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A school provides milk to its students in cylindrical glasses, each with a diameter of 7 cm. If each glass is filled with milk to a height of 12 cm, how many litres of milk are needed to serve 1600 students? Use 1000 cm³ = 1 litre.
The diameter is 7 cm, giving radius r = 3.5 cm and height h = 12 cm. Volume per glass is πr²h = (22/7) × (7/2) × (7/2) × 12 = 22 × 7 × 3 = 462 cm³. For 1600 students, the total milk needed is 1600 × 462 = 739,200 cm³. Using the conversion 1000 cm³ = 1 litre, this equals 739,200 / 1000 = 739.2 litresRead more
The diameter is 7 cm, giving radius r = 3.5 cm and height h = 12 cm. Volume per glass is πr²h = (22/7) × (7/2) × (7/2) × 12 = 22 × 7 × 3 = 462 cm³. For 1600 students, the total milk needed is 1600 × 462 = 739,200 cm³. Using the conversion 1000 cm³ = 1 litre, this equals 739,200 / 1000 = 739.2 litres of milk.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 14 Math of Space: Surface Area and Volume Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-14/
See lessThe surface area of a sphere of radius 5 cm is five times the area of the curved surface of a cone of radius 4 cm. Find the height and the volume of the cone.
Surface area of the sphere is 4π(5)² = 100π cm². Curved surface area of the cone is πrl = 4πl. Given 100π = 5(4πl), we get l = 5 cm. Using l² = r² + h², h² = 25 − 16 = 9, so h = 3 cm. Hence, cone volume is (1/3)π(4)²(3) = 16π cm³. For more NCERT Solutions of Class 9 Maths Ganita Manjari PartRead more
Surface area of the sphere is 4π(5)² = 100π cm². Curved surface area of the cone is πrl = 4πl. Given 100π = 5(4πl), we get l = 5 cm. Using l² = r² + h², h² = 25 − 16 = 9, so h = 3 cm. Hence, cone volume is (1/3)π(4)²(3) = 16π cm³.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 14 Math of Space: Surface Area and Volume Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-14/
See lessTake Earth to be a perfect sphere with radius 6370 km. Take Jupiter to be a perfect sphere with radius 69,900 km. Take the Sun to be a perfect sphere with radius 6,95,700 km. Compute approximately: (i) the ratio of the volume of Jupiter to the volume of the Earth; (ii) the ratio of the volume of the Sun to the volume of the Earth. (Calculator can be used.)
Since sphere volume is proportional to r³, the volume ratio between two spheres equals the cube of the ratio of their radii. (i) Ratio of Jupiter to Earth is (69,900/6,370)³ ≈ 1321 : 1. (ii) Ratio of Sun to Earth is (6,95,700/6,370)³ ≈ 1,303,000 : 1. For more NCERT Solutions of Class 9 MathsRead more
Since sphere volume is proportional to r³, the volume ratio between two spheres equals the cube of the ratio of their radii. (i) Ratio of Jupiter to Earth is (69,900/6,370)³ ≈ 1321 : 1. (ii) Ratio of Sun to Earth is (6,95,700/6,370)³ ≈ 1,303,000 : 1.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 14 Math of Space: Surface Area and Volume Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-14/
See lessShow that the volume of a sphere is equal to 2/3 of the volume of the smallest cylinder which encloses it.
Consider a sphere of radius r, which has volume (4/3)πr³. The smallest cylinder enclosing this sphere must have a base radius equal to r and a height equal to the sphere's diameter, 2r. The cylinder's volume is πr²(2r) = 2πr³. Comparing volumes gives [(4/3)πr³] / [2πr³] = 2/3. For more NCERTRead more
Consider a sphere of radius r, which has volume (4/3)πr³. The smallest cylinder enclosing this sphere must have a base radius equal to r and a height equal to the sphere’s diameter, 2r. The cylinder’s volume is πr²(2r) = 2πr³. Comparing volumes gives [(4/3)πr³] / [2πr³] = 2/3.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 14 Math of Space: Surface Area and Volume Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-14/
See lessSuppose the Earth is perfectly spherical. A string is wrapped tightly around the equator of the Earth. Another string, 1 metre longer, is placed around the Earth so that it forms a larger circle, staying the same distance above the ground everywhere. How high above the ground is the second string? Repeat this for: (i) the Moon (ii) Jupiter (iii) a volleyball. Are you surprised by the three answers? Why or why not?
Let original circumference be C = 2πR and new circumference be C + 1 = 2π(R + h). Subtracting gives 2πh = 1, so the height above ground is h = 1 / (2π) ≈ 0.159 m ≈ 16 cm. This value remains identical for the Moon, Jupiter and a volleyball because radius cancels out completely. The result is counter-Read more
Let original circumference be C = 2πR and new circumference be C + 1 = 2π(R + h). Subtracting gives 2πh = 1, so the height above ground is h = 1 / (2π) ≈ 0.159 m ≈ 16 cm. This value remains identical for the Moon, Jupiter and a volleyball because radius cancels out completely. The result is counter-intuitive but mathematically expected since height depends solely on the additional string length.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 14 Math of Space: Surface Area and Volume Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-14/
See less