KIITEE Mock Test - 7 - JEE MCQ

speed of electron of the hydrogen atom in the ground state,

v₁ = 2.2 x 10⁶ m/s

Since

∴

[for third excited state. n₂ = 4]

⇒

5.5 x 10⁵ m/s

Let us suppose that original length simple pendulum is L and time period is now if the acceleration due to gravity is g.
Now the frequency should be 
Given the ratio of frequencies,it means,

(A + B) · (A - B) = 0

A·A + B·A - A·B - B·B = 0

|A| + BAcosθ - AB cosθ - |B|² = 0

|A|² = |B|²

Hence. |A| = |B|

α = I_C/I_E = 0.98 (given)

The ratio of collector current to the base current at constant collector-emitter voltage is given by

β = f_C/f_a

These are related as β = α/1-α

⇒ β = 0.98/1-0.98 = 0.98/0.02 = 49

The heat of neutralisation of strong acid and strong base is -57.33 kJ. MgO is a weak base, while HCl is a strong acid. So, the heat of neutralisation of MgO and HCl is less than -57.33 kJ because MgO requires some heat in ionisation. Thus, the net released amount of heat is decreased.

For simple cubic cells,

Packing effedency =

∴ Percentage of unoccupied volume in Scc = 100 - 52.4 = 47.6%

Mohr's salt is (NH₄)₂SO₄.FeSO₄.6H₂O.
The equation is 5Fe²⁺ + MnO^-₄ + 8H⁺ 5Fe³⁺ + Mn²⁺ + 4H₂O
Total change in oxidation number of iron = (+3) - (+2)
So, equivalent weight of Mohr's salt
= 
= 
= 392

In crystallography, the tetragonal crystal system is one of the 7 crystal systems.
Tetragonal crystal lattices result from stretching a cubic lattice along one of its lattice vectors, so that the cube becomes a rectangular prism with a square base a  and height c, which is different from a.

At constant pressure, q_p = ΔU -W.

q_p = ΔU - pΔV

q_p = U₂ - U₁ - (-p(V₂ - V₁))

q_p = (U₂ + pV₂) - (U₁ + pV₁)

q_p = H₂ - H₁

q_p = ΔH

As we know, ΔH = ΔU + pΔV

ΔH = ΔU + W

Putting the value of ΔH in (1)

q_p = ΔU + W

Other corrected options are as follows:

• For adiabatic process ΔU = W

• For an isochoric process ΔU = q_v

• For an isothermal process q = -W

There are 4-metameric ethers that are represented by the molecular formula C₄H₁₀. Which are as follows

Anisole reacts with HI to give phenol and methyl iodide. First, anisole on protonation forms methylphenyl oxonium ion. Since the bond between O-CH₃ is weaker than the bond between O-C₆H₅, (because of a partial double bond is character between oxygen and carbon of the benzene ring) it breaks more readily.
Hence, the I^- ion attacks and breaks the O-CH₃ bond to give CH₃I and phenol.

The phenols formed do not react with halides because nucleophilic substitution on aromatic ring is difficult.

Electrochemical equivalent = Equivalent weight/96500

gm per coulomb. Its unit is kgC^-1.

⇒ 2 moles of sodium metal reacts to give 1 mole of ethane.

Also,

1 mole of Na metal = 23 g

∴ 2 motesol Na metal = 23x 2 = 46g

Thus, 46 g of sodium is required to prepare one mole of ethane from methyl chloride by Wurtz reaction.

p + q = - a, p.q = b
Now, p² q + q² p = pq(p + q)
= b(-a) = - ab
p²q . q²p = (pq)³ = b³
The required equation is
x² - (p²q + q²p) x+ (p²q . q²p) = 0
x² + abx + b³ = 0

Given, pair of lines, x² - 4pxy + 8y² = 0

Which is in the form of ax² + 2hxy + by² = 0

According to the question,

Sum of the slopes = 3 x product of the slopes

⇒ m₁ + m₂ = 3 x (m₁ m₂)

⇒

⇒ 4p = 3 ⇒ p = 3/4

According to the question, we get

Let E₁, E₂ and A be the events defined as follows
E₁ = six occurs
E₂= six does not occurs
A = the man reports that it is a six. We have.

Now, P(A/E₁)= Probability that the man reports that there is a six on the die given then 6 has occured on die
= Probability the man speaks truth

=3/4

P(A/E₂) = Probability that the man reports that there is six is on die given that six has not occured on die
= Probability that the man does not speak truth

We have to find P(E₁/A) i.e. the probability that there is six on the die given that the man has reported that there is six. By Baye's rule, we have

We know that, a_n = S_n - S_n-1

= 5(2ⁿ - 1) - 5(2^n-1 - 1)

= 5(2ⁿ - 2^n-1)

= 5(2^n-1)

1. Start with
   sin⁻¹(1−x) − 2 sin⁻¹ x = π/2

2. Let θ = sin⁻¹ x,
   so x = sin θ and θ∈[−π/2, π/2].
   Then the equation becomes
   sin⁻¹(1−sin θ) − 2θ = π/2

3. Take sine of both sides:
   1 − sin θ = sin(π/2 + 2θ) = cos 2θ = 1 − 2 sin² θ

4. So
   1 − sin θ = 1 − 2 sin² θ
   ⇒ 2 sin² θ − sin θ = 0
   ⇒ sin θ (2 sin θ − 1) = 0
   ⇒ sin θ = 0 or sin θ = 1/2
   ⇒ x = 0 or x = 1/2

5. Check in the original:
   • x=0 ⇒ sin⁻¹(1) − 2 sin⁻¹(0) = π/2 − 0 = π/2 ✓
   • x=1/2 ⇒ sin⁻¹(1/2) − 2 sin⁻¹(1/2) = π/6 − 2·(π/6) = −π/6 ≠ π/2 ✗

Therefore the only solution is
 x = 0

Put sin^-1 (0.4) = θ ⇒ sin θ = 0.4

∴ E = sin(3θ) = 3 sinθ - 4sin³θ

= 3(0.4) - 4(0.4)³

= 1.2 -4(0.064)

= 1.2 - 0.256

= 0.944

A =

Therefor, |A| =

= (1 + 2i)(1 - 2i) + i²

= 1 - (2i)² + (-1) = 1 + 4 -1 = 4

∴ A(adj A) = |A| = 4i

N₁ + N₂ + N₃ + N₄ + N₅  =  Total ways – {when no odd}

Total ways = ⁹C₅

Number of ways when no odd number is selected is 0 (∵ only available even are 2,4,6,8)