JEE Main Practice Mock Test - 2 Free Online Test 2026

Applying energy conservation

Given, galvanometer resistance, R_G = 40Ω
Shunt resistance, R_λ = 2Ω
Reading = 10 div/mA
Number of divisions, n = 50

∴ Galvanometer current, I_G = 50 / 10 = 5 mA

Let shunt current be I.

Since,
I / (R_G + R_λ) = I_G / R_λ

I = (5 × (40 + 2)) / 2

I = 105 mA

The escape velocity for the surface of earth is or

So, v_es(p) = 12v_es(e)

The pressure due to surface tension = 4T / R

The pressure due to electrostatic forces = σ² / 2ε₀

Just before the bubble bursts,

4T / R = σ² / 2ε₀ or R = 8Tε₀ / σ²

A→B  ⇒ constant pressure(linear)
B→C  ⇒ constant temperature(linear)
C→A  ⇒ constant volume(hyperbola)

Here,
T₁ = 6000 K, λ₁ = 4800 Å
T₂ = 3000 K, λ₂ = ?
According to Wien's displacement law,
∴ λ₂ / λ₁ = T₁ / T₂
⇒ λ₂ = (T₁ / T₂) × λ₁
= (6000 / 3000) × 4800
= 9600 Å

The magnetic field at the origin due to the current in the wire consists of contributions from three segments, but only the part of the wire parallel to the Y-axis contributes significantly here.

The magnetic field due to a semi-infinite straight current-carrying wire at a perpendicular distance a from the wire is given by:
B = (μ₀ I) / (4π a)

For the given geometry, the direction of the magnetic field vector is along (-î + k̂) direction, normalized by √2.

Thus, the magnetic field at the origin is:

B = (μ₀ I) / (4π a) × ( (-î + k̂) / √2 ) = (μ₀ I) / (8π a) (-î + k̂)

Therefore, option (c) is correct.

In the forward biased mode, we apply voltage in a direction opposite to that of barrier potential, that is, p-side to the positive terminal and n-side to the negative terminal of the battery. Thus, electrons in the n-side, holes in the p-side are pushed towards the junction which results in increased diffusion.
In reverse biased mode, we apply voltage in a direction of that of barrier potential. The electrons in n-side, holes in p-side pushed away from the junction that results in the drift current.

Initial angular moment about 'O' is zero and also no torque about 'O'.
So ω = 0
∴   Not possible

Option D is correct.

Use the formula g = GM/R², where G and M are constants and R is the original radius.

For the new radius take R' = R/2.

The new acceleration is g' = GM/(R')² = GM/(R/2)² = GM/(R²/4) = 4·GM/R².

Since g = GM/R², we get g' = 4g.

Therefore the required acceleration is 4g, so option D is correct.

Induced electric field,

Torque due to field about centre of ring,

The ring starts rotating when,
τ due to electric field = τ due to friction.
τ₁ = (μmg)r
Solving,

= 4 s

As work done in stretching an elastic body per unit volume is given by


So

Now according to given problem heat produced H = (75/100)  W = 3/4.W



i.e. 

Assertion (A) Electronic configuration of boron (Z = 5) is [He] 2s²2p¹. So, in first ionisation ( 1IE₁ or Δ_iH₁ ) removal will take place from unpaired p¹-electron. Whereas that of Be will be from paired 2s²-electrons which requires more energy.
Electronic configuration of Be(Z = 4) : [He]2s² So, the Assertion is a correct statement.
Reason (R) s-orbital is being symmetrical in shape (spherical), it shields nuclear force (nuclear charge) strongly. So, 2p¹-electron of B is experiences lesser nuclear attractive force for ionisation. As a result,
IE₁ or Δ_iH₁ : B < Be
So, the Reason is correct explanation for Assertion.

Stability Pb⁺² > Sn⁺² > Ge⁺²

In PbI₄, Pb⁺⁴ is less stable.

Acidic nature:

CO₂ > SiO₂ > GeO₂ > SnO₂ > PbO₂

GeO₂, SnO₂, PbO₂ = amphoteric

B.E. = N - 4 > B - 4 > As - 4

Initially aldehyde reacts with Grignard reagent to form a secondary alcohol.  Which undergo nucleophilic substitution reaction.

