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A cylindrical rod of length h is melted and cast into a cone of base radius twice that of the cylinder. What is the height of the cone?

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3 h 4

4 h 3

h 2

answer is A.

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Detailed Solution

Let the base radius of the cylinder be

r 1

, height be h and the base radius of the cone be

r 2

and the height be H.
The base radius of the cone is twice the base radius of the cylinder. This means

r 2 = 2 r 1

Volume of the cylindrical rod is

π r 2 h

⇒ π × r 1 2 × h = π r 1 2 h

Volume of The cone is

13 π r 2 h

⇒ 13 π × r 2 2 × H ​ r 2 = 2 r 1 ​ ⇒ 13 π × 2 r 1 2 × H = 43 π r 1 2 H ​

To find the height of the cone, equate the volume of the cone with the volume of the cylinder.

V o l u m e c o n e = V o l u m e c y l i n d e r ​ ⇒ 43 π r 1 2 H = π r 1 2 h ​ ⇒ H = π r 1 2 h π r 1 2 × 34 ​ ⇒ H = h × 34 ​ ∴ H = 3 h 4 ​

Therefore, the height of the cone is

3 h 4

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