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:
Answer & explanation
Correct answer: option 4
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.