Question

A wooden block of dimensions of 6 cm ⨯ 6 cm ⨯ 15 cm floats with the long axis oriented horizontally and the of the square faces are oriented so that they point up vertically, straight up out of the water like a diamond shape. If the wooden block is submerged to a depth 7.96 cm (the bottom point to the water level), then: (a) What is the submerged volume of the block?(b) What is the mass of the block
(c) What is the density of the block?

Answer 1

Explanation:

(a)  Formula to calculate volume of the submerged wooden block is as follows.

            $V_{sub} = l \times w \times d$

It is given data of the wooden block is as follows.

          depth = 7.96 cm,       length (l) = 6 cm

         width (w) = 4 cm,      

So, we will calculate the volume of the submerged wooden block as follows.

           $V_{sub} = l \times w \times d$

                       = $6 \times 6 \times 7.96$

                       = 286.56  $cm^{3}$

Hence,  the submerged volume of the block is 286.56  $cm^{3}$.

(b)   Expression for the buoyant force acting on the wooden block is as follows.

            $F_{B} = \rho_{w} g V_{sub}$

And, expression for the force of gravity of the wooden block is as follows.

            $F_{g} = m_{b}g$

As the wooden block is floating on the water hence, buoyant force is balanced by the weight of the block.

                 $F_{g} = F_{B}$

Hence, mass of the wooden block will be calculated as follows.

         $F_{g} = F_{B}$

       $m_{b}g = \rho_{w}gV_{sub}$

          $m_{b} = \rho_{w}V_{sub}$

                      = $997 kg/m^{3} \times 286.56 cm^{3}$

                      = $997 kg/m^{3} \times 286.56 \times 10^{-6} m^{3}$

                      = 0.02857 kg

Therefore, mass of the given block is 0.02857 kg

(c)   Expression for the density of the block is as follows.

             $\rho_{b} = \frac{m_{b}}{V_{b}}$

Now, expression for the total volume of the wooden block is as follows.

             $V_{b} = l \times w \times h$

Hence, density of the given block is as follows.

              $\rho_{b} = \frac{m_{b}}{V_{b}}$

                         = $\frac{m_{b}}{lwh}$

                         = $\frac{0.02857 kg}{4 \times 4 \times 15}$

                         = $1.19 \times 10^{-4} kg/cm^{3}$

Therefore, density of the given block is $1.19 \times 10^{-4} kg/cm^{3}$.