Question

If the height and the radius of a cone are doubled, the volume of the cone becomes
(1) 2 times
(2)4 times
(3) 6 times
(4)8 times​

Answer 1

❍ Let the Height and Radius of the cone be h and r respectively.

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$\underline{\bf{\dag} \:\mathfrak{As\;we\;know\: that\: :}}$⠀

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$\star\;\boxed{\sf{\pink{Volume_{\:(cone)} = \dfrac{1}{3}\pi r^2h}}}$

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$\bigstar\underline{\textsf{\;According to given Question :}}$

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• The Height and Radius of the cone are doubled. Therefore, h becomes 2h and Radius becomes 2r.

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Now,

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$:\implies\sf Volume_{\:(cone)} = \dfrac{1}{3} \pi r^2 h \\\\\\:\implies\sf Volume_{\:(cone)} = \dfrac{1}{3} \pi \times (2r)^2 \times 2h\\\\\\:\implies\sf Volume_{\:(cone)} = \dfrac{1}{3} \pi \times 4r \times 2h\\\\\\:\implies{\underline{\boxed{\frak{\pink{ \: \mathcal{V}olume_{\:(cone)} = 8 \times \dfrac{1}{3} \pi r^2 h \: \: }}}}}\;\bigstar$

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$\therefore{\underline{\sf{Hence, \; the \: Volume\; of \; cone\; becomes\; 8 \; times\; \bf{ Option\; d)}.}}}$

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$\qquad\qquad\boxed{\bf{\mid{\overline{\underline{\purple{\bigstar\: Formulae\:related\:to\:cone :}}}}}\mid}$

• $\sf Area\:of\:base = \bf{\pi r^2}$

• $\sf Curved\:surface\:area\:of\:cone = \bf{\pi rl}$

• $\sf Total\:surface\:area\:of\:cone = Area\:of\:base + CSA = \pi r^2 + \pi rl = \bf{\pi r(r + l)}$

• $\sf Volume\:of\:cone = \bf{\dfrac{1}{3} \pi r^2 h}$

Answer 2

Given :-

• Height and radius of cone is double

To Find :-

Volume

Solution :-

As we know that

$\sf \: Volume = \dfrac{1}{3}\pi {r}^{2} h$

Let the Radius be R and Height be H

Volume = ⅓ × π × R² × H

Now,

When radius and Height Doubled

Volume = ⅓ × π × 2R² × 2H

Volume = ⅓ × π × 4R × 2H

Volume = 8 times