Question

a cylindrical tub of radius of 16 cm contains water to a depth of 30 cm a sperical iron ball is dropped into the tub and thus level of water is raised by 9 cm what is radius of ball

Answer 1

Hi ,

Let the radius of the cylidrical tub

R = 16 cm

Height of the water raised ( h ) = 9cm

radius of the iron ball = r cm

We know that ,
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Volume of the cylider = πR²h

Volume of the Sphere = ( 4 / 3)πr³

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Volume of water raised in the tank

after immersion of the iron ball

equals to volume of the iron ball.

Volume of the iron ball = volume of

the water raised in the tank

(4/3)πr³ = πR² h

r³ = ( πR² h ) / [ (4/3 )π ]

r³ =(3 R² h) / 4

r³ = ( 3 × 16 × 16 × 9 ) / 4

r³ = ( 3 × 4 × 16 × 9 )

r³ = ( 3 × 3 × 3 × 4 × 4 × 4 )

r³ = 12 × 12 × 12

r = 12 cm

Therefore ,

radius of the iron ball = r = 12cm

I hope this helps you.

:)

Answer 2

Answer:

Step-by-step explanation:

Let the radius of the cylidrical tub

R = 16 cm

Height of the water raised ( h ) = 9cm

radius of the iron ball = r cm

Volume of the cylider = πR²h

Volume of the Sphere = ( 4 / 3)πr³

Volume of water raised in the tank

after immersion of the iron ball

equals to volume of the iron ball.

Volume of the iron ball = volume of

the water raised in the tank

(4/3)πr³ = πR² h

r³ = ( πR² h ) / [ (4/3 )π ]

r³ =(3 R² h) / 4

r³ = ( 3 × 16 × 16 × 9 ) / 4

r³ = ( 3 × 4 × 16 × 9 )

r³ = ( 3 × 3 × 3 × 4 × 4 × 4 )

r³ = 12 × 12 × 12

r = 12 cm

Therefore ,

radius of the iron ball = r = 12cm

Hope it helps:-)

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