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:

$5\frac{25}{27}cm$

The correct answer is Option (4) → $5\frac{25}{27}cm$

1. Identify the Formulas

• Volume of a Sphere: $V_{sphere} = \frac{4}{3} \pi r^3$
• Volume of a Cylinder (Rise): $V_{rise} = \pi R^2 h$

(Where $R$ is the radius of the jar and $h$ is the height the water rises)

2. Calculate the Volume of the Spherical Ball

Given the radius of the ball ($r$) is 10 cm:

$V_{sphere} = \frac{4}{3} \pi (10)^3 = \frac{4}{3} \pi (1000) = \frac{4000}{3} \pi$

3. Set Up the Equation for Water Rise

The volume of the water that rises forms a small "cylindrical" shape inside the jar with radius ($R$) = 15 cm. Let the increase in height be $h$.

$V_{rise} = \pi (15)^2 h = 225 \pi h$

Since $V_{rise} = V_{sphere}$:

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

4. Solve for $h$

First, cancel $\pi$ from both sides:

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

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

Now, simplify the fraction. Both 4000 and 225 are divisible by 25:

• $4000 \div 25 = 160$
• $225 \div 25 = 9$

$h = \frac{160}{3 \times 9} = \frac{160}{27}=5\frac{25}{27}$

Conclusion

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