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The number of solid spheres, each of diameter 6cm that could be moulded to form a solid metal cylinder of height 45 cm and diameter 4 cm, is
Volume of one sphere = (4/3)π(3³) = 36π cm³. Volume of cylinder = π(2²)(45) = 180π cm³. Number of spheres = 180π / 36π = 5. 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 vRead more
Volume of one sphere = (4/3)π(3³) = 36π cm³.
Volume of cylinder = π(2²)(45) = 180π cm³.
Number of spheres = 180π / 36π = 5.
This question related to Chapter 12 Mathematics Class 10th NCERT. From the Chapter 12. Surface Areas and Volumes. Give answer according to your understanding.
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See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/
A circus tent is cylindrical to a height of 4m and conical above it. If its diameter is 105 m and its slant height is 40m, the total area of the canvas required in m² is
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:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/
The material of a cone is converted into the shape of a cylinder of equal radius. If height of the cylinder is 5 cm, then height of the cone is
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.
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See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/
If the radius of the base of a right circular cylinder is halved, keeping the height the same, then the ratio of the volume of the cylinder thus obtained to the volume of the original cylinder is
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:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/
A hollow sphere of internal and external diameter 4 cm and 8cm respectively is melted into a cone of base diameter 8 cm. The height of the cone is
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:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/