Question

a solid 50 gf in air and 44 gf when completely immersed in water. calculate upthrust, the volume of the solid and relative density of the solid​

Answer 1

Explanation:

Let m be the mass of the object.

Actual weight of the object, W = 50 gf = 50g N

Apparent weight of the object, Wa = 44 gf = 44g N

Let the buoyancy be B.

Now,

Wa = W – B

=> B = 50g – 44g

=> B = 6g N

This is the upthrust on the object.

Let ‘d’ be the density of the object, and ‘V’ be its volume.

Weight, W = Vdg = 50g

=> Vd = 50

The buoyancy is given by,

B = V(density of water)g

=> B = V(1000)g = 6g

=> V = 6/1000 = 0.006 m3

This is the volume of the object.

Now, Vd = 50

=> d = 50/V = 50/0.006

=> d = 8333.33 kg/m3

This is the density of the solid.

Answer 2

Given, weight of solid in air W1 = 50 gf and weight of solid in water w2 = 44 gf

(i) Upthurst = loss in weight when immersed in water = W1 - W2 = 50 - 44 = 6 gf

(ii) Weight of water displaced = upthurst = 6 gf

Since density of water is 1 g/m³, therefore volume of water displaced = 6cm³

But the solid displaces water equal to its own volume, therefore volume of solid = 6cm³.

(iii) R.D. of solid = Weight of solid in air / Weight in air - Weight in water

= W1 / W1 - W2 = 50 / 50 - 44 = 50 / 6 = 8.33