Question

a body of mass 4.5 kg displaces 200cm3 of water when fully immersed inside it. Calculate a) volume of the body b) upthrust on body c) apparent weight of body in water

Answer 1

Answer:

Explanation:volume of body is equal to the water displaced by it so it is 200 cm³

Answer 2

Answer:

A body of mass 4.5 kg displaces 200cm3 of water when fully immersed inside it.

a) volume of the body =

b) upthrust on body     =  $1.96\ N$

c) apparent weight of body in water = $3.52 \ kg$

Explanation:

We can solve this problem using Archimede's principle.

Archimedes's principle states that, When a body is fully or partially immersed in a fluid, the volume of the body immersed will be equal to the volume of the fluid displaced.

Let $v _b$ be the volume of the body and $v_w$ be the volume of water displaced.

Then,

$v_b = v_w$    

In the question, it is given that,

$v_w =200\ cm^{3} = 0.0002\ m^{3}$

Mass of the body (m) = $4.5\ kg$

The true weight of the body  = mg = $4.5\times10=45\ kg$

So,

1) The volume of the body,

        $v_b = v_w =200\ cm^{3} = 0.0002\ m^{3}$

2) The upthrust (B) on the body is given by,

$\mathrm{B}=\mathrm{V}_{\mathrm{b}} \rho_{\mathrm{w}} \mathrm{g}$           ..(1)

Where

$\rho_{\mathrm{w}}$  - Density of the water    

g     - Acceleration due to gravity

$\rho_{\mathrm{w}} = 1000\ kg/m^{3}$      ,        $g = 9.8\ m/s^{2}$

Substituting these values into the above equation,

$\mathrm{B}= 0.0002\times 1000 \times 9.8=1.96 \mathrm{~N}$

Upthrust on the body = $1.96\ N$

3) The apparent weight ($W_a$) of the body in the water is given by the expression,

   $\begin{aligned}\text { Apparent weight } &=\text { true weight }-\text { upthrust } \\&=4.5 \times 10 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}^{2}-9.8 \mathrm{~N} \\&=(45-9.8) \mathrm{N} \\&=35.2 \mathrm{~N} \\=3.52 \ kg\end{aligned}$