JEE Main Practice Mock Test - 7 Free Online Test 2026

Refer to Fig.

Refer to Fig.

So, it is 60^∘ upstream.

Total upward buoyant force on the block is
U = buoyant force due to oil + buoyant
force due to water
= ρ₀(l − x)l²g + ρ_wxl²g
For floating U = weight of the block = ρ_bl³g
∴ ρ_bl³g = ρ₀(l − x)l²g + ρwxl²g

So the correct choice is (c).

Given:

• Charge q = 10⁻¹² C
• Velocity v = 10⁵ m/s along x̂
• Force F = 10⁻¹⁰ N along ŷ

Magnetic force:

For: 
Compute:

Magnitude:
F = qvB sin(θ), maximum when θ = 90°

Minimum magnetic field B:

If t is the thickness of the glass sheet, the fringes are displaced by an amount given by

In order to bring the adjacent minimum to the centre of the screen (i.e. to bring the first dark fringe the central bright fringe), the fringes must be displaced by half the fringe width, i.e.

AB and BC are in series

AD and DC are in series

ABC and ADC are in parallel

or

Hence,

When the key, K is inserted, the current starts growing and after some time it acquires a steady value. At this stage, no current flows through the capacitor (because an ideal capacitor offers an infinite resistance to a steady current). All the current flows through the inductor (because an ideal inductor offers zero resistance to a steady current). Now, the network of resistors is a balanced wheatstone bridge. Hence, no current flows through the resistance 2R. Therefore, this resistance can be ignored. The net resistance between points X and Y = resistance of the parallel combination of 2 R and 2 R = = R

Hence, the current in the circuit is

I =

Now, the terminal voltage of cell A is

2r_A = R + r_A + r_B or R = r_A – r_B

So, the correct choice is (a).

We are given:

• Inductance (L) = 0.20 H
• EMF (E) = 1.6 V
• Resistance (R) = 4.0 Ω

We are asked to find the initial rate of growth of current when the switch is just closed.

Concept:

For an RL circuit, the initial rate of change of current is given by:

(di/dt)₀ = E / L (This is only true at t = 0, when the inductor initially resists any change in current)

But in a real RL circuit, with resistance, the correct formula is:

(di/dt)₀ = E / L — only if R = 0

When R ≠ 0, the correct expression is:

But this only gives the maximum rate at the very beginning. Actually, the current obeys:

Then:

So yes, (di/dt) at t = 0 = E / L

Step-by-step Calculation:

Final Answer: D: 8.0 A/sor 8.0As^-1

Let d be the distance between the moon and the point at force on mass m is zero then

⇒60R − d = 9d
⇒60R = 10d
⇒d = 6R

In new situation we have

As the length of magnet is halved , Magnetic moment M’= m(l) = M/2

Resultant magnetic Moment

Let 'x' be the length of the 4^th rod and the centre of the mass of all the rods lies at origin then X_cm = 0

The frequency at which a wire vibrates is given by:

Here, g is the acceleration due to gravity and l is the length of the wire.

Let’s express the fundamental frequency of the vibration at T₁​=30^∘C as follows,

Let’s express the fundamental frequency of the vibration at T₂​=10^∘C as follows

Dividing equation (2) by equation (1), we get,

We can express the linear expansion of the wire due to change in temperature as,

Here, α is the linear expansion coefficient and ΔT is the change in temperature.

Using equation (3), we can rewrite the above equation as,

Substituting 1KHz for f_1​, 1.001KHz for f₂​ and −20^∘C for ΔT in the above equation, we get,

Given: C_p - C_v = 4150 Jkg-1K_-1 and C_p/C_v = 1.4 Or C_p = 1.4 C_v.

Therefore,

1.4 C_v - C_v = 4150

Or C_V = 4150/0.4 = 10375 J kg^-1 K^-1

Moment of inertia (I) = mr²

[I] = [ML²]

Moment of force (C) = rF

[C] = [r][F] = [L][MLT^-2] or [C] = [ML²T^-2]

Moment of inertia and moment of a force do not have identical dimensions.

Tension=[MLT⁻²]

Surface Tension = [ML⁰T⁻²]

Clearly these two have different dimension.

F.B.D. of man and plank are

For plank to be at rest, applying Newton's second law to plank along the incline

Mgsinα = f

and applying Newton's second law to man along the incline.

mgsinα + f = ma

Direction: Down the incline (same as gravity component).

a = gsinα(1 + M/m) down the incline

If = constant then

So should be parallel to so coefficient should be in same ratio. So

So α = −1

= gcosθ (perpendicular to plane)

Let ρ the density of weight.

σ be the density of liquid.

and V be the volume of the weights.

From (1) and (2)

From (1) and (3)

Poles are not actually at the end of the magnet they are little inside from the end this makes the magnetic length of the magnet little less than the Geometric length.

