If the radius of the base is doubled, keeping the height constant, what is the ratio of the volume of the large cone to the smaller cone?

4 : 1

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3 : 1

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8 : 1

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None of these

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Solution

The correct option is A $4 : 1$
Here, $r_{2} = 2 r_{1}$ & $h_{1} = h_{2}$ .
We know, volume of cone $V = \frac{1}{3} π r^{2} h$ .
Then, the ratio of volume,
$\frac{V_{2}}{V_{1}} = \frac{\frac{π}{3} r_{2}^{2} h_{2}}{\frac{π}{3} r_{1}^{2} h_{1}}$
$\frac{V_{2}}{V_{1}} = \frac{\frac{π}{3} (2 r_{1})^{2} h_{1}}{\frac{π}{3} r_{1}^{2} h_{1}}$
$\Rightarrow V_{2} : V_{1} = 4 : 1$ .

Hence, option $A$ is correct.

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If the radius of the base is doubled, keeping the height constant, what is the ratio of the volume of the large cone to the smaller cone?

The correct option is A 4:1Here, r2=2r1 & h1=h2.We know, volume of cone V=13πr2h.Then, the ratio of volume,V2V1=π3r22h2π3r21h1V2V1=π3(2r1)2h1π3r21h1⇒V2:V1=4:1 .Hence, option A is correct.