JEE Main Part Test - 4 Free Online Test 2026

Key Idea: Like charges repel each other while unlike charges attract each other.

From Coulomb's law, the force of attraction/repulsion between two point charges q1 and q2 placed a distance r apart is given by

When similar charges are taken

q₁ = 3 × 10^-6 C, q₂ = 8 × 10^-6 C

When an additional charge - 6 × 10⁶ C is given to each charge, then


Dividing Eq. (ii) by Eq. (i), we get

Negative sign indicates, force is attractive.

Here, F_CA = F_CB = F_C

Hence, the resultant force on the charge at C is

The charge inside the closed surface is given by
q = net electric flux pass through the surface × ε₀
= (4 × 10³ - 8 × 10³) ε₀
Therefore, q = -4 × 10³ ε₀ coulomb

• The electric field is always perpendicular to the surface of a conductor.
• On the surface of a metallic solid sphere, the electric field is perpendicular to the surface and directed towards the centre of the sphere.

Hence, the correct option is (d).

The correct option is (c). In a non-uniform electric field, a dipole experiences a force, which gives it a translational motion, and a torque, which gives it a rotational motion.

The charge enclosed by the Gaussian surface surrounding the cavity is zero. Hence, the electric field is also zero. So, the assertion is true. Charges in a conductor reside only at its surface. So, in the cavity there is no charge. So, the reason is also true and properly explains the assertion.

Gravitational force is the dominating force in the universe so assertion is false. Gravitational force is weaker than Coulombic force so, reason is false

Given potential,
V = 3x² + 5
Electric field intensity,

At x = -2,
E = -6 (-2) = +12V/m

If Q is the initial charge on capacitor C₁, then the initial energy is given by U_i = Q²/2C₁.

As the two capacitors are connected together and the charge is distributed equally, the charge on each capacitor is Q/2.

Since the potential difference (in a parallel connection) across the two capacitors is also the same, it follows that their capacitance are equal (since C = Q/V).

Thus, C₁ = C₂ = C (say)

Also, Q₁ = Q₂ = Q/2

Therefore, the final energy stored in the two capacitors is

The charge Q remains unchanged as the battery is disconnected. The capacitance C decreases if the separation between the plates is increased.
Now, energy stored U = Q²/2C
Since Q remains the same and C is decreased, U will increase.
Also, work has to be done to increase the separation between the plates of a charged capacitor.
Hence, option d is correct.

To get equivalent capacitance 6 μF, out of the three, 4 μF capacitances, two are connected in series and the third one is connected in parallel.

Before connections,
C₁ = 3µF, V₁ = 300V
Charge on C₁,
Q₁ = 900μC
C₂ = 2µF, V₁ = 200V
Charge on C₂,
Q₂ = 400μC
When the positive terminal of C₁ is connected with the negative terminal of C₂, common potential of the system is:

Substituting the values, we get
V = 100V
New charge on positive plate of C₁,
Q'₁ = 3 x 100 = 300µC
So, charge Q1 - Q'₁ = 600 µC has flown from the positive plate of the C₁ to the negative plate of C₂.

As insulator plate is passed between the plates of the capacitor, its capacity increases first and then decreases as the plate slips out. As a result, positive charge on plate A increase first and then decreases, hence, current in outer circuit flows from B to A and then from A to B.

By Gauss’s law:
∮ E ⋅ dA = Q_enc / ε₀

If ∮ E ⋅ dA = 0, then Q_enc = 0, so the reason is true. However, the assertion is false: zero flux does not imply E = 0 everywhere on the surface (e.g., a dipole with no net charge produces non-zero field).

Thus, both are not correct, and the correct answer is D: Assertion is false, Reason is true.

Though the net charge on the conductor is still zero but due to induction the negatively charged region is nearer to the rod as compared to the positively charged region. That is why the conductor gets attracted towards the rod

• When an external electric field is applied to the conductor, the free electrons in the conductor move in an opposite direction to that of the applied electric field.
• This movement of electrons induces another electric field inside the conductor which opposes the original external electric field.
• This continues until the induced electric field cancels out the external field. 

The external electric field polarizes the dielectric and an electric field is produced.
The net electric field inside the dielectric decreases due to polarization.

Potential difference across PQ, i.e. potential difference across the resistance of 20Ω, is

∴ V = 0.16 × 20 = 3.2 V

Let the voltage supply be V volts.

Let the current flowing initially be i amperes.

Now, by Ohm's law, i = V/90

Let the final resistance be R ohms.

As the final current is 90% of the initial value,

Or R = 100 ohms

Let the resistance added in series be r ohms.

Now, r + 90 = 100

Or r = 10

Thus, 10 ohms resistance ought to be connected in series to the given resistor.

The above arrangement is a balanced wheatstone bridge, so no current will pass through the 10 ohm resistance.
Or

Or 
 ohms is the equivalent resistance.

In the circuit I = 9/3 = 3A
V_C - V_A = 2 × 1.5 = 3 .... (I)
V_c - V_B = 4 × 1.5 = 6 .....  (II)
Eqⁿ(II) - Eqⁿ(I)
V_A - V_B = 6 - 3 = 3 Volt

Given:

Equating the two expressions for current, i,

This equation simplifies to:
100 (5 + R_G) =  1050 x 5 + R_G x 5
Solving for the resistance of the galvanometer, R_G:
95 R_G = 4750
R_G = 50Ω
So, the resistance of the galvanometer is 50Ω.

For SHE E°_SHE = 0.00 V
Oxidation at anode (left)

Reduction at cathode (right) 
Net

This is the type of the cell in which electrodes at different pressures are dipped in same electrolyte and connectivity is made by a salt-bridge.

Reaction Quotient (Q) 

∵ 

This is a type of concentration cell using hydrogen electrode as anode and cathode.

Thermal stability ∝ bond dissociation energy 
(where, E = group 16 elements)
On moving down the group, bond dissociation energy decreases due to increase in bond length. Thus, the order of bond dissociation energy or thermal stability is 
H₂O > H₂S > H₂Se > H₂Te > H₂PO

Ozone is thermodynamically unstable w.r.t oxygen since, its decomposition into oxygen results in liberation of heat.

This is in accordance with Henry's law.
'At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to partial pressure of that gas in equilibrium with that liquid.'