1
JEE Main 2018 (Online) 15th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A charge $$Q$$ is placed at a distance $$a/2$$ above the center of the square surface of edge a as shown in the figure.

JEE Main 2018 (Online) 15th April Morning Slot Physics - Electrostatics Question 177 English
The electric flux through the square surface is
A
$${Q \over {{ \in _0}}}$$
B
$${Q \over {2{ \in _0}}}$$
C
$${Q \over {3{ \in _0}}}$$
D
$${Q \over {6{ \in _0}}}$$
2
JEE Main 2018 (Online) 15th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The velocity-time graphs of a car and a scooter are shown in the figure. (i) The difference between the distance travelled by the car and the scooter in $$15$$ $$s$$ and (ii) the time at which the car will catch up with the scooter are, respectively.

JEE Main 2018 (Online) 15th April Morning Slot Physics - Motion in a Straight Line Question 89 English
A
$$112.5$$ $$m$$ and $$22.5$$ $$s$$
B
$$337.5$$ $$m$$ and $$25$$ $$s$$
C
$$112.5$$ $$m$$ and $$15$$ $$s$$
D
$$225.5$$ $$m$$ and $$10$$ $$s$$
3
JEE Main 2018 (Online) 15th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
In a screw gauge, $$5$$ complete rotations of the screw cause it to move a linear distance of $$0.25$$ $$cm.$$ There are $$100$$ circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of $$4$$ main scale divisions and $$30$$ circular scale divisions. Assuming negligible zero error, the thickness of the wire is :
A
$$0.4300$$ $$cm$$
B
$$0.2150$$ $$cm$$
C
$$0.3150$$ $$cm$$
D
$$0.0430$$ $$cm$$
4
JEE Main 2018 (Online) 15th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Two electrons are moving with non-relativistic speed perpendicular to each other. If corresponding de Broglie wavelength are $${\lambda _1}$$ and $${\lambda _2},$$ their de Broglie wavelength in the frame of reference attached to their center of masses :
A
$${\lambda _{CM}} = {\lambda _1} = {\lambda _2}$$
B
$${\lambda _{CM}} = {{2{\lambda _1}{\lambda _2}} \over {\sqrt {\lambda _1^2 + \lambda _2^2} }}$$
C
$${1 \over {{\lambda _{CM}}}} = {1 \over {{\lambda _1}}} + {1 \over {{\lambda _2}}}$$
D
$${\lambda _{CM}} = \left( {{{{\lambda _1} + {\lambda _2}} \over 2}} \right)$$
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