JEE Main Mock Test - 8 - JEE MCQ

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.

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

Given that, R₁ = 1 kΩ,

Zener breakdown voltage V_z = 6 V

Let I₁ and I₂ be the current passing through R₁ and the Zener diode, then using the equation,

I₁ = V_z / R₁ = 6 / 1000 A

According to the given condition, Zener current:

I_z = 5I₁ = 30/1000 A

Let I be the current drawn from the source battery (Vₛ = 30V), then

I = (Vₛ - V_z) / R = I₁ + Iₓ

(30 - 6) / R = 6/1000 + 30/1000

⇒ R = 2000/3 Ω

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.

Diode with no biasing ⇒ I_drift = i_diffusion  

Diode in forward bias ⇒İ_diffusion >> i_drift

Diode in reverse bias without breakdown

⇒ İ_drift> i_diffusion

Diode in reverse bias with breakdown

⇒ i_drift>> i_diffusion

Given:
Number of turns across the primary coil, Nₚ = 50
Number of turns across the secondary coil, Nₛ = 1500
Magnetic flux linked with the primary coil, ϕ = ϕ₀ + 4t

∴ Voltage across the primary coil,

Vₚ = dϕ/dt = d/dt (ϕ₀ + 4t) = 4 volt

Also,

Vₛ / Vₚ = Nₛ / Nₚ

∴ Vₛ = (1500 / 50) × 4 = 120 V

⇒ k = 2

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

Here, magnetic field, B = 0.025 T
Radius of the loop, r = 2 cm = 2 × 10⁻² m
Constant rate of shrinking of the loop,
dr/dt = 1 × 10⁻³ m/s
The linked magnetic flux is:
ϕ = B A cos(θ) = B (πr²) cos(0°) = B (πr²)
Using Faraday's law, the induced emf is of the magnitude:
|ε| = dϕ/dt = d/dt (Bπr²) = (πB) 2r (dr/dt)
Substituting values:
|ε| = 0.025 × π × 2 × 2 × 10⁻² × 1 × 10⁻³
|ε| = π × 10⁻⁶ V = π μV

For total internal reflection at A:
4 sin 30° = μ₂ sin 90°
Therefore, μ₂ = 2
Case - I
If μ₂ < 2, then total internal reflection always takes place, and in this situation, the angle of deviation is 120°.

Case - II
If μ₂ > 2, then for the given angle of incidence, no total internal reflection takes place at A, so the light strikes on interface B for total internal reflection.

At A,
4 sin 30° = μ₂ sin θ ........(Equation i)
Now at interface B,
μ₂ sin θ = 2 sin α ........(Equation ii)
From both equations,
α = 90°
Hence, the deviation is 60°.

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

Given, mass of disc, M = 4 kg
Radius of disc, R = 0.4 m
Initial angular velocity of disc, ω₁ = 30rads⁻¹
Mass of point-masses, m = 0.25 kg
Now, moment of inertia of disc, I₁ = 1/2MR²
Moment of inertia of disc and two point mass system,
I₂ = 1/2MR² + 2mR²
Let ω₂ be the final angular velocity.
According to conservation of angular momentum,
I₁ω₁ = I₂ω₂ ⇒ω₂ = I₁ω₁ / I₂
Substituting the values, we get

ω₂ = 24rad/s
Hence, the final angular velocity of system is 24rad/s.

Bulk modulus, K = 2 × 10⁹ N/m²

Change in volume, ΔV = 0.1% of initial volume

= 0.1% of V = (V × 0.1) / 100 = V × 10⁻³

∴ K = pV / ΔV

2 × 10⁹ = (p × V) / (V × 10⁻³)

p = 2 × 10⁹ × 10⁻³

= 2 × 10⁶ N/m²

Comparing with P = 2 × 10^K

2 × 10⁶ = 2 × 10^K

K =  6K

Thus, the value of K is 6.

= 2

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).

Problem includes the concept of the Arrhenius equation for different rate constants of a common reaction. To solve this problem, the student is advised to write the Arrhenius equation for every rate constant. Then, put the value in

k = (k₁k₂ / k₃)^(2/3)

to get the value of Eₐ.

