Question

a body if weight 3,5kgf displase 1 litter of water when fully immersed calculate the volume of body and upthrust acting on the body​

Answer 1

Answer:

The volume of the body is 0.001 m³ and the upthrust acting on the body is 9.8 N.

Explanation:

The weight of the body, W = 3.5 Kg

The volume of water displaced by the body = 1 L = 0.001 m³

By Archimede's principle, the volume of the fluid displaced by a body when it is immersed in a fluid will be equivalent to the volume of the body.

∴ The volume of the body, V = 0.001 m³

The upthrust acting on the body, F = Vρg

Where V is the volume of the body, ρ is the density of the fluid and g is the acceleration due to gravity.

The density of water, ρ = 1000 Kg/m³

∴ F = 0.001 × 1000 × 9.8

  F = 9.8 N

Answer 2

Answer:

A body if weight 3,5kgf displace 1 litter of water when fully immersed

(i)  Volume of the body =  $1 \ L = 0.001\ m^{3}$

(ii)  Upthrust = $9.8 \ N$

Explanation:

According to the Archimedes principle,

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,

$v_b = v_w$

Given,

$v_w = 1 \ L = 0.001\ m^{3}$

So, The volume of the body,

$v_b = v_w = 1\ L$

The upthrust B is given by,

$\mathrm{B}=\mathrm{V}_{\mathrm{b}} \rho_{\mathrm{w}} \mathrm{g}$

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,

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