Water is transported on a level road in the cylindrical container of diameter 0.5 m and height 0.8 m. The maximum acceleration of vehicle is 5 m/s². What is the initial allowable height of water (cm) in the tank, if there is no spill?
Try it first: pick an option, then read the explanation.
Asked in SSC JE 29-01-2018 (Evening Shift)
Detailed Explanation
When an open container carrying liquid undergoes uniform linear horizontal acceleration, the liquid undergoes rigid-body motion. The free liquid surface tilts at an angle \(\\theta\) such that the surface normal aligns with the resultant of gravity and the fictitious inertial force.
1. Given data:
- Tank diameter \(d = 0.5\text{ m}\)
- Tank height \(H = 0.8\text{ m}\)
- Horizontal acceleration \(a = 5\text{ m/s}^2\)
- Gravitational acceleration \(g = 9.81\text{ m/s}^2\)
2. Calculate the tilt angle of the free surface:
\[ \tan\theta = \frac{a}{g} = \frac{5}{9.81} \approx 0.50968 \]
3. By conservation of liquid volume, the free surface pivots about the tank's central axis. The total difference in liquid level across the container diameter is:
\[ x = d \tan\theta = 0.5\text{ m} \times 0.50968 = 0.2548\text{ m} \]
4. The maximum liquid rise at the rear wall above the initial still level is:
\[ \Delta h_{\text{rise}} = \frac{x}{2} = \frac{0.2548}{2} = 0.1274\text{ m} = 12.74\text{ cm} \]
5. To prevent spilling over the top edge of the tank (\(H = 0.80\text{ m}\)):
\[ h_{\text{initial}} = H - \Delta h_{\text{rise}} = 0.80\text{ m} - 0.1274\text{ m} = 0.6726\text{ m} \approx 67.3\text{ cm} \]
Why “67.3” Is Correct
Under a horizontal acceleration of \(5\text{ m/s}^2\), the water tilts with slope \(a/g = 5/9.81 = 0.5097\). At the rear boundary, the liquid climbs by \((d/2)\tan\theta = 0.25 \times 0.5097 = 0.1274\text{ m}\). Subtracting this rise from the tank height of \(0.80\text{ m}\) leaves an initial filling limit of \(67.3\text{ cm}\).
Why the Other Options Are Wrong
- A. 50.5: 50.5 cm results from mistakenly subtracting the full level difference x (0.255 m) from the height: 0.80 − 0.255 ≈ 54.5 cm with additional errors.
- C. 92.7: 92.7 cm exceeds the total physical height of the container (0.8 m = 80 cm) and is impossible.
- D. 100: 100 cm is completely impossible because the container height is only 80 cm.
Key Point
Rigid body acceleration: tan θ = a/g; Rise at rear lip = (d/2) · (a/g); Max depth = H − (d/2)·(a/g).
Key Concepts for Students
- In steady horizontal acceleration, isobars remain parallel to the tilted free surface.
- Because volume is conserved in a symmetrical tank, the surface pivots about its geometric centreline.
- Pressure distribution remains hydrostatic along any direction perpendicular to the tilted free surface.
Practice more: this is question 19 in Hydrostatic Pressure MCQs – Part 2. Try the whole set and check every answer instantly.