1
JEE Main 2017 (Online) 8th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
There is a uniform electrostatic field in a region. The potential at various points on a small sphere centred at $$P,$$ in the region, is found to vary between the limits 589.0 V to 589.8 V. What is the potential at a point on the sphere whose radius vector makes an angle of 60o with the direction of the field ?
A
589.5 V
B
589.2 V
C
589.4 V
D
589.6 V
2
JEE Main 2017 (Online) 8th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Let the refractive index of a denser medium with respect to a rarer medium be n12 and its critical angle be θC . At an angle of incidence A when light is travelling from denser medium to rarer medium, a part of the light is reflected and the rest is refracted and the angle between reflected and refracted rays is 90o. Angle A is given by :
A
$${1 \over {{{\cos }^{ - 1}}\left( {\sin {\theta _C}} \right)}}$$
B
$${1 \over {{{\tan }^{ - 1}}\left( {\sin {\theta _C}} \right)}}$$
C
$${\cos ^{ - 1}}\,\left( {\sin {\theta _C}} \right)$$
D
$${\tan ^{ - 1}}\,\left( {\sin {\theta _C}} \right)$$
3
JEE Main 2017 (Online) 8th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A single slit of width b is illuminated by a coherent monochromatic light of wavelength $$\lambda $$. If the second and fourthminima in the diffraction pattern at a distance 1 m from the slit are at 3 cm and 6 cm respectively from the central maximum, what is the width of the central maximum ? (i.e. distance between first minimum on either side of the central maximum)
A
1.5 cm
B
3.0 cm
C
4.5 cm
D
6.0 cm
4
JEE Main 2017 (Online) 8th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Magnetic field in a plane electromagnetic wave is given by

$$\overrightarrow B $$ = B0 sin (k x + $$\omega $$t) $$\widehat j\,T$$

Expression for corresponding electric field will be :
Where c is speed of light.
A
$$\overrightarrow E $$ = B0 c sin (k x + $$\omega $$t) $$\widehat k$$ V/m
B
$$\overrightarrow E $$ = $${{{B_0}} \over c}$$ sin (k x + $$\omega $$t) $$\widehat k$$ V/m
C
$$\overrightarrow E $$ = $$-$$ B0 c sin (kx +$$\omega $$t) $$\widehat k$$ V/m
D
$$\overrightarrow E $$ = B0 c sin (kx $$-$$$$\omega $$t) $$\widehat k$$ V/m
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