Chapter 19, Problem 60A

A circus elephant weighing 18,800 N is taken on board a barge of length 6.2 m and breadth 3.0 m, which floats in a river. Show that the barge sinks 10 cm when the elephant gets on board.

To determine

To show: Barge will sink at a depth of 10 cm when the elephant gets on board.

Explanation of Solution

Given:

Weight of elephant, $F_g=mg=18800\text{N}$

Barge length is 6.2 m and breadth is 3 m.

Formula Used:

The density is given as,

  $\rho =\frac{m}{V}$

Where $m$ is mass and $V$ is volume.

Calculation:

To support the additional weight of elephant, the volume of water displaced is,

  $\begin{array}{l}{V}_{\text{displaced}}=\frac{m}{{\rho }_{\text{water}}}=\frac{mg}{g{\rho }_{\text{water}}}\\ \Rightarrow V_{\text{displaced}}=\frac{\text{weight}}{g{\rho }_{\text{water}}}\end{array}$

Taking $g=10\text{N/kg}$ and $1000{\text{kg/m}}^3$

On substituting the values in above equation,

  $\begin{array}{l}{V}_{\text{displaced}}=\frac{18800\text{N}}{10\text{N/kg}\times 1000{\text{kg/m}}^3}\\ \Rightarrow V_{\text{displaced}}=1.88{\text{m}}^3\end{array}$

Now, this volume will be equal to the volume of the barge.

So,

  $\begin{array}{l}1.88{\text{m}}^3=6.2\text{m}\times 3\text{m}\times \text{depth}\\ \Rightarrow \text{depth}=\frac{1.88{\text{m}}^3}{6.2\text{m}\times 3\text{m}}=0.1\text{m}=10\text{cm}\end{array}$

Conclusion:

Hence, it is proved that the barge sinks 10 cm when the elephant gets on board.