The weight of a stone in air is 25 N and the weight of the stone when submerged in water is 20 N. What is the relative density of the stone?

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
5

The relative density = weight of the body in $\frac{a i r}{l o s s}$ of weight in water
So, given weight in air is 25 N
Loss of weight in water = 25 - 20 = 5 N
This loss is nothing but the weight of displaced i.e., equal volume of liquid.
Hence, specific gravity or relative density $= \frac{25}{5} = 5$ (no unit).

Suggest Corrections

Similar questions

Q. The weight of an object in air is

25 N

and the weight of that object when submerged in water is

20 N

? What is the relative density of the object?

Q. A body weight

50 N

in air and when immersed in water weight only

25 N

. Find the relative density:

A piece of stone of mass 15.1 g is first immersed in a liquid and it weighs 10.9 gf. Then on immersing the piece of stone in water, it weighs 9.7 gf. Calculate :
(a) the weight of the piece of stone in air,
(b) the volume of the piece of stone,
(c) the relative density of stone,
(d) the relative density of the liquid.

Q. A stone weight 392.4 N in air and 196.2 N IN water. Compute the volume of stone and its specific gravity.

The following observations were noted while performing an experiment. Find the relative density of the salt solution with the help of the observation noted.
(I) Weight of the stone in the air = 272
(II) Weight of the stone in the water = 192 g
(III) Weight of the stone in the salt solution = 176 g

Join BYJU'S Learning Program