Question

c) A body of mass 5 kg displaces 500 cm of water when fully immersed inside it
Calculate (i) the volume of body (ii) the upthrust on body and (ii) the apparent weight of
body in water.​

Answer 1

Answer:

ANSWER

(i) :

Volume of body is equal to the volume of water displaced.

V

b

=V

w

=1000 cm

3

=0.001 m

3

(ii) :

Upthrust B=V

b

ρ

w

g=0.001×1000×10=10 N

(iii) :

True weight W=3.5×10=35 N

Apparent weight W

a

=W−B=35−10=25 N or 2.5 kg

Answer 2

Answer:

A body of mass 5 kg is displaced 1000 cm3 of water when  fully immersed in it.

i)  Volume of the body    =   $0.0005\ m^3$

ii) The  upthrust on the body   =    $4.9\ N$

iii) The apparent weight of the body in water   =   $3.92\ kg$

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,

$v_b = v_w$    

From the question,

$v_w =500\ cm^{3} = 0.0005\ m^{3}$

Mass of the body  = $5\ kg$

The true weight of the body = mg =  $5\times 10 = 50\ kg$

where

g - Acceleration due to gravity  ($g = 9.8\ m/s^{2}$)  

So,

The volume of the body,

 $v_b = v_w =500\ cm^{3} = 0.0005\ m^{3}$

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      

$\rho_{\mathrm{w}} = 1000\ kg/m^{3}$      

Substituting these values into equation(1),

$\mathrm{B}= 0.0005\times 1000 \times 9.8=4.9 \mathrm{~N}$

Upthrust on the body = $4.9\ N$

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

   $\begin{aligned}\text { Apparent weight } &=\text { true weight }-\text { upthrust } \\&=4.9 \times 10 \mathrm{~kg} \cdot \ -9.8 \mathrm{~N} \\&=(49-9.8) \mathrm{N} \\&=39.2 \mathrm{~N} \\\end{aligned}$

Thus,

The apparent weight of the body = $\frac{39.2}{10} = 3.92\ kg$