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  1. (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/

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  2. 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/

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  3. (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/

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  4. 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/

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  5. 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/

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