Geometric length is the actual length of the magnet.

So when the relationship between these two measured it comes to be,

Magnetic length = 0.8× Geometric length

If the effective decay constant is λ, then

Now,

We are given:

• Mass of each astronaut: M = 100 kg
• Initial separation: r = 100 m
• They move 1 cm = 0.01 m closer due to gravitational attraction
• They are initially at rest relative to each other
• No external forces (isolated system in space)
• We are to calculate time taken to move 1 cm closer

Step 1: Use Newton’s Law of Gravitation

The gravitational force between them:

This force causes acceleration toward each other. Since both astronauts are the same mass, each experiences half of the relative acceleration.

Let’s calculate the relative acceleration (i.e., how fast the gap is closing):

Where:

• G = gravitational constant = 6.674 × 10⁻¹¹ N·m²/kg²

Step 2: Use Kinematic Equation

Since they start from rest and accelerate toward each other, and the displacement is s = 0.01 m, we use:

Now plug in the values:

• s = 0.01 m
• r = 100 m
• G = 6.674 × 10⁻¹¹
• M = 100 kg

Step 3: Convert seconds to days

Final Answer: Nearest integer = 1 day

Answer: 1

The effective focal length of the silvered lens is given by:

, which gives F = 15/2 cm.

The silvered lens behaves like a concave mirror.

Using the spherical mirror formula , we have

, which gives v = -12 cm. The negative sign indicates that the image is formed to the left of the lens.

Mass of water = 55 kg, mass of drum = 5 kg, total = 60 kg.
Height lifted, h = 20 m, g = 10 m/s²

Work done lifting water and drum:

W₁ = 60 × 10 × 20 = 12000 J

Each rope: length = 20 m, mass = 20 × 0.5 = 10 kg
Total rope mass = 10 + 10 = 20 kg
Center of mass of ropes rises by 10 m:

W₂ = 20 × 10 × 10 = 2000 J

Total work:

W = 12000 + 2000 = 14000 J = 1.4 × 10⁴ J

Thus, c = 1.4

Corrected Answer: c = 1.4 (corrected from c = 1)

The change in length ∆l is proportional to ll and ∆T. Stated mathematically

Δl = αlΔT

Where, α is called the coefficient of linear thermal expansion for the material.

Given, ΔT = 100°C

l = 10 m

∴ Δl=10 × 100 × 10 × 10⁻⁶

= 10⁻² m = 1 cm

During charging voltage across capacitor with time is given by

V_c =  V(1 − e^−t/T)

V = V_C + V_R

Here, V is the applied potential.

Given that V_C = 3V_R = 3(V−V_C)

Here τ = CR =10 s

⇒ t = 2 × 10 × 0.693 = 13.86 s ≃ 14 s

Reaction:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) — exothermic reaction

Analysis:
Adding H₂:
According to Le Chatelier’s Principle, adding more H₂ shifts the equilibrium toward the formation of NH₃ to reduce the added H₂.

Exothermic reaction:
Since the forward reaction releases heat, increasing temperature would shift the equilibrium toward the reactants.

Pressure consideration:
The reaction goes from 4 moles (1 N₂ + 3 H₂) to 2 moles (2 NH₃).
Therefore, the total pressure decreases as the equilibrium shifts forward.

Conclusion:
Option A: Incorrect – Pressure does not increase; it decreases as fewer moles of gas are formed.
Option B: Incorrect – Equilibrium does shift, so it's not unchanged.
Correct: Option C is correct.

We are given:
Glycine is a diprotic molecule (acts as both acid and base).
Ka₁ = 4.5 × 10⁻³, Ka₂ = 1.7 × 10⁻¹⁰
Concentration = 0.01 M

In water, glycine mostly exists as a zwitterion, and its pH ≈ average of pKa₁ and pKa₂:

But in very dilute solutions, the pH slightly shifts toward neutral due to water's ionization.
So, for 0.01 M glycine, the pH ≈ 7.06 (experimentally observed).
Final Answer: B: 7.06 

When Al(OH)₃ is added to NaOH, it dissolves due to the formation of a soluble complex ion.

Reaction:
Al(OH)₃ + OH⁻ → [Al(OH)₄]⁻

This happens because Al(OH)₃ is amphoteric — it reacts with both acids and bases. In a basic solution like NaOH, it forms a tetrahydroxoaluminate complex.

Correct complex ion: [Al(OH)₄]⁻

This matches: Final Answer: C: Al(H₂O)₂(OH)₄⁻
(Since water molecules can also coordinate, it's the same as [Al(OH)₄]⁻ with water ligands)

HO − SO₂ − OH + PCl₅ → Cl − SO₂ − O + POCl₃ + HCl

LHS   (equality occurs if x = y = π/3)