The weight of body in air is 100 N. How much will it weight in water, if it displaces 400 cc of water? (g = 10m/s²).

Concept:

Weight (W):

• The weight of an object is defined as the force of gravity on the object and may be calculated as the mass times the acceleration of gravity, i.e.,
• W = mg

Where m = mass of the object and g = acceleration due to gravity

• Since the weight is a force, so its SI unit is the Newton.

Archimedes principle:

• The principle of Archimedes states “When a body is immersed in a liquid, an upward thrust, equal to the weight of the liquid displaced, acts on it.”

• Thus, when a solid is fully immersed in a liquid, it loses weight which is equal to the weight of the liquid it displaces.

\(Apparent\;weight = Actual\;weight - boyant\;force = mg - \rho gV\)

Where m is mass of the object, ρ is the density of the fluid

Hence if there is no gravity, there is no Upward Thrust.

The upward force applied to the object is called the buoyant force.

Here,

ρ = density of liquid

g = acceleration due to gravity

V = volume of liquid displaced by the object

Calculation:

Given,

Weight in air = 100 N

Also,

Density of water = 1000 L/m³

So, when the body is kept in water it displaces water and loss in weight

Given volume displace by body V = 400 cm³ = 400 × 10^-6 m³ = 4 × 10^-4 m³

\(Apparent\;weight = Actual\;weight - boyant\;force = mg - \rho gV\)

\(\Rightarrow Apparaent\;weight = 100\;N - \left( {1000 \times 10 \times 4 \times {{10}^{ - 4}}} \right) = 100 - 4 = 96\;N\)

So when the body is kept in water

Its apparent weight becomes 96 Newton.