source: ncert_exercises
topic: geometry
class_level: class_9
chapter: ch11
exercise: ex11.2
difficulty: intermediate
Class 9 | Ch11: Surface Areas and Volumes | Exercise 11.2 (Sphere and Hemisphere)
Key Formulas — Sphere (radius=r): Surface Area = 4πr² Volume = (4/3)πr³ Hemisphere: CSA=2πr², TSA=3πr², Volume=(2/3)πr³
Q1. Find surface area of spheres with radius: (i) 10.5 cm (ii) 5.6 cm (iii) 14 cm. Answer: SA=4πr². (i) 4×22/7×10.5²=4×22/7×110.25=1386 cm². ✅ (ii) 4×22/7×5.6²=4×22/7×31.36=394.24 cm². ✅ (iii) 4×22/7×196=2464 cm². ✅
Q2. Find surface area of spheres with diameter: (i) 14 cm (ii) 21 cm (iii) 3.5 m. Answer: r=d/2. (i) r=7: SA=4×22/7×49=616 cm². ✅ (ii) r=10.5: SA=4×22/7×110.25=1386 cm². ✅ (iii) r=1.75: SA=4×22/7×3.0625=38.5 m². ✅
Q3. Find total surface area of hemisphere of radius 10 cm. Answer: TSA=3πr²=3×22/7×100=942.86 cm². ✅
Q4. The radius of spherical balloon increases from 7 cm to 14 cm. Find ratio of surface areas. Answer: SA∝r². Ratio=(7/14)²=1:4. ✅
Q5. Brass sphere of radius 4.2 cm is melted and recast as cylinder of radius 6 cm. Find height. Answer: Volume of sphere=4/3×π×4.2³=4/3×π×74.088. Volume of cylinder=π×36×h. 4/3×74.088=36×h → h=74.088×4/(3×36)=2.744 cm≈2.74 cm. ✅
Q6. Diameter of moon=1/4 of earth's diameter. Find ratio of volumes. Answer: V∝r³∝d³. Ratio=(1/4)³=1:64. ✅
Q7. Hollow sphere of inner radius 3 cm and outer radius 5 cm. Find volume of metal. Answer: V=4/3×π×(R³-r³)=4/3×π×(125-27)=4/3×π×98=410.67 cm³. ✅
Q8. Tank (hemisphere + cylinder) of base radius 4.5 m, height of cylindrical part 2.1 m. Find water it can hold. Answer: V=2/3×π×r³+π×r²×h=π×r²×(2r/3+h)=π×20.25×(3+2.1)=π×20.25×5.1≈323.6 m³. In litres=323600 L. ✅
Q9. Right circular cylinder just encloses a sphere of radius r. Find: (i) SA of sphere (ii) CSA of cylinder (iii) ratio. Answer: Cylinder has radius r and height 2r (to enclose sphere). (i) SA of sphere=4πr². (ii) CSA of cylinder=2πr×2r=4πr². (iii) Ratio=4πr²:4πr²=1:1. ✅