Question

One sphere of diameter 18cm is dropped in a cylindrical vessel of diameter 36cm, partially filled with water. If the sphere is completely immersed in water, find the rise in the level of water.

Answer 1

Hint: Assume that the initial height of the water inside the cylinder by “h” and let its level rise by d. Use the fact that the increase in the height of the water is due to occupancy of volume by the sphere. Hence find a linear equation in d. Solve for d to get the value of height rise in the column of water. Use volume of cylinder $=\pi {{r}^{2}}h$ and volume of a sphere $=\dfrac{4}{3}\pi {{r}^{3}}$.

Complete step-by-step answer:
Let the initial height of the water column inside the cylinder be “h” and let its level rose by d after the sphere is immersed in it. Let us denote the radius of the cylinder by r and that of the sphere by 2r.
Initial volume of the cylinder $=\pi {{r}^{2}}h$
Final volume of the cylinder $=\pi {{r}^{2}}\left( h+d \right)$
Volume of the sphere $=\dfrac{4}{3}\pi {{\left( r' \right)}^{3}}$
But the final volume of the cylinder is equal to the sum of the initial volume of the cylinder and the volume of the sphere.

Hence we have
$\pi {{r}^{2}}\left( h+d \right)=\pi {{r}^{2}}h+\dfrac{4}{3}\pi {{\left( r' \right)}^{3}}$
Dividing both sides of the equation by $\pi $ we get
${{r}^{2}}\left( h+d \right)={{r}^{2}}h+\dfrac{4}{3}{{\left( r' \right)}^{3}}$
Subtracting ${{r}^{2}}h$ from both sides, we get
$\begin{align}
  & {{r}^{2}}\left( h+d \right)-{{r}^{2}}h=\dfrac{4}{3}{{\left( r' \right)}^{3}} \\
 & \Rightarrow {{r}^{2}}d=\dfrac{4}{3}{{\left( r' \right)}^{3}} \\
\end{align}$
Dividing both sides by ${{r}^{2}}$, we get
$d=\dfrac{4{{\left( r' \right)}^{3}}}{3{{r}^{2}}}$
Substituting $r'=\dfrac{18}{2}=9$ and $r=\dfrac{36}{2}=18$, we get
$d=\dfrac{4{{\left( 9 \right)}^{3}}}{3{{\left( 18 \right)}^{2}}}=3$
Hence the rise in level of water = 3cm.

Note: [1] Calculation trick: This method minimises the risk of calculation mistake.
Observe that the radius of the cylinder is twice the radius of the sphere.
Also, since there is a dcm rise in level of the cylinder, net change in volume $=\pi {{r}^{2}}d$
But change in volume is due to immersion of the sphere.
Hence we have
$\pi {{r}^{2}}d=\dfrac{4}{3}\pi {{\left( r' \right)}^{3}}$
Since r = 2r’, we have
$\begin{align}
  & \pi {{\left( 2r' \right)}^{2}}d=\dfrac{4}{3}\pi {{\left( r' \right)}^{3}} \\
 & \Rightarrow d=\dfrac{r'}{3}=\dfrac{9}{3}=3 \\
\end{align}$
[2] In questions of this type do not make the mistake of substituting the value of $\pi $ as this will cause errors in calculations and hence yield incorrect answers.