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1
JEE Main 2020 (Online) 7th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The electric field of a plane electromagnetic wave is given by
$$\overrightarrow E = {E_0}{{\widehat i + \widehat j} \over {\sqrt 2 }}\cos \left( {kz + \omega t} \right)$$

At t = 0, a positively charged particle is at the point (x, y, z) = $$\left( {0,0,{\pi \over k}} \right)$$.
If its instantaneous velocity at (t = 0) is $${v_0}\widehat k$$ , the force acting on it due to the wave is :
A
parallel to $$\widehat k$$
B
parallel to $${{\widehat i + \widehat j} \over {\sqrt 2 }}$$
C
antiparallel to $${{\widehat i + \widehat j} \over {\sqrt 2 }}$$
D
zero
2
JEE Main 2020 (Online) 7th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A mass of 10 kg is suspended by a rope of length 4 m, from the ceiling. A force F is applied horizontally at the mid point of the rope such that the top half of the rope makes an angle of 45o with the vertical. Then F equal : (Take g = 10 ms–2 and the rope to be massless)
A
100 N
B
75 N
C
90 N
D
70 N
3
JEE Main 2020 (Online) 7th January Evening Slot
Numerical
+4
-0
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The sum of two forces $$\overrightarrow P $$ and $$\overrightarrow Q $$ is $$\overrightarrow R $$ such that $$\left| {\overrightarrow R } \right| = \left| {\overrightarrow P } \right|$$ . The angle $$\theta $$ (in degrees) that the resultant of 2$${\overrightarrow P }$$ and $${\overrightarrow Q }$$ will make with $${\overrightarrow Q }$$ is , ..............
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4
JEE Main 2020 (Online) 7th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
In a Young's double slit experiment, the separation between the slits is 0.15 mm. in the experiment, a source of light of wavelengh 589 nm is used and the interference pattern is observed on a screen kept 1.5 m away. The separation between the successive bright fringes on the screen is :
A
4.9 mm
B
5.9 mm
C
6.9 mm
D
3.9 mm

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