Question

A cylindrical tube of radius 12 cm contains water upto a depth of 20 cm. A spherical shaped ball of iron is dropped into the tube and the level of water is raised by 6.75 cm. What is the radius of the spherical ball ?

Answer 1

$\huge\mathfrak{Solution:}$

Let radius of iron sphere be r cm

Therefore, volume of the sphere, V1 = 4/3 × pi × r^3 cm^3

Radius of the cylindrical tube = R = 12 cm

Water level raised in the tube = h = 6.75 cm

Volume of water displaced by the iron sphere = V2

Therefore, V2 = pi × R^2 × h

= pi × 12 × 12 × 6.75 cm^3

Clearly, volume of the ball = volume of the water displaced by the ball

i.e., V1 = V2

Therefore, 4/3 × pi × r^3

= pi × 12 × 12 × 6.75

or r^3 = 12 × 12 × 6.75 × 3 / 4

= 3 × 81 × 3

= 9 × 9 × 9

= (9)^3

r = cube root (9)^3

= 9 cm

Hence, radius of iron ball = 9 cm.

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Answer 2

$\huge\mathfrak{Solution:}$

Let radius of iron sphere be r cm

Therefore, volume of the sphere, V1 = 4/3 × pi × r^3 cm^3

Radius of the cylindrical tube = R = 12 cm

Water level raised in the tube = h = 6.75 cm

Volume of water displaced by the iron sphere = V2

Therefore, V2 = pi × R^2 × h

= pi × 12 × 12 × 6.75 cm^3

Clearly, volume of the ball = volume of the water displaced by the ball

i.e., V1 = V2

Therefore, 4/3 × pi × r^3

= pi × 12 × 12 × 6.75

or r^3 = 12 × 12 × 6.75 × 3 / 4

= 3 × 81 × 3

= 9 × 9 × 9

= (9)^3

r = cube root (9)^3

= 9 cm

Hence, radius of iron ball = 9 cm.

________________

HOPE IT HELPS ❤❤