 # RS Aggarwal Solutions Class 9 Mathematics Solutions for Volume And Surface Area Of Solids Exercise 15D in Chapter 15 - Volume And Surface Area Of Solids

Question 20 Volume And Surface Area Of Solids Exercise 15D

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

Consider r cm as the radius of ball and R cm as the radius of cylindrical tub

So we get

4/3 πr3 = π R2h

By substituting the values

4/3 × π × r3 = π × 122 × 6.75

On further calculation

r3 = (π × 122 × 6.75)/ 4/3 × π

So we get

r3 = 2916/ 4 = 729

By taking cube root

r = 9cm

Therefore, the radius of the ball is 9cm.

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