Question

A rectangular wooden block of weight W floats with exactly one-half of its volume below the waterline. The wooden block is removed from the water bath and placed in an unknown liquid. In this liquid, only one-third of the wooden block is submerged this unknown liquid is more dense than water. What is the density of the unknown liquid rhounknown?

Answer 1

The given question is incomplete. The complete question is as follows.

A rectangular wooden block of weight W floats with exactly one-half of its volume below the waterline. The wooden block is removed from the water bath and placed in an unknown liquid. In this liquid, only one-third of the wooden block is submerged. Mass of the wooden block = 20g What is the density of the unknown liquid?

Explanation:

The given data is as follows.

Gravitational acceleration (g) = 9.81 $m/s^{2}$

Density of water ($\rho_{1}$) = 1000 $kg/m^{3}$

Density of the unknown liquid ($\rho_{2}$) = ?

Mass of the wooden block = m = 20 g = 0.02 kg      (as 1 kg = 1000 g)

Weight of the wooden block = W = mg

Volume of the wooden block = V

Since, it is given that wooden block floats in water with exactly one-half of its volume below the water.

Therefore, volume of the wooden block submerged in water will be as follows.

      $Vs_{1} = \frac{V}{2}$

Now, buoyancy force on the wooden block from water is as follows.

       $F_{1} = \rho_{1}Vs_{1}g$

As weight of the wooden block is supported by the buoyancy force of water.

So,           W = $F_{1}$

The wooden block floats in water with exactly one-third of its volume below the unknown liquid.

Volume of the wooden block submerged in unknown liquid = $Vs_{2} = \frac{V}{3}$

Buoyancy force on the wooden block from the unknown liquid = $F_{2} = \rho_{2}Vs_{2}g$

Hence, weight of the wooden block is supported by the buoyancy force of the unknown liquid.

                       W = $F_{2}$

Therefore,    $F_{1} = F_{2}$

Also,     $\rho_{1}Vs_{1}g = \rho_{2}Vs_{2}g$

            $\rho_{1}(\frac{V}{2}) = \rho_{2}(\frac{V}{3})$

              $\frac{\rho_{1}}{2} = \frac{\rho_{2}}{3}$

          $\frac{1000}{2} = \frac{\rho_{2}}{3}$

            $\rho_{2} = 1500 kg/m^{3}$

Now, we will convert $kg/m^{3}$ to $g/cm^{3}$

               $\rho_{2} = \frac{1500}{1000} g/cm^{3}$

                $\rho_{2} = 1.5 g/cm^{3}$

Thus, we can conclude that density of the unknown liquid is $1.5 g/cm^{3}$ .

Answer 2

The density of the unknown liquid is equal to 1.5 gram per cubic centimeter.

What is buoyancy?

Buoyancy can be defined as a force which is created by the water displaced by an object. Thus, buoyancy is directly proportional to the volume of water that is being displaced by an object.

Mathematically, the buoyancy of an object is given by this formula;

$F_b =\rho gh$

Note: Volume of wooden block is $\frac{V}{3}$ after submerssion in water.

Also, volume of wooden block is $\frac{V}{2}$ after submerssion in unknown liquid.

$F_{bw}=F_{bl}\\\\\frac{\rho_w gh}{2} =\frac{\rho_l gh}{3} \\\\3\rho_w gh=2\rho_l gh\\\\3\rho_w =2\rho_l \\\\\rho_l=\frac{3}{2} \\\\\rho_l=1.5\;g/cm^3$

Read more on density here: brainly.com/question/3173452