Given: - Total volume of solid = 3 × Volume of cone, - Height of cone = h. Volumes - Volume of cone = (1/3)πr²h, - Volume of cylinder = πr²H (H = height of cylinder), - Total volume = Volume of cone + Volume of cylinder. Equation for total volume Total volume = 3 × Volume of cone: (1/3)πr²h + πr²H =Read more
Given:
– Total volume of solid = 3 × Volume of cone,
– Height of cone = h.
Volumes
– Volume of cone = (1/3)πr²h,
– Volume of cylinder = πr²H (H = height of cylinder),
– Total volume = Volume of cone + Volume of cylinder.
Equation for total volume
Total volume = 3 × Volume of cone:
(1/3)πr²h + πr²H = 3 × (1/3)πr²h.
Cancel πr²:
H = (2/3)h.
This question related to Chapter 12 Mathematics Class 10th NCERT. From the Chapter 12. Surface Areas and Volumes. Give answer according to your understanding.
Given: - Cone is cut at the midpoint of its height. Smaller cone dimensions - Height and radius of smaller cone are half of the original cone. Volume ratio - Volume of smaller cone = (1/8) × Volume of original cone. Final Answer: d) 1:8. For more please visit here: https://www.tiwariacademy.in/ncertRead more
Given:
– Cone is cut at the midpoint of its height.
Smaller cone dimensions
– Height and radius of smaller cone are half of the original cone.
Volume ratio
– Volume of smaller cone = (1/8) × Volume of original cone.
The sphere's volume is 288π cm³. When submerged, it displaces an equal volume of water in the cylinder. Using the formula for the cylinder's volume, πr²h = 288π, and substituting r = 8 cm, we get: 64πh = 288π → h = 288 / 64 = 4.5 cm. This question related to Chapter 12 Mathematics Class 10th NCERT.Read more
The sphere’s volume is 288π cm³. When submerged, it displaces an equal volume of water in the cylinder. Using the formula for the cylinder’s volume, πr²h = 288π, and substituting r = 8 cm, we get:
64πh = 288π → h = 288 / 64 = 4.5 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.
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.
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.
A solid consists of a circular cylinder with an exact fitting right circular cone placed at the top. The height of the cone is h. If the total volume of the solid is 3 times the volume of the cone, then the height of the circular cylinder is
Given: - Total volume of solid = 3 × Volume of cone, - Height of cone = h. Volumes - Volume of cone = (1/3)πr²h, - Volume of cylinder = πr²H (H = height of cylinder), - Total volume = Volume of cone + Volume of cylinder. Equation for total volume Total volume = 3 × Volume of cone: (1/3)πr²h + πr²H =Read more
Given:
– Total volume of solid = 3 × Volume of cone,
– Height of cone = h.
Volumes
– Volume of cone = (1/3)πr²h,
– Volume of cylinder = πr²H (H = height of cylinder),
– Total volume = Volume of cone + Volume of cylinder.
Equation for total volume
Total volume = 3 × Volume of cone:
(1/3)πr²h + πr²H = 3 × (1/3)πr²h.
Simplify:
πr²H = (3/3)πr²h – (1/3)πr²h,
πr²H = (2/3)πr²h.
Cancel πr²:
H = (2/3)h.
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/
If a cone is cut into two parts by a horizontal plane passing through the mid point of its axis, the ratio of the volume of the upper part and the cone is
Given: - Cone is cut at the midpoint of its height. Smaller cone dimensions - Height and radius of smaller cone are half of the original cone. Volume ratio - Volume of smaller cone = (1/8) × Volume of original cone. Final Answer: d) 1:8. For more please visit here: https://www.tiwariacademy.in/ncertRead more
Given:
– Cone is cut at the midpoint of its height.
Smaller cone dimensions
– Height and radius of smaller cone are half of the original cone.
Volume ratio
– Volume of smaller cone = (1/8) × Volume of original cone.
Final Answer: d) 1:8.
For more please visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/
A sphere of radius 6cm is dropped into a cylindrical vessel partly filled with water. The radius of the vessel is 8cm. If the sphere is submerged completely, then the surface of the water rises by
The sphere's volume is 288π cm³. When submerged, it displaces an equal volume of water in the cylinder. Using the formula for the cylinder's volume, πr²h = 288π, and substituting r = 8 cm, we get: 64πh = 288π → h = 288 / 64 = 4.5 cm. This question related to Chapter 12 Mathematics Class 10th NCERT.Read more
The sphere’s volume is 288π cm³. When submerged, it displaces an equal volume of water in the cylinder. Using the formula for the cylinder’s volume, πr²h = 288π, and substituting r = 8 cm, we get:
64πh = 288π → h = 288 / 64 = 4.5 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/
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.
For more please visit here:
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/