JEE Exam  >  JEE Tests  >  Mock Tests for JEE Main and Advanced 2024  >  JEE Main Physics Test- 8 - JEE MCQ

JEE Main Physics Test- 8 - JEE MCQ


Test Description

25 Questions MCQ Test Mock Tests for JEE Main and Advanced 2024 - JEE Main Physics Test- 8

JEE Main Physics Test- 8 for JEE 2024 is part of Mock Tests for JEE Main and Advanced 2024 preparation. The JEE Main Physics Test- 8 questions and answers have been prepared according to the JEE exam syllabus.The JEE Main Physics Test- 8 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Main Physics Test- 8 below.
Solutions of JEE Main Physics Test- 8 questions in English are available as part of our Mock Tests for JEE Main and Advanced 2024 for JEE & JEE Main Physics Test- 8 solutions in Hindi for Mock Tests for JEE Main and Advanced 2024 course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt JEE Main Physics Test- 8 | 25 questions in 60 minutes | Mock test for JEE preparation | Free important questions MCQ to study Mock Tests for JEE Main and Advanced 2024 for JEE Exam | Download free PDF with solutions
1 Crore+ students have signed up on EduRev. Have you? Download the App
JEE Main Physics Test- 8 - Question 1

Two charges of magnitude +Q and two charges of magnitude -q are placed at the opposite corners of a square of side length ‘a’. If the resultant force on a charge Q is zero than

JEE Main Physics Test- 8 - Question 2

Work done in carrying a charge ‘Q’ from ‘A’ to ‘B’ (as shown in figure) on a circle of radius ‘r’ with a charge ‘Q’ at the centre is 

Detailed Solution for JEE Main Physics Test- 8 - Question 2
A circle around a point charge represents an equipotential surface (V= Kq/r throughout the perimeter), and work done on an equipotential surface is zero.
JEE Main Physics Test- 8 - Question 3

Two spheres having radii of 5 cm and 10 cm bear identical charge of 6.6 nC. Find the potential of spheres after they have been connected by a conductor, assume that the spheres are at large distance from each other

JEE Main Physics Test- 8 - Question 4

An ideal electrical cell does 5 joule of work in carrying 10 coulomb of charge around a closed circuit. The heat loss or energy consumption in the circuit outside the battery is

JEE Main Physics Test- 8 - Question 5

A charged particle of mass 0.1 g is in equilibrium in an electric field E = 98 N/C. The charge on the particle is

JEE Main Physics Test- 8 - Question 6

An electric charge ‘q’ is lying at the open end of a cylindrical vessel. The flux coming out of the cylinder will be

Detailed Solution for JEE Main Physics Test- 8 - Question 6
If you put a another cylinder on it then charge will completely in side and flux will distribute half half so flux =q/2εₒ
JEE Main Physics Test- 8 - Question 7

On moving a charge of 20 coulombs by 2cm, 2J of work is done, then the potential difference between the points is

Detailed Solution for JEE Main Physics Test- 8 - Question 7
Va - Vb =W/q.
W= work done :2Jq= charge : 20C Va - Vb = 2/20 = 0.1
JEE Main Physics Test- 8 - Question 8

The capacitance of a parallel plate condenser does not depend upon 

JEE Main Physics Test- 8 - Question 9

A semi-circular are of radius ‘a’ is charged uniformly and the charge per unit length is . The electric field at its centre is

JEE Main Physics Test- 8 - Question 10

Which of the following cannot be the units of electric intensity?

JEE Main Physics Test- 8 - Question 11

The unit of permittivity of free space ∈0  is

Detailed Solution for JEE Main Physics Test- 8 - Question 11

Answer :- a

Solution :- By coulomb's law the electrostatic force

F = 1/4πε0(q1q2/r2)

⇒ε0 = 1/4πF(q1q2/r2)

Substituting the unit for q, r,and F , we abtain unit of

ε0 = coulomb×coulomb newton−(metre)2

= (coulumb)2 newton −(meter)2

=C2N−m2

JEE Main Physics Test- 8 - Question 12

Ampere hour is a unit of 

JEE Main Physics Test- 8 - Question 13

Six resistors each of value 10, are connected to a 5 V battery as shown in the figure. The reading of ammeter'A' is 

Detailed Solution for JEE Main Physics Test- 8 - Question 13

where ampere is placed it should be 2A as 0.5A in each resistor and 4 resistors ahead so total current will be 4x0.5=2A going through the ammeter. 

