Math, asked by Shubhamjagtap664, 1 year ago

2 metallic right circular cones having their height 4.1cm and 4.3cm respectively and the radii of their bases 2.1cm each have been melting together and recasted into a shpere determine the diameter of the sphere so formed

Answers

Answered by BEJOICE
24

sum \: of \: the \: volume \: of \: two \: cones \\  = volume \: of \: sphere \\  \frac{1}{3} \pi \times  {2.1}^{2}  \times (4.1 + 4.3) =  \frac{4}{3} \pi {r}^{3}  \\  {r}^{3}  =  \frac{ {2 .1}^{2}  \times 8.4}{4}  =  {2.1}^{3}  \\ r = 2.1 \\ diameter \:  = 2r = 4.2 \: cm
Answered by wifilethbridge
7

The diameter of the sphere so formed is 6.66 cm

Step-by-step explanation:

Cone 1 :

Height of cone = 4.1 cm

Radius of cone = 2.1 cm

Volume of cone = \frac{1}{3} \pi r^2 h =\frac{1}{3} \times \frac{22}{7} \times 2.1^2 \times 4.1 =18.942 cm^3

Cone 2 :

Height of cone = 4.3 cm

Radius of cone = 2.1 cm

Volume of cone = \frac{1}{3} \times \frac{22}{7} \times 2.1^2  \times 4.3 =19.866 cm^3

Volume of sphere = \frac{4}{3} \pi r^3 =\frac{4}{3} \times \frac{22}{7} \times r^3

Two cones are melted to form sphere

So, volume of two cones will be equal to the volume of sphere

So, \frac{1}{3} \times \frac{22}{7} \times 2.1^2 \times 4.1 +\frac{1}{3} \times \frac{22}{7} \times 2.1^2 \times 4.3 =\frac{4}{3} \times \frac{22}{7} \times r^3\\\\ 2.1^2 \times 4.1 + 2.1^2 \times 4.3 =r^3\\\\ \sqrt[3]{2.1^2 \times 4.1 + 2.1^2 \times 4.3 }=r\\\\3.33=r

So, radius of sphere = 3.33 cm

Diameter of sphere =3.33 \times 2 =  6.66

Hence the diameter of the sphere so formed is 6.66 cm

#Learn more :

2 metallic circular cones having their heights 4.1 cm and 4.3 cm and the radii of their bases 21 cm each have been melted together and recast into a sphere. Find the diameter of the sphere.

https://brainly.in/question/2487748

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