A metallic metre stick translates in a direction making an angle of `60^(@)` with its length.The plane of motion is perpendicular to a uniform magnetic field of `0.1 T` that exists in the space.Find the `emf` induced between the ends of the rod if the speed of translation is `0.2 m//s`.

Explanation

To solve the problem, we need to find the electromotive force (emf) induced between the ends of a metallic meter stick that is translating in a magnetic field. We will use the formula for induced emf in a moving conductor in a magnetic field.

### Step-by-Step Solution:

1. **Identify the given values:**
- Magnetic field strength, \( B = 0.1 \, \text{T} \)
- Speed of translation, \( v = 0.2 \, \text{m/s} \)
- Length of the rod, \( L = 1 \, \text{m} \) (since it is a meter stick)
- Angle with the direction of motion, \( \theta = 60^\circ \)

2. **Recall the formula for induced emf:**
The induced emf (\( E \)) in a rod moving in a magnetic field is given by:
\[ E = B \cdot v \cdot L \cdot \sin(\theta) \]

3. **Substitute the known values into the formula:**
\[ E = 0.1 \, \text{T} \cdot 0.2 \, \text{m/s} \cdot 1 \, \text{m} \cdot \sin(60^\circ) \]

4. **Calculate \( \sin(60^\circ) \):**
\[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \]

5. **Substitute \( \sin(60^\circ) \) into the equation:**
\[ E = 0.1 \cdot 0.2 \cdot 1 \cdot \frac{\sqrt{3}}{2} \]

6. **Perform the multiplication:**
\[ E = 0.1 \cdot 0.2 \cdot \frac{\sqrt{3}}{2} = 0.02 \cdot \frac{\sqrt{3}}{2} \]

7. **Calculate the final value:**
\[ E = \frac{0.02\sqrt{3}}{2} = 0.01\sqrt{3} \, \text{V} \]

8. **Approximate \( \sqrt{3} \):**
\[ \sqrt{3} \approx 1.732 \]
\[ E \approx 0.01 \cdot 1.732 = 0.01732 \, \text{V} \]

Thus, the induced emf between the ends of the rod is approximately:
\[ E \approx 0.01732 \, \text{V} \text{ or } 1.732 \times 10^{-2} \, \text{V} \]

### Final Answer:
The induced emf is approximately \( 1.732 \times 10^{-2} \, \text{V} \).