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JEE Main Physics Test- 4 - JEE MCQ


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25 Questions MCQ Test Mock Tests for JEE Main and Advanced 2024 - JEE Main Physics Test- 4

JEE Main Physics Test- 4 for JEE 2024 is part of Mock Tests for JEE Main and Advanced 2024 preparation. The JEE Main Physics Test- 4 questions and answers have been prepared according to the JEE exam syllabus.The JEE Main Physics Test- 4 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Main Physics Test- 4 below.
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JEE Main Physics Test- 4 - Question 1

Figure shows some of the electric field lines corresponding to an electric field. The figure suggests that 

Detailed Solution for JEE Main Physics Test- 4 - Question 1

At point A and C, electric field lines are dense and equally spaced, so EA = EC
While at B they are far apart.
∴ EA = E> EB

JEE Main Physics Test- 4 - Question 2

When the separation between two charges is increased, the electric potential energy of the charges 

Detailed Solution for JEE Main Physics Test- 4 - Question 2

Potential  energy between two charges 
Now if r increases and q1 and q2 are of same sign then
 decrases. if q1 and q2 are of opposite  increases.
Therefore depending on signs of q1 and q may increase or decrease.

JEE Main Physics Test- 4 - Question 3

An electric dipole is placed in a uniform electric field. The net electric force on the dipole 

Detailed Solution for JEE Main Physics Test- 4 - Question 3
As dipole is combination of equal positive and negative charges, when an electric field is generated in the region of the dipole, torque can act on the dipole, but there will be no force on the dipole.
JEE Main Physics Test- 4 - Question 4
Electric charge are distributed in a small volume. The flux of the electric field through a spherical surface of radius 10cm surrounding the total charge is 25V-m The flux over a concentric sphere of radius 20cm will be
JEE Main Physics Test- 4 - Question 5

A charge q is placed at the centre of the open end of a cylindrical vessel in figure. The flux of the electric field through the surface of the vessel is 

Detailed Solution for JEE Main Physics Test- 4 - Question 5

when a charge Q is placed at the centre of the open end of a cylindrical vessel then only half of the charge will contribute to the flux,because half will lie inside the surface and half will outside the surface so flux=Q/2ε0

JEE Main Physics Test- 4 - Question 6
Mark the correct option 
JEE Main Physics Test- 4 - Question 7

Two capacitors each having capacitance C and breakdown voltage V are joined in series. The capacitance and the breakdown voltage of the combination will be 

Detailed Solution for JEE Main Physics Test- 4 - Question 7

JEE Main Physics Test- 4 - Question 8

A dielectric slab is inserted between the plates of an isolated capacitor. The force between the plates will 

Detailed Solution for JEE Main Physics Test- 4 - Question 8
Here,the capacitor is isolated. It is already charged or else there would be no force. But when the dielectric is inserted, the charge remains fixed. If the capacitor were connected to a voltage source, then current would flow to keep the voltage fixed and then the force would increase.The voltage goes down even though there is no current in or out.The permittivity of any material is greater than the permittivity of free space. Therefore,When we insert the dielectric slab the capacitance rises.

 
JEE Main Physics Test- 4 - Question 9

Two metal spheres of capacitances C1 & C2   carry some charges. They are put in contact and then separated. The final charges Q1  and Q2  on them will satisfy

JEE Main Physics Test- 4 - Question 10

Two resistances R and 2R are connected in parallel in an electric circuit. The thermal energy developed in R and 2 R are in the ratio 

Detailed Solution for JEE Main Physics Test- 4 - Question 10

Thermal energy = V2t/R
In parellel arrangement V is constant
So, for a time t
Thermal energy ∝ 1/R
Ratio of their thermal energy = R₂ : R₁ = 2R : R = 2 : 1

JEE Main Physics Test- 4 - Question 11
A uniform wire of resistance 50  is cut into 5 equal parts. These parts are now connected in parallel. The equivalent resistance of the combination is
Detailed Solution for JEE Main Physics Test- 4 - Question 11
Each part resistance is 10 ohm as R is directly proportional to length now 10 ohms 5 resistance are connected in parallel. Therefore 2 is equivalent resistance.
JEE Main Physics Test- 4 - Question 12
The net resistance of an ammeter should be small to ensure that 
Detailed Solution for JEE Main Physics Test- 4 - Question 12
Based on basic construction and implementation of ammeter,it is in connected in series in a circuit and should not appreciably change the current to be measured.
JEE Main Physics Test- 4 - Question 13

The equivalent capacitance between the points A and B of five identical capacitors each of capacity C, will be 

