# 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

(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

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A piece of iron of and volume 100 cm3is totally immersed in water. Calculate (a) the weight of the iron piece in air (b) the upthrust and (c) apparent weight in water. (ans. (a) 7.8N (b) 1N (c) 6.8 N)

8) density= 7.8*103kg/m3

volume= 100cm3= 0.0001 m3

mass= density*volume

= 7.8*103kg/m3 *0.0001 m3

= 7.8 *103kg/m3* 10-4 m3

=7.8*10-1kg

..a)weight= mg

= 7.8*10-1kg*10

=7.8N ..

.........

b) upthrust= vdg

where,

v= volume

d= denstiy

and g = accn due to gravity

..

=1000kg/m3 * 0.0001 m3 *10 N

=1N .

c)

density=1000kg/m3

volume= 100cm3= 0.0001 m3

mass= density*volume

= 1000kg/m3 *0.0001 m3

=103kg/m3* 10-4 m3

=10-1kg

..a)weight= mg

= 10-1kg*9.8

=9.8N

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