(i) The large dough ball has volume (4/3)π(6)³ cm³. An individual chapati dough ball typically has a radius of about 1.5 cm, giving volume (4/3)π(1.5)³ cm³. Dividing volumes gives (6 / 1.5)³ = 4³ = 64 chapatis. (ii) Treating the halved fruit as a concentric hemisphere, let outer radius be R and inneRead more
(i) The large dough ball has volume (4/3)π(6)³ cm³. An individual chapati dough ball typically has a radius of about 1.5 cm, giving volume (4/3)π(1.5)³ cm³. Dividing volumes gives (6 / 1.5)³ = 4³ = 64 chapatis. (ii) Treating the halved fruit as a concentric hemisphere, let outer radius be R and inner cavity radius be r. The edible flesh volume equals (2/3)π(R³ − r³).
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)
The ink reservoir inside a ballpoint pen refill forms a right circular cylinder. Measuring a typical standard refill, the internal radius is approximately r = 0.1 cm (diameter 2 mm) and the column of ink has a height of about h = 10 cm. The volume of ink is given by V = πr²h ≈ 3.14 × (0.1)² × 10 = 0Read more
The ink reservoir inside a ballpoint pen refill forms a right circular cylinder. Measuring a typical standard refill, the internal radius is approximately r = 0.1 cm (diameter 2 mm) and the column of ink has a height of about h = 10 cm. The volume of ink is given by V = πr²h ≈ 3.14 × (0.1)² × 10 = 0.314 cm³, equivalent to roughly 0.31 millilitres.
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)
Let the side length of the cube be a. The volume of the cube is given by V = a³, which means that a = V¹ᐟ³. The total surface area of the cube is given by S = 6a². Substituting the expression for a into the surface area formula gives S = 6(V¹ᐟ³)² = 6V²ᐟ³. Thus, S = 6∛(V²). For more NCERT SoluRead more
Let the side length of the cube be a. The volume of the cube is given by V = a³, which means that a = V¹ᐟ³. The total surface area of the cube is given by S = 6a². Substituting the expression for a into the surface area formula gives S = 6(V¹ᐟ³)² = 6V²ᐟ³. Thus, S = 6∛(V²).
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)
Let the side length of the cube be a. The total surface area is given by the formula S = 6a². Solving for the side length a gives a² = S/6, so a = √(S/6) = (S/6)¹ᐟ². The volume of the cube is V = a³. Substituting a gives V = ((S/6)¹ᐟ²)³ = (S/6)³ᐟ² = S√S/(6√6). For more NCERT Solutions of ClasRead more
Let the side length of the cube be a. The total surface area is given by the formula S = 6a². Solving for the side length a gives a² = S/6, so a = √(S/6) = (S/6)¹ᐟ². The volume of the cube is V = a³. Substituting a gives V = ((S/6)¹ᐟ²)³ = (S/6)³ᐟ² = S√S/(6√6).
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)
Solve: (i) A ball of chapati dough of radius 6 cm is prepared. Estimate how many chapatis can be made from it? (ii) Cut a coconut/muskmelon in half and find out the approximate volume of edible coconut flesh/fruit by taking the necessary measurements.
(i) The large dough ball has volume (4/3)π(6)³ cm³. An individual chapati dough ball typically has a radius of about 1.5 cm, giving volume (4/3)π(1.5)³ cm³. Dividing volumes gives (6 / 1.5)³ = 4³ = 64 chapatis. (ii) Treating the halved fruit as a concentric hemisphere, let outer radius be R and inneRead more
(i) The large dough ball has volume (4/3)π(6)³ cm³. An individual chapati dough ball typically has a radius of about 1.5 cm, giving volume (4/3)π(1.5)³ cm³. Dividing volumes gives (6 / 1.5)³ = 4³ = 64 chapatis. (ii) Treating the halved fruit as a concentric hemisphere, let outer radius be R and inner cavity radius be r. The edible flesh volume equals (2/3)π(R³ − r³).
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 the volume of ink in a new ball point pen. Take the necessary measurements and make approximations, as needed.
The ink reservoir inside a ballpoint pen refill forms a right circular cylinder. Measuring a typical standard refill, the internal radius is approximately r = 0.1 cm (diameter 2 mm) and the column of ink has a height of about h = 10 cm. The volume of ink is given by V = πr²h ≈ 3.14 × (0.1)² × 10 = 0Read more
The ink reservoir inside a ballpoint pen refill forms a right circular cylinder. Measuring a typical standard refill, the internal radius is approximately r = 0.1 cm (diameter 2 mm) and the column of ink has a height of about h = 10 cm. The volume of ink is given by V = πr²h ≈ 3.14 × (0.1)² × 10 = 0.314 cm³, equivalent to roughly 0.31 millilitres.
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 lessWhat is the change in volume when: (i) The length of a cuboid with dimensions l, w, h is increased by 1 unit? (a) 1 cubic unit (b) (lwh + 1) cubic unit (c) wh cubic units (d) lw cubic units (e) lh cubic units (ii) The radius of a cylinder with dimensions r, h is increased by 1 unit? (a) 1 cubic unit (b) πr²h cubic units (c) πr² cubic units (d) 2πrh + 2πh cubic units (e) 2πrh + πh cubic units (iii) The radius of a sphere is decreased by 1 unit?
(i) New volume is (l + 1)wh = lwh + wh. Change in volume is (lwh + wh) − lwh = wh cubic units, which is option (c). (ii) New volume is π(r + 1)²h = π(r² + 2r + 1)h = πr²h + 2πrh + πh. Subtracting original volume πr²h yields 2πrh + πh cubic units, corresponding to option (e). (iii) Change in volume iRead more
(i) New volume is (l + 1)wh = lwh + wh. Change in volume is (lwh + wh) − lwh = wh cubic units, which is option (c).
(ii) New volume is π(r + 1)²h = π(r² + 2r + 1)h = πr²h + 2πrh + πh. Subtracting original volume πr²h yields 2πrh + πh cubic units, corresponding to option (e).
(iii) Change in volume is (4/3)πr³ − (4/3)π(r − 1)³ = (4/3)π(3r² − 3r + 1) cubic units.
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 lessGiven a cube with volume V, express its total surface area S in terms of V.
Let the side length of the cube be a. The volume of the cube is given by V = a³, which means that a = V¹ᐟ³. The total surface area of the cube is given by S = 6a². Substituting the expression for a into the surface area formula gives S = 6(V¹ᐟ³)² = 6V²ᐟ³. Thus, S = 6∛(V²). For more NCERT SoluRead more
Let the side length of the cube be a. The volume of the cube is given by V = a³, which means that a = V¹ᐟ³. The total surface area of the cube is given by S = 6a². Substituting the expression for a into the surface area formula gives S = 6(V¹ᐟ³)² = 6V²ᐟ³. Thus, S = 6∛(V²).
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 lessGiven a cube with total surface area S, express its volume V in terms of S.
Let the side length of the cube be a. The total surface area is given by the formula S = 6a². Solving for the side length a gives a² = S/6, so a = √(S/6) = (S/6)¹ᐟ². The volume of the cube is V = a³. Substituting a gives V = ((S/6)¹ᐟ²)³ = (S/6)³ᐟ² = S√S/(6√6). For more NCERT Solutions of ClasRead more
Let the side length of the cube be a. The total surface area is given by the formula S = 6a². Solving for the side length a gives a² = S/6, so a = √(S/6) = (S/6)¹ᐟ². The volume of the cube is V = a³. Substituting a gives V = ((S/6)¹ᐟ²)³ = (S/6)³ᐟ² = S√S/(6√6).
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