1
JEE Main 2017 (Online) 8th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
JEE Main 2017 (Online) 8th April Morning Slot Physics - Current Electricity Question 317 English
A 9 V battery with internal resistance of 0.5 $$\Omega $$ is connected across an infinite network as shown in the figure. All ammeters A1, A2, 3 and voltmeter V are ideal.
Choose correct statement.
A
Reading of A1 is 2 A
B
Reading of A1 is 18 A
C
Reading of   V   is  9 V
D
Reading of   V   is   7 V
2
JEE Main 2017 (Online) 8th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The V-I characteristic of a diode is shown in the figure. The ratio of forward to reverse bias resistance is :

JEE Main 2017 (Online) 8th April Morning Slot Physics - Semiconductor Question 202 English
A
10
B
10$$-$$6
C
106
D
100
3
JEE Main 2017 (Online) 8th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A small circular loop of wire of radius a is located at the centre of a much larger circular wire loop of radius b. The two loops are in the same plane. The outer loop of radius b carries an alternating current I = Io cos ($$\omega $$t). The emf induced in the smaller inner loop is nearly :
A
$${{\pi {\mu _o}{I_o}} \over 2}.{{{a^2}} \over b}\,\omega \sin \left( {\omega t} \right)$$
B
$${{\pi {\mu _o}{I_o}} \over 2}.{{{a^2}} \over b}\,\omega \cos \left( {\omega t} \right)$$
C
$$\pi {\mu _o}{I_o}\,{{{a^2}} \over b}\omega \sin \left( {\omega t} \right)$$
D
$${{\pi {\mu _o}{I_o}\,{b^2}} \over a}\omega \cos \left( {\omega t} \right)$$
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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