Question

A measuring jar of internal diameter 10 cm is partially filled with water. Four equal spherical balls of diameter 2 cm each is dropped in it and they sink down in water completely. What will be the change in the level of water in the jar?
$
  {\text{A}}{\text{. }}\dfrac{4}{{75}}cm \\
  {\text{B}}{\text{. }}\dfrac{{16}}{{75}}cm \\
  {\text{C}}{\text{. }}\dfrac{{32}}{{75}}cm \\
  {\text{D}}{\text{. none of these}} \\
 $

Answer 1

Hint: - Here we go through by calculating the volume of all the spheres and equating it by the volume of the jar by taking the height of the jar as the rise height of water. To find out the level of water in the jar.

Complete step-by-step answer:
Here in the question it is given that a measuring jar of internal diameter 10 cm is partially filled with water.
It means the radius of jar R=$\dfrac{{10}}{2} = 5cm$.
Now for a spherical ball it is given that diameter is 2cm it means the radius of the spherical ball, r=1cm.
As the balls dropped in the jar and they sank down in water completely.
By this the water rises some height, suppose water rises by h cm.
It means that the volume of water raised in the jar = Volume of 4 balls
As we know that the formula of volume of cylinder is $\pi {R^2}h$ and the volume of sphere is$\dfrac{4}{3}\pi {r^3}$.
By applying these formula we can write,
$
   \Rightarrow \pi \times {5^2} \times h = 4\left( {\dfrac{4}{3} \times \pi \times {1^3}} \right) \\
   \Rightarrow 25h = \dfrac{{16}}{3} \\
  \therefore h = \dfrac{{16}}{{75}}cm \\
 $
Hence option B is the correct answer.

Note: - Whenever we face such a type of question the key concept for solving the question is always equate the volume. As the water rises by the same amount of the volume that the sphere is put in the jar filled with water.