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

Answer:

Step-by-step explanation:

Given radius of the cylinder, r = 12 cm It is also given that a spherical iron ball is dropped into the cylinder and the water level raised by 6.75 cm .

Hence volume of water displaced = volume of the iron ball Height of the raised water level, h = 6.75 m

Volume of water displaced = πr2h = π × 12 × 12 × 6.75 cm3

⇒ Volume of iron ball = π × 12 × 12 × 6.75 cm3 → (1)

But, volume of iron ball =From (1) and (2) we get

⇒ r3 = 33.

∴ r = 9 .

Thus the radius of the iron ball is 9 cm

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

Therefore, r = cube root (9)^3 = 9 cm

Hence, radius of iron ball = 9 cm

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