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1
JEE Main 2018 (Online) 16th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
In the given circuit, the current through zener diode is :

JEE Main 2018 (Online) 16th April Morning Slot Physics - Semiconductor Question 204 English
A
5.5 mA
B
6.7 mA
C
2.5 mA
D
3.3 mA
2
JEE Main 2018 (Online) 16th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
In a circuit for finding the resistance of a galvanometer by half deflection method, a 6 V battery and a high resistance of 11 k$$\Omega $$ are used. The figure of merit of the galvanometer is 60 $$\mu A/$$division. In the absence of shunt resistance, the galvanometer produces a deflection of $$\theta $$ = 9 divisions when current flows in the circuit. The value of the shunt resistance that can cause the deflection of $$\theta /2,$$ is closest to :
A
500 $$\Omega $$
B
220 $$\Omega $$
C
55 $$\Omega $$
D
110 $$\Omega $$
3
JEE Main 2018 (Online) 16th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A power transmission line feeds input power at 2300 V to a step down transformer with its primary windings having 4000 turns, giving the output power at 230 V. If the current in the primary of the transformer is 5 A, and its efficiency is 90%, the output current would be ;
A
50 A
B
45 A
C
25 A
D
20 A
4
JEE Main 2018 (Online) 16th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Unpolarized light of intensity I is incident on a system of two polarizers, A followed by B. The intensity of emergent light is I/2. If a third polarizer C is placed between A and B, the intensity of emergent light is reduced to I/3. The angle between the polarizers A and C is $$\theta $$. Then :
A
cos$$\theta $$ = $${\left( {{2 \over 3}} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}$$
B
cos$$\theta $$ = $${\left( {{2 \over 3}} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 4$}}}}$$
C
cos$$\theta $$ = $${\left( {{1 \over 3}} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}$$
D
cos$$\theta $$ = $${\left( {{1 \over 3}} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 4$}}}}$$

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