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

Object true weight is given as

$mg = 123 N$

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

$m* 9.8 = 123$

$m = \frac{123}{9.8} = 12.55 kg$

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

$F_b = p_{liq}Vg$

$41 = 1000*V*9.8$

$V = \frac{41}{1000*9.8}$

$V = 4.18 * 10^{-3} m^3$

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

$density = \frac{m}{V}$

$density = \frac{12.55}{4.18*10^{-3}$

$density = 3002.4 kg/m^3$

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

Answer 2

Answer:

3000 kg/m^3

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

$V = \frac{41}{1000 g}$       .... (1)

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

$123 = \frac{41}{1000 g} \times d\times g$      from (1)

d = 3000 kg/m^3