Question

A hemishpere of lead of radius 9cm is melted and cast into a right circular cone of height 72cm.Find the radius of the base of the cone

Answer 1

i have first substituted the volume of both the given solids. we are taking volume because the concept described here is about melting. always be careful in calculation. especially du

Answer 2

Answer:

• Volume of cone = 4.5 cm

Step-by-step explanation:

Given:

• Radius of hemisphere = 9 cm
• Height of Circular cone = 72 cm

To find:

• Radius of the base of cone.

Now, it is given that, a hemisphere is melted and cast into right circular cone. So,

$\implies \sf Volume\;of\;hemisphere = Volume\;of\;cone\\ \\ \\ \implies \sf \dfrac{2}{3}\pi r^{3}=\dfrac{1}{3}\pi r^{2}h\\ \\ \\ \implies \sf \dfrac{2}{3}\pi \times 9\times 9\times 9=\dfrac{1}{3}\pi r^{2}\times 72\\ \\ \\ \implies \sf 2\times3\times9\times 9=24\times r^{2}\\ \\ \\ \implies \sf r^{2}=486/24\\ \\ \\ \implies \sf r^{2}=20.25\\ \\ \\ \implies \sf r=\sqrt{20.25}\\ \\ \\ \implies \sf r=4.5\;cm$

Hence, Volume of cone = 4.5 cm