Q. A solid metallic sphere of diameter 28 cm is melted and recast into a number of smaller cones, each of diameter 14/3 cm and height 3 cm. Find the number of cones so formed.
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Given: Diameter of sphere = 21cm
Radius of sphere, R = 21/2=10.5 cm
Now,
Diameter of small cone= 7cm
Radius of small cones, r = 3.5 cm
Height of small cones, h = 3 cm
Let the numbers of small cones those can be formed by melting the metallic sphere = n.
∴ n × Volume of a small cone = Volume of the sphere.
Thus, numbers of small cones can be formed by melting the given metallic sphere is 126.
Radius of sphere, R = 21/2=10.5 cm
Now,
Diameter of small cone= 7cm
Radius of small cones, r = 3.5 cm
Height of small cones, h = 3 cm
Let the numbers of small cones those can be formed by melting the metallic sphere = n.
∴ n × Volume of a small cone = Volume of the sphere.
Thus, numbers of small cones can be formed by melting the given metallic sphere is 126.
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⛦Hᴇʀᴇ Is Yoᴜʀ Aɴsᴡᴇʀ⚑
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➧ Diameter➙ 21 cm
➧ So, Radius➙ 21 / 2 cm
➧ Radius of cone
➾ 7/2 cm
➧ Height of cone
➾ 3 cm
➧ Volume of sphere =
No. of cones x Volume of cone
4/3 x 22/7 x 21/2 x 21/2 x 21/2 =
No.of cones x 1/3 x 22/7 x 7/2 x 7/2 x 3
➧ That implies,
11 x 441 = no. of cones x 77/2
4851 = no. of cones x 77/2
4851 x 2/77 = no. of cones
➧ So, No. of cones
➾ 126 ...✔
_________
Thanks...✊
▬▬▬▬▬▬▬▬▬▬▬▬☟
➧ Diameter➙ 21 cm
➧ So, Radius➙ 21 / 2 cm
➧ Radius of cone
➾ 7/2 cm
➧ Height of cone
➾ 3 cm
➧ Volume of sphere =
No. of cones x Volume of cone
4/3 x 22/7 x 21/2 x 21/2 x 21/2 =
No.of cones x 1/3 x 22/7 x 7/2 x 7/2 x 3
➧ That implies,
11 x 441 = no. of cones x 77/2
4851 = no. of cones x 77/2
4851 x 2/77 = no. of cones
➧ So, No. of cones
➾ 126 ...✔
_________
Thanks...✊
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