1. 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.

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  2. Given: - Height of original cone = H = 30 cm, - Volume of smaller cone = (1/27) × Volume of original cone. For similar cones: (h/H)³ = 1/27 ⇒ h/H = 1/3. Height of smaller cone: h = (1/3) × 30 = 10 cm. Height above the base: H - h = 30 - 10 = 20 cm. Final Answer: c) 20 cm. For more please visit here:Read more

    Given:
    – Height of original cone = H = 30 cm,
    – Volume of smaller cone = (1/27) × Volume of original cone.

    For similar cones:
    (h/H)³ = 1/27 ⇒ h/H = 1/3.

    Height of smaller cone:
    h = (1/3) × 30 = 10 cm.

    Height above the base:
    H – h = 30 – 10 = 20 cm.

    Final Answer: c) 20 cm.

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  3. Given: - Height of original cone = H = 30 cm, - Volume of smaller cone = (1/27) × Volume of original cone. For similar cones: (h/H)³ = 1/27 ⇒ h/H = 1/3. Height of smaller cone: h = (1/3) × 30 = 10 cm. Height above the base: H - h = 30 - 10 = 20 cm. Final Answer: c) 20 cm. This question related to ChRead more

    Given:
    – Height of original cone = H = 30 cm,
    – Volume of smaller cone = (1/27) × Volume of original cone.

    For similar cones:
    (h/H)³ = 1/27 ⇒ h/H = 1/3.

    Height of smaller cone:
    h = (1/3) × 30 = 10 cm.

    Height above the base:
    H – h = 30 – 10 = 20 cm.

    Final Answer: c) 20 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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  4. Class 6 Mathematics Chapter 5 MCQ assesses understanding of basic geometric concepts like angles, shapes, and measurements, enhancing spatial reasoning, foundational math skills, and preparation for advanced geometry in higher classes. For Practice MCQ visit here: https://www.tiwariacademy.in/ncert-Read more

    Class 6 Mathematics Chapter 5 MCQ assesses understanding of basic geometric concepts like angles, shapes, and measurements, enhancing spatial reasoning, foundational math skills, and preparation for advanced geometry in higher classes.

    For Practice MCQ visit here:
    https://www.tiwariacademy.in/ncert-solutions-class-6-maths-chapter-1/

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  5. Class 6 Mathematics Chapter 2 MCQ evaluates knowledge of whole numbers, their properties, and operations, strengthening numerical skills, logical thinking, and foundational math concepts essential for advanced arithmetic and problem-solving in higher classes. For Practice MCQ visit here: https://wwwRead more

    Class 6 Mathematics Chapter 2 MCQ evaluates knowledge of whole numbers, their properties, and operations, strengthening numerical skills, logical thinking, and foundational math concepts essential for advanced arithmetic and problem-solving in higher classes.

    For Practice MCQ visit here:
    https://www.tiwariacademy.in/ncert-solutions-class-6-maths-chapter-2/

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