Question

150 spherical marbles each of diameter 1.4cm are dropped in a cylindrical vessel of diameter 7cm containing some water, which are completely immersed in water. Find the level of rising water in the vessel.

Answer 1

Hint: Assume that the rise in the level of water is d. Use the fact that Volume of all sphere = Area of base of cylinder times the rise in the level of water. Hence form a linear equation in d and solve for d. The value of d will be the rise in the level of water.

Complete step-by-step answer:

Let d be the rise in the level of water. Let r be the radius of the base of the cylinder and x the radius of the spheres.
We know that Volume of all spheres = Area of base of cylinder times the rise in the level of water.
Now the base of the cylinder is circular.
So area of the base $=\pi {{r}^{2}}$
So, we have
The volume of all the spheres $=\pi {{r}^{2}}d$
Also, we know that volume of a sphere $\dfrac{4}{3}\pi {{r}^{3}}$
Substituting r = x, we get
The volume of one spherical ball $=\dfrac{4}{3}\pi {{x}^{3}}$
Hence the volume of 150 spherical balls $150\times \dfrac{4}{3}\pi {{x}^{3}}=200\pi {{x}^{3}}$
Hence, we have
$200\pi {{x}^{3}}=\pi {{r}^{2}}d$
Put x =0.7 and r = 3.5, we get
$\begin{align}
  & 200{{\left( 0.7 \right)}^{3}}={{\left( 3.5 \right)}^{2}}d \\
 & \Rightarrow d=\dfrac{68.6}{12.25}=5.6 \\
\end{align}$
Hence the level of water will rise by 5.6cm
Note: Volume of spherical balls + Volume of water = Volume of the cylinder occupied finally.
Let the water be initially at height h and let its level rise by d. Hence final height = h+d.
So, we have
The volume of spherical balls $+\pi {{r}^{2}}h=\pi {{r}^{2}}\left( h+d \right)$=
Hence the volume of spherical balls $=\pi {{r}^{2}}\left( h+d \right)-\pi {{r}^{2}}h=\pi {{r}^{2}}d$
Hence the volume of spherical balls = area of the base of cylinder times the level of rising of water, which is the result we used in solving the question above.