JEE Main Practice Mock Test - 4 Free Online Test 2026

According to the EM Wave Spectrum, wavelength range for, 
(a) Gamma-rays is fm.
(b) X-rays is A⁰
(c) Microwave is in mm
(d) A. M. radio wave is in m

We know that, angular momentum,

mvr = nh / 2π ⇒ 2πr = n (h / mv)

According to de Broglie's wavelength, λ = h / mv

⇒ 2πr = nλ

Where r is the radius of the orbit, λ is the de Broglie wavelength, and n represents integral quantisation for standing wave (electron).

∴ λ = 2πr / n = (4 × 10⁻⁹) / 2 = 2 × 10⁻⁹ m

If momentum is Zero ie, if p = 0,then kinetic energy
K = p² / 2m = 0
But potential energy cannot be zero, thus a body can have energy without momentum.

Here, λ = 3 mm = 3 × 10⁻³ m, E_₀ = 66 Vm⁻¹

∴ B_₀ = E_₀ / c = 66 / (3 × 10⁸) = 2.2 × 10⁻⁷ T

As the electromagnetic wave is propagating along the x-axis and the electric field oscillation is along the y-direction, the magnetic field oscillation is along the z-direction, using the relation for a harmonic wave:

E_y = E_₀ cos (2π / λ) (ct - x)

E_y = E_₀ cos (2πc / λ) (t - x / c)

∴ E_y = 66 cos (2π × 3 × 10⁸ / 3 × 10⁻³) (t - x / c)

= 66 cos (2π × 10¹¹) (t - x / d)

and B_z = B_₀ cos (2πc / λ) (t - x / c)

= 2.2 × 10⁻⁷ cos (2π × 10¹¹) (t - x / c)

After taking the mean of velocities, we get mean velocity.

The root-mean-square speed is the measure of the speed of particles in a gas, defined as the square root of the mean velocity-squared of the molecules in a gas.

Then if,

u_a = (u₁ + u₂ + ... ) / n

For one molecule,

u_a = u₁ / 1 and u_rms = √( u₁² / 1 )

⇒ u_a = u_rms

The emf induced in the second coil is

Where M is the mutual inductance of two coils, d_i / d_t is the rate of change of current in the first coil.

But 

For maximum value of emf induced, cosωt = 1
∴ The maximum value of the emf induced is ε_max  = Mi_mω
Here, M = 0.005H, i_m = 10A,ω = 100πrads⁻¹
ε_max = (0.005H)(10A)(100π) = 5πV

C₁, C₂ and C₃ are in series

or 

or 

All the capacitors in upper branch are in series so the charge on each capacitor is Q′ = 6 / 11CV

Also charge on capacitor C₄ is Q = 4CV

∴ 

Net electric field will be zero at origin.
At any co-ordinate (0, y)


For maximum electric field, dE / dy = 0
Solving, 

⇒ R = 5 Ω

In Statement I:
The decrease in mechanical energy in Case I will be:
ΔME₁ = (1/2) mv²
However, the decrease in mechanical energy in Case II will be:
ΔME₂ = (1/2) mv² - mgh
∴ ΔU₂ < ΔU₁, meaning Statement I is correct.
In Statement II:
The coefficient of friction will not change, which makes this statement incorrect.

The pressure exerted by the photons is given by:
P = 0.75 × (I/c) + 0.25 × (2I/c)
(P = pressure, c = speed of light, I = intensity)
P = 1.25 × (I/c)
We know that force can be computed using:
F = P × A
(F = force, P = pressure, A = area)
F = 1.25 × (50 × 1 / (3 × 10⁸))
F = 20.83 × 10⁻⁸ N

Magnetic fields at P :

In case I. B₁ = 

In case II. B₂ = 

If R₁ and R₂ be the corresponding radii,

Change in length, 

At equilibrium
2Tcosθ = mg

⇒x = 5.92 cm

By Energy Conservation:
mg (3l / 2) = (1/2) × (m(3l)² / 3) × ω²

ω = √(g / l)

Velocity of Centre of Mass of Rod BC:
V_cm = √(g / l) × 2l

V_cm = 2√(gl)

If the time taken by the center of mass of rod BC from the breaking position to line PQ is t,

8l = (1/2) × g × t²

t = 4√(l / g)

Angular Displacement:
θ = ω × t = 4 radians

  [l = Change in length, L = Original length] (where Y is Young's modulus of elasticity)
Since Y, L and A remain same.

