If by dropping a stone in a quiet lake a wave moves in circle at a speed of 3.5cm/sec, then the rate of increase of the enclosed circular region when the radius of the circular wave is 10 cm, is `(pi=(22)/(7))`

Options

A. 220sq.cm/sec
B. 110sq.cm/sec
C. 35sq.cm/sec
D. 350sq.cm/sec

Explanation

To solve the problem, we need to find the rate of increase of the enclosed circular region (area) when the radius of the circular wave is 10 cm. We are given the speed of the wave (dr/dt) as 3.5 cm/sec.

### Step-by-Step Solution:

1. **Identify the formula for the area of a circle**:
The area \( A \) of a circle is given by the formula:
\[ A = \pi r^2 \]

2. **Differentiate the area with respect to time**:
To find the rate of change of the area with respect to time, we differentiate \( A \) with respect to \( t \):
\[ \frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = \pi \cdot 2r \cdot \frac{dr}{dt} \]
This simplifies to:
\[ \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \]

3. **Substitute the given values**:
We know:
- \( r = 10 \) cm (the radius of the circular wave)
- \( \frac{dr}{dt} = 3.5 \) cm/sec (the rate of increase of the radius)

Substituting these values into the equation:
\[ \frac{dA}{dt} = 2 \cdot \pi \cdot 10 \cdot 3.5 \]

4. **Use the value of \( \pi \)**:
We are given \( \pi = \frac{22}{7} \). Substituting this value:
\[ \frac{dA}{dt} = 2 \cdot \frac{22}{7} \cdot 10 \cdot 3.5 \]

5. **Calculate the expression**:
First, calculate \( 2 \cdot 10 = 20 \):
\[ \frac{dA}{dt} = 20 \cdot \frac{22}{7} \cdot 3.5 \]
Now calculate \( 3.5 = \frac{7}{2} \):
\[ \frac{dA}{dt} = 20 \cdot \frac{22}{7} \cdot \frac{7}{2} \]
The \( 7 \) in the numerator and denominator cancels out:
\[ \frac{dA}{dt} = 20 \cdot \frac{22}{2} = 20 \cdot 11 = 220 \text{ cm}^2/\text{sec} \]

6. **Final answer**:
The rate of increase of the enclosed circular region when the radius is 10 cm is:
\[ \frac{dA}{dt} = 220 \text{ cm}^2/\text{sec} \]