JEE Main Practice Mock Test - 8 Free Online Test 2026

Since the pass axes of two polarizers P₁ and P₂ were kept such that the incident unpolarized beam of intensity I₀ gets completely blocked. This means that both polarizers were kept cross (perpendicular of their pass axes) to each other. When the third polarizer P₃ is introduced between P₁ and P₂ such that the angle between the pass axes of P₁ and P₃ is 60°, i.e., θ₁ = 60°.

Intensity of polarized light emerging from the first polaroid,
I₁ = I₀ / 2

Intensity of polarized light (I₃) emerging from polaroid P₃ is given by the law of Malus, i.e.,

I₃ = I₁ cos² θ₁
= (I₀ / 2) cos² 60°
= (I₀ / 2) * (1 / 4)
I₃ = I₀ / 8

Now, the angle between the pass axis of P₃ and P₂,
θ₂ = 90° - 60° = 30°

Thus, the intensity of polarized light emerging from the second polaroid P₂:

I₂ = I₃ cos² 30°
= (I₀ / 8) * (3 / 4)
= 3I₀ / 32

According to the problem, the mass of gases is equal, so the number of moles will not be equal, i.e., μ_A ≠ μ_B.

From the ideal gas equation P_V = μRT,

⇒ (P_AV_A) / μ_A = (P_BV_B) / μ

[As the temperature of the container is equal]

From this relation, it is clear that if P_A = P_B, then

V_A / V_B = μ_A / μ_B ≠ 1, i.e., V_A ≠ V_B

Similarly, if V_A = V_B, then P_A / P_B = μ_A / μ_B ≠ 1, i.e., P_A ≠ P_B

Option B is correct: A-4, B-3.

Angular momentum has dimension given by r × p, where p (linear momentum) has dimension [M L T⁻¹] and r has dimension [L]. Therefore angular momentum has dimension [M L² T⁻¹].

Torque has dimension given by r × F, where force [F] = [M L T⁻²] and r is [L]. Hence torque has dimension [M L² T⁻²].

Tension is a force; its dimension is the same as that of force: [M L T⁻²].

The gravitational constant G appears in F = G m₁ m₂/r². Rearranging gives G = F r²/m², so the dimension of G is [M⁻¹ L³ T⁻²].

From the above, angular momentum = [M L² T⁻¹] (matches 4) and torque = [M L² T⁻²] (matches 3). Therefore the pairing A-4, B-3 is correct.

The solution is based on the analysis of the shown circuit.

• In the given circuit, all resistors are of 1 ohm, including the middle resistor.
• Since the circuit is symmetrical, the current through resistor "x" is zero. This is because there is no potential difference across it.
• The current distribution in symmetrical circuits often leads to zero current through the central or connecting resistors when arranged in a bridge-like configuration.

Thus, the correct statement is that the current through resistor "x" is zero.

Given Work Functions:
W_Na = 2.75 eV
W_Cu = 4.65 eV
W_Au = 51 eV
The frequency range of visible light is 4 × 10¹⁴ Hz to 8 × 10¹⁴ Hz, so we calculate the maximum and minimum energy of photons in visible light.

Using E = hν,

Maximum Photon Energy:
E_max = (6.63 × 10⁻³⁴ × 8 × 10¹⁴) / (1.6 × 10⁻¹⁹)
= 331 eV

Minimum Photon Energy:
Similarly,
E_min = (6.63 × 10⁻³⁴ × 4 × 10¹⁴) / (1.6 × 10⁻¹⁹)

Now, comparing the work functions of the mentioned metals with E_max, we see that only sodium can emit photoelectrons.

Further, a sodium device can also work with photons of energy higher than 2.75 eV, such as X-rays and UV rays.

[OA→ circular path]

[AB→ circular path]

[BC→ parabolic path]

From B to C,

Co-ordinate of pt. C

∴ x = 2 & y = 2 So 2 + 2 = 4

Time required to empty the tank,

∴  t₂ = 2t

Refer figure, A and B are the extreme position and O is the mean position.
For SHM, acceleration and force is always directed towards the mean position. It is given that the direction A to B is positive.

