Question

An object weigh 40N in air ,weigh 20N when submerged in water,and 30N when submerged in a liquid of unknown liquid density.what is the density of unknown of liquid?

Answer 1

Answer:

The density is  $\rho_u =500 kg /m^3$

Explanation:

From the question we are told that

    The weight in air is  $W_a = 40 \ N$

     The weight in water is  $W_w = 20 \ N$

     The weight in a unknown liquid is  $W_u = 30 \ N$

Now according to Archimedes principle the weight of the object in water is mathematically represented as

       $W_w = W_a -m _w g$

Where  $m_w$ is he mass of the water displaced

 substituting value

       $m_w g = 40 -20$

      $m_w g = 20 \ N --- (1)$

Now according to Archimedes principle the weight of the object in unknown  is mathematically represented as

       $W_u = W_a -m _u g$

Where  $m_u$ is he mass of the unknown liquid  displaced

 substituting value

       $m_u g = 40 -30$

      $m_u g = 10 \ N ---(2)$

dividing equation 2 by equation 1

      $\frac{m_ug}{m_wg} = \frac{10}{20}$

     $\frac{m_u}{m_w} = \frac{1}{2}$

=>  $m_u = 0.5 m_w$

Now since the volume of water and liquid displaced are the same then

      $\rho _u = 0.5 \rho_w$

This because

         $density = \frac{mass}{volume}$

So if  volume is constant

         mass = constant * density

Where $\rho_u$ is the density of the liquid

     and  $\rho_ w$ is the density of water which is a constant with a value $\rho_w = 1000 kg/m^3$

So

        $\rho_u = 1000*0.5$

        $\rho_u =500 kg /m^3$