A cylindrical jar, whose base has a radius of 15 cm, is filled with water upto 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 (in cm).

Options

A. `5""(17)/(27)`
B. `5""(5)/(7)`
C. `5""(8)/(9)`
D. `5""(25)/(27)`

Answer

To find the increase in the water level when a solid iron spherical ball is submerged in a cylindrical jar, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the dimensions of the cylindrical jar and the spherical ball:**
- Radius of the cylindrical jar (r) = 15 cm
- Height of water in the jar (h) = 20 cm
- Radius of the spherical ball (R) = 10 cm

2. **Calculate the volume of the spherical ball:**
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi R^3 \]
Substituting the radius of the sphere:
\[ V = \frac{4}{3} \pi (10)^3 = \frac{4}{3} \pi (1000) = \frac{4000}{3} \pi \text{ cm}^3 \]

3. **Calculate the increase in water level in the cylindrical jar:**
The volume of water displaced by the submerged sphere will be equal to the volume of the sphere. The volume \( V \) of water in a cylinder is given by:
\[ V = \pi r^2 h' \]
where \( h' \) is the increase in the height of the water level. Setting the volume of the sphere equal to the volume of the cylindrical water displaced:
\[ \frac{4000}{3} \pi = \pi (15^2) h' \]
Simplifying this equation:
\[ \frac{4000}{3} = 225 h' \]

4. **Solve for \( h' \):**
\[ h' = \frac{4000}{3 \times 225} = \frac{4000}{675} = \frac{4000 \div 25}{675 \div 25} = \frac{160}{27} \text{ cm} \]

5. **Convert \( h' \) to a mixed fraction:**
To express \( \frac{160}{27} \) as a mixed fraction:
- Divide 160 by 27, which gives 5 remainder 25.
- Thus, \( \frac{160}{27} = 5 \frac{25}{27} \) cm.

### Final Answer:
The increase in the level of water is \( 5 \frac{25}{27} \) cm.

---