Maths!!!! A spherical ball of lead 3cm in radius is melted and recasted in 3 spherical balls if the radius of the two balls are 1.5cm and 2cm respectively . Find the radius of third ball
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Radius of ball which is melted = 3cm
Radius of first formed ball = 1.5 cm
Radius of second formed ball = 2 cm
Let Radius of third formed ball = r cm
Volume of large ball = sum of volume of 3 formed balls
⇒4π/3 × 3³ = 4π/3(1.5³ + 2³ + r³)
⇒ 3³ =1.5³ + 2³ + r³
⇒ 27 = 3.375 + 8 + r³
⇒ r³ = 27 - 8 - 3.375 = 15.625
⇒ r = ∛15.625
⇒ r = 2.5cm
∴Radius of third ball is 2.5cm
Hope this helps!
Radius of first formed ball = 1.5 cm
Radius of second formed ball = 2 cm
Let Radius of third formed ball = r cm
Volume of large ball = sum of volume of 3 formed balls
⇒4π/3 × 3³ = 4π/3(1.5³ + 2³ + r³)
⇒ 3³ =1.5³ + 2³ + r³
⇒ 27 = 3.375 + 8 + r³
⇒ r³ = 27 - 8 - 3.375 = 15.625
⇒ r = ∛15.625
⇒ r = 2.5cm
∴Radius of third ball is 2.5cm
Hope this helps!
Answered by
12
conservation of mass principle, as the density is constant:
volume of 3 cm ball = sum of volumes of small ones.
4π/3 * 3³ = 4π/3 (1.5³ + 2³ + R³)
Solving that R = 2.5 cm
volume of 3 cm ball = sum of volumes of small ones.
4π/3 * 3³ = 4π/3 (1.5³ + 2³ + R³)
Solving that R = 2.5 cm
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