JEE Main Mock Test - 10 Free Online Test 2026

In terms of mass-energy interconversion, a chemical reaction is conceptually equivalent to a nuclear reaction. The difference in chemical (not nuclear) binding energies of atoms and molecules on the two sides of a reaction can be traced to the energy emitted or consumed in a chemical reaction.

A chemical reaction's mass defects, on the other hand, are about a million times smaller than those in a nuclear reaction.

Since, h = 1 / 2at² ⇒a should be same in both cases, because h and t are same in both cases as given.

In (i) F₁ - mg = ma
⇒ F₁ = mg + ma

In (ii) 2F₂ - mg = ma
⇒ F₂ = (mg + ma) / 2

∴ F₁ > F₂.

The escape velocity for the surface of earth is or

From triangle,  sinθ = h / l

Putting this value in (1), we get, required expression.

L.C. = M.S.D − V.S.D
20(VSD) = 18 (MSD)
⇒ VSD = 18/20 MSD
∴ L.C. = (1 − 18/20) MSD
= 2/ 20 × 1 mm
= 0.1 mm

The wavelength of spectral line in the Balmer series is given by: 1/λ = R [1/2² - 1/n²]
For the first line of the Balmer series, n = 3:
⇒ 1/λ₁ = R [1/2² - 1/3²] = 5R/36
For the second line, n = 4:
⇒ 1/λ₂ = R [1/2² - 1/4²] = 3R/16
∴ λ₂ / λ₁ = 20/27
⇒ λ₂ = (20/27) × 6561 = 4860

on putting values V = √10 m/s
Ans. (1) After C is brought in contact with A, charge on A is q/2 and on C is q/2. Now C is in contact with B.
The charge on B is q + q/2 / 2 = 3q / 4
∴ 

Uniform current is flowing. Current enclosed in the first amperian path is,

πR² = (I r₁²) / R²

∴ B = (μ₀ × current) / path
= (μ₀ I r₁²) / (2π r₁ R²)
= (μ₀ I r₁) / (2π R²)

Magnetic induction at a distance,

r₂ = (μ₀ I) / (2π r₂)

∴ B₁ / B₂ = (r₁ r₂) / R²
= (a / 2) * (2a / a²)
= 1

Induced emf is given by ε = Δφ / Δt
Current, I = Q / Δt = Δφ / Δt × 1 / R [where Q is the total charge passed in time Δt]
Thus, Q = Δφ / R

As here initially,
P_I = P₀ = 1 × 10⁵ N/m²,  V_I = V₀ = 2.4 × 10^-3 m³,
T₁ = T₀ = 300 K
​And finally,

= 2 × 10⁵ N/m²
V_F = V₀ + Ax = 2.4 × 10^-3 + 0.1 × 8 × 10^-3
= 3.2 × 10^-3 m³

So from gas equation PV = nRT, i.e.,
We have, 

T_F = 800 K
As, 
So, 

i.e.,  ΔW = 80 + 40 = 120 J
Note : 80 J of work is done against atmosphere and 40 J against spring.
and as 

Hence total heat supplied,
ΔQ = ΔU + ΔW = 600 + 120 = 720 J
Note : In this problem none of the variables F, P, V, T, U or Q is constant.

When the rod is bent in the form of a square, then each side has a resistance of 1/4 Ω.
As shown R₁, R₂ and R₃ are connected in series, so, their equivalent resistance,





= 3/16 Ω

From the graph given in the question, we can see that the current is rising up to time T₂ from 0, and then decaying up to time T from T₂.
Now, the voltage induced in an inductor is given as:
V = -L (dI/dt) ⇒ |V| = L (dI/dt) ....(1)
If we observe the slope from t = 0 to t = T₂, dI/dt is positive and constant, while L is already constant. This shows that the voltage value remains unchanged from t = 0 to t = T₂.
Similarly, if we observe the slope from t = T₂ to t = T, dI/dt is negative but constant, which shows that the voltage value remains unchanged from t = T₂ to t = T.
Hence, the graph between voltage and time is as shown below.

∴ V_A  –  V_B = 0 J

Forms the carbocation is SN1 mechanism, which gets the stability by delocalization of positive charge through two benzene rings.

First ionization energy for nitrogen is greater than oxygen.

This is due to stable configuration of nitrogen (halffilled 2p-orbital).

Due to screening effect the valency electron experiences less attraction towards the nucleus. This brings decrease in the nuclear charge (Z) actually present on the nucleus. The reduced nuclear charge is termed effective nuclear charge and its magnitude increases in a period when we move from left to right.

The order of increase of energy can be calculated from (n + l) rule. If two orbitals have same value of (n + l), the orbital with lower value of n will be filled first.
(i) For n = 4,l = 1,(n + l) = 4 + 1 = 5
(ii) For n = 4,l = 0,(n + l) = 4 + 0 = 4
(iii) For n = 3,l = 2,(n + l) = 3 + 2 = 5
(iv) For n = 3,l = 1,(n + l) = 3 + 1 = 4
Therefore correct order is
(iv) < (ii) < (iii) < (i).

