A small bob tied at one end of a thin string of length 1m is describing a vertical circle so that the maximum and minimum tension in the string are in the ratio 5: 1. The velocity of the bob at the height position is ____ m/s. (Take `g=10m//s^(2))`

Answer

To solve the problem, we need to find the velocity of the bob at the highest point of its vertical circular motion, given that the maximum and minimum tensions in the string are in the ratio of 5:1.

### Step-by-Step Solution:

1. **Identify the Forces at the Highest and Lowest Points**:
- At the highest point (Point B), the forces acting on the bob are its weight (mg) acting downwards and the tension (T_min) in the string also acting downwards.
- At the lowest point (Point A), the forces are the weight (mg) acting downwards and the tension (T_max) acting upwards.

2. **Write the Equations for Tension**:
- At the highest point (B):
\[ T_{min} + mg = \frac{mv_2^2}{r} \]
Rearranging gives:
\[ T_{min} = \frac{mv_2^2}{r} - mg \quad \text{(Equation 1)} \]
- At the lowest point (A):
\[ T_{max} - mg = \frac{mv_1^2}{r} \]
Rearranging gives:
\[ T_{max} = \frac{mv_1^2}{r} + mg \quad \text{(Equation 2)} \]

3. **Use the Ratio of Tensions**:
- Given that \( T_{max} : T_{min} = 5 : 1 \), we can write:
\[ T_{max} = 5T_{min} \]
- Substitute the expressions from Equations 1 and 2 into this ratio:
\[ \frac{\frac{mv_1^2}{r} + mg}{\frac{mv_2^2}{r} - mg} = 5 \]

4. **Cross Multiply and Simplify**:
- Cross multiplying gives:
\[ \frac{mv_1^2}{r} + mg = 5\left(\frac{mv_2^2}{r} - mg\right) \]
- Expanding and rearranging:
\[ \frac{mv_1^2}{r} + mg = \frac{5mv_2^2}{r} - 5mg \]
\[ \frac{mv_1^2}{r} + 6mg = \frac{5mv_2^2}{r} \]

5. **Solve for v1**:
- Rearranging gives:
\[ mv_1^2 + 6mgr = 5mv_2^2 \]
- Dividing through by m:
\[ v_1^2 + 6gr = 5v_2^2 \]

6. **Use Conservation of Energy**:
- The total mechanical energy at the lowest point (A) and the highest point (B) must be equal:
\[ \frac{1}{2}mv_1^2 = \frac{1}{2}mv_2^2 + mg(2r) \]
- Simplifying gives:
\[ v_1^2 = v_2^2 + 4gr \]

7. **Substitute v1^2 into the Tension Equation**:
- Substitute \( v_1^2 \) from the energy equation into the tension equation:
\[ (v_2^2 + 4gr) + 6gr = 5v_2^2 \]
- This simplifies to:
\[ 10gr = 4v_2^2 \]
- Rearranging gives:
\[ v_2^2 = \frac{10gr}{4} = \frac{5gr}{2} \]

8. **Substitute g and r**:
- Given \( g = 10 \, \text{m/s}^2 \) and \( r = 1 \, \text{m} \):
\[ v_2^2 = \frac{5 \times 10 \times 1}{2} = 25 \]
- Therefore, \( v_2 = \sqrt{25} = 5 \, \text{m/s} \).

### Final Answer:
The velocity of the bob at the highest position is **5 m/s**.