Inverted Cone and Lead Shots Problem | Class 10 Maths

Quick answer: The water that flows out equals the total volume of the lead shots. The cone holds $\frac{1}{3}\pi(5)^2(8) = \frac{200\pi}{3}$ cm³, so one-fourth of it is $\frac{50\pi}{3}$ cm³. Each shot has volume $\frac{4}{3}\pi(0.5)^3 = \frac{\pi}{6}$ cm³. So the number of shots is $\frac{50\pi/3}{\pi/6} =$ 100.

Syllabus note: This is from the NCERT exercise on conversion of solids from one shape to another. That section was removed from the rationalised NCERT Class 10 Surface Areas and Volumes chapter (from 2023–24). The volume formulas used are still in the syllabus, so it remains good practice.

The idea: displaced water

When a solid sinks completely in a vessel filled to the brim, the volume of water that overflows is equal to the volume of the solid. With many identical solids:

$\text{water that flows out} = N \times \text{volume of one solid}$

The problem

A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its open top is 5 cm. It is filled with water to the brim. When lead shots, each a sphere of radius 0.5 cm, are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots.

  \------- 5 cm -------/
   \      water       /
    \  o o o (shots) /
     \              /  8 cm
      \            /
       \__________/
        (vertex)

Step-by-step solution

Step 1: Volume of the cone.

$V_{\text{cone}} = \frac{1}{3}\pi R^2 H = \frac{1}{3} \times \pi \times 25 \times 8 = \frac{200\pi}{3} \text{ cm}^3$

Step 2: Volume of water that flows out.

$\frac{1}{4} \times \frac{200\pi}{3} = \frac{50\pi}{3} \text{ cm}^3$

Step 3: Volume of one lead shot. With $r = 0.5 = \frac{1}{2}$ cm:

$V_{\text{shot}} = \frac{4}{3}\pi\left(\frac{1}{2}\right)^3 = \frac{4}{3} \times \frac{1}{8}\pi = \frac{\pi}{6} \text{ cm}^3$

Step 4: Number of shots.

$N \times \frac{\pi}{6} = \frac{50\pi}{3}$

$N = \frac{50}{3} \times 6 = 100$

Answer: 100 lead shots were dropped into the vessel.

Quick check with one formula

Cancelling $\pi$ and $\frac{1}{3}$:

$N = \frac{\frac{1}{4} \cdot \frac{1}{3}\pi R^2 H}{\frac{4}{3}\pi r^3} = \frac{R^2 H}{16 r^3} = \frac{25 \times 8}{16 \times 0.125} = \frac{200}{2} = 100$

Summary table

Exam tips

• Write the key idea in words: "volume of water that flows out = total volume of lead shots".
• Keep $\pi$ as a symbol; it cancels at the end, so there is no need to use 22/7.
• Change 0.5 to $\frac{1}{2}$ before cubing: $\left(\frac{1}{2}\right)^3 = \frac{1}{8}$.
• Use the one-line formula to check your answer.

Common mistakes

• Using the whole volume of the cone instead of one-fourth, which gives 400 instead of 100.
• Using the diameter instead of the radius for the shots.
• Forgetting the $\frac{1}{3}$ in the cone formula.
• Cubing 0.5 incorrectly: $(0.5)^3 = 0.125$, not 0.25.

Frequently Asked Questions

How many lead shots are dropped into the cone?

100 lead shots. The water that flows out is 50π/3 cm³ and each shot has a volume of π/6 cm³.

Why is the volume of water that flows out equal to the volume of the shots?

When a solid sinks completely in a full vessel, it pushes out a volume of water equal to its own volume.

What is the volume of a cone of radius 5 cm and height 8 cm?

It is (1/3) × π × 25 × 8 = 200π/3 cm³, or about 209.5 cm³.

What is the volume of a sphere of radius 0.5 cm?

It is (4/3) × π × (0.5)³ = π/6 cm³, or about 0.52 cm³.

Is this question in the current Class 10 syllabus?

It comes from the "conversion of solids" section, which was removed from the rationalised NCERT Class 10 book, though the cone and sphere volume formulas are still in the syllabus.