1
JEE Main 2018 (Online) 15th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A monochromatic beam of light has a frequency $$v = {3 \over {2\pi }} \times {10^{12}}Hz$$ and is propagating along the direction $${{\widehat i + \widehat j} \over {\sqrt 2 }}.$$
It is polarized along the $$\widehat k$$ direction. The acceptable form for the magnetic field is :
A
JEE Main 2018 (Online) 15th April Morning Slot Physics - Electromagnetic Waves Question 141 English Option 1
B
JEE Main 2018 (Online) 15th April Morning Slot Physics - Electromagnetic Waves Question 141 English Option 2
C
JEE Main 2018 (Online) 15th April Morning Slot Physics - Electromagnetic Waves Question 141 English Option 3
D
JEE Main 2018 (Online) 15th April Morning Slot Physics - Electromagnetic Waves Question 141 English Option 4
2
JEE Main 2018 (Online) 15th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A body of mass $$M$$ and charge $$q$$ is connected to spring of spring constant $$k.$$ It is oscillating along $$x$$-direction about its equilibrium position, taken to be at $$x=0,$$ with an amplitude $$A$$. An electric field $$E$$ is applied along the $$x$$-direction. Which of the following statements is correct ?
A
The new equilibrium position is at a distance $${{qE} \over {2k}}$$ from $$x=0.$$
B
The total energy of the system is $${1 \over 2}m{\omega ^2}{A^2} + {1 \over 2}{{{q^2}{E^2}} \over k}.$$
C
The total energy of the system is $${1 \over 2}m{\omega ^2}{A^2} - {1 \over 2}{{{q^2}{E^2}} \over k}.$$
D
The new equilibrium position is at a distance $${{2qE} \over k}$$ from $$x=0.$$
3
JEE Main 2018 (Online) 15th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
One mole of an ideal monoatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, $${27^ \circ }C.$$ The work done on the gas will be :
A
$$300$$ $$R$$
B
$$300$$ $$R$$ $$ln$$ $$6$$
C
$$300$$ $$R$$ $$ln$$ $$2$$
D
$$300$$ $$R$$ $$ln$$ $$7$$
4
JEE Main 2018 (Online) 15th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A thin uniform tube is bent into a circle of radius $$r$$ in the vertical plane. Equal volumes of two immiscible liquids, whose densities are $${\rho _1}$$ and $${\rho _2}$$ $$\left( {{\rho _1} > {\rho _2}} \right),$$ fill half the circle. The angle $$\theta $$ between the radius vector passing through the common interface and the vertical is :
A
$$\theta = {\tan ^{ - 1}}\pi \left( {{{{\rho _1}} \over {{\rho _2}}}} \right)$$
B
$$\theta = {\tan ^{ - 1}}{\pi \over 2}\left( {{{{\rho _1}} \over {{\rho _2}}}} \right)$$
C
$$\theta = {\tan ^{ - 1}}\left( {{{{\rho _1} - {\rho _2}} \over {{\rho _1} + {\rho _2}}}} \right)$$
D
$$\theta = {\tan ^{ - 1}}{\pi \over 2}\left( {{{{\rho _1} + {\rho _2}} \over {{\rho _1} - {\rho _2}}}} \right)$$
JEE Main Papers
2023
2021
EXAM MAP