1
JEE Main 2019 (Online) 12th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The figure shows a square loop L of side 5 cm which is connected to a network of resistances. The whole setup is moving towards right with a constant speed of 1 cm s-1. At some instant, a part of L is in a uniform magnetic field of 1 T, perpendicular to the plane of the loop. If the resistance of L is 1.7 $$\Omega $$, the current in the loop at that instant will be close to : JEE Main 2019 (Online) 12th April Morning Slot Physics - Electromagnetic Induction Question 87 English
A
115 $$\mu $$A
B
170 $$\mu $$A
C
60 $$\mu $$A
D
150 $$\mu $$A
2
JEE Main 2019 (Online) 12th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
At 40o C, a brass wire of 1 mm radius is hung from the ceiling. A small mass, M is hung from the free end of the wire. When the wire is cooled down from 40oC to 20oC it regains its original length of 0.2 m. The value of M is close to : (Coefficient of linear expansion and Young’s modulus of brass are 10–5 /oC and 1011 N/m 2 , respectively; g= 10 ms–2 )
A
1.5 kg
B
0.5 kg
C
9 kg
D
0.9 kg
3
JEE Main 2019 (Online) 12th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
To verify Ohm's law, a student connects the voltmeter across the battery as, shown in the figure. The measured voltage is plotted as a function of the current, and the following graph is obtained : JEE Main 2019 (Online) 12th April Morning Slot Physics - Current Electricity Question 220 English
If V0 is almost zero, identify the correct statement :
A
The value of the resistance R is 1.5 $$\Omega $$
B
The emf of the battery is l.5 V and its internal resistance is 1.5 $$\Omega $$
C
The emf of the battery is l.5 V and the value of R is 1.5 $$\Omega $$
D
The potential difference across the battery is 1.5 V when it sends a current of 1000 mA
4
JEE Main 2019 (Online) 12th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
An electromagnetic wave is represented by the electric field $$\overrightarrow E = {E_0}\widehat n\sin \left[ {\omega t + \left( {6y - 8z} \right)} \right]$$ . Taking unit vectors in x, y and z directions to be $$\widehat i,\widehat j,\widehat k$$ , the direction of propagation $$\widehat s$$, is :
A
$$\widehat s = {{3\widehat i - 4\widehat j} \over 5}$$
B
$$\widehat s = {{ - 4\widehat k + 3\widehat j} \over 5}$$
C
$$\widehat s = \left( {{{ - 3\widehat j + 4\widehat k} \over 5}} \right)$$
D
$$\widehat s = {{4\widehat j - 3\widehat k} \over 5}$$
JEE Main Papers
2023
2021
EXAM MAP
Medical
NEETAIIMS
Graduate Aptitude Test in Engineering
GATE CSEGATE ECEGATE EEGATE MEGATE CEGATE PIGATE IN
Civil Services
UPSC Civil Service
Defence
NDA
Staff Selection Commission
SSC CGL Tier I
CBSE
Class 12