A spherical ball of lead 3 cm is radius is melted and recast into three sperical balls. The radius of two of these are 1.5 cm and 2 cm respectively the radius of third ball
Answers
Given that radius of the spherical ball r = 3cm.
We know that volume of the sphere = 4/3pir^3
= 4/3 * 22/7 * (3)^3
= 4/3 * 22/7 * 27
= 4 * 22 * 9/7
= 792/7 cm^3.
= 113.1428cm^3
Given radius of the 1st sphere = 1.5cm
Volume of the 1st sphere = 4/3 pir^3
= 4/3 * 22/7 * (1.5)^3
= 4/3 * 22/7 * 3.375
= 14.1428cm^3
Given radius of the 2nd sphere = 2cm.
the volume of the 2nd sphere = 4/3 pir^3
= 4/3 * 22/7 * (2)^3
= 33.5238cm^3
Volume of the two small spheres = 14.1428 + 33.5238
= 47.6666cm^3
Volume of the third sphere = 113.1428 - 47.6666
= 65.4762cm^3.
Let the radius of the third be r cm.
4/3pir^3 = 65.4762
4/3 * 22/7 * r^3 = 65.4762
r^3 = 65.4762 * 7/22 * 3/4
= 65.4762 * 21/88
= 1375.0002/88
= 15.62500
r = 2.5cm.
Therefore the radius of the third ball = 2.5cm.
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Answer:
The diameter of the third ball is 5 cm.
SOLUTION :
Given :
Let the Radius of the third ball be r3
Radius of the spherical ball , R = 3 cm
Radius of the first ball, r1 = 1.5 cm
Radius of the second ball, r2 = 2 cm
Volume of the spherical ball, V = 4/3πR³
Volume of the spherical ball is equal to the volume of the 3 small spherical balls.
Volume of the spherical ball, V = Volume (V1) of first ball + Volume (V2) of second ball + Volume of third ball (V3)
4/3πR³ = 4/3πr1³ + 4/3πr2³ + 4/3πr3³
4/3πR³ = 4/3π (r1³ + r2³ + r3³)
R³ = (r1³ + r2³ + r3³)
3³ = (1.5³ + 2³ + r3³)
27 = 3.375 + 8 + r3³
27 = 11.375 + r3³
r3³ = 27 - 11.375
r3³ = 15.625
r3 = ³√15.625
r3 = ³√ 2.5 × 2.5 × 2.5
r3 = 2.5 cm
Radius of the third ball = 2.5 cm
Diameter of the third ball = 2 × r3 = 2 × 2.5 = 5 cm
Diameter of the third ball = 5 cm
Hence, the diameter of the third ball is 5 cm.
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