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Question

There is a uniform electric field of strength $$10^{3}V/m$$ along $$y-axis$$. A body of mass $$1 g$$ and charge $$10^{-6}C$$ is projected into the field from origin along the positive $$x-axis$$ with a velocity $$10 m/s$$. Its speed in m/s after $$10s$$ is (neglect gravitation) :



A
10
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B
52
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C
102
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D
20
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Solution

The correct option is C $$10\sqrt{2}$$
Velocity in $$x$$ direction remains constant i.e, $$v_{x}=10\ m/s$$
Now acceleration is only in y direction
$$a=\dfrac{qE}{m}$$
$$a=\dfrac{10^{3}\times 10^{-6}}{1\times 10^{-3}}$$

$$\therefore$$ Velocity in y direction $$v_{y}=a\times t$$
$$v_{y}=10m/s$$
$$\therefore v=10\hat{x}+10\hat{y}$$

$$\left | v \right |=10\sqrt{2}$$

Physics
NCERT
Standard XII

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