Question

Some spherical balls of diameter $2.8\,\text{cm}$ are dropped into a cylindrical container containing some water and are fully submerged. The diameter of the container is $14\,\text{cm}$. Find how many balls have been dropped in it if the water rises by $11.2\,\text{cm}$.

• A. 50
• B. 150
• C. 250
• D. 350

Hint:

Use displacement: rise in cylinder volume $=\pi R^2 h$ equals $n$ times sphere volume $\frac{4}{3}\pi r^3$. The $\pi$ cancels, keeping arithmetic clean.

Answer 1

The Correct Option is B

Step 1: Volumes displaced.
Rise in water gives cylinder volume increase:
Container radius $R=\dfrac{14}{2}=7\,\text{cm}$, rise $h=11.2\,\text{cm}$.
\[ V_{\text{rise}}=\pi R^2 h=\pi\cdot 7^2\cdot 11.2=\pi\cdot 49\cdot 11.2=548.8\,\pi\ \text{cm}^3. \] Step 2: Volume of one sphere.
Ball radius $r=\dfrac{2.8}{2}=1.4\,\text{cm}$.
\[ V_{\text{sphere}}=\frac{4}{3}\pi r^3=\frac{4}{3}\pi(1.4)^3=\frac{4}{3}\pi(2.744)=\frac{1372}{375}\pi\ \text{cm}^3. \] Step 3: Number of balls.
Let $n$ be the number of balls. Then \[ n=\frac{V_{\text{rise}}}{V_{\text{sphere}}} =\frac{548.8\,\pi}{\tfrac{1372}{375}\pi} =\frac{548.8\times 375}{1372} =\frac{\tfrac{2744}{5}\times 375}{1372} =\frac{2744}{1372}\times \frac{375}{5} =2\times 75=150. \] \[ \boxed{150} \]

Answer 2

A quicker route notices that the container's radius is exactly five times the ball's radius, which lets the volume ratio be simplified before plugging in numbers.

Setting up the ratio. Container radius \(R=7\) cm, ball radius \(r=1.4\) cm, so \(R=5r\). The rise in water volume equals the total volume of the \(n\) balls: \[ \pi R^2 h = n\times\frac{4}{3}\pi r^3. \] Substituting \(R=5r\): \[ \pi(5r)^2h = n\cdot\frac{4}{3}\pi r^3 \ \Rightarrow\ 25r^2h = \frac{4}{3}n\,r^3 \ \Rightarrow\ n=\frac{75h}{4r}. \]

Substituting values. With \(h=11.2\) cm and \(r=1.4\) cm: \[ n=\frac{75\times11.2}{4\times1.4}=\frac{840}{5.6}=150. \]

1. Option 50: Too small — would correspond to a water rise of only about \(3.7\) cm, not \(11.2\) cm; ruled out.
2. Option 150: Matches the value obtained from \(n=75h/(4r)\) exactly.
3. Option 250: Would require a rise of about \(18.7\) cm, more than what's given; ruled out.
4. Option 350: Far too large for an \(11.2\) cm rise; ruled out.

The radius-ratio shortcut confirms the same count without needing to expand \(\left(\tfrac{2.8}{2}\right)^3\) directly.

So the correct answer is 150.