Question

A solid metallic object is shaped like a double cone. Radius of tge base of both cones is same but their

Answer 1

I think your question is incomplete. Complete it

Answer 2

The quantity of water it will displace is $\dfrac{1}{3} \pi r^{2} (h+H) cm^{3}.$

Step-by-step explanation:

"The complete quwestion"

A solid metallic object is shaped like a double cone.Radius of base of both cones is same but their height are different. If this cone is immersed in water,find the quantity of water it will displace.

Let height of cone = h and H and radius of both cone = r cm

To find, the quantity of water it will displace = ?

Quantity of water, which will displace = The volume of double cone

$=\dfrac{1}{3} \pi r^{2} h+\dfrac{1}{3} \pi r^{2} H$

$=\dfrac{1}{3} \pi r^{2} (h+H) cm^{3}$

Hence, the quantity of water it will displace is $\dfrac{1}{3} \pi r^{2} (h+H) cm^{3}.$