1
JEE Main 2019 (Online) 10th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A ray of light AO in vacuum is incident on a glass slab at angle 60° and refracted at angle 30° along OB as shown in the figure. The optical path length of light ray from A to B is: JEE Main 2019 (Online) 10th April Morning Slot Physics - Geometrical Optics Question 180 English
A
2a + 2b/$$\sqrt 3$$
B
2a + 2b/3
C
2a + 2b
D
2$$\sqrt3$$/a + 2b
2
JEE Main 2019 (Online) 10th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A uniformly charged ring of radius 3a and total charge q is placed in xy-plane centred at origin. A point charge q is moving towards the ring along the z-axis and has speed u at z = 4a. The minimum value of u such that it crosses the origin is :
A
$$\sqrt {{2 \over m}} {\left( {{2 \over {15}}{{{q^2}} \over {4\pi {\varepsilon _0}a}}} \right)^{1/2}}$$
B
$$\sqrt {{2 \over m}} {\left( {{1 \over {15}}{{{q^2}} \over {4\pi {\varepsilon _0}a}}} \right)^{1/2}}$$
C
$$\sqrt {{2 \over m}} {\left( {{1 \over {5}}{{{q^2}} \over {4\pi {\varepsilon _0}a}}} \right)^{1/2}}$$
D
$$\sqrt {{2 \over m}} {\left( {{4 \over {15}}{{{q^2}} \over {4\pi {\varepsilon _0}a}}} \right)^{1/2}}$$
3
JEE Main 2019 (Online) 10th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Two wires A & B are carrying currents I1 & I2 as shown in the figure. The separation between them is d. A third wire C carrying a current I is to be kept parallel to them at a distance x from A such that the net force acting on it is zero. The possible values of x are : JEE Main 2019 (Online) 10th April Morning Slot Physics - Magnetic Effect of Current Question 160 English
A
$$x = \left( {{{{I_1}} \over {{I_1} + {I_2}}}} \right)d\,and\,x = \left( {{{{I_2}} \over {{I_1} - {I_2}}}} \right)d$$
B
$$x = \left( {{{{I_2}} \over {{I_1} + {I_2}}}} \right)d\,and\,x = \left( {{{{I_2}} \over {{I_1} - {I_2}}}} \right)d$$
C
$$x = \left( {{{{I_1}} \over {{I_1} - {I_2}}}} \right)d\,and\,x = \left( {{{{I_2}} \over {{I_1} + {I_2}}}} \right)d$$
D
$$x = \pm {{{I_1}d} \over {{I_1} - {I_2}}}$$
4
JEE Main 2019 (Online) 10th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A particle of mass m is moving along a trajectory given by
x = x0 + a cos$$\omega $$1t
y = y0 + b sin$$\omega $$2t
The torque, acting on the particle about the origin, at t = 0 is :
A
Zero
B
+my0a $$\omega _1^2$$$$\widehat k$$
C
$$ - m\left( {{x_0}b\omega _2^2 - {y_0}a\omega _1^2} \right)\widehat k$$
D
m (–x0b + y0a) $$\omega _1^2$$$$\widehat k$$
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