JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2021
- Hard
- MCQ
An inclined plane making an angle of \(30^{\circ}\) with the horizontal is placed in a uniform horizontal electric field \(200 \, \frac{ N }{ C }\) 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 (in \(s\) )taken by the body to reach the bottom. \(\left[ g =9.8 \,m / s ^{2}, \sin 30^{\circ}=\frac{1}{2}\right.\); \(\left.\cos 30^{\circ}=\frac{\sqrt{3}}{2}\right]\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(1.3\)
Step-by-step Solution
Detailed explanation
\(FBD\)
here \(N =9.8 \cos 30+1 \sin 30\)
\(\approx 9 \,N\)
so \(a=\frac{9.8 \sin 30-1 \cos 30-\mu \,N}{1}\)
\(a=2.233\, m / s ^{2}\)
By \(S=u t+\frac{1}{2} a t^{2}\)
\(=\frac{1}{2}(2.233) t ^{2}\)
\(\sin 30^{\circ}\)
\(t \approx 1.3\, sec\)
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