Question

A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. spherical iron ball is dropped into the tub and thus the level of water is raised by 6.75 cm. What is the radius of the ball?
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Answer 1

Answer:

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Step-by-step explanation:

Solution:-

Radius of the cylinder = 12 cm

Height of the cylinder = 6.75 cm

Let 'r' be the radius of the spherical iron ball.

Volume of the cylinder = volume of the spherical iron ball

πr²h = 4/3πr³

22/7*12*12*6.75 = 4/3*22/7*r³

3 × 12 × 12 × 6.75 = 4r³

2916 = 4r³

r³ = 2916/4

r³ = 729

r = 9 cm

So, the radius of the spherical iron ball is 9 cm.

Answer 2

Step-by-step explanation:

Volume of cylinder Vc=πr2h

Thus, volume of water in cylinder V1=π×144×20=2880πcm3

After dropping the spherical ball in cylinder new volume 

V2=144×20π+34πr3=π×144×26.75

⇒34r3=144(6.75)

r3=729∴r=9cm