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JEE MAIN 2021
26.02.2021_S2
Question
An inclined plane making an angle of $30^{\circ}$ with the horizontal is placed in a uniform horizontal electric field $200 \frac{\mathrm{~N}}{\mathrm{C}}$ as shown in the figure. A body of mass 1 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 bottc $$ \left[\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2, \sin 30^{\circ}=\frac{1}{2} ; \cos 30^{\circ}=\frac{\sqrt{3}}{2}\right] $$
Question Image
Select the correct option:
A
0.92 s
B
0.46 s
C
2.3 s
D
1.3 s
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Solution Image
Question Tags
JEE Main
Physics
Easy
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