A thin cylindrical rod of length `2.5m` and radius `5mm` is firmly fixed at upper end when lower end is twiced, the shear angle is found to be `0.06^(@)` then angle of twising is

Options

A. `10^(@)`
B. `20^(@)`
C. `30^(@)`
D. `40^(@)`

Explanation

To find the angle of twisting (θ) of a thin cylindrical rod, we can use the relationship between the shear angle (φ), the radius (r), and the length (L) of the rod. The formula we will use is:

\[ \phi = \frac{G \theta}{L} \]

Where:
- φ is the shear angle (0.06 degrees)
- G is the modulus of rigidity (not provided, but we will see it cancels out)
- θ is the angle of twisting we want to find
- L is the length of the rod (2.5 m)

### Step-by-Step Solution:

1. **Convert the given values to consistent units**:
- Length (L) = 2.5 m = 2500 mm
- Radius (r) = 5 mm
- Shear angle (φ) = 0.06 degrees (we will convert this to radians for calculations)

To convert degrees to radians:
\[ \text{Radians} = \text{Degrees} \times \frac{\pi}{180} \]
\[ \phi = 0.06 \times \frac{\pi}{180} \approx 0.00105 \text{ radians} \]

2. **Use the relationship between shear angle and angle of twisting**:
The angle of twisting (θ) can be related to the shear angle (φ) as follows:
\[ \phi = \frac{r \theta}{L} \]
Rearranging this gives:
\[ \theta = \frac{\phi \cdot L}{r} \]

3. **Substitute the known values into the equation**:
\[ \theta = \frac{0.00105 \cdot 2500}{5} \]

4. **Calculate θ**:
\[ \theta = \frac{0.00105 \cdot 2500}{5} = \frac{2.625}{5} = 0.525 \text{ radians} \]

5. **Convert radians back to degrees**:
To convert radians back to degrees:
\[ \text{Degrees} = \text{Radians} \times \frac{180}{\pi} \]
\[ \theta = 0.525 \times \frac{180}{\pi} \approx 30.1 \text{ degrees} \]

### Final Answer:
The angle of twisting (θ) is approximately **30 degrees**.