1
JEE Main 2018 (Online) 15th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Truth table for the following digital circuit will be :
JEE Main 2018 (Online) 15th April Evening Slot Physics - Semiconductor Question 166 English
A
JEE Main 2018 (Online) 15th April Evening Slot Physics - Semiconductor Question 166 English Option 1
B
JEE Main 2018 (Online) 15th April Evening Slot Physics - Semiconductor Question 166 English Option 2
C
JEE Main 2018 (Online) 15th April Evening Slot Physics - Semiconductor Question 166 English Option 3
D
JEE Main 2018 (Online) 15th April Evening Slot Physics - Semiconductor Question 166 English Option 4
2
JEE Main 2018 (Online) 15th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A plane polarized light is incident on a polariser with its pass axis aking angle $$\theta $$ with x-axis, as shown in the figure. At four different values of $$\theta ,\,\theta $$ = 8o, 38o, 188o and 218o, the observed intensities are same.
What is the angle between the direction of polarization and x-axis ?
JEE Main 2018 (Online) 15th April Evening Slot Physics - Wave Optics Question 111 English
A
98o
B
128o
C
203o
D
45o
3
JEE Main 2018 (Online) 15th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
JEE Main 2018 (Online) 15th April Evening Slot Physics - Electromagnetic Induction Question 104 English
A copper rod of mass m slides under gravity on two smooth parallel rails, with separation l and set at an angle of $$\theta $$ with the horizontal. At the bottom rails are joined by a resistance R. There is a uniform magnetic field B normal to the plane of the rails, as shown in the igure. The terminal speed of the copper rod is :
A
$${{mg\,R\,\tan \,\theta } \over {{B^2}\,{l^2}}}$$
B
$${{mg\,R\,\cot \,\theta } \over {{B^2}\,{l^2}}}$$
C
$${{mg\,R\,\sin \,\theta } \over {{B^2}\,{l^2}}}$$
D
$${{mg\,R\,\cos \,\theta } \over {{B^2}\,{l^2}}}$$
4
JEE Main 2018 (Online) 15th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A capacitor C1 = 1.0 $$\mu $$F is charged up to a voltage V = 60 V by connecting it to battery B through switch (1). Now C1 is disconnected from battery and connected to a circuit consisting of two uncharged capacitors $${C_2} = 3.0\mu F$$ and C3 = 6.0 $$\mu $$F through switch (2), as shown in the figure. The sum of final charges on C2 and C3 is :
JEE Main 2018 (Online) 15th April Evening Slot Physics - Capacitor Question 121 English
A
40 $$\mu $$C
B
36 $$\mu $$C
C
20 $$\mu $$C
D
54 $$\mu $$C
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