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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
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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?
Answer:
ANSWER:
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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