Question

A cylindrical pipe of radius $1.4\,\text{m}$ has water flowing out at $2.5\,\text{m/s}$ into a cuboidal tank of dimensions $28\,\text{m}\times 11\,\text{m}\times 25\,\text{m}$. The flow completely occupies the pipe's cross-section. What percentage of the tank is filled up in $8$ min $20$ s?

• A. 66.66%
• B. 100%
• C. 86%
• D. 75%

Hint:

For steady flow, use $Q=Av$ and $V=Qt$. Converting time to seconds keeps units consistent; then compare $V$ to the tank capacity for the percentage.

Answer 1

The Correct Option is B

Step 1: Compute flow rate from the pipe.
Cross-sectional area \(A=\pi r^2=\pi(1.4)^2=\pi\cdot 1.96\). 
Speed \(v=2.5\,\text{m/s}\). 
Volumetric flow rate \(Q=Av=1.96\pi\times 2.5=4.9\pi\ \text{m}^3/\text{s}\). 

Step 2: Volume delivered in the given time.
Time \(t=8\ \text{min}\ 20\ \text{s}=500\ \text{s}\). 
Volume \(V_{\text{in}}=Qt=4.9\pi\times 500=2450\pi\ \text{m}^3\). 

Step 3: Tank volume and fill percentage.
Tank volume \(V_T=28\times 11\times 25=7700\ \text{m}^3\). 
Fill fraction \(=\dfrac{2450\pi}{7700}=\dfrac{7\pi}{22}\approx 0.9996\). 
Percentage \(\approx 99.96\%\ \approx 100\%\). \[ \boxed{100\%} \]

Answer 2

Rather than carrying \(\pi\) symbolically to the end, this problem becomes exact if \(\pi\) is approximated as \(\dfrac{22}{7}\) from the start — a natural choice since the pipe radius is \(1.4=\dfrac{7}{5}\) m.

Cross-sectional area of the pipe. \[ A=\pi r^2=\frac{22}{7}\times(1.4)^2=\frac{22}{7}\times1.96=6.16\ \text{m}^2. \]

Flow rate and volume delivered. \[ Q=A\times v=6.16\times2.5=15.4\ \text{m}^3/\text{s}. \] Time \(t=8\ \text{min}\ 20\ \text{s}=500\ \text{s}\), so \[ V_{\text{in}}=Q\times t=15.4\times500=7700\ \text{m}^3. \]

Comparing with tank capacity. Tank volume \(=28\times11\times25=7700\ \text{m}^3\), which is exactly equal to the volume delivered.

1. Option 66.66%: Would mean only two-thirds of the tank filled, but the delivered volume equals the full tank volume; ruled out.
2. Option 100%: Matches exactly — \(7700\ \text{m}^3\) delivered against a \(7700\ \text{m}^3\) tank.
3. Option 86%: Would require a volume of only about \(6622\ \text{m}^3\), well short of the computed \(7700\ \text{m}^3\); ruled out.
4. Option 75%: Would require \(5775\ \text{m}^3\), also short of the computed value; ruled out.

The delivered volume exactly matches the tank's capacity, so the tank is completely filled.

So the correct answer is 100%.