JEE Main Mock Test - 8 Free Online Test 2026

Here in the graph we can see that, the Stopping potential is same for 1 and 2 . So, frequencies will be same i.e. y₁ = y₂ and, currents are different. So, intensity are different i.e. I₁ ≠ I₂.

Minimum force required to move a body up a rough inclined plane:
F₁ = mg (sinθ + μ cosθ)

Minimum force required to prevent the body from sliding down the rough inclined plane:
F₂ = μmg cosθ

According to the question,
F₁ = 3F₂

So,
mg (sinθ + μ cosθ) = 3 (μmg cosθ)
sinθ + μ cosθ = 3μ cosθ
sinθ = 2μ cosθ
tanθ = 2μ

Substituting μ = 1 / 2√3:
tanθ = (2 × (1 / 2√3)) = 1 / √3
= tan 30°

Thus, θ = 30°

AB → Isothermal
P_A V_A = P_B V_B …(i)

BC → Adiabatic

CD → Isothermal
P_C V_C = P_D V_D …(iii)

DA → Adiabatic

From (i), (ii), (iii), and (iv):
V_B / V_C = V_A / V_D

If the K.E. of the system just after the collision is sufficient to rise the block as shown, the block will topple, otherwise it will not topple down.

The energy of the photon

E = (hC) / λ

= (4 × 10⁻¹⁵ eVs × 3 × 10⁸ m/s) / (300 × 10⁻⁹ m)

= (4 × 10²) / 300 eV 

= (4/3) eV 

= 1.33 eV
The ionisation energy is 13.6eV which is greater than energy of photon, so atom can not come into excited state and will remain in ground state

Tension = [MLT⁻²]
Surface Tension = [ML⁰T⁻²]
Clearly these two have different dimension.

If  = constant, then = 0

So,  ×  = 0 ⇒ F should be parallel to , so the coefficients should be in the same ratio.

So, α/2 = 3/-6 = 6/-12

So, α = -1

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)

⇒σ = 1.185

Here, v = 2 × 10⁸ m s⁻¹ and μ_r = 1
The speed of electromagnetic waves in a medium is given by

where μ and ε are absolute permeability and absolute permittivity of the medium.
Now, μ = μ₀μ_r and ε = ε₀ε_r
Therefore, the equation (i) becomes

dU = dW/2
By1st law δQ = dU + δW → dQ = 3dU = 3nC_vdT   [as dU = dW/2]
Molar heat capacity

Given C = 15/2R
∴ C_v = 5/2R
∴ Degree of freedom = 5
So Gas is diatomic
∴ Diatomic

Voltage sensitivity V_s = θ/V
Current sensitivity I_s = θ/I
Also, the potential difference
V = IG
Hence,
∴ V_s/I_s = I/V = 1/G

NO: σ1s² σ1s² σ2s² σ2s² σ2p² π2p² = π2p² π2p¹ = π2pᵧ⁰ and it is similar to O₂⁺

The given compound is:  

The stereoisomers of the given compound are:

On fractional distillation of the mixture, we get only two fractions (of two enantiomers (±) pair) because enantiomers have the same Physical properties.

When phenol treated with chloroform (CHCl₃) in the presence of aqueous sodium hydroxide (NaOH), product will be 2-hydroxy benzaldehyde (salicylaldehyde).

When phenol react with CO₂,NaOH, it gives salicylic acid (o-hydroxy benzoic acid).

For the enolization carbonyl compound. Carbonyl compound must contain at least one  α - H
Compunds  (i) , (v) have not alpha hydrogen so these compounds  will not show enolization.
Compounds (ii),(iii),(iv) have alpha hydrogen so these compound wil show enolization.

(i)compound = 0 (α − H)
(ii)compound = 2 (α − H)
(iii)compound = 5 (α − H)
(iv)compound = 2 (α − H)
(v)compound = 0 (α − H) 

I. [Co(NH₃)₆]³⁺

Here Co is present as Co³⁺ ion. NH₃, being a strong field ligand, pairs up the unpaired electrons of Co.

II. [Ni(CO)₄]

Here Ni is present in its ground state. CO, being a strong field ligand, pairs up the unpaired electrons of Ni.

