1.

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

Answer»

Given, 

the radius of the cylindrical tube, r = 12 cm 

Level of water raised in tube, h = 6.75 cm 

Volume of cylinder = πr2 h 

= π (12)2 × 6.75 cm3 

= π x 122 x 6.75 cm3----------(i) 

let r be the radius of a spherical shell balls 

volume of the sphere = \(\frac{4}{3}πr^3\)-------(ii) 

∵ volume of cylinder = volume of spherical ball 

∴ π x 122 x 6.75

= \(\frac{4}{3}πr^3\)

⇒ r3 = \(\frac{{12}^2\times{6.75}}{\frac{4}{3}\times{π}}\)

⇒ r = 9 cm 

∴ radius of spherical ball, r = 9 cm



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