A mass of 100 grams is attached to the end of a rubber string 49 cm. long and having an area of cross section 20 sq. mm. The string is whirled round, horizontally at a constant speed of 40 r.p.s in a circle of radius 51 cm. Find Young's modulus of rubber.

Explanation

To find Young's modulus of the rubber string, we will follow these steps:

### Step 1: Convert all units to SI units
- Mass (m) = 100 grams = 0.1 kg
- Length of the rubber string (L) = 49 cm = 0.49 m
- Area of cross-section (A) = 20 mm² = 20 × 10⁻⁶ m²
- Radius (r) = 51 cm = 0.51 m
- Frequency (f) = 40 r.p.s = 40 Hz

### Step 2: Calculate the angular velocity (ω)
The angular velocity (ω) can be calculated using the formula:
\[ \omega = 2\pi f \]
Substituting the value of frequency:
\[ \omega = 2\pi \times 40 = 80\pi \text{ rad/s} \]

### Step 3: Calculate the centripetal force (F)
The centripetal force (F) acting on the mass can be calculated using the formula:
\[ F = m \cdot r \cdot \omega^2 \]
Substituting the values:
\[ F = 0.1 \cdot 0.51 \cdot (80\pi)^2 \]
Calculating \( (80\pi)^2 \):
\[ (80\pi)^2 = 6400\pi^2 \]
Now substituting this back:
\[ F = 0.1 \cdot 0.51 \cdot 6400\pi^2 \]

### Step 4: Calculate the extension (ΔL)
The extension (ΔL) of the rubber string can be calculated as:
\[ \Delta L = L' - L \]
Where \( L' \) is the new length when the string is stretched. Since the problem states the rubber string is being whirled, we can assume the effective length under tension is the same as the original length, hence:
\[ \Delta L = 0.51 - 0.49 = 0.02 \text{ m} \]

### Step 5: Calculate Young's modulus (Y)
Young's modulus (Y) is given by the formula:
\[ Y = \frac{F \cdot L}{A \cdot \Delta L} \]
Substituting the values we have:
\[ Y = \frac{(0.1 \cdot 0.51 \cdot 6400\pi^2) \cdot 0.49}{(20 \times 10^{-6}) \cdot 0.02} \]

### Step 6: Simplify and calculate Y
Now we can simplify and calculate:
1. Calculate the force \( F \) first.
2. Substitute \( F \) into the Young's modulus formula.
3. Perform the final calculation to find \( Y \).

After performing the calculations, we find:
\[ Y \approx 3.95 \times 10^{10} \text{ dyne/cm}^2 = 3.95 \times 10^{9} \text{ N/m}^2 \]

### Final Answer
Thus, the Young's modulus of rubber is approximately \( 3.95 \times 10^{9} \text{ N/m}^2 \).

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