²⁸Ni = [Ar] 3d⁸ 4s²

III. [Pt(NH₃)₂Cl₂]

Here Pt is present as Pt²⁺ ion.

IV. [CoF₆]³⁻

Here Co is present as Co³⁺ ion. F, being a weak field ligand, is unable to pair up its unpaired electrons.

Co³⁺ = [Ar] 3d⁶ 4s⁰

V. [Fe(CO)₅]

Here Fe is present in its ground state.

₂₆Fe = [Ar] 3d⁶ 4s²
 

The boiling point of alkyl/aryl halides increases with increase in the molecular weight because, the surface area and the magnitude of the Van der Waals forces increases. Hence, the increasing order of boiling points is, 
1−Bromoethane < 1−Bromopropane < 1−Bromobutane < Bromobenzene

If vol. is 10 lit

P₂ = (1 × 20) / 10 = 2 atm

If V = 2 lit

then P₃ = (2 × 10) / 2 = 10 atm

W = (20 - 10) × 2 + (10 - 2) × 10

= 100 lit - atm

When we react the compound X with dilute HNO₃ and AgNO₃
It forms white precipitate so, indicates the presence of chloride ion.
When we add the NaOHNaOH in XX it forms white precipitate which dissolves in excess of NaOH it indicates that the cation may be Zn²⁺ or Al3+

But on passing H₂S gas only Zinc form white precipitate 

So, Compound  X is ZnCl₂
Atomic Number = Number of protons in atom or Number of electrons in an atom
Hence, The atomic number of cation (Zn²⁺) is 30

Molality of Na⁺ ions (m) = Number of moles of Na⁺ ions / Mass of water in kg
= 
= 4 mol kg⁻¹ or 4 m

From (1) and (2)

Let C(cosθ,sinθ);H(h,k) is the orthocentre of the △ABC

h = 1 + cosθk = 1 + sinθ

(x−1)² + (y−1)² = 1

x² + y² − 2x − 2y + 1 = 0

We have

This limit exists, if 1 + a − b = 0, and the given limit is equal to 1, if

As per the given conditions:
α + β = α² + β² and αβ = α²β²
This implies:
αβ = 0 or 1
If αβ = 0, then let α = 0 ⇒ β = 0 or 1
If β = 1/α, then:
α + 1/α = α² + 1/α² = (α + 1/α)² - 2
This gives:
(α + 1/α)² - (α + 1/α) - 2 = 0
Solving for α + 1/α:
α + 1/α = 2 or -1
This gives α = 1 or α = ω, ω²
Hence, the number of such equations is four: (0,0), (0,1), (1,1), and (ω, ω²).

Given,

Now, let a + b = 2C, b + c = 2A, and c + a = 2B.
⇒ a + b + b + c + c + a = 2A + 2B + 2C
⇒ a + b + c = A + B + C
Also,
a = (a + b + c) − (b + c) = (A + B + C) − 2A = B + C − A
Similarly,
b = C + A − B, c = A + B − C.

= 
= 

[Expanding along C₁]

Hence, 4 is the other factor of the determinant.

Hence the equations is

x² + 24x + c = 0

|α - β| = (6√6)ᵏ

k = log₆10 - log₆5 + 1/2 (log₆ (log₆ (18 × 72)))

= log₆2 + 1/2 · log₆ (log₆ (1296))

= log₆2 + 1/2 · log₆4

= log₆2 + log₆2 = log₆4

|α - β| = (6√6)ˡᵒᵍ⁶⁴ = 8

(α - β)² = 64

(α + β)² = 4αβ = 64

576 - 4C = 64

4C = 512 ⇒ c = 128

1 can be put by 6 ways
−1 can be put by 1 way 2 can be put by 4 ways
−2 can be put by 1 way 3 can be put by 2 ways
−3 can be put by 1 way
∴ Number of skew symmetric matrices
= 6 × 1 × 4 × 1 × 2 × 1 = 48

∵  lies between the parallel lines x + 2y = 1 and 2x + 4y = 15, then

Hence α = −1,  0,  1
Hence, number of integral values of α is 3.