1
JEE Main 2019 (Online) 9th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Expression for time in terms of G(universal gravitional constant), h (Planck constant) and c (speed of light) is proportional to :
A
$$\sqrt {{{h{c^5}} \over G}} $$
B
$$\sqrt {{{{c^3}} \over {Gh}}} $$
C
$$\sqrt {{{Gh} \over {{c^5}}}} $$
D
$$\sqrt {{{Gh} \over {{c^3}}}} $$
2
JEE Main 2019 (Online) 9th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Two point charges q1$$\left( {\sqrt {10} \mu C} \right)$$ and q2($$-$$ 25 $$\mu $$C) are placed on the x-axis at x = 1 m and x = 4 m respectively. The electric field (in V/m) at a point y = 3 m on y-axis is,
[take $${1 \over {4\pi { \in _0}}}$$ = 9 $$ \times $$ 109 Nm2C$$-$$2]
A
$$\left( {63\widehat i - 27\widehat j} \right) \times {10^2}$$
B
$$\left( { - 63\widehat i + 27\widehat j} \right) \times {10^2}$$
C
$$\left( {81\widehat i - 81\widehat j} \right) \times {10^2}$$
D
$$\left( { - 81\widehat i + 81\widehat j} \right) \times {10^2}$$
3
JEE Main 2019 (Online) 9th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A particle having the same charge as of electron moves in a ciurcular path of radius 0.5 cm under the influence of a magnetic field 0f 0.5 T. If an electric field of 100 V/m makes it to move in a straight path, then the mass of the particle is (Given charge of electron = 1.6 $$ \times $$ 10$$-$$19C)
A
9.1 $$ \times $$ 10$$-$$31 kg
B
1.6 $$ \times $$ 10$$-$$27 kg
C
1.6 $$ \times $$ 10$$-$$19 kg
D
2.0 $$ \times $$ 10$$-$$24 kg
4
JEE Main 2019 (Online) 9th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The energy associated with electric field is (UE) and with magnetic field is (UB) for an electromagnetic wave in free space. Then :
A
$${U_E} = {{{U_B}} \over 2}$$
B
$${U_E} > {U_B}$$
C
$${U_E} < {U_B}$$
D
$${U_E} = {U_B}$$
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