Question

A piece of iron has density 7.8/$cm^3$ and volume ${10^{ - 4}}{m^3}$. It is totally immersed in the water to calculate the weight of iron in the water.
A) 7.8N
B) 8N
C) 6.8N
D) 6N

Answer 1

Hint: We know that a body immersed in a liquid is felt lighter than when it is in air. Any object placed inside a liquid suffers an apparent loss of weight.

Complete step by step answer:
Given,
The density of iron, ${\rho _i} = 7.8c{m^{ - 3}}$ $ \Rightarrow 7800kg{m^{ - 3}}$
The Volume of iron, $V = {10^{ - 4}}{m^3}$
Then the density of water, ${\rho _w} = 1000kg{m^{ - 3}}$
Initially, we need to calculate the weight of the iron in the air.
Weight of iron in air, ${W_a} = {\rho _i}Vg$
$\Rightarrow {W_a} = 7800 \times {10^{ - 4}} \times 10 = 7.8N$
Now an iron body is totally immersed in water, then we need to find the weight of the iron in the water. When an iron body is immersed inside water then there upward force is acting which is buoyancy force.
We know that,
Weight of iron in water = (weight in air) - (buoyancy force)
$\Rightarrow {W_w} = {W_a} - {\rho _w}Vg$
$\Rightarrow {W_w} = 7.8 - 1000 \times {10^{ - 4}} \times 10$
$\Rightarrow {W_w} = 6.8N$

$\therefore $ The weight of iron body in water is $6.8N$.

Additional information:
According to Archimedes’ principle, when a body is immersed fully or partially in a fluid, it experiences an upward force that is equal to the weight of the fluid displaced by it. The loss of weight of the body is equal to the resultant upthrust. The resultant upthrust on a body is called buoyancy. The resultant upthrust acting on a body is equal to the weight of the displaced liquid.
For a body to float in equilibrium on the surface of a liquid, the following two conditions must be satisfied.

Note:
- The Archimedes principle has many applications. It is used in designing ships and submarines.
- Lactometers, which are used to determine the purity of a sample of milk, and hydrometers used for determining the density of the liquids.
- The weight of the floating body is equal to the weight of the liquid displaced. The center of gravity of the floating body and the center of gravity of the displaced liquid must lie in the same vertical line.