Acceleration
a = F/m = 4/20 = 1/5ms⁻²
(Displacement in n^th second is given by S_nth = u + a/2(2n − 1)
Distance covered by body in 3rd second
= 1/2 × 1/5 × (2 × 3 − 1) = 5/10 = 1/2m
∴  W = 4 × 1/2 = 2J

When an electron is moving in a circular path with angular velocity, ω, then-current produced in this process, I = eω / 2π, and strength of magnetic field due to the circular current-carrying wire, B = μ₀I / 2R, here μ₀ is the permeability of in free space. It means, , here the value of , e = 1.6 × 10⁻¹⁹ C, v = 2.19 × 10⁶ m s⁻¹ and R = 0.529 × 10⁻¹⁰ m,
 

The constituents of nucleic acids are nitrogenous bases, sugar and phosphoric acid. The sugar present in DNA is D(−)−2-deoxyribose and the sugar present in RNA is D(−) ribose. Due to these D(−)-sugar components, DNA and RNA molecules are chiral molecules.

(a) [Fe(CN)₆]³⁻ has d²sp³ (inner d-complex) hybridisation with one electron unpaired.

(b) The Fe²⁺ ion in [Fe(CN)₆]⁴⁻ has an outer electronic configuration of 3d⁶. The complex is a low-spin complex. It contains 0 unpaired electrons with a magnetic moment of 0 BM. Therefore, under the influence of the octahedral crystal field, the possible electronic arrangement of Fe(II) ion is t₂g⁶, e₉⁰.

(c) [Ni(CN)₄]²⁻ has a square planar geometry formed by dsp² hybridisation. [Ni(CN)₄]²⁻ is diamagnetic, so Ni²⁺ ion has 3d⁸ outer configuration with two unpaired electrons.

For the formation of the square planar structure by dsp² hybridisation, two unpaired d-electrons are paired up due to energy made available by the approach of ligands, making one of the 3d-orbitals empty.

∴Cl⁻ is a weak ligand so, there is no pairing of electrons.

Number of unpaired  = 2

Hence, the correct option is (d).

Ionisation isomers in coordination chemistry is when complex compound having same molecule formula but gives different ions in aqueous solution on ionisation.

[Co(NH₃)₅SO₄]Br and [Co(NH₃)₅Br]SO₄

[Co(NH₃)₅SO₄]Br  ⇌  [Co(NH₃)₅SO₄]⁺ + Br⁻

[Co(NH₃)₅Br]SO₄  ⇌  [Co(NH₃)₅Br]²⁺ + SO₄²⁻

Due to inert pair effect Pb has four electrons in its valence shell but it shows +2 oxidation state. In other words due to inert pair effect +2 oxidation state is more stable than +4 of Pb.

So, both Assertion and Reason are true and Reason is the correct explanation of Assertion.

This product is highly stable conjugate di-ene system with benzene ring also so it is formed here as major product.

Acidic nature ∝ − I (Electron withdrawing groups)
More the −I group at α 'C' of carboxylic acid greater the acidic strength.

P is sp³ hybridised in P₄. so, p character is 75% in it.

Given function:
f(x) = P e^(2x) + Q e^x + R x

Differentiation:
f'(x) = 2P e^(2x) + Q e^x + R

Given:
31 = 2P e^{² ln 2} + Q e ^{ln 2} + R
8P + 2Q + R = 31 ...(i)

Also, -1 = P + Q ...(ii)

15P + 6Q = 39 ...(iii)

Solving (i), (ii), and (iii), we get P = 5, Q = -6, R = 3.

