JEE MAIN - Physics (2021 - 26th February Evening Shift - No. 14)
An inclined plane making an angle of 30$$^\circ$$ with the horizontal is placed in a uniform horizontal electric field $$200{N \over C}$$ as shown in the figure. A body of mass 1 kg and charge 5 mC is allowed to slide down from rest at a height of 1 m. If the coefficient of friction is 0.2, find the time taken by the body to reach the bottom.
[g = 9.8 m/s2; $$\sin 30^\circ = {1 \over 2}$$; $$\cos 30^\circ = {{\sqrt 3 } \over 2}$$]
_26th_February_Evening_Shift_en_14_1.png)
[g = 9.8 m/s2; $$\sin 30^\circ = {1 \over 2}$$; $$\cos 30^\circ = {{\sqrt 3 } \over 2}$$]
0.46 s
0.92 s
1.3 s
2.3 s
Explanation
_26th_February_Evening_Shift_en_14_2.png)
f = mg sin$$\theta$$ $$-$$ ($$\mu$$N + qE cos$$\theta$$)
F = mg sin$$\theta$$ $$-$$ $$\mu$$(mg cos$$\theta$$ + qE sin$$\theta$$) $$-$$ qE cos$$\theta$$
F = 1 $$\times$$ 10 $$\times$$ sin30
$$-$$ 0.2 (1 $$\times$$ 10 $$\times$$ cos30$$^\circ$$ + 200 $$\times$$ 5 $$\times$$ 10$$-$$3 sin30$$^\circ$$)
$$-$$ 200 $$\times$$ 5 $$\times$$ 10$$-$$3 cos30$$^\circ$$
$$ \Rightarrow $$ F = 2.3 N
$$a = {F \over m} = {{2.3} \over 1} = 2.3$$ m/sec2
$$t = \sqrt {{{25} \over 9}} = \sqrt {{{2 \times 2} \over {2.3}}} = 1.3$$ sec
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