The structure of A is

The isomerism in the reaction is called keto-enol tautomerism. Enol form tautomerises into keto form.

 (Aliphatic amines) Most basic due to +I effect of methyl group. Methoxy group is electron donating group and increase electron density by (+M) effect on NH₂ group.

-NO₂ group withdraws electron density from N of −NH₂ and decrease basic strength.

 Most basic because NH₂ connected with sp³ C while in i,ii&iii Connected with sp² C & basic strength  ∝∝ +M/+I
basic strength ∝ 1/-M ∝ 1/+1
NO₂ exert -M so least basic while -OCH₃ exert +M so more basic.

M effect is a stronger effect than the I effect.
Aliphatic amines are more basic than aromatic amines because in aliphatic amines lone pair of electrons on nitrogen is localized and easy to donate, while in aromatic amines lone pair of electrons on nitrogen atom is delocalized and participate in resonance.

% of N in the compound = 1%

∴ mass of N in the compound = (1/100) × 2.8 g

Number of moles of N in the compound = (1/100) × (2.8/14) = 0.2/100 mol

Number of mmol of NH₃ formed = (0.2/100) × 1000 = 2

Total number of mmol of H₂SO₄ added = 60 × (1/20) = 3

meq of H₂SO₄ = 3 × 2 = 6

Let the volume of NaOH used be V mL

meq of H₂SO₄ = meq of NH₃ + meq of NaOH

6 = 2 + (1/20 × VₘL)

⇒ 4 × 20 = VₘL

⇒ 80 mL = VₘL

Reaction:

3Fe₂O₃ (s) → 2Fe₃O₄ + (1/2) O₂

Given data:

• 3 mol of Fe₂O₃ corresponds to 0.5 mol of O₂
• Mass of 3 moles Fe₂O₃ = 3 × 160 g
• Mass of (1/2) mol O₂ = (1/2) × 32 g

Mass loss calculation:

• Loss of 16 g O₂ corresponds to 480 g Fe₂O₃
• Loss of 0.04 g O₂ leads to Fe₂O₃ loss:
  0.04 × (480 / 16) = 1.2 g Fe₂O₃

Percentage by mass of Fe:
(0.84 / 1.5) × 100 = 56%

We have, 

⇒ A = 2(π + 1)

Hence, f(x) = (π + 1) sin x + 2(π + 1)

∴ fₘₐₓ = 3(π + 1) = M

and fₘᵢₙ = (π + 1) = m

⇒ M / m = 3

In the integral J, substitute x + 1 = t

⇒ dx = dt and x² + 2x = (t² - 1)

Now,  and 

Hence,

x² + y² + 4x + 6y − 3 = 0
has centre C(−2,−3) and radius =4
∴r₂ = 4 + OC / 2 & r₁ = 4 − OC / 2

Put z = x + iy

(x + iy) + i(|z|) = i(x - iy) + 1

⇒ x = y + 1 & y + |z| = x

⇒ x - y = 1 & x - y = |z|

⇒ |z| = 1

Variance of (3xᵢ + 2) = 9 var(xᵢ)
= 9 ( (Σ xᵢ²) / 5 - ( (Σ xᵢ) / 5 )² )
= 9(5 - 1)
= 36
⇒ S.D = 6

∴  Coefficient of x⁵ in the given expression
= Coefficient of x⁵ in 
= Coefficient of x⁵ in 
= ³¹C₆ − ²¹C₆

Given equation of hyperbola is

Equation of directrix will be x = ±a / e,
Where, 
i.e. k≥  −1
 is the equation of directrix


⇒ k = 2 as k = −3 is rejected
So equation of hyperbola will be 
∴  (√5,−2  satisfy the given hyperbola

Let

Let e₁ & e₂ be the eccentricities of the ellipse & hyperbola.
Since, b = a * e₁ ...(1)
b² = a² (1 − e₁²) ...(2)
a² = b² (e₂² − 1) ...(3)
From (1) & (2):
2e₁² = 1 ⇒ e₁ = √(1/2)
From (1) & (3):
2 = e₂² − 1 ⇒ e₂ = √3
Therefore, √3 * λ = √3
⇒ λ = 1

Given, 
and 
Let 
So, 

On comparing we get,

On solving equation (i), (ii) and (iii) we get,

So,  = 29