The boiling point of an azeotropic mixture of water and ethyl alcohol is less than that of theoretical value of water and alcohol mixture. Hence, the mixture shows positive deviation from Raoult's law.
Positive deviations from Raoult's law are noticed when
(i) Exp. vaue of vapour pressure of mixture is more than calculated value.
(ii) Exp. value of boiling point of mixture is less than calculated value.
(iii) ΔH_mixing = +ve
(iv) ΔV_mixing = +ve

Here X is 2, 4, 6-tribromofluorobenzene

When an organic compound is present in an aqueous medium, it is separated by shaking it with an organic solvent in which it is more soluble than in water. The organic solvent and the aqueous solution should be immiscible with each other so that they form two distinct layers which can be separated by a separatory funnel.

The hybridisation of the central atom can be decided based on the number of atomic orbitals used in the hybridisation. This can be found using the formula:
H = 1 / 2 [V.E. + M − C + A]
Where,
H = Number of hybrid orbitals;
V.E.= Number of valence electrons on the central atom;
M = Number of monovalent atoms attached to the central atom;
C = Magnitude of charge if the given species is a cation;
A = Magnitude of charge if the given species is an anion.
Based on this, the hybridisation of NO⁻³, NO⁺²  and NH⁺⁴ can be found out as:

Form Raoult's law, 

Putting the known information, we get

M = 100

The elevation in the boiling point is given by:
T′_b − T_b = ΔT_b = i × K_b × m
Where:

• i = Van't Hoff factor for the solute
• K_b = Molal elevation constant
• m = Molality of the solution

Now,
100.16 − 100 = 0.16 = 1 × K_b × 0.1 ...(1)
And
T′_b − 100 = ΔT_b = 1 × Kb × 0.5 ...(2)
From equations (1) and (2):
T′_b − 100 = 0.8
T′_b = 100.80°C

Δ_rG^∘ = 0−77.1 × 2 = −154.2 kJ/mol

= 
ΔG = Δ_rG^∘ + RT ln Q

= −129.5 kJ/mol

Benzoic acid is monobasic acid, so its valence factor is 1 and molar conductivity of benzoic acid is equal to its equivalent conductivity.
By Kohlrausch's law:

Let tan⁻¹x = θ

⇒ y = tan⁻¹x + c

The roots of the equation x² − 9x + 8 = 0 are 1 and 8.
Let y = [sin2α + sin4α + sin6α + ... ∞] logₑ2

According to question
2^{(tan 2α)} = 8 = 2³ ⇒ tan ²α = 3
⇒ tan α = √3 ⇒ α = π/3
∴ sin⁻¹(sin 2π/3) = π - 2π/3 = π/3 = α

A² − 4A + 10I = A

⇒ 9 − 3k = −3, −6 + 2k = 2
And 4 + k² − 4k = k
⇒ k2 − 5k + 4 = 0 ⇒ k = 4, 1
But, k = 1 does not satisfy the Eq (1)1.

The equation of the curve is y = 2x⁴ − x² = (2x² − 1)x²
The curve is symmetrical about the y-axis.
Also, it is a polynomial of degree four having roots 0, 0, ±1√2
x = 0 is repeated root. Hence, the graph touches x-axis at (0, 0) and intersects the x-axis at

So,  x= ±1/2 are the points of local minima
Thus, the graph of the curve is shown in the diagram

Here, y ≤ 0, as x varies from x = −1/2 to x = 1/2
∴ The required area
= 2 Area OCDO

= 7/120 sq. units

Given, f(x + y, x − y) = xy
Let, x + y = p, x − y = q
Then, 


= 
= 0

Dividing the given differential equation by y² cos x:

- (1 / y²) (dy/dx) + sec x * (1 / y) = 1 - sin x

Rewriting:

d/dx (1/y) + sec x * (1/y) = 1 - sin x

This is a linear differential equation with integrating factor (I.F.):

I.F. = exp ∫ sec x dx = sec x + tan x

Solution:

(sec x + tan x) / y = ∫ ((1 - sin x)(1 + sin x) dx) / cos x

= ∫ cos x dx = sin x + c

Given initial condition: x = 0, y = 1
⇒ c = 1

Thus,

y = (sec x + tan x) / (sin x + 1)

Finally, solving for y(π/3):

y(π/3) = 2

Given, f(1) = 1/3 and  for all x ≥ 1
Using Newton-Leibniz formula.
Differentiating both sides,

Integrating both sides,

∴ 3f(2) = 8