a spherical ball of radius 6 cm is recast into 27 spherical balls of equal radii find the percentage change in the total surface area
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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.
= 113.1428
Given radii of the 1st sphere = 1.5cm
The volume of the 1st sphere = 4/3 pir^3
= 4/3 * 22/7 * (1.5)^3
= 4/3 * 22/7 * 3.375
= 14.1428.
Given radii of the 2nd sphere = 2cm.
the volume of the 2nd sphere = 4/3 pir^3
= 4/3 * 22/7 * (2)^3
= 33.5238.
The volume of the two small spheres = 14.1428 + 33.5238
= 47.6666.
The 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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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.
= 113.1428
Given radii of the 1st sphere = 1.5cm
The volume of the 1st sphere = 4/3 pir^3
= 4/3 * 22/7 * (1.5)^3
= 4/3 * 22/7 * 3.375
= 14.1428.
Given radii of the 2nd sphere = 2cm.
the volume of the 2nd sphere = 4/3 pir^3
= 4/3 * 22/7 * (2)^3
= 33.5238.
The volume of the two small spheres = 14.1428 + 33.5238
= 47.6666.
The 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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