Question

A piece of iron of density 7.8 * 103kg/m³ and volume 100 cm³ is totally immersed in water. Calculate

(a) The weight of the iron piece in air

(b) The upthrust

(c) Apparent weight in water

Answer 1

(a)

we know that

weight = m.g

here

m = mass = density x volume

or as per given

density = 7.8 x 10³ kg/m³

volume = 100 cm³ = 100 x 10^-6 m³

m = (7.8 x 10³ kg/m³) x (100 x 10^-6 m³)

or

m = 0.78 kg

thus, the weight will be

W = m.g = 0.78 x 9.81 = 7.65N

(b)

now, the upthrust is given as

F = ρgV

here

ρ = density of liquid (here water) = 1000 kg/m³

V = volume of immersed object or displaced liquid = 100 x 10^-6 m³

thus,

F = 1000 x 9.81 x 100 x 10^-6 m³

or

upthrust

F = 0.98 N

(c)

Now the apparent weight of the body will be

W' = real weight - upthrust

or

W' = W - F

thus,

W' = 7.65 - 0.98

so, we get

W' = 6.67 N