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 $E = \frac{6 m g}{5 q} a n d g = 10 m s^{- 2}$

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

answer is 24.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

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 $q E l + 14 m g l = m u^{2} + 4 m g l - 4 q E l$

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

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

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

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

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

Watch 3-min video & get full concept clarity

No courses found

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test