NEET Physics: Topic-wise Test- 4 Free Online Test 2026

A.Statement A is wrong.
B.m= m0​​/√(1− v2/c2) ​​
specific charge q/m=q/m0√(1​−(v2/C2))​​
so, as v increases the specific charge decreases
C. None of the fundamental particles that form the Standard Model have charge without mass. The Standard Model describes all energy and matter that makes up the universe, except gravitation. And gravitation does not have charge, it has mass.
D.Yes. Repulsion is observed only when two bodies have like charges that means the bodies must be charged. Therefore, repulsion is sure test for electrification than attraction.
 

The charge in the middle experiences force along the line. The equilibrium is stable along the line connecting charges while The equilibrium is unstable along the line perpendicular to the line of charges  
∴ Only option B is correct (given consider equilibrium only along line joining charges).
 

If the potential energy of the system is minimum, it will be stable equilibrium, i.e , d²U/dx²​>0 and when potential energy is maximum then it will be unstable equilibrium, i.e, d²U/dx²<0. 
As along y direction no electric field, potential energy is minimum and it will be stable equilibrium along y-axis.
Along x-axis potential energy is maximum due to all charges situated along x-axis.so it will be unstable equilibrium.
 

(A) electric field due to positive charge is away from it and electric field due to negative charge is towards it so there cannot be a point on AB where electric field can be zero
(D) negative charge will experience a repulsion force due to −ve charge and attraction force due to +ve charge, so if pushed away from charges then a net attractive force and if pushed towards AB then a net repulsive force will be there so there will be oscillations.
 

If field is straight then the charge moves along the straight line .if the fielss is a curve then the electrostatic force is not sufficient to change the direction along the curved field line .

The acceleration due to gravity is insignificant compared with the acceleration caused by the electric field, so the gravitational field can be ignored. Hence, when a particle of mass m and charge q is thrown in a region where uniform gravitational field and electric field are present, before entering the electric field, the charge follows a straight line path (since, no net force due to any of the field). As soon as it enters the field, the charge begins to follow a parabolic path (since, constant force is always in the same direction). As soon as it leaves the field, the charge follows a straight line path  (since, no net force due to any of the field). The path of a particle may be straight line or may be parabola.

The dipole will have some distance along the electric field, so, option (1) is correct.

Let P is the observation point at a distance r from –2q and at (L + r) from +8q.

Given Now, Net EFI at P = 0

​

​

∴ P is at x = L + L = 2L from origin

∴ Correct Option is (iii) 2L

Combination of Capacitors in Series
C(effective) = C/3 = 1

When equal charges are given to two spheres of different radii, the potential will be more or the smaller sphere as per the equation, Potential = Charge / Radius.
Since potential is inversely proportional to radius, the smaller radius will have higher potential and vice versa.

For series combination,
1/C₁ = (1/C)+(1/C)+(1/C) [ Since all capacitors have equal capacitances]=3/C Or
C₁ = C/3−−−(i)
For parallel combination,
C₂ = C + C + C = 3 C−−−(ii)
Now,C₁ / C₂ = (C/3) / 3C = (C/3) × (1/3C) = 1/9

C1= 20×10µf

and C2= 30×10µf

in series Ceq = C1C2/(C1+C2)

Ceq = 20×10^(-6)×30×10^(-6)/20×10^(-6)+30^×10(-6)

Ceq= 12×10^(-6)f

As we know that Q = CV

Putting the values of C and V= 40V, we get

Q = (12 * 10^-6) * 40

= 480µC

Let C_p be the equivalent capacitance of parallel
C_p=C₁+C₂
16= C₁+C₂……..(1)
Let Cs be the equivalent of capacitance in series
1/C_s=(1/C₁) + (1/C₂)
Or, C_s=C₁C₂/C₁+C₂
Or,3=C₁C₂/16
Or,C₂C₁=48
C₂=48/C₁……..(ii)
16=C1+48/C1
(C₁²-16C₁+48)=0
(C₁-4)(C₁-12)=0
C₁=4 ; C₁=12
If,
C₁=4
C₂=12
If,
C₂=4
C₁=12
So, the answer is, Either 12μf or4μf

Since the electric field inside the conductor is zero and has no tangential component on its surface, therefore, no work is done in moving a test charge within the conductor or on its surface. It means the potential difference between any two points inside or on the surface is zero. Hence, electrostatic potential is constant throughout the volume of the charged conductor and has the same value on its surface as inside it.

On an irregularly shaped conductor, the surface charge density is greatest at the locations where the radius of curvature of the surface is smallest i.e, charge density is inversely proportional to radius of curvature of surface. Since the radius of curvature is least at sharp portion and maximum at flat portion, so, charge density is high at sharp portions and less at flat portions.

When there is potential difference across the conductor, the electric field is set up. Due to this charge will flow across the conductor. But when conductors have the same potential the charge will not flow from one conductor to the other.

If a charge q is placed outside than the electric field lines incident on the conducting sphere , if some charge is developed due to thee lines than opposite surface becomes oppositely charge and the total charge becomes zero

The deflection in galvanometer will not be changed due to interchange of battery and the galvanometer.

Let x be the resistance shunted with S for the bridge to be balanced.
For a balance Wheatstone's bridge


or S' = 2Ω 
From Figure 

x = 6Ω

The bridge will be balanced when the shunted resistance is of the value of 3Ω

For balanced Wheatstone's bridge
P/Q = R/S.....(i)
Power dissipation in resistance R with voltage V is V²/R.
∴ 
From equation (i)

⇒ 
Using (i), we get

∴ 

Wire of resistance =1Ω
Keeping in view that, ABCD is a square where each side is uniform wire of resistance 1Ω. If a uniform wire of resistance 1Ω is connected across AE and a potential difference is applied across A and C, the points B and E will be equipotential; a point E on CD is CE​/ ED =√2/1
∴    CE:ED= √2​:1

A battery is of emf E is being charged from a charger and hence current inside the cell is from anode to cathode 
I=V−E/r​
V=E+Ir
Therefore, when a cell is being charged the potential difference across its terminals is greater than emf of a cell. Also in charging the positive terminal is connected to anode of the cell and negative terminal to cathode.
Here, positive terminal of the battery is connected to terminal A of charger and negative terminal of the battery is connected to terminal B of charger.
Hence, Potential difference across points A and B must be more than E, A must be at higher potential than B and In battery, current flows from positive terminal to the negative terminal
 

When a resistor is added parallel to other resistor in circuit it results in the reduction of overall resistance, since there are multiple pathways by which charge can flow. adding another resistor in a separate branch provides another pathway by which to direct charge through the main area of resistance within the circuit. This decreased resistance resulting from increasing the number of branches will have the effect of increasing the rate at which charge flows i.e. the current.