A small bob tied at one end of a thin string of length 1 m is describing a vertical circle so that the maximum and minimum tension in the string is in the ratio 5 : 1. The velocity of the bob at the highest position is ______ m s$$^{-1}$$. (Take $$g = 10$$ m s$$^{-2}$$)

For a bob of mass $$m$$ tied to a string of length $$L = 1$$ m describing a vertical circle, the tension at any point depends on the speed and the component of gravity along the string.

At the lowest point, the tension is maximum: $$T_{max} = \frac{mv_{bot}^2}{L} + mg$$.

At the highest point, the tension is minimum: $$T_{min} = \frac{mv_{top}^2}{L} - mg$$.

Using energy conservation between the top and bottom of the circle (height difference $$= 2L$$):

$$\frac{1}{2}mv_{bot}^2 = \frac{1}{2}mv_{top}^2 + mg(2L)$$

$$v_{bot}^2 = v_{top}^2 + 4gL$$

Given $$\frac{T_{max}}{T_{min}} = 5$$:

$$\frac{\frac{mv_{bot}^2}{L} + mg}{\frac{mv_{top}^2}{L} - mg} = 5$$

$$\frac{v_{bot}^2 + gL}{v_{top}^2 - gL} = 5$$

Substituting $$v_{bot}^2 = v_{top}^2 + 4gL$$:

$$\frac{v_{top}^2 + 4gL + gL}{v_{top}^2 - gL} = 5$$

$$v_{top}^2 + 5gL = 5v_{top}^2 - 5gL$$

$$10gL = 4v_{top}^2$$

$$v_{top}^2 = \frac{10 \times 10 \times 1}{4} = 25$$

$$v_{top} = 5$$ m s$$^{-1}$$

The velocity of the bob at the highest position is $$5$$ m s$$^{-1}$$.