A cylindrical jar having a base of radius 15 cm, is filled with water up to a height of 20 cm. If a solid iron spherical ball of radius 10 cm is dropped in the jar to submerge completely in the water, then the increase in the level of water is:

When a ball is dropped into water in a jar, it displaces water equal to its volume. This displaced water causes the water level to rise.

Given:

• Cylindrical jar with base radius $R = 15$ cm
• Initial water height $= 20$ cm
• Spherical iron ball with radius $r = 10$ cm
• Ball completely submerged

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The volume of the spherical ball is:

$V_{sphere} = \frac{4}{3}\pi r^3$

$V_{sphere} = \frac{4}{3}\pi (10)^3$

$V_{sphere} = \frac{4}{3}\pi \times 1000$

$V_{sphere} = \frac{4000\pi}{3}$ cm³

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The base area of the cylinder is:

$A_{base} = \pi R^2$

$A_{base} = \pi (15)^2$

$A_{base} = 225\pi$ cm²

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The displaced water spreads across the circular base, creating a rise in water level.

The volume of water displaced equals the volume of the ball, which also equals the volume of the new water column formed:

$\text{Volume of ball} = \text{Base Area} \times \text{Height rise}$

$\frac{4000\pi}{3} = 225\pi \times h$

Dividing both sides by $\pi$:

$\frac{4000}{3} = 225h$

Solving for $h$:

$h = \frac{4000}{3 \times 225}$

$h = \frac{4000}{675}$

Simplifying by dividing both numerator and denominator by $25$:

$h = \frac{160}{27}$ cm

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The increase in the level of water is $\frac{160}{27}$ cm or $5\frac{25}{27}$ cm.