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JEE Main 2021 (26 Feb, Shift 2) · Motion of Charges in an Electric Field

Motion of Charges in an Electric Field question from JEE Main 2021 (26 Feb, Shift 2)

Q28JEE Main 2021 · 26 FEB · SHIFT 2Medium

Motion of Charges in an Electric Field

An inclined plane making an angle of 30∘^\circ with the horizontal is placed in a uniform horizontal electric field 200 NC200\,\frac{N}{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∘=12\sin 30^\circ = \frac{1}{2}; cos⁡30∘=32\cos 30^\circ = \frac{\sqrt{3}}{2}]

Figure

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Answer: (c)

30°mgqENf = 0.2Nslides down
  1. qE=5×10−3×200=1 NqE = 5 \times 10^{-3} \times 200 = 1\text{ N} (horizontal, towards the incline); mg=9.8 Nmg = 9.8\text{ N}.
  2. Normal reaction: N=mgcos⁡30∘+qEsin⁡30∘=8.49+0.5=8.99 NN = mg\cos 30^\circ + qE\sin 30^\circ = 8.49 + 0.5 = 8.99\text{ N}, so friction =0.2N=1.80 N= 0.2N = 1.80\text{ N}.
  3. Net force down the slope: mgsin⁡30∘−qEcos⁡30∘−0.2N=4.9−0.87−1.80=2.24 Nmg\sin 30^\circ - qE\cos 30^\circ - 0.2N = 4.9 - 0.87 - 1.80 = 2.24\text{ N}, so a=2.24 m/s2a = 2.24\text{ m/s}^2.
  4. Slope length s=1sin⁡30∘=2 ms = \frac{1}{\sin 30^\circ} = 2\text{ m}; t=2sa=42.24≈1.3 st = \sqrt{\frac{2s}{a}} = \sqrt{\frac{4}{2.24}} \approx 1.3\text{ s}
  5. So the answer is (c) 1.3 s.
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