A right circular cone is cut by a plane parallel to its base in such a way that the slant heights of the original and the smaller cone thus obtained are in the ratio $2 : 1$ . If $V_{1}$ and $V_{2}$ are respectively the volumes of the original cone and of the new cone, then what is $V_{1} : V_{2}$ ?

2 : 1

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3 : 1

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4 : 1

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8 : 1

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Solution

The correct option is B $8 : 1$
R.E.F image
$s i n θ = \frac{r_{1}}{l} = \frac{r_{2}}{2 l_{1}} \Rightarrow r_{1} = \frac{r_{2}}{2}$
$∴ V_{2} = \frac{π}{3} r_{2}^{2} l_{2} = \frac{π}{3} (4 r_{1}^{2}) (2 l_{1})$
$\Rightarrow V_{2} = 8 V_{1}$

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A right circular cone is cut by a plane parallel to its base in such a way that the slant heights of the original and the smaller cone thus obtained are in the ratio 2:1. If V1 and V2 are respectively the volumes of the original cone and of the new cone, then what is V1:V2?

The correct option is B 8:1R.E.F imagesinθ=r1l=r22l1⇒r1=r22∴V2=π3r22l2=π3(4r21)(2l1)⇒V2=8V1

The correct option is B $8 : 1$
R.E.F image
$s i n θ = \frac{r_{1}}{l} = \frac{r_{2}}{2 l_{1}} \Rightarrow r_{1} = \frac{r_{2}}{2}$
$∴ V_{2} = \frac{π}{3} r_{2}^{2} l_{2} = \frac{π}{3} (4 r_{1}^{2}) (2 l_{1})$
$\Rightarrow V_{2} = 8 V_{1}$