Question

An object's true weight is 123 N. When it is completely submerged in water, its

apparent weight is 82.0 N. If the density of water is 1000 kg/m3
, then what is the density of the object?

Answer 1

3000 kg/m^3

Step-by-step explanation:

True weight = weight in air = 123 N

Apparent weight = weight in water = 82 N

Loss in weight of the object = true weight - apparent weight

Loss in weight = 123 - 82 = 41 N

According to the Archimedes principle, the loss in weight of the object is equal to the buoyant force acting on the object.

Let V be the volume of the object an d be the density of the object.

Buoyant force = volume of the object x density of water x gravity

41 = V x 1000 x g

`V = (41)/(1000 g)` .... (1)

Now, true weight = Volume of the object x density of object x gravity

`123 = (41)/(1000 g) * d* g` from (1)

d = 3000 kg/m^3

Answer 2

Object true weight is given as

`mg = 123 N`

now we know that g = 9.8 m/s^2

`m* 9.8 = 123`

`m = (123)/(9.8) = 12.55 kg`

now when it is complete submerged in water its apparent weight is given as 82 N

apparent weight = weight - buoyancy force

apparent weight = 82 N

weight = 123 N

now we have

82 = 123 - buoyancy force

buoyancy force = 123 - 82 = 41 N

now we also know that buoyancy force is given as

`F_b = p_(liq)Vg`

`41 = 1000*V*9.8`

`V = (41)/(1000*9.8)`

`V = 4.18 * 10^(-3) m^3`

now as we know that mass of the object is 12.55 kg

its volume is 4.18 * 10^-3 m^3

now we know that density will be given as mass per unit volume

`density = (m)/(V)`

`density = 3002.4 kg/m^3`

so here density of object is 3002.4 kg/m^3