From a right circular cylinder of radius $10$ cm and height $21$ cm a right circular cone of same base radius is removed. If the volume of the remaining portion is $4400 {cm}^{3}$ , then the height of the cone removed is

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Solution

The correct option is C $21$ cm
Let the height of cone be 'h', then
$π r^{2} H = \frac{1}{3} π r^{2} h = 4400 ㎤$
$\Rightarrow π r^{2} [21 - \frac{h}{3}] = 4400 \Rightarrow 63 - h = \frac{3 \times 4400 \times 7}{22 \times 10 \times 10} = 42$
$\Rightarrow h = 63 - 42 = 21 ㎝$

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From a right circular cylinder of radius 10 cm and height 21 cm a right circular cone of same base radius is removed. If the volume of the remaining portion is 4400 cm3, then the height of the cone removed is

The correct option is C 21 cmLet the height of cone be 'h', then πr2H=13πr2h=4400㎤⇒πr2[21−h3]=4400⇒63−h=3×4400×722×10×10=42⇒h=63−42=21㎝

From a right circular cylinder of radius $10$ cm and height $21$ cm a right circular cone of same base radius is removed. If the volume of the remaining portion is $4400 {cm}^{3}$ , then the height of the cone removed is

The correct option is C $21$ cm
Let the height of cone be 'h', then
$π r^{2} H = \frac{1}{3} π r^{2} h = 4400 ㎤$
$\Rightarrow π r^{2} [21 - \frac{h}{3}] = 4400 \Rightarrow 63 - h = \frac{3 \times 4400 \times 7}{22 \times 10 \times 10} = 42$
$\Rightarrow h = 63 - 42 = 21 ㎝$