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JEE Main Physics Test- 5 - JEE MCQ


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25 Questions MCQ Test Mock Tests for JEE Main and Advanced 2024 - JEE Main Physics Test- 5

JEE Main Physics Test- 5 for JEE 2024 is part of Mock Tests for JEE Main and Advanced 2024 preparation. The JEE Main Physics Test- 5 questions and answers have been prepared according to the JEE exam syllabus.The JEE Main Physics Test- 5 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Main Physics Test- 5 below.
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JEE Main Physics Test- 5 - Question 1

An electron moving with kinetic energy 6.6 * 10-14J   enters a magnetic field 4 * 10-3T  at right angle to it.The radius of its circular path will be nearest to

JEE Main Physics Test- 5 - Question 2

A closely wound flat circular coil of 25 turns of wire has diameter of 10 cm which carries current of 4 A, the flux density at the centre of a coil will be

Detailed Solution for JEE Main Physics Test- 5 - Question 2

Flux density = 2 * π * N * i/10r
or,  = 12.56G or 1.25 * 10-3T 

JEE Main Physics Test- 5 - Question 3

An electron is travelling along the x-direction. It encounters a magnetic field in the y-direction. Its subsequent motion will be

Detailed Solution for JEE Main Physics Test- 5 - Question 3
Because a force of q(v×B) is acted on e⁻ toward +z axis that will change the direction of e⁻ and then it will move in a circle in x-z plane
JEE Main Physics Test- 5 - Question 4
The magnetic field due to a straight conductor of uniform cross-section of radius a and carrying a steady current is represented by
JEE Main Physics Test- 5 - Question 5

A cyclotron is operating at a frequency of  12 * 106 Hz. Mass of deuteron is 3.3 * 10-27 kg  and charge on deuteron is 1.6 * 10-19  induction of the neccessary magnetic field is 

JEE Main Physics Test- 5 - Question 6

The strength of the transverse magnetic field required to bend photoelectrons of maximum kinetic energy with in a circle of radius 50 cm when light of wavelength 4000Å is incident on barium emitter is (work function of barium is 2.5 eV)

Detailed Solution for JEE Main Physics Test- 5 - Question 6

key idea: Body performing circular motion is always acted upon by a centripetal force provided by magnetic field. If the kinetic energy of photoelectrons emitted from the metal surface is E1 and W the work function them from Einstein's photoelectric equation, we have Eh = hv -W where v is frequency and h the Planck's constant 
∴  

Putting the numercial values from the question, we have 
∴ 
 

A body performing circular motion is acted upon by a force directed towards its centre  Also the magnetic force  On equating.
We have  

JEE Main Physics Test- 5 - Question 7
The semicircular portion (radius R) of a wire carrying current i is placed in a uniform magnetic field B. The magnetic field is perpendicular to the plane of the circle. The magnetic of the force on the wire is 
JEE Main Physics Test- 5 - Question 8
A & B are two concentric circular conductors of centre O and carrying currents  and  as shown in the diagram. If the ratio of their radii is 1:2 and ratio of the flux densities at O due to A and B is 1:3, then the value of  will be 
JEE Main Physics Test- 5 - Question 9
Domain formation is the necessary feature of
JEE Main Physics Test- 5 - Question 10
The direction of the null points is on the equation line of a bar magnet, when the north pole of the magnet is pointing 
JEE Main Physics Test- 5 - Question 11
Two magnets each of magnetic moment M are placed so as to form a cross at right angles to each other. The magnetic moment of the system will be
JEE Main Physics Test- 5 - Question 12

At a certain place, the horizontal component of earth’s magnetic field is  times of its vertical component. the angle of dip at that place is 

JEE Main Physics Test- 5 - Question 13

A magnet makes 30 oscillations per minute in the earth’s magnetic field. If the magnetic field is doubled. The period of oscillation of magnet is

