Question

The internal and external diameters of hollow hemispherical shell are 6 cm and 10 cm respectively it is melted and recast into a solid cone of base diameter 14 centimetre find the height of the cone so formed

Answer 1

(1/3)*pi*7^2 *h = (4/3)*pi*(5^3 - 3^3)
=> 49h = 4*(125-27)
=> 49h = 4*98
=> h = 8
height of the cone so formed = 8 cm

Answer 2

Answer:

4 cm

Step-by-step explanation:

Internal diameter of  hemisphere = 6 cm

So, internal radius r = Diameter/2 = 6/2 = 3 cm

External diameter of hemisphere R = 10 cm

So, external radius = 10/2 = 5cm

Volume of hemisphere = $\frac{2}{3} \pi r^{3}$

volume of given hemisphere =  $\frac{2}{3} \pi (R^3-r^3)$

                                                  =  $\frac{2}{3} \pi (5^3-3^3)$

                                                  =  $\frac{2}{3} \pi (125-27)$

                                                  =  $\frac{2}{3} \pi 98$

Diameter of cone = 14 cm

Radius of cone = 14 /2 = 7 cm

Volume of cone = $\frac{1}{3} \pi r^{2} h$

Volume of given cone =  $\frac{1}{3} \pi 7^{2} h$

                                     = $\frac{1}{3} \pi 49 \times h$

Now since we are given that Hemisphere is recast into cone So, volume will remain same .

$\frac{2}{3} \pi 98= \frac{1}{3} \pi 49 \times h$

$2 \times 98=  49 \times h$

$\frac{196}{49}= h$

$4= h$

Thus the height of the cone is 4 cm.