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Question

In the uniform electric field of E=1×104NC1, an electron is accelerated from rest. The velocity of the electron when it has travelled a distance of 2×102m is nearly ________ ms1 (em of electron 1.8×1011Ckg1)

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 8.5×106
Given : Distance covered S=2×102 m
Acceleration of the electron a=eEm
a=1.8×1011×104=1.8×1015 m/s2
Initial speed of the electron is zero i.e. u=0
Using v2u2=2aS
v20=2(1.8×1015)(2×102) v=8.5×106 m/s

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