A piece of iron of density 7.8 * 103kg/m^{3} and volume 100 cm^{3} is totally immersed in water. Calculate

(a) The weight of the iron piece in air

(b) The upthrust

(c) Apparent weight in water

**(a)**

we know that

weight = m.g

here

m = mass = density x volume

or as per given

density = 7.8 x 10^{3} kg/m^{3}

volume = 100 cm^{3} = 100 x 10^{-6} m^{3}

m = (7.8 x 10^{3} kg/m^{3}) x (100 x 10^{-6} m^{3})

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/m^{3}

V = volume of immersed object or displaced liquid = 100 x 10^{-6} m^{3}

thus,

F = 1000 x 9.81 x 100 x 10^{-6} m^{3}

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**

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