JEE Main Physics Test- 8 - Question 14

The potential difference between points ‘A’ and ‘B( -)' in the figure if R=0.7 is

Detailed Solution for JEE Main Physics Test- 8 - Question 14
Ans) Applying Kirchhoff's first law
Va - 6 - 2(3) + 9 - (0.7)(3) =Vb
Vb - Va = 9- 6 - 6 - 2.1
=- 3 - 2.1
= - 5.1
JEE Main Physics Test- 8 - Question 15

If 2% of the main current is to be passed through the galvanometer of resistance G. the resistance of shunt required is

Detailed Solution for JEE Main Physics Test- 8 - Question 15
Let total current through the circuit be I.
Then, 0.02I current is to be passed through galvanometer of resistance G.
Let the resistance of shunt is Rs . 
∵ Shunt are in parallel combination with Galvanometer.
∴ 0.02I * G = (1 - 0.02)I * Rs 
⇒0.02G = 0.98 * Rs 
⇒Rs = 0.02G/0.98 = G/49 

Hence answer is G/49
JEE Main Physics Test- 8 - Question 16

The thermo e.m.f. of a thermocouple is given by e = 2164 t - 6.2 t2 . The neutral temperature and a temperature of inversion are

JEE Main Physics Test- 8 - Question 17

The relation between Faraday constant F, electron charge e, and Avogadro's Number N is

JEE Main Physics Test- 8 - Question 18

In the circuit as shown in Fig. The heat produced by 6 resistance due to current flowing in it is 60 calories per second. The heat generated across 3 ohm resistance per second will be 

JEE Main Physics Test- 8 - Question 19

A thin bar magnet of length 2l and breadth 2b, pole strength p and magnetic moment M is divided into four equal parts with length and breadth of each part being half of the original magnet. Then the pole strength of each part is

JEE Main Physics Test- 8 - Question 20

Susceptibility is positive and small for a ........substance: 

*Answer can only contain numeric values
JEE Main Physics Test- 8 - Question 21

A battery of internal resistance 4 ohm is connected to the network of resistance as shown. In the order that the maximum power can be delivered to the network, the value of R in ohm should be :-


Detailed Solution for JEE Main Physics Test- 8 - Question 21

Condition for maximum power is
r = R

R = 2

*Answer can only contain numeric values
JEE Main Physics Test- 8 - Question 22

A capacitor of capacitance 5 μF is connected as shown in the fig. The internal resistance of the cell is 0.5 Ω. The amount of charge on the capacitor plate is :- (μC) :-


*Answer can only contain numeric values
JEE Main Physics Test- 8 - Question 23

Consider two concentric non-conducting uniformly charged spherical shells of radius 3m and 6m having charges 10 µC and 200 µC respectively. Find the interaction potential energy between two spheres in joules.


Detailed Solution for JEE Main Physics Test- 8 - Question 23

*Answer can only contain numeric values
JEE Main Physics Test- 8 - Question 24

Diagram shows a part of circuit. Find VB–VA when current I is 5A and it is decreasing at a rate of 10+3 A/s ?


Detailed Solution for JEE Main Physics Test- 8 - Question 24

*Answer can only contain numeric values
JEE Main Physics Test- 8 - Question 25

A current of 4 A flows in a coil when connected to a 12 V d.c. source. If the same coil is connected to a 12 V, 50 rad/s, a.c. source, a current of 2.4 A flows in the circuit. Determine the reactance (in Ω) of coil.


Detailed Solution for JEE Main Physics Test- 8 - Question 25

357 docs|148 tests
Information about JEE Main Physics Test- 8 Page
In this test you can find the Exam questions for JEE Main Physics Test- 8 solved & explained in the simplest way possible. Besides giving Questions and answers for JEE Main Physics Test- 8, EduRev gives you an ample number of Online tests for practice

Up next

Download as PDF

Up next

Download the FREE EduRev App
Track your progress, build streaks, highlight & save important lessons and more!