Math, asked by vishnusjvn21, 1 year ago

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

Answers

Answered by sam2001envy
8
Given, radius of base of cylindrical tub = 12cm
Depth of the tub = 20cm
Let the radius of the spherical ball be "R"
Now, volume of the spherical ball = Volume of water raised
Height by which water is raised = 6.75cm
Hence, 4/3*22/7*R³ = 22/7*12*12*6.75
⇒4/3R³ = 144*6.75
⇒R³=144*6.75*3/4  ⇒ R³ = 729
∴ R = ∛729 = 9cm
Hence, the radius of the ball is 9cm.

Answered by mathsdude85
3
Answer:

The radius of the spherical ball is 9 cm.

Step-by-step explanation:

Given :  

Radius of a cylindrical tube (R) =12 cm

Level of water raised in the cylindrical tube (h) = 6.75 cm

Let ‘r’ be the radius of the spherical ball.

Volume of water raised in the cylindrical tube is equal to the volume of spherical ball.

Volume of water raised in the cylindrical tube = Volume of spherical ball

πR²h = 4/3πr³

R²h = 4/3r³

12² × 6.75 = 4/3 r³

r³ = (12 × 12 × 6.75 × 3) / 4

r³ = (3 × 12 × 6.75 × 3)  

r³ = 729

r = ³√729 = 9

r = 9 cm

Hence, The radius of the spherical ball is 9 cm.

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