General term of (a + b/x²)¹¹ is

T₍ᵣ₊₁₎ = C₍ᵣ₎¹¹ (ax)¹¹⁻ʳ × (b/x²)ʳ

⇒ T₍ᵣ₊₁₎ = C₍ᵣ₎¹¹ a¹¹⁻ʳ bʳ x¹¹⁻³ʳ

For coefficient of x⁻⁷,

11 - 3r = -7
⇒ 3r = 18
⇒ r = 6

So,

T₍₆₊₁₎ = C₆¹¹ a¹¹⁻⁶ b⁶ x⁻⁷

T₍ᵣ₊₁₎ = (C₆¹¹ a⁵ b⁶) x⁻⁷

So, L = C₆¹¹ a⁵ b⁶

Similarly, the general term of (bx² + a/x)¹¹ is

T₍ᵣ₊₁₎ = C₍ᵣ₎¹¹ (bx²)¹¹⁻ʳ × (a/x)ʳ

⇒ T₍ᵣ₊₁₎ = C₍ᵣ₎¹¹ b¹¹⁻ʳ × aʳ × x⁽²²⁻³ʳ⁾

So, for the coefficient of x⁷,

22 - 3r = 7 ⇒ r = 5

Hence,

M = C₅¹¹ b⁶ a⁵

So, L + M = 2 × C₆¹¹ a⁵ b⁶ = 924a⁵b⁶

And, the general term of (ax² + b/x)¹² is

T₍ᵣ₊₁₎ = C₍ᵣ₎¹² (ax²)¹²⁻ʳ × (b/x)ʳ

T₍ᵣ₊₁₎ = C₍ᵣ₎¹² a¹²⁻ʳ bʳ x⁽²⁴⁻³ʳ⁾

For the coefficient of x⁶,

24 - 3r = 6 ⇒ r = 6

So, the coefficient of x⁶ is

= C₆¹² a⁶ b⁶ = 924 a⁶ b⁶

Hence,

L + M = (1/a) × [Coefficient of x⁶ in (ax² + b/x)¹²]

Let y = (x + 1)(x + 2) ...... (x + n)
⇒ log y = log (x + 1) + log (x + 2) + ...... + log (x + n)
⇒ (1/y) (dy/dx) = (1 / (x+1)) + (1 / (x+2)) + ...... + (1 / (x+n))
⇒ dy/dx = [(x + 1) ⋅ (x + 2) ...... (x + n)] [ (1 / (x+1)) + (1 / (x+2)) + ...... + (1 / (x+n))

Given limit can be written as

Using L'Hospital' rule,


= 

Let A(0, -3), B(-2√3, 3), and C(2√3, 3)
Let, D, E & F are the mid-point of sides AB, BC and CA respectively.

∴ ΔDEF is an equilateral triangle.

Given equation of hyperbola is

Here, a² = cos²α and b² = sin²α
(i.e., comparing with standard equation )
We know, foci = (±ae, 0) where

⇒Foci = (±1, 0)
whereas vertices are (±cos α, 0)
⇒ae = 1
⇒Eccentricity, e = 1/cosα
Hence, foci remains constant with change in 'α'.

Here is the corrected text without LaTeX:
Set S = {1, 2, 3, ..., 12}
Given that A ∪ B ∪ C = S and A ∩ B = B ∩ C = A ∩ C = ϕ
Set S is partitioned into 3 equal parts A, B, C, each containing 4 elements.
Thus, the number of ways to partition is:
¹²C₄ × ⁸C₄ × ⁴C₄ = (12! / (4! 8!)) × (8! / (4! 4!)) × (4! / (4! 0!)) = 12! / (4!)³.

Let 'l' and 'b' be the length and breadth, respectively, of the rectangle.
∴ Area A = l × b = b² × b = b³
∴ dA/dt = 3b² (db/dt)
⇒ 48 = 3b² (db/dt) ...(i)
But l = b²
⇒ dl/dt = 2b (db/dt) ...(ii)
⇒ dl/dt = 2b × (1 / 3b²) × dA/dt
= (2/3b) × dA/dt
⇒ dl/dt | b = 4.5 = (2 × 48) / (3 × 4.5) = 192 / 27 cm/sec
Hence, 27λ = 192

Differentiating with respect to x, we get

⇒ 4λ = 1