What's your question?
  1. CSA_cylinder = 2πrh = 2π(52.5)(4) = 420π m². CSA_cone = πrl = π(52.5)(40) = 2100π m². Total area = 420π + 2100π = 2520π m². Using π ≈ 22/7: Total area = 2520 × (22/7) = 7920 m². This question related to Chapter 12 Mathematics Class 10th NCERT. From the Chapter 12. Surface Areas and Volumes. Give ansRead more

    CSA_cylinder = 2πrh = 2π(52.5)(4) = 420π m².
    CSA_cone = πrl = π(52.5)(40) = 2100π m².
    Total area = 420π + 2100π = 2520π m².
    Using π ≈ 22/7: Total area = 2520 × (22/7) = 7920 m².
    This question related to Chapter 12 Mathematics Class 10th NCERT. From the Chapter 12. Surface Areas and Volumes. Give answer according to your understanding.

    For more please visit here:
    https://www.tiwariacademy.in/ncert-solutions/class-10/maths/

    See less
    • 20
  2. Equating volumes: (1/3)h_cone = h_cylinder. Substitute h_cylinder = 5: (1/3)h_cone = 5 → h_cone = 15 cm. This question related to Chapter 12 Mathematics Class 10th NCERT. From the Chapter 12. Surface Areas and Volumes. Give answer according to your understanding. For more please visit here: https://Read more

    Equating volumes: (1/3)h_cone = h_cylinder.
    Substitute h_cylinder = 5: (1/3)h_cone = 5 → h_cone = 15 cm.
    This question related to Chapter 12 Mathematics Class 10th NCERT. From the Chapter 12. Surface Areas and Volumes. Give answer according to your understanding.

    For more please visit here:
    https://www.tiwariacademy.in/ncert-solutions/class-10/maths/

    See less
    • 30
  3. New radius = r/2. V_new = π(r/2)²h = (1/4)πr²h. Ratio = V_new : V_original = (1/4) : 1 = 1:4. This question related to Chapter 12 Mathematics Class 10th NCERT. From the Chapter 12. Surface Areas and Volumes. Give answer according to your understanding. For more please visit here: https://www.tiwariaRead more

    New radius = r/2.
    V_new = π(r/2)²h = (1/4)πr²h.
    Ratio = V_new : V_original = (1/4) : 1 = 1:4.
    This question related to Chapter 12 Mathematics Class 10th NCERT. From the Chapter 12. Surface Areas and Volumes. Give answer according to your understanding.

    For more please visit here:
    https://www.tiwariacademy.in/ncert-solutions/class-10/maths/

    See less
    • 27
  4. Volume of hollow sphere: V_sphere = (4/3)π(R³ - r³) = (4/3)π(4³ - 2³) = (224/3)π. Volume of cone: V_cone = (1/3)πr²h = (1/3)π(4²)h. Equating volumes: (224/3)π = (1/3)π(16)h → 224 = 16h → h = 14 cm. This question related to Chapter 12 Mathematics Class 10th NCERT. From the Chapter 12. Surface Areas aRead more

    Volume of hollow sphere:
    V_sphere = (4/3)π(R³ – r³) = (4/3)π(4³ – 2³) = (224/3)π.

    Volume of cone:
    V_cone = (1/3)πr²h = (1/3)π(4²)h.

    Equating volumes:
    (224/3)π = (1/3)π(16)h → 224 = 16h → h = 14 cm.
    This question related to Chapter 12 Mathematics Class 10th NCERT. From the Chapter 12. Surface Areas and Volumes. Give answer according to your understanding.

    For more please visit here:
    https://www.tiwariacademy.in/ncert-solutions/class-10/maths/

    See less
    • 21
  5. The correct answer is (B) Mango. Mango is a large, perennial tree that belongs to the genus Mangifera. It produces sweet, edible fruits and has a long lifespan. In contrast, rose is a shrub, tulsi is a herb and tomato is a small plant. For more visit here: https://www.tiwariacademy.in/ncert-solutionRead more

    The correct answer is (B) Mango. Mango is a large, perennial tree that belongs to the genus Mangifera. It produces sweet, edible fruits and has a long lifespan. In contrast, rose is a shrub, tulsi is a herb and tomato is a small plant.

    For more visit here:
    https://www.tiwariacademy.in/ncert-solutions/class-6/science/

    See less
    • 29