Question

A rectangular tank $25$cm long and $20$cm wide contains $4.5$liters of water. When a metal cube is lowered in the tank, the water level rises to a height of $11$cm. Find the length of the edge of the cube?

Answer 1

Hint: Try to find the height of the water level using the formula of the volume of a cuboid. Then find the volume of water when the metal cube is lowered in the tank. Change in the volume is the volume of the metal cube.

Complete step by step solution:
It is given that a rectangular tank, $25$cm long and $20$cm wide contains$4.5$liters of water.

First, assume that the height of the water level is $\left( h \right)$

Now, covert the given quantity of the water, $4.5$liters in cubic centimeters using the relation given as:

$1{\text{ litre}} = 1000{\text{ cubic cm}}$
So, $4.5$liters of water in term of cubic centimeters is given as:
${\text{4}}{\text{.5 litre}} = \left( {4.5 \times 1000} \right){\text{ cubic cm}}$
${\text{4}}{\text{.5 litre}} = 4500{\text{ c}}{{\text{m}}^3}$
So, there are $4500{\text{ c}}{{\text{m}}^3}$of water present in the rectangular tank. Now, find the height of the water level using the formula for the volume of a cuboid. The formula is given as:

Volume of water$ = \left( {{\text{length of the tank}} \times {\text{width of the tank}}} \right) \times {\text{height of the water level}}$
Substitute the value of the water as $4500$, length and width of the tank as $25$ and$20$ respectively, and put height as the height of the water level. So, the above equation is given as:
$4500 = \left( {25 \times 20} \right) \times {\text{h}}$
Now, solve the equation for the value of $h$:
$4500 = 500 \times {\text{h}}$
$ \Rightarrow {\text{h}} = \dfrac{{4500}}{{500}}$
$ \Rightarrow {\text{h}} = 9$
Thus, the height of the water level is $9$cm.

Now, it is given that a metal cube is lowered in a tank, and the water level rises to the height of $11$cm.
It means that the increase in the volume of the water is equal to the volume of the metal cube.
When the height of the water level is $11$cm, then the volume of the water is given as:
Volume of water$ = \left( {{\text{length of the tank}} \times {\text{width of the tank}}} \right) \times {\text{height of the water level}}$
Volume of water$ = \left( {25 \times 20} \right) \times 11$
Volume of water$ = 500 \times 11$
Volume of water$ = 5500$
After lowering the metal cube, the volume of the water is $5500{\text{ c}}{{\text{m}}^3}$.

Now, find the change in the volume of the liquid.
Change in the volume is given as:
$ = \left( {{\text{Volume of liquid with 11cm height}}} \right) - \left( {{\text{Volume of liquid with 9cm height}}} \right)$
$ = \left( {5500} \right) - \left( {{\text{4500}}} \right)$
$ = 1000{\text{ c}}{{\text{m}}^3}$
It means that the change in the volume of the water is $1000{\text{ c}}{{\text{m}}^3}$, thus the volume of the metal cube is $1000{\text{ c}}{{\text{m}}^3}$.
Now, assume that the side length of the metal cube is expressed as$\left( s \right)$. Then the volume of the metal cube is given as:
The volume of the metal cube$ = {\left( s \right)^3}$
Since, the volume of the cube is analyzed as $1000{\text{ c}}{{\text{m}}^3}$, so the above equation is given as:

$1000 = {\left( s \right)^3}$
We can express $1000$as${\left( {10} \right)^3}$then the above equation is expressed as:
$
  {\left( {10} \right)^3} = {s^3} \\
   \Rightarrow s = 10 \\
$

Therefore, the length of the edge of the cube is $10{\text{ cm}}$

Note: The volume of the water is $4500{\text{ c}}{{\text{m}}^3}$ and when a metal cube is lowered in the tank, then the volume of water is increased and its value is $5500{\text{ c}}{{\text{m}}^3}$. So, the change in the volume of the water is the volume of the metal cube.