1
JEE Main 2020 (Online) 2nd September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A charged particle (mass m and charge q)
moves along X-axis with velocity V0. When it
passes through the origin it enters a region having uniform electric field
$$\overrightarrow E = - E\widehat j$$ which extends upto x = d.
Equation of path of electron in the region x > d is JEE Main 2020 (Online) 2nd September Morning Slot Physics - Electrostatics Question 138 English
A
y = $${{qEd} \over {mV_0^2}}\left( {x - d} \right)$$
B
y = $${{qEd} \over {mV_0^2}}\left( {{d \over 2} - x} \right)$$
C
y = $${{qEd} \over {mV_0^2}}x$$
D
y = $${{qE{d^2}} \over {mV_0^2}}x$$
2
JEE Main 2020 (Online) 2nd September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language
An amplitude modulated wave is represented by the expression
vm = 5(1 + 0.6 cos 6280t) sin(211 × 104 t) volts
The minimum and maximum amplitudes of the amplitude modulated wave are, respectively
A
$${3 \over 2}$$ V, 5 V
B
$${5 \over 2}$$ V, 8 V
C
5 V, 8 V
D
3 V, 5 V
3
JEE Main 2020 (Online) 2nd September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A plane electromagnetic wave, has
frequency of 2.0 $$ \times $$ 1010 Hz and its energy density is 1.02 $$ \times $$ 10–8 J/m3 in vacuum. The amplitude of the magnetic field of the wave is close to
( $${1 \over {4\pi {\varepsilon _0}}} = 9 \times {10^9}{{N{m^2}} \over {{C^2}}}$$ and speed of light
= 3 $$ \times $$ 108 ms–1)
A
190 nT
B
150 nT
C
160 nT
D
180 nT
4
JEE Main 2020 (Online) 2nd September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A uniform cylinder of mass M and radius R is to be pulled over a step of height a (a < R) by applying a force F at its centre ‘O’ perpendicular to the plane through the axes of the cylinder on the edge of the step (see figure). The minimum value of F required is : JEE Main 2020 (Online) 2nd September Morning Slot Physics - Rotational Motion Question 120 English
A
$$Mg\sqrt {1 - {{\left( {{{R - a} \over R}} \right)}^2}} $$
B
$$Mg\sqrt {1 - {{{a^2}} \over {{R^2}}}} $$
C
$$Mg{a \over R}$$
D
$$Mg\sqrt {{{\left( {{R \over {R - a}}} \right)}^2} - 1} $$
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