Mathematics — Standard (041) — CBSE Class 10 board question

Given: Height of cone (h) = 8 cm, radius (r) = 5 cm, radius of each lead shot = 0.5 cm

Step 1: Volume of water in the cone (= volume of cone)

$$V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 5 \times 5 \times 8 = \frac{2200}{21} \approx \frac{2200}{21} \text{ cm}^3$$

Step 2: Volume of water that flows out (one-fourth)

$$V_{\text{water out}} = \frac{1}{4} \times \frac{2200}{21} = \frac{550}{21} \text{ cm}^3$$

Step 3: Volume of each lead shot (sphere, r = 0.5 cm)

$$V_{\text{shot}} = \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (0.5)^3 = \frac{4}{3} \times \frac{22}{7} \times \frac{1}{8} = \frac{11}{21} \text{ cm}^3$$

Step 4: Number of lead shots

The volume of lead shots dropped = volume of water displaced = $\dfrac{550}{21}$ cm³

$$n = \frac{550/21}{11/21} = \frac{550}{11} = \boxed{100}$$

The number of lead shots dropped into the vessel is 100.

Source: Chapter 12, Section 12.3 (Volume of a Combination of Solids), Exercise 12.2, Q.5

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• The key concept: the volume of lead shots that sink into the vessel equals the volume of water displaced (which flows out). So: n × V(one shot) = ¼ × V(cone).
• Examiners award marks at each step — writing the formula, substituting correctly, computing V(cone), V(water out), V(shot), and the final division. Don't skip steps.
• Use $\pi = \frac{22}{7}$ throughout (as instructed in the exercise) so fractions cancel neatly to give exactly 100.
• A common error is forgetting the $\frac{1}{4}$ — the shots displace only the water that flows out, not the entire cone's volume.