Question

A cylindrical jar whose base has a radius of 15 cm is filled with water up to a height of 20 cm. A solid iron spherical ball of radius 10 cm is dropped in the jar to submerge completely in water. Find the increase in the level of water.

• A. \(5 \dfrac{17}{27}\) cm
• B. \(5 \dfrac{5}{7}\) cm
• C. \(5 \dfrac{8}{9}\) cm
• D. \(5 \dfrac{25}{27}\) cm

Hint:

To find the increase in liquid level when a solid is submerged, equate the volume of the solid to the volume of liquid displaced in the container's shape.

Answer 1

The Correct Option is D

Volume of sphere \(= \dfrac{4}{3} \pi r^3 = \dfrac{4}{3} \pi (10)^3 = \dfrac{4}{3} \pi \times 1000 = \dfrac{4000}{3} \pi \) cm\(^3\)
Let the rise in water level be \( h \) cm.
Volume of water displaced \(= \) volume of cylinder part raised
Volume \(= \pi r^2 h = \pi (15)^2 h = 225\pi h \) cm\(^3\)
Now equate the two:
\[ 225\pi h = \dfrac{4000}{3} \pi \Rightarrow h = \dfrac{4000}{3 \times 225} = \dfrac{4000}{675} = \dfrac{80}{13.5} = \dfrac{160}{27} \] Now convert \( \dfrac{160}{27} \) to mixed fraction:
\( 160 \div 27 = 5 \) remainder \(25\)
\[ \therefore h = 5 \dfrac{25}{27} \text{ cm} \]