Detailed Solution for JEE Main Physics Test- 4 - Question 13
The two identical branches capacitors have capacitance equal to C/2 each.
adding these in parallel will give you C/2 +C/2+C = 2C.
JEE Main Physics Test- 4 - Question 14

The electric field intensity at a point situated 4 meters from a point charge is 200 N/C. If the distance is reduced to 2 meters, the field intensity will be

JEE Main Physics Test- 4 - Question 15

If an electron is brought towards an another electron, the electric potential energy of the system 

JEE Main Physics Test- 4 - Question 16
Two identical conductors of a copper and aluminium are placed in an identical electric field. What is the magnitude of induced charge in the aluminium ? 
Detailed Solution for JEE Main Physics Test- 4 - Question 16
As aluminum and copper are metals, their mobile electrons move under the influence of the external field until they reach the surface of the metal and collect there. External electric fields induce surface charges on metal objects that exactly cancel the field within. Since the field applied is same in both case, therefore induced charge will be the same. Since both are metals so equal amount of charge will induce on them.
JEE Main Physics Test- 4 - Question 17

The electric field intensity at a point situated 4 meters from a point charge is 200 N/C. If the distance is reduced to 2 meters, the field intensity will be

Detailed Solution for JEE Main Physics Test- 4 - Question 17
200=KQ/16-------equ-1
E=KQ/4-------------equ-2
on dividing1by 2
E= 800
JEE Main Physics Test- 4 - Question 18

An electric dipole is placed at an angle of 30° with an electric field intensity 2 × 105 N/C. It experiences a torque equal to 4 N m. The charge on the dipole, if the dipole length is 2 cm, is

JEE Main Physics Test- 4 - Question 19

Shown in the figure is a distribution of charges. The flux of electric field due to these charges through the surface S is

Detailed Solution for JEE Main Physics Test- 4 - Question 19

Flux flowing through a body depends on the net charge enclosed inside the surface, here net charge inside is zero so flux is also zero. And the third charge is outside so it will not effect the flux.

JEE Main Physics Test- 4 - Question 20

Ten identical cells each of emf E and internal resistance r are connected in series to form a closed circuit. An ideal voltmeter connected across the three cells will read

Detailed Solution for JEE Main Physics Test- 4 - Question 20

No. of cells = 10
Emf = E
Internal resistance = r
Let I = current through the ckt

Now potential (V) of across each cell will be 
V = E -Ir

We see there being no load resistance internal resistance play no effect on the observed emf
Thus the pd across 3 cells will be 
E + E + E = 3Ir = 3E

*Answer can only contain numeric values
JEE Main Physics Test- 4 - Question 21

A dipole consists of two particles one with charge +1µC and mass 1kg and the other with charge –1µC and mass 2kg separated by a distance of 3m. for small oscillations about its equilibrium position, the angular frequency, when placed in a uniform electric field of 20kV/m is :-


Detailed Solution for JEE Main Physics Test- 4 - Question 21

Restoring torque

τ = – PE sin θ
τ = – PE θ, for small θ

*Answer can only contain numeric values
JEE Main Physics Test- 4 - Question 22

A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge –3Q, the new potential difference between the same two surface is xV, then x is :-


Detailed Solution for JEE Main Physics Test- 4 - Question 22


*Answer can only contain numeric values
JEE Main Physics Test- 4 - Question 23

A potentiometer is used for the comparison of e.m.f. of two cells E1 and E2. For cell E1 the no deflection point is obtained at 20 cm and for E2 the no deflection point is obtained at 30 cm. The ratio of their e.m.f.’s is N/3 then N is:


Detailed Solution for JEE Main Physics Test- 4 - Question 23

Ratio will be equal to the ratio of no deflection lengths i.e

*Answer can only contain numeric values
JEE Main Physics Test- 4 - Question 24

The equivalent resistance of the arrangement of resistances shown in adjoining figure between the points A and B is N × 4 ohm then N is:


Detailed Solution for JEE Main Physics Test- 4 - Question 24



*Answer can only contain numeric values
JEE Main Physics Test- 4 - Question 25

In the circuit shown here C1 = 6μ F, C2 = 3μ F and battery B = 20V. The switch sis first closed. It is then opened and afterwards sis closed. What is the charge finally on C2.


Detailed Solution for JEE Main Physics Test- 4 - Question 25

After closing S1, charge on C1 is q=6×20=120μC.

Now S1 is opened.

On closing S2, charge q will be distributed between C1 and C2 according to their capacitances.

So charge on C2 is q

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