(i) CH₃−CH₂−COOH
(ii) CH₃−COOH
(iii) C₆H₅−COOH
(iv) C₆H₅−CH₂COOH
Die to −I effect (iii) is most acidic and order will be (iii) > (iv) > (ii) > (i).

(A) Manganese show pale purple colour in flame test.
(B) 
(C) Both Cu²⁺ and Mn²⁺ form precipitate with H₂ S in basic medium.
(D) E^o_Cu2+ |Cu = +0.34 V, E^o_{Mn2 +} |Mn = −1.18 V

Bond order of He₂⁺ → 0.5
O₂⁻ → 1.5
C₂ → 2
NO → 2.5

Only aldehydes can be synthesized by Stephens reaction but not ketones. Alkane nitriles are reduced with stannous chlorides and hydrochloric acid in the presence of ether to form corresponding aldimines. Aldimines give aldehydes on hydrolysis but not ketones. This reaction is known as Stephen's reaction.

As we can see, 1 mL contains 70 mg of Na⁺. So, 50 mL contains:
50 × (70/1000) = 3.5 g
Gram atomic mass of sodium = 23 g
Number of moles of sodium = 3.5 / 23 = 0.152
Applying the POAC concept, the number of moles of Na⁺ is equal to the number of moles of NaNO₃.
Molar mass of NaNO₃ = 85 g/mol
So, mass of NaNO₃ = 0.152 × 85 = 12.92 g

Given Equilibrium Reactions:

1. NO + NO₃ ⇌ 2NO₂ ; Kc

   Initial: 1 - x/2, 3 - x/2, x

2. NO + NO₃ ⇌ 2NO₂ ; Kc

   Initial: 3 - x, 3 - x, 2x

Equilibrium Expression:

(x² / ((1 - x/2)(3 - x/2))) = (4x² / (3 - x)²) = (4x² / ((2 - x)(6 - x)))

Solving for x:

(3 - x)² = (2 - x)(6 - x)

9 + x² - 6x = 12 - 2x - 6x + x²

9 - 6x = 12 - 8x

2x = 3

x = 1.5

Calculating Kc:

Kc = (4 × 1.5 × 1.5) / (1.5 × 1.5) = 4

We have r̅ = (11î - 2ĵ - 8k̅) + λ(10î - 4ĵ - 11k̅)

So, coordinates of any point on this line are [(10λ + 11), (-11λ - 2), (-11λ - 8)]

Let P = (2, -1, 5) and let M be the foot of the perpendicular.

∴ The direction ratios of PM are (10λ + 9), (-4λ - 1), (-11λ - 13)

Since PM is perpendicular to the given line, we write

(10λ + 9)(10) + (-4λ - 1)(-4) + (-11λ - 13)(-11) = 0

100λ + 90 + 16λ + 4 + 121λ + 143 = 0
⇒ 237λ = -237
⇒ λ = -1

Now, M = (-10 + 11, 4 - 2, 11 - 8) = (1, 2, 3) ​

∴ Not differentiable at x = 1

∴ f¹(x) is discontinuous at x = 1

Also, f¹(3/2) + f¹(0) = -2(3/2) + 1 + 0 = -2

We can reduce the function by substituting
x = cosθ ⇒ θ = cos⁻¹x

As θ = cos⁻¹x

In the first quadrant, tan⁻¹ (tan (θ/2)) = θ/2

= cos(2 × (θ/2)) - 2 cos⁻¹ (sin (θ/2))

As sin⁻¹ x + cos⁻¹ x = π/2

y = cosθ - 2(π/2 - sin⁻¹ (sin (θ/2)))

As sin⁻¹ (sin (θ/2)) = θ/2 in the first quadrant,

y = cosθ - π + θ

y = x - π + cos⁻¹ x

Let, [x² - 1/2] = [x² + 1/2 - 1] = α ∈ I

⇒ [x² + 1/2] - 1 = α

∴ y = 2sin⁻¹(α + 1) + cos⁻¹α, where α = {-1,0}

At α = -1  
y = 2sin⁻¹(0) + cos⁻¹(-1) = π  

At α = 0  
y = 2sin⁻¹(1) + cos⁻¹(0) = 3π/2  

Given equation: ln(x + y) = 2xy ........(1)
For x = 0:
ln(0 + y) = 2(0) ⋅ y = 0
⇒ ln y = 0
⇒ y = 1
Now, differentiating (1), we get,

At point (0, 1), we get,

Now, from AM ⩾ GM


≥ 1 + 2a^-1

k² + 4a² = r² ....(i)
h² + (k − 2b)² = r² ....(ii)
Hence, from (i) and (ii):
k² + 4a² = h² + (k − 2b)²
⇒ 4a² = h² + 4b² − 4bk
⇒ 4a² = x² + 4b² − 4by
Thus, the locus is: 4a² = x² + 4b² − 4by, which represents a parabola.