Question

c. A body of mass 3.5 kg displaces 1000 cm3

of water when fully immersed inside it. Calculate: (i)

the volume of body, (ii) the upthrust on body and (iii) the apparent weight of body in water​

Answer 1

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 3.5 kg displaces 1000 cm3  of water when fully immersed inside it.

(i)  The volume of the body =    $0.001m^3$

(ii) The upthrust on body     =    $9.8\ N$

(iii) The apparent weight of the body in water  =  $2.52\ kg$

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,

$v_w =1000\ cm^{3} = 0.001\ m^{3}$

Mass of the body  = $3.5\ kg$

The true weight of the body = mg =  $3.5\times 10 = 35 \ kg$

where

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

So,

1) The volume of the body,

        $v_b = v_w =1000\ cm^{3} = 0.001\ m^{3}$

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      

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

Substituting these values into the above equation,

$\mathrm{B}= 0.001\times 1000 \times 9.8=9.8 \mathrm{~N}$

Upthrust on the body = $9.8\ N$

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

   $\begin{aligned}\text { Apparent weight } &=\text { true weight }-\text { upthrust } \\&=3.5 \times 10 \mathrm{~kg} \cdot \ -9.8 \mathrm{~N} \\&=(35-9.8) \mathrm{N} \\&=25.2 \mathrm{~N} \\=2.52 \ kg\end{aligned}$