Question

If the radius of the base of a cone is doubled keeping the height same. What is the ratio of the volume of the larger cone to the smaller cone?

Answer 1

Solution :-

The radius of smaller cone is r

The height of smaller cone is h

But,

According to the question,

The radius of new/larger cone is doubled

that is 2r .

The height of the new/larger cone = h

[ Height of both the cones are same ]

Now,

As we know that,

The volume of cone = 1/3πr^2h

Therefore,

The volume of smaller cone = 1/3πr^h

The volume of larger cone

= 1/3π(2r)^2h = 1/3π4rh = 4/3πr^2h

Now, We have to calculate the ratio of the volume of both the cone

Therefore,

Volume of larger cone / Volume of smaller cone

= 4/3πr^h / 1/3πr^h

= 4 : 1

Hence, The ratio of the volume of the larger cone to the smaller. cone is 4 : 1$.$

Answer 2

Given :

• Radius of the base of a cone is doubled keeping the height same.

To find :

• Ratio of the volume of the larger cone to the smaller cone.

Knowledge required :-

• Formula to calculate volume of cone :-

⠀⠀⠀⠀⠀Volume of cone = $\boldsymbol{ \dfrac{1}{3}\pi {r}^{2}h}$

Where,

• Take π = 22/7
• r = radius
• h = height

Let,

• Height of the larger and smaller cone = x
• Radius of the smaller cone = r
• Radius of the larger cone = 2r

⠀⠀⠀⇒ Volume of smaller cone = 1/3 πr²h

⠀⠀⠀⇒ Volume of larger cone = 1/3 π(2r)²h

⠀⠀⠀⇒ Volume of larger cone = 1/3 π4r²h

Ratio = Volume of larger cone/Volume of smaller cone

⠀⠀⠀⇒ $\sf{ \dfrac{ \tfrac{1}{3} \small{\pi 4{r}^{2}h} }{ \tfrac{1}{3} \small{\pi {r}^{2}h}}}$

⠀⠀⠀⇒ $\sf{ \dfrac{ \tfrac{1}{3} \small{\pi h( 4{r}^{2})} }{ \tfrac{1}{3}\small{\pi h({r}^{2})}}}$

⠀⠀⠀⇒ $\sf{ \dfrac{ \not\tfrac{1}{3} \small{ \not\pi \not h( {4r}^{2})} }{ \not \tfrac{1}{3}\small{ \not\pi \not h({r}^{2})}}}$

⠀⠀⠀⇒ $\sf{ \dfrac{ 4{r}^{2} }{ {r}^{2}}}$

⠀⠀⠀⇒ $\sf{ \dfrac{ 4 \not{r}^{2} }{ \not{r}^{2}}}$

⠀⠀⠀⇒ $\sf{ \dfrac{4}{1}}$

Ratio of the volume of the larger cone to the smaller cone = 4 : 1