Answer :
The Volume of the cone will increase EIGHT times.
1st case,
consider the
diameter=2x cm⇒ radius=x cm
height= h cm
Volume of the 1st Cone=[tex] \frac{1}{3} \pi r^{2} h[/tex]
Putting the value we get,
Volume of 1st cone= [tex] \frac{1}{3} \pi x×h[/tex]
2nd case, (since it gets double)
Consider the
Diameter= 4x⇒ radius= 2x cm
height= 2h cm
Putting the value we get,
Volume of 2nd cone= [tex] \frac{1}{3} \pi 4 x^{2} ×2h[/tex]
Now divide the volume of 2nd Cone by 1st cone,
the answer remains is 8.
Hence, the volume increases by 8 times.
1st case,
consider the
diameter=2x cm⇒ radius=x cm
height= h cm
Volume of the 1st Cone=[tex] \frac{1}{3} \pi r^{2} h[/tex]
Putting the value we get,
Volume of 1st cone= [tex] \frac{1}{3} \pi x×h[/tex]
2nd case, (since it gets double)
Consider the
Diameter= 4x⇒ radius= 2x cm
height= 2h cm
Putting the value we get,
Volume of 2nd cone= [tex] \frac{1}{3} \pi 4 x^{2} ×2h[/tex]
Now divide the volume of 2nd Cone by 1st cone,
the answer remains is 8.
Hence, the volume increases by 8 times.