Detailed Solution for JEE Main Physics Test- 5 - Question 13


n1 = 30 oscillation/min
n1 = 1/2  oscillation/s
T1 = 2s

JEE Main Physics Test- 5 - Question 14
A moving conductor coil produces an induced emf. This is in accordance with 
JEE Main Physics Test- 5 - Question 15
The varying current in a coil changes from 10 A to zero in 0.5 s. If the average emf induced in the coil is 220 V, then the self-inductance of the coil will be 
Detailed Solution for JEE Main Physics Test- 5 - Question 15
Avg E= L(∆i/∆t) 220=L (10/0.5) 220=L(20) L=11H
JEE Main Physics Test- 5 - Question 16
Two circular coils can be arranged in any of the three situations shown in the figure. Their mutual inductance will be 
JEE Main Physics Test- 5 - Question 17

In the circuit shown in the figure, what is the value of  just after pressing the key K?

JEE Main Physics Test- 5 - Question 18
A uniform but time-varying magnetic field B(t) exists in cylindrical region of radius a and is directed into the plane of the paper, as shown. The magnitude of the induced electric field at point P at a distance r from the centre of the circular region 
Detailed Solution for JEE Main Physics Test- 5 - Question 18
Construct a concentric circle of radius r. The induced electric field (E) at any point on the circle is equal to that at P. For this circle 

JEE Main Physics Test- 5 - Question 19
A device for generating and alternating current of a desired frequency is known as 
JEE Main Physics Test- 5 - Question 20
In an AC circuit containing only capacitance, the current
*Answer can only contain numeric values
JEE Main Physics Test- 5 - Question 21

A long solenoid has 100 turns/meter and a current of 3.5 A is flowing through it. If it is filled with material of relative permeability (μr = 20) then magnetic field within the solenoid will be xπ × 10–4, find x?


Detailed Solution for JEE Main Physics Test- 5 - Question 21

B = µ0µRni

*Answer can only contain numeric values
JEE Main Physics Test- 5 - Question 22

A bar magnet is freely suspended in such a way that, when it oscillates in the horizontal plane. It makes 20 oscillations per minute at a place, where dip angle is 30° and 15 oscillations per minute at a place, where dip angle is 60°. Ratio of total earth's magnetic field at these two places is x : 9√3, find x :-


Detailed Solution for JEE Main Physics Test- 5 - Question 22

*Answer can only contain numeric values
JEE Main Physics Test- 5 - Question 23

A square loop ABCD carrying a current i, is placed near and coplanar with a long straight conductor XY carrying a current i, the net force on the loop will be  Find the value of k.


Detailed Solution for JEE Main Physics Test- 5 - Question 23

*Answer can only contain numeric values
JEE Main Physics Test- 5 - Question 24

A square metal wire loop of side 10 cm and resistance 1Ω moved with a constant velocity v in a uniform magnetic field B = 2T as shown in figure. The magnetic field is perpendicular to the plane of the loop and directed into the paper. The loop is connected to a network of resistors, each equal to 3Ω. What should be the speed of the loop in m/s so as to have a steady current of 1mA in the loop?


Detailed Solution for JEE Main Physics Test- 5 - Question 24

The given network PQRS is balanced Wheatstone's bridge. Hence the resistance connected in the diagonal arm PR becomes ineffective. The net effective resistance of the network is

The total resistance of the loop + network is
R = 1 + 3 = 4Ω
Induced current in the network, 

v = 2 × 10–2 m/s = 2 cm/s

*Answer can only contain numeric values
JEE Main Physics Test- 5 - Question 25

An AC source of angular frequency ω is fed across a resistor R and a capacitor C in series. The current registered is I. If now the frequency of source is changed to ω/3 (but maintaining the same voltage), the current in the circuit is found to be halved. The ratio of reactance to resistance at the original frequency ω is   then find the value of x.


Detailed Solution for JEE Main Physics Test- 5 - Question 25

At angular frequency ω, the current in RC circuit is given by

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