A circular disc of area `(0.49 pi) m^2` rolls down a length of 1 76 km. The number of revolutions it makes, is :

Options

A. 300
B. 400
C. 600
D. 4000

Answer

To solve the problem of how many revolutions a circular disc of area \(0.49\pi \, m^2\) makes when it rolls down a distance of \(1.76 \, km\), we can follow these steps:

### Step 1: Find the radius of the circular disc.
The area \(A\) of a circle is given by the formula:
\[ A = \pi r^2 \]
We are given that the area is \(0.49\pi \, m^2\). Setting the two equal gives:
\[ \pi r^2 = 0.49\pi \]
Dividing both sides by \(\pi\):
\[ r^2 = 0.49 \]
Taking the square root of both sides:
\[ r = \sqrt{0.49} = 0.7 \, m \]

### Step 2: Convert the distance from kilometers to meters.
The distance rolled by the disc is given as \(1.76 \, km\). To convert this to meters:
\[ 1.76 \, km = 1.76 \times 1000 \, m = 1760 \, m \]

### Step 3: Calculate the circumference of the circular disc.
The circumference \(C\) of a circle is given by the formula:
\[ C = 2\pi r \]
Substituting the radius we found:
\[ C = 2\pi(0.7) = 1.4\pi \, m \]

### Step 4: Determine the number of revolutions.
The number of revolutions \(n\) made by the disc when it rolls a distance \(d\) is given by:
\[ n = \frac{d}{C} \]
Substituting the values we have:
\[ n = \frac{1760}{1.4\pi} \]
Using \(\pi \approx \frac{22}{7}\):
\[ n = \frac{1760}{1.4 \times \frac{22}{7}} = \frac{1760 \times 7}{1.4 \times 22} \]
Calculating \(1.4 \times 22 = 30.8\):
\[ n = \frac{1760 \times 7}{30.8} \]
Calculating \(1760 \times 7 = 12320\):
\[ n = \frac{12320}{30.8} \approx 400 \]

Thus, the number of revolutions the disc makes is:
\[ \boxed{400} \]