Two solid spheres of radii 2 cm and 4 cm are melted and recast into a cone of height 8 cm. Find the radius of the cone so formed.
Please give step by step solution.
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Let the radius of first sphere be r1 , second sphere be r2 and radius of the cone be r3 , then -
4/3πr1^3 + 4/3πr2^3 = 1/3πr3^3h
4/3π(r1^3 + r2^3) = 1/3πr3^3h
4(8+64) = 8r3^3
72 = 2r3^3
r3^3 = 36
r3 = 6
Thus , radius of the cone = 6 cm.
4/3πr1^3 + 4/3πr2^3 = 1/3πr3^3h
4/3π(r1^3 + r2^3) = 1/3πr3^3h
4(8+64) = 8r3^3
72 = 2r3^3
r3^3 = 36
r3 = 6
Thus , radius of the cone = 6 cm.
rockingstar16:
Thanks
Answered by
0
Answer: 2 cm
Step-by-step explanation:
Let the radius of first sphere be r1 , second sphere be r2 and radius of the cone be r3 , t
4/3πr1^3 + 4/3πr2^3 = 1/3πr3^3h
4/3π(r1^3 + r2^3) = 1/3πr3^3h
4(8+64) = 8r3^3
72 = 2r3^3
r3^3 = 36
r3 = 6^1/3 = 1.9=2cm approx
radius of the cone =2 cm
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