Question

Marbles of diameter 1.4 cm are dropped into a beaker containing some water and fully submerged. The diameter of the beaker is 7 cm. Find how many marbles have been dropped in it if water rises by 5.6 cm.

Answer 1

Hint: This problem is based on a principle from science known as Archimedes’ principle which states that the volume of liquid displaced by submerging an object into the liquid is equal to the volume of the object itself. We are given the diameter of the marble. We are also provided with the diameter of the beaker. According to the aforementioned Archimedes’ principle, we can say that the change in volume of water will be equal to the total volume of the marbles submerged in the water. The beaker is cylindrical in shape and we know that the volume of a cylinder is given as $\dfrac{\pi {{d}^{2}}h}{4}$ . The marbles are spherical in shape and volume of a sphere is given as $\dfrac{\pi {{d}^{3}}}{6}$.

Complete step by step answer:
It is given to us that marbles of diameter 1.4 cm are dropped into a beaker containing some water and fully submerged.
We know that marbles are spherical in shape and the volume of a sphere is given by the relation ${{V}_{s}}=\dfrac{\pi {{d}^{3}}}{6}$.

Thus, we will find the volume of one marble of 1.4 cm diameter.
\[\Rightarrow {{V}_{s}}=\dfrac{\pi {{\left( 1.4 \right)}^{3}}}{6}c{{m}^{3}}\]
Let us assume that n marbles are dropped into the beaker filled with water.
Therefore, cumulative volume of n marbles will be \[{{V}_{m}}=n\times \dfrac{\pi {{\left( 1.4 \right)}^{3}}}{6}c{{m}^{3}}\].
It is given that the diameter of the beaker is 7cm. Let the height of water level be h.
The beaker is of cylindrical shape and we know that the volume of a beaker is given as ${{V}_{b}}=\dfrac{\pi {{d}^{2}}h}{4}$.

Thus, the volume of the beaker before the marbles were dropped are given as
$\Rightarrow {{V}_{b1}}=\dfrac{\pi {{\left( 7 \right)}^{2}}h}{4}c{{m}^{3}}$

After dropping n marbles, the height increases by 5.6 cm. Therefore, the final height will be (h + 5.6) and the volume will be given as:
$\Rightarrow {{V}_{b2}}=\dfrac{\pi {{\left( 7 \right)}^{2}}\left( h+5.6 \right)}{4}c{{m}^{3}}$
Therefore, the change in volume will be ${{V}_{b2}}-{{V}_{b1}}$.
$\Rightarrow {{V}_{b2}}-{{V}_{b1}}=\dfrac{\pi {{\left( 7 \right)}^{2}}\left( 5.6 \right)}{4}c{{m}^{3}}$
With the help of Archimedes’ principle, we can say that the change in volume of water will be equal to the total volume of the marbles submerged in the water.
\[\begin{align}
  & \Rightarrow {{V}_{b2}}-{{V}_{b1}}={{V}_{m}} \\
 & \Rightarrow \dfrac{\pi {{\left( 7 \right)}^{2}}\left( 5.6 \right)}{4}c{{m}^{3}}=n\times \dfrac{\pi {{\left( 1.4 \right)}^{3}}}{6}c{{m}^{3}} \\
\end{align}\]
We will solve the above equation to find the value of n.
\[\begin{align}
  & \Rightarrow \dfrac{\left( 7 \right)\times \left( 7 \right)\times \left( 5.6 \right)}{2}=n\times \dfrac{\left( 1.4 \right)\times \left( 1.4 \right)\left( 1.4 \right)}{3} \\
 & \Rightarrow 2.8\times 3=n\times \left( 0.2 \right)\times \left( 0.2 \right)\times \left( 1.4 \right) \\
 & \Rightarrow 2\times 3=n\times \left( 0.04 \right) \\
 & \Rightarrow \dfrac{6}{0.04}=n \\
 & \Rightarrow n=150 \\
\end{align}\]

Therefore, the number of marbles is 150.

Note: It is very common for mathematical problems to be based on scientific principles. Students are advised to be well aware of such common principles which are true in physical forms. Students are also advised to be careful while finding the volumes as the formula for volume with diameter is different from the formula for volume with radius.