Question

The weight of a body in air is $100N $. How much will it weigh in water, if it displaces \[400cc\] of water?
(A) $96N $
(B) $94N $
(C) $98N $
(D) None of these

Answer 1

Hint The weight of a body in air is given. The body displaces a certain amount of water when we put it in the water. We have to find the weight of the body in water. The volume of water displaced is given. We know the density of water. Use the density formula to find the mass of the body in water.

Complete step by step answer
Weight and mass are different from each other.
The earth’s gravitational force (acceleration due to gravity) affects the heaviness of the body. If we measure the heaviness of the body on earth and if we measure the heaviness of the body where there is no gravity in space at the same time. The heaviness of the body on earth will be larger than the heaviness of the body in space.
The measure of the heaviness of the body under gravitational force is known as weight
The measure of the heaviness of the body without gravitational force is known as mass
So, we can say weight of a body is the product of mass of the body and acceleration due to gravity.
 $ \Rightarrow {\text{Weight = m}} \times {\text{g}} $
Where,
m is the mass
g is the acceleration due to gravity
Given,
The weight of a body in air, ${W_a} = 100N $
The volume of water displaced by the given body, \[v = 400cc\]
The weight of the body in water, ${W_w} = ? $
The mass of the water displaced can be found from the density equation.
The density is determined by the ratio of mass and volume
 $ \Rightarrow \sigma = \dfrac{m}{v} $
 $ \Rightarrow m = \sigma \times v $
 $ \Rightarrow m = 1 \times 400 $
 $ \Rightarrow m = 400g $
 $ \Rightarrow 1kg = 1000g $ then,
 $ \Rightarrow m = \dfrac{{400}}{{1000}} $
 $ \Rightarrow m = \dfrac{2}{5}kg $
We know that,
 $ \Rightarrow {\text{Weight = m}} \times {\text{g}} $
The mass of the body in water, $m = \dfrac{2}{5}kg $
Acceleration due to gravity, $g = 9.8{\text{m/s}} $
 $ \Rightarrow {\text{Weight = }}\dfrac{2}{5} \times 9.8 $
 $ \Rightarrow {\text{Weight = 3}}{\text{.92N}} $
The weight of the water displaced is $3.92N $
According to Archimedes’ principle, the weight of the body in water is derived when the weight of water displaced is subtracted from the weight of the body in air.
We know that the weight of a body in air is $100N $.
 $ \Rightarrow {W_w} = 100 - 3.92 $
 $ \Rightarrow {W_w} = 96.08N $
The weight of the body in water, ${W_w} = 96.08N $

Hence the correct answer is option (A) $96N $

Note Weight is expressed in terms of kilogram. In this problem we use newton as the unit of weight because in question the weight of the body in air is expressed in newton and in the given options the answers are expressed in newton. Since the gravity acts on the body, its weight is expressed in newton.