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5. A solid metallic hemisphere with radius r is melted and cast into a solid right circular cone with the radius of the base = r. What is the ratio of their curved surface areas ?5/?2 2/?5 3/2 5/2

Question

  1. A solid metallic hemisphere with radius r is melted and cast into a solid right circular cone with the radius of the base = r. What is the ratio of their curved surface areas ?5/?2 2/?5 3/2 5/2
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Solution

Step 1: The volume of a hemisphere is given by the formula (2/3)πr³.

Step 2: The volume of a cone is given by the formula (1/3)πr²h, where h is the height of the cone.

Step 3: Since the hemisphere is melted and cast into a cone of the same radius, their volumes are equal. Therefore, we can set the two volume formulas equal to each other and solve for h:

(2/3)πr³ = (1/3)πr²h

Solving for h gives h = 2r.

Step 4: The curved surface area of a hemisphere is given by the formula 2πr².

Step 5: The curved surface area of a cone is given by the formula πrl, where l is the slant height of the cone.

Step 6: In a right circular cone, the slant height l can be found using the Pythagorean theorem: l = √(r² + h²) = √(r² + (2r)²) = √5r.

Step 7: Substituting l = √5r into the formula for the curved surface area of a cone gives πr√5r = √5πr².

Step 8: The ratio of the curved surface area of the hemisphere to the cone is therefore (2πr²) / (√5πr²) = 2/√5 = 2√5/5.

So, the ratio of their curved surface areas is 2√5/5.

This problem has been solved

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