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

Given: Total length of capsule = 14 mm, Diameter = 4 mm → Radius (r) = 2 mm

Length of cylindrical part = 14 − 2r − 2r = 14 − 2(2) − 2(2) = 14 − 4 − 4 = 6 mm

(Each hemisphere contributes a length equal to its radius.)

---

Surface Area:

TSA of capsule = CSA of cylinder + CSA of 2 hemispheres

$$= 2\pi r h + 2 \times 2\pi r^2$$

$$= 2 \times \frac{22}{7} \times 2 \times 6 + 2 \times 2 \times \frac{22}{7} \times 2 \times 2$$

$$= \frac{528}{7} + \frac{352}{7} = \frac{880}{7} \approx 160 \text{ mm}^2$$

---

Volume:

Volume of capsule = Volume of cylinder + Volume of 2 hemispheres

$$= \pi r^2 h + 2 \times \frac{2}{3}\pi r^3 = \pi r^2 h + \frac{4}{3}\pi r^3$$

$$= \frac{22}{7} \times 4 \times 6 + \frac{4}{3} \times \frac{22}{7} \times 8$$

$$= \frac{528}{7} + \frac{704}{21} = \frac{1584}{21} + \frac{704}{21} = \frac{2288}{21} \approx 109.0 \text{ mm}^3$$

Surface Area ≈ 160 mm²; Volume ≈ 109 mm³

Source: Chapter 12, Section 12.2 & 12.3

---

• The capsule consists of one cylinder + two hemispheres. The two hemispheres together form one complete sphere.
• The cylindrical height = total length − diameter of both hemispheres = 14 − 2(2) − 2(2) = 6 mm. Students commonly make errors here.
• Surface area = CSA of cylinder + CSA of 2 hemispheres (= full sphere surface area $4\pi r^2$). The flat circular ends are not visible/counted because hemispheres cover them.
• Volume = Volume of cylinder + Volume of full sphere ($\frac{4}{3}\pi r^3$).
• Note: The textbook Exercise 12.1 Q.6 uses diameter = 5 mm; this question uses diameter = 4 mm — use r = 2 mm throughout.