A ball of mass $m$ with a charge $q$ can rotate in a vertical plane at the end of a string of length l = 3.6 m in a uniform electrostatic field whose lines of force are directed upwards. What horizontal velocity, in $m s^{- 1}$ , must be imparted to the ball in the upper position so that the tension in the string in the lower position of the ball is 15 times than the weight of the ball? Given $E = \frac{6 m g}{5 q} a n d g = 10 m s^{- 2}$

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answer is 24.

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Detailed Solution

The Corresponding situation is shown in figure

Here by using work energy theorem between position B and A we have

$\frac{1}{2} m u^{2} + 2 m g l - 2 q E l = \frac{1}{2} m v^{2}$ ...............(1)

At position A we have $T - q E = \frac{m v^{2}}{l} + m g$

Since $T = 15 m g$

$\Rightarrow q E l = m v^{2} - 14 m g l {∵ a t t h e l o w e s t p o int, T = 15 m g}$

From equation (1), we have $q E l + 14 m g l = m u^{2} + 4 m g l - 4 q E l$

$\Rightarrow u^{2} = \frac{5 q E}{m} + \frac{10 m g}{m}$

$\Rightarrow u = \sqrt{\frac{5 q E}{m} + 10 g l}$

$\Rightarrow u = \sqrt{\frac{5 l}{m} (q E + 2 m g)}$

$\Rightarrow u = \sqrt{\frac{5 l}{m} (\frac{6 m g}{5} + 2 m g)} = \sqrt{\frac{5 l}{m} (\frac{16 m g}{5})}$

$\Rightarrow u = 4 \sqrt{g l} = 4 \sqrt{(10) (3.6)} = (4) (6) = 24 m s^{- 1}$

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