First slide

Electric field

Question

The bob of a simple pendulum has a mass of 40 g and a positive charge of 4x10^-6 C. It makes 20 oscillation in 45 s. A vertical electric field pointing upward and of magnitude 2.5x10⁴ NC^-1 is switched on. How much time, in seconds will it now take to complete 20 oscillations? $Take g = 10 m / s^{2}$

Difficult

Solution

$Initially T = 2 π \sqrt{\frac{L}{g}} = \frac{45}{20} s$ ……………………..(1)
$Finally T^{'} = 2 π \sqrt{\frac{L}{(g - \frac{qE}{m})}}$ ……………………………….(2)

Dividing equation (1) and (2), we get
$\frac{T}{T^{'}} = \sqrt{\frac{g - \frac{qE}{m}}{g}}$
$\Rightarrow \frac{T}{T^{'}} = \sqrt{1 - \frac{q E}{m g}} = \sqrt{1 - \frac{4 \times 10^{- 6} \times 2.5 \times 10^{4}}{40 \times 10^{- 3} \times 10}} \Rightarrow T^{'} = 2.598 s$
For 20 oscillation, time taken is t = 20x2.598
$\Rightarrow$ t = 51.96 s

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