Chapter 12 · Question 5

A vessel is in the form of an inverted cone. Its height is $8$ cm and the radius of its top, which is open, is $5$ cm. It is filled with water up to the brim. When lead shots, each of which is 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 dropped in the vessel.

Answer Revealed

Direct Answer:

The vessel is a cone with radius

r = 5

cm and height

h = 8

cm, filled completely with water. Volume of water in the cone

= \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \times 5^2 \times 8 = \frac{1}{3}\pi \times 25 \times 8 = \frac{200}{3}\pi

^3

. When lead shots are dropped, one-fourth of the water overflows: Volume of water that flows out

= \frac{1}{4} \times \frac{200}{3}\pi = \frac{50}{3}\pi

^3

. This outflow volume equals the total volume of all lead shot spheres. Volume of one lead shot (sphere)

= \frac{4}{3}\pi R^3

, where

R = 0.5

cm. So volume of one shot

= \frac{4}{3}\pi \times (0.5)^3 = \frac{4}{3}\pi \times 0.125 = \frac{0.5}{3}\pi = \frac{1}{6}\pi

^3

. Let

n

be the number of shots. Then:

n \times \frac{1}{6}\pi = \frac{50}{3}\pi

. Cancel

\pi

n \times \frac{1}{6} = \frac{50}{3} \implies n = \frac{50}{3} \times 6 = 100

. Thus,

100

lead shots were dropped.

Simple Explanation

The inverted cone holds

\frac{200}{3}\pi

^3

of water. When lead balls are dropped, they push out one-quarter of the water, which is

\frac{50}{3}\pi

^3

. Each tiny lead ball (radius

0.5

cm) has a volume of

\frac{4}{3}\pi \times 0.125 = \frac{1}{6}\pi

^3

. The number of balls needed to match the overflow volume:

\frac{50}{3}\pi \div \frac{1}{6}\pi = 100

. So

100

lead shots were dropped in.

Exam-Ready Structure

This is a classic conversion-of-solids problem using Archimedes' principle: the volume of water displaced equals the volume of the solid submerged. 1. Calculate the volume of water in the inverted cone: The cone has radius

r = 5

cm and height

h = 8

cm. Volume

V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \times 25 \times 8 = \frac{200}{3}\pi

^3

. This is the initial volume of water. 2. Overflow condition: When lead shots are fully submerged, one-fourth of the water flows out. Volume of water displaced

= \frac{1}{4} \times V_{\text{cone}} = \frac{1}{4} \times \frac{200}{3}\pi = \frac{50}{3}\pi

^3

. 3. By the principle of displacement, this displaced volume of water equals the total volume of all lead shots. Volume of one lead shot (sphere, radius

R = 0.5

cm)

= \frac{4}{3}\pi R^3 = \frac{4}{3}\pi \times (0.5)^3 = \frac{4}{3}\pi \times 0.125 = \frac{0.5}{3}\pi = \frac{\pi}{6}

^3

. 4. Let the number of lead shots be

n

. Then:

n \times \frac{\pi}{6} = \frac{50}{3}\pi

. Cancel

\pi

\frac{n}{6} = \frac{50}{3} \implies n = \frac{50}{3} \times 6 = 50 \times 2 = 100

. 5. Therefore,

100

lead shots were dropped. 6. Key principle: When a solid is immersed in a liquid, it displaces a volume of liquid equal to its own volume. This converts the geometric problem of finding how many spheres fit into the overflow volume.

Key Points

Cone volume $= \frac{1}{3}\pi (5)^2 \times 8 = \frac{200}{3}\pi$ cm $^3$
Water displaced $= \frac{1}{4} \times \frac{200}{3}\pi = \frac{50}{3}\pi$ cm $^3$
One lead shot (sphere $r = 0.5$ cm) volume $= \frac{4}{3}\pi (0.5)^3 = \frac{\pi}{6}$ cm $^3$
Number of shots $n$ : $n \times \frac{\pi}{6} = \frac{50}{3}\pi \implies n = 100$
Conversion principle: volume of submerged solid equals volume of displaced liquid

Common Mistakes

Multiplying by $\frac{1}{4}$ incorrectly — one-fourth of the water flows OUT, so the volume that flows out is $\frac{1}{4}$ of the cone's volume, not $\frac{3}{4}$
Using diameter ( $1$ cm) instead of radius ( $0.5$ cm) when computing the volume of a lead shot

Simple Explanation

Exam-Ready Structure

Key Points

Common Mistakes

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