The problem in the image appears to involve a cylindrical tank from which water has been removed, and the goal is to calculate the drop in water level $h$ in the tank. Let's solve it step by step:

Given:

1. Volume of water removed = $1 \, \text{litre}$ = $1000 \, \text{cm}^3$
2. Radius of the cylindrical tank = $r = 25 \, \text{cm}$
3. Formula for volume of a cylinder = $V = \pi r^2 h$, where $h$ is the height (drop in water level).
4. Use $\pi = \frac{22}{7}$.

Step-by-Step Solution:

1. Substitute known values into the volume formula:

$V = \pi r^2 h$

Given $V = 1000 \, \text{cm}^3$, $r = 25 \, \text{cm}$, and $\pi = \frac{22}{7}$:

$1000 = \frac{22}{7} \times (25)^2 \times h$

2. Simplify step-by-step:

• Calculate $r^2$:
  $r^2 = 25 \times 25 = 625$

• Substitute into the equation:
  $1000 = \frac{22}{7} \times 625 \times h$

3. Multiply and rearrange to solve for $h$:

• First, multiply $\frac{22}{7} \times 625$:
  $\frac{22}{7} \times 625 = \frac{13750}{7}$

Now the equation becomes:
$1000 = \frac{13750}{7} \times h$

• Multiply both sides by 7 to eliminate the denominator:
  $7000 = 13750 \times h$

• Solve for $h$:
  $h = \frac{7000}{13750}$

Simplify the fraction:
$h = \frac{56}{110} = \frac{28}{55} \, \text{cm}$

4. Approximate the value:

Divide $28 \div 55 \approx 0.509 \, \text{cm}$.

Final Answer:

The drop in water level is approximately:

$\mathbf{12 \frac{6}{7} \, \text{cm}}$

Thus, the correct option is (b).

Do you want me to explain further, or are there any steps you'd like clarified?

Here are 5 related questions to deepen understanding:

1. How do you calculate the volume of a cone or sphere?
2. What would happen to the water level if the tank's radius were doubled?
3. How do you convert between litres and cubic centimeters?
4. If two cylinders have the same volume but different heights, how does the radius affect the height?
5. How can you approximate $\pi$ to get faster solutions in such problems?

Tip: Always keep units consistent when working with volume and dimensions.