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Question

In the uniform electric field of $$E = 1 \times {10}^{4} N{C}^{-1}$$, an electron is accelerated from rest. The velocity of the electron when it has travelled a distance of $$2 \times {10}^{-2} m$$ is nearly ________ $$m{s}^{-1}$$  $$(\dfrac{e}{m}$$ of electron $$\simeq 1.8 \times {10}^{11} C {kg}^{-1}$$)


A
1.6×106
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B
0.85×106
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C
0.425×106
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D
8.5×106
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Solution

The correct option is D $$8.5 \times {10}^{6}$$
Given : Distance covered   $$S = 2\times 10^{-2}$$ m
Acceleration of the electron    $$a = \dfrac{eE}{m} $$
$$\therefore$$  $$a = 1.8\times 10^{11}\times 10^4 = 1.8\times 10^{15}$$  $$m/s^2$$
Initial speed of the electron is zero  i.e. $$u = 0$$
Using    $$v^2  -u^2 = 2aS$$
$$\therefore$$    $$v^2 - 0 = 2(1.8\times 10^{15})(2\times 10^{-2})$$          $$\implies v = 8.5\times 10^{6}$$  $$m/s$$

Physics

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