R.D =
Since weight is proportional to mass
R.D
When a body is immersed in water, it displaces its own volume of water. Upthrust on the body is equal to the weight of this displaced volume of water, which is also equal to loss of weight of the body. Hence 'weight of equal volume of water' can be replaced by upthrust or loss of weight in water.
R.D
or R.D
Numerical :
A solid weighs 50 gf in air and 44 gf when completely immersed in water. Calculate (i) R.D of the solid (ii) upthrust (iii) density of the solid in cgs and SI units.
Suggested answer :
(i) R.D
= 8.33
(ii) upthrust = loss of weight = 6 gf
(iii) density in cgs unit 8.33 g cm-3
In SI unit 8.33 x 103 kg m-3
When a solid body is immersed in a liquid and then in water, the volume of displaced liquid is the same as the volume of displaced water which is equal to the volume of the solid.
R.D of liquid
Since the volume of displaced liquid = Volume of displaced water,
R.D of liquid
Numericals :
1. A body weighs 20gf in air, 18.2 gf in a liquid and 18.0 gf in water. Calculate (i) the relative density of the body (ii) and relative density of the liquid.
Suggested answer :
(i) R.D of the solid
= 10
(ii) R.D of liquid
= 0.9
2. A solid weighs 32 gf in air and 28.8 gf in water. Find how much will it weigh in a liquid of R.D 0.9.
Suggested answer :
R.D of liquid
0.9
0.9 x 3.2 = x
x = 2.88 gf
Weight of solid in liquid = 32 - 2.88 = 29.12 gf
Tie the solid firmly with a thread and suspend it from the hook as shown in figure. Lower the solid in water as shown and find its weight (W2 gf).
Weight of the solid in air: W1 gf
Apparent loss of weight of solid = (W1 - W2) gf
Record the observations as shown below:
Weight of the sinker in water = W2 gf
Loss of weight of the sinker in water = (W1 - W2) gf
Choose a sinker and find its weight in water by suspending it in water as shown in figure (a) below.
Remove the cork and tie it together with the sinker and suspend it in water as shown in figure (c) and find the weight of the cork together with the sinker in water.
Record your observations as shown below:
Weight of sinker in water = x gf
Weight of sinker in water + cork in air = y gf
Weight of cork in air = (y - x) gf
Weight of cork + sinker in water = z gf
Upthrust on cork in water = (y - z) gf
The upthrust on cork in water also represents the weight of the water displaced by the cork. Hence, relative density of cork
Numericals :
Following observations are taken while finding the R.D of cork.
Weight of sinker in water = 12.6 gf
Weight of sinker in water + cork in air = 13.7 gf
Weight of sinker and cork, both in water = 10.5 gf
Find the R.D and density of cork.
Suggested answer :
Weight of cork in air = 13.7 - 12.6
= 1.1 gf
Weight of cork in water = 10.5 - 12.6
= -2.1 gf
Since cork floats in water, upthrust on it is more than its own weight as it is made to sink with the force of the sinker, hence weight of cork in water is -ve.
R.D of cork
= 0.33
density = 0.33 g cm-3
با تشکر از سرکارخانم مهندس ترکاشوند.