Question

Three pipes are connected to an inverted cone, with its base at the top. Two inlet pipes, $A$ and $B$, can fill the empty cone in $8$ hours and $12$ hours, respectively. The outlet pipe $C$ (at the bottom) can empty a filled cone in $4$ hours. When the cone is completely filled with water, all three pipes are opened. Two of the pipes remain open for $20$ hours continuously and the third pipe remains open for a lesser time. As a result, the height of water comes down to $50%$. Which of the following options would be possible?

• A. Pipe A was open for 19 hours.
• B. Pipe A was open for 19 hours 30 minutes.
• C. Pipe B was open for 19 hours 30 minutes.
• D. Pipe C was open for 19 hours 50 minutes.
• E. The situation is not possible.

Hint:

For conical tanks, volume $\propto h^3$. Convert height change to volume change first, then set up a net-rate equation for each case.

Answer 1

The Correct Option is C

Cone volume–height relation

If height reduces to \(50\%\), volume reduces to \(\left(\tfrac{1}{2}\right)^3 = \tfrac{1}{8}\). Hence, net volume removed \(= \tfrac{7}{8}\) of the full cone.

Pipe rates (fraction of full-cone per hour)

\[ A:\ +\tfrac{1}{8}, \qquad B:\ +\tfrac{1}{12}, \qquad C:\ -\tfrac{1}{4}. \]

Exactly two pipes run the full \(20\) hours; the third runs for \(t(<20)\) hours. Consider the three possibilities.

(i) Pipe A is the shorter-time pipe (B and C run 20h)

\[ \Delta V = 20\Big(\tfrac{1}{12} - \tfrac{1}{4}\Big) + t\Big(\tfrac{1}{8}\Big) = -\tfrac{10}{3} + \tfrac{t}{8} = -\tfrac{7}{8}. \] \[ \Rightarrow\ t = \tfrac{59}{3} = 19\text{ h }40\text{ m}. \] Not among the given options.

(ii) Pipe B is the shorter-time pipe (A and C run 20h)

\[ \Delta V = 20\Big(\tfrac{1}{8} - \tfrac{1}{4}\Big) + t\Big(\tfrac{1}{12}\Big) = -\tfrac{5}{2} + \tfrac{t}{12} = -\tfrac{7}{8}. \] \[ \Rightarrow\ t = \tfrac{39}{2} = 19\text{ h }30\text{ m}. \] This matches option (C).

(iii) Pipe C is the shorter-time pipe (A and B run 20h)

\[ \Delta V = 20\Big(\tfrac{1}{8} + \tfrac{1}{12}\Big) + t\Big(-\tfrac{1}{4}\Big) = \tfrac{25}{6} - \tfrac{t}{4} = -\tfrac{7}{8}. \] \[ \Rightarrow\ t = \tfrac{121}{6} \approx 20.17 > 20, \] impossible since the “shorter” time would exceed 20h.

Final Answer

The only feasible case is when Pipe B was open for \[ \boxed{19 \ \text{h}\ 30 \ \text{m (Option C)}.} \]