Question

The height of a cone is 30 cm .A small cone is cut off at the top by a plane parallel to the base . If its volume be 1/27 of the volume of the given cone , then the height above the base at which the section has been made is

10 cm b) 15 cm c) 20 cm d) 25 cm

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Answer 1

$\sf \large \red{\underline{ Question:-}}$

• The height of a cone is 30 cm .A small cone is cut off at the top by a plane parallel to the base . If its volume be 1/27 of the volume of the given cone , then the height above the base at which the section has been made is 10 cm b) 15 cm c) 20 cm d) 25 cm

$\\\\\sf \large \red{\underline{Given:-}}$

• The height of a cone is 30 cm .

• If its volume be

$\\\\\sf \large \red{\underline{To \: Find:-}}$

• find the hight of Base $\frac{1}{27}$

$\\\\\sf \large \red{\underline{Solution :- }}$

$\sf \to\frac{h}{r} = \frac{30}{3} \\ \\ \sf \to \: h= 10r$

$\sf \boxed{ \sf \red{Volume \: of \: small \: Cone \: = \frac{1}{3} πr^2h }}$

$\bf \underline{putting \: all \: values : }$

$\sf \to \: \frac{1}{3} π r^2 \times 10r \\ \\ </p><p> \sf \to \: \frac{1}{3} \times 10 \times r^3$

Now it is given that :

$\sf \to \frac{1}{3} π \times 10 \times r^3 = \frac{1}{27} \times \frac{1}{3} π \times 9 \times 30 \\ \\ </p><p> \sf \to \: r^3 = 1 \\ \\ \sf \to \: r = 1 \\ \\ </p><p> \sf \to h = 10r = 10$

So, the cone must be cut 30 - 10 = 20 cms from the base.

Answer 2

Answer:

$$\begin{lgathered}\sf \large \red{\underline{ Question:-}}\\\\\end{lgathered}$$

The height of a cone is 30 cm .A small cone is cut off at the top by a plane parallel to the base . If its volume be 1/27 of the volume of the given cone , then the height above the base at which the section has been made is 10 cm b) 15 cm c) 20 cm d) 25 cm

$$\begin{lgathered}\\\\\sf \large \red{\underline{Given:-}}\\\\\end{lgathered}$$

The height of a cone is 30 cm .

If its volume be

$$\begin{lgathered}\\\\\sf \large \red{\underline{To \: Find:-}}\\\\\end{lgathered}$$

find the hight of Base $$\frac{1}{27}$$

$$\begin{lgathered}\\\\\sf \large \red{\underline{Solution :- }}\\\\\end{lgathered}$$

$$\begin{lgathered}\sf \to\frac{h}{r} = \frac{30}{3} \\ \\ \sf \to \: h= 10r \\ \\ \\\end{lgathered}$$

$$\begin{lgathered}\sf \boxed{ \sf \red{Volume \: of \: small \: Cone \: = \frac{1}{3} πr^2h }} \\ \\\end{lgathered}$$

$$\bf \underline{putting \: all \: values : }$$

$$\begin{lgathered}\sf \to \: \frac{1}{3} π r^2 \times 10r \\ \\ \sf \to \: \frac{1}{3} \times 10 \times r^3\end{lgathered}$$

Now it is given that :

$$\begin{lgathered}\sf \to \frac{1}{3} π \times 10 \times r^3 = \frac{1}{27} \times \frac{1}{3} π \times 9 \times 30 \\ \\ \sf \to \: r^3 = 1 \\ \\ \sf \to \: r = 1 \\ \\ \sf \to h = 10r = 10 \\\end{lgathered}$$

So, the cone must be cut 30 - 10 = 20 cms from the base.