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We have a cylinder with a base radius of r cm and height h cm. A square pyramid is fitted inside it. The square base of the pyramid lies on the base of the cylinder, its corners on the boundary of the cylinder. The apex of the pyramid lies on the top of the cylinder. Find the ratio of the volume of this pyramid to the volume of the cylinder.
The square base of the pyramid is inscribed inside the circular base of radius r, making its diagonal equal to the diameter 2r. The area of this square base is (1/2) × d² = (1/2)(2r)² = 2r². The pyramid has height h, giving volume V_pyramid = (1/3) × base area × h = (2/3)r²h. The cylinder volume isRead more
The square base of the pyramid is inscribed inside the circular base of radius r, making its diagonal equal to the diameter 2r. The area of this square base is (1/2) × d² = (1/2)(2r)² = 2r². The pyramid has height h, giving volume V_pyramid = (1/3) × base area × h = (2/3)r²h. The cylinder volume is V_cylinder = πr²h. Thus, the ratio of their volumes is [(2/3)r²h] / [πr²h] = 2 / (3π) or 2 : 3π.
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 lessFind (i) the curved surface area and (ii) the total surface area of a hemisphere of radius 21 cm.
Given the radius of the hemisphere r = 21 cm. (i) The curved surface area of a hemisphere is given by CSA = 2πr². Using π = 22/7, CSA = 2 × (22/7) × 21 × 21 = 2 × 22 × 3 × 21 = 2772 cm². (ii) The total surface area includes the curved surface and circular base: TSA = 3πr² = 3 × (22/7) × 21 × 21 = 41Read more
Given the radius of the hemisphere r = 21 cm. (i) The curved surface area of a hemisphere is given by CSA = 2πr². Using π = 22/7, CSA = 2 × (22/7) × 21 × 21 = 2 × 22 × 3 × 21 = 2772 cm². (ii) The total surface area includes the curved surface and circular base: TSA = 3πr² = 3 × (22/7) × 21 × 21 = 4158 cm².
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 lessMetal spheres, each of radius 2 cm, are packed into a rectangular box of internal dimensions 16 cm × 8 cm × 8 cm. When 16 spheres are packed, the box is filled with preservative liquid. Find the volume of this liquid. Round your answer to the nearest integer.
The internal volume of the rectangular box is 16 × 8 × 8 = 1024 cm³. Each metal sphere has a radius of 2 cm, so the volume of 16 spheres is 16 × (4/3)πr³ = 16 × (4/3) × (22/7) × 8 = 11264 / 21 ≈ 536.38 cm³. The volume of preservative liquid needed is 1024 − 536.38 = 487.62 cm³, which rounds to the nRead more
The internal volume of the rectangular box is 16 × 8 × 8 = 1024 cm³. Each metal sphere has a radius of 2 cm, so the volume of 16 spheres is 16 × (4/3)πr³ = 16 × (4/3) × (22/7) × 8 = 11264 / 21 ≈ 536.38 cm³. The volume of preservative liquid needed is 1024 − 536.38 = 487.62 cm³, which rounds to the nearest integer as 488 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 lessThe radius of a sphere is increased by 10%. Show that the volume increases by approximately 33.1%.
Let the original radius of the sphere be r, so its original volume is V = (4/3)πr³. When the radius increases by 10%, the new radius becomes r' = 1.1r. The new volume is V' = (4/3)π(1.1r)³ = 1.331 × (4/3)πr³ = 1.331V. The fractional increase is (V' − V) / V = 0.331, which corresponds to a 33.1% incrRead more
Let the original radius of the sphere be r, so its original volume is V = (4/3)πr³. When the radius increases by 10%, the new radius becomes r’ = 1.1r. The new volume is V’ = (4/3)π(1.1r)³ = 1.331 × (4/3)πr³ = 1.331V. The fractional increase is (V’ − V) / V = 0.331, which corresponds to a 33.1% increase in volume.
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 radius of a sphere is increased by x%. The volume of the sphere increases by 72.8%. Find the value of x.
Let the initial radius be r and initial volume be V = (4/3)πr³. An increase in volume of 72.8% gives new volume V' = 1.728V. Because volume is proportional to the cube of the radius, (r')³ = 1.728r³. Taking the cube root gives r' = 1.2r = r + 0.2r, which corresponds to a 20% increase. Thus, x = 20.Read more
Let the initial radius be r and initial volume be V = (4/3)πr³. An increase in volume of 72.8% gives new volume V’ = 1.728V. Because volume is proportional to the cube of the radius, (r’)³ = 1.728r³. Taking the cube root gives r’ = 1.2r = r + 0.2r, which corresponds to a 20% increase. Thus, x = 20.
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/
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