Question

Diameter of the base of a water-filled inverted right circular cone is $26$ cm. A cylindrical pipe, $5$ mm in radius, is attached to the surface of the cone at a point. The perpendicular distance between the point and the base (the top) is $15$ cm. The distance from the edge of the base to the point is $17$ cm, along the surface. If water flows at the rate of $10$ meters per minute through the pipe, how much time would elapse before water stops coming out of the pipe?

• A. $<4.5$ minutes
• B. $\ge 4.5$ minutes but $<4.8$ minutes
• C. $\ge 4.8$ minutes but $<5$ minutes
• D. $\ge 5$ minutes but $<5.2$ minutes
• E. $\ge 5.2$ minutes

Hint:

For draining from a side hole, the level drops to the hole’s height. In similar solids Cones), volume scales with the cube of linear dimensions, so a quick ratio gives the drained volume before using the flow rate.

Answer 1

The Correct Option is D

Step 1: Find cone dimensions from the given point.
Let base radius $R=13$ cm Diameter $26$ cm), height $H$ cm, and slant height $L$ cm. The point on the surface is $17$ cm along the generator from the rim and is $15$ cm below the top plane, so \[ \frac{H}{L}=\frac{15}{17}⇒ L=\frac{17}{15}H. \] But $L^2=R^2+H^2$; hence \[ \left(\frac{17}{15}\right)^2=1+\frac{R^2}{H^2} ⇒ \frac{R}{H}=\frac{8}{15}⇒ H=\frac{15}{8} 13=\frac{195}{8}=24.375\ \text{cm}. \] Step 2: Volume that drains until the level reaches the hole.
Water stops when the level falls to the hole, i.e., the new water height is $H-15=9.375$ cm. Since cross-sections are similar, volume scales as (height)$^3$. If $V$ is the full cone volume, \[ V=\frac{1}{3}\pi R^2H,\qquad V_{\text{remain}}=V\left(\frac{H-15}{H}\right)^3 =V\left(\frac{5}{13}\right)^3. \] Thus drained volume \[ V_{\text{drain}}=V\left(1-\frac{125}{2197}\right). \] Numerically (in m$^3$): $R=0.13,\ H=0.24375$, \[ V=\frac{1}{3}\pi(0.13)^2(0.24375)\approx 0.0043138, V_{\text{drain}}\approx 0.0040684\ \text{m}^3. \] Step 3: Time from flow rate.
Pipe radius $r=5\text{ mm}=0.005$ m, speed $v=10$ m/min.
Volumetric flow $Q=v \pi r^2=10 \pi(0.005)^2\approx 7.85398\times 10^{-4}\ \text{m}^3/\text{min}$.
Time \[ t=\frac{V_{\text{drain}}}{Q} \approx \frac{0.0040684}{7.85398\times 10^{-4}} \approx 5.18\ \text{minutes}. \] \[ \boxed{t\approx 5.18\ \text{minutes}} \] which lies in $\,[5,\ 5.2)\,$ minutes.