Loss in weight of the solid when immersed in liquid = 25 – 21 = 4 gf
But the weight of liquid displaced is equal to the loss in weight of the solid when immersed in liquid.
∴ V × 0.8 = 4
(or) V = 4/0.8
= 5 cm3
(ii) Density or solid = (Mass)/(Volume)
= 25/5
= 5 g cm–3.
Hence relative density of solid = 5 g cm–3.
4.

Relative density of silver is 10.8. What is the density of silver in S.I. unit ?

Sol:

When the liquid medicine is heated, it will expand and its volume will increase. The amount of liquid medicine leaking at 40°C will be equal to the increase in its volume on heating to 40°C. So, in this problem we have to calculate the increase in volume of the liquid medicine on heating. We know that the formula for the coefficient of cubical expansion of a liquid is :

R.D. = (Density of silver) / (Density of water)
∴ Density of silver = R.D × density of water
= 10.8 × 103 kg m–3.