Putting k = Ae^(-Eₐ/RT)

Comparing coefficients

e^{(1/RT { (2/3) (Eₐ₁ + Eₐ₂) - (2/3) Eₐ₃ })} = e^(-Eₐ/RT)

Eₐ = (2/3) [Eₐ₁ + Eₐ₂ - Eₐ₃]

Eₐ = (2/3) [200 + 90 - 80]

= (2/3) [210]

= 140 kJ/mol

The electrolysis of CuCl₂ solution is taking place in the presence of Cu electrode.
So, at anode, the oxidation of Cu (s) and at cathode, the reduction of Cu⁺² ions will occur.
The electrode reactions are as follows:
At anode: Cu (s)→Cu⁺² (aq) + 2e
At cathode: Cu⁺² (aq) + 2e⁻ → Cu (s)
From the reactions, it is understood that one mole of copper is consumed at the anode and one mole is deposited at the anode.
The number of moles of copper consumed at the anode is equal to the number of moles of copper deposited at the cathode.
If mass of the cathode is increased by 3.15 g, then at anode, loss of mass of 3.15 g of Cu will take place.

• In the 14th group of the periodic table (Group 14), the maximum inert pair effect is exhibited by the element lead (Pb).
• The inert pair effect refers to the tendency of heavier elements, particularly those in Groups 13 through 16 of the periodic table, to preferentially retain their valence electrons in the s-orbital rather than participating in chemical bonding. This results in a lower oxidation state than would be expected based on the group number.
• In lead (Pb), the inert pair effect is particularly pronounced due to the increasing energy difference between the s and p orbitals as you move down the group. Lead has a configuration of [Xe]4f¹⁴5d¹⁰6s²6p². The 6s and 6p orbitals are farther from the nucleus compared to the 5d orbital, resulting in weaker interactions with other atoms. As a result, lead tends to exhibit an oxidation state of +2 instead of the expected +4.

For a particular reversible reaction at temperature T, ΔH and ΔS were found to be both positive. If T_e is the temperature at equilibrium, the reaction would be spontaneous when T > T_e.
At equilibrium, ΔG = 0
ΔG = ΔH − T_eΔS
0 = ΔH − T_eΔS
ΔH = T_eΔS
T_e = ΔH / ΔS
For a spontaneous reaction, ΔG must be negative, which is possible only if:
ΔH − TΔS < 0
ΔH < TΔS or T > ΔH / ΔS
Thus, T > T_e.

The correct extracted text is:

f₂(x) = f(f(x)) = f(x) = x

f₃(x) = f(f₂(x)) = f(x) = x

x³ - 25x² + 175x - 375 = 0

(x - 5)(x² - 20x + 75) = 0

(x - 5)²(x - 15) = 0 ⇒ x = 5, 15

We have,

Put,

Put, 3x − 2 = t ⇒ x = t + 2 / 3

We have

(OMₙ₋₁)² = (OPₙ)² + (PₙMₙ₋₁)² = 2(OPₙ)² = 2aₙ (say)

Also, (OPₙ₋₁)² = (OMₙ₋₁)² + (Pₙ₋₁Mₙ₋₁)²

⇒ aₙ₋₁² = 2aₙ² + 1/2 aₙ₋₁ ²
⇒ aₙ = 1/2 aₙ₋₁

Since, each next term of the sequence is obtained by dividing the previous term by 2, this gives a GP with common ratio 1/2.

⇒ OPₙ = aₙ = 1/2 aₙ₋₁ = 1/2² aₙ₋₂ = ……… = 1/2ⁿ OP = (1/2)ⁿ

(As P lies on x + y = 1 and is the midpoint of AB, P = (1/2, 1/2))

⇒ OP = 1/2 units

Since, 3(3)² + 5(5)² − 32 > 0.
So, the given point lies outside the ellipse.
Hence, two real tangents can be drawn from the point to the ellipse.

Given two circles x² + y² = 1 and (x − 9)² + (y − 12)² = 4
Here, the centres and radii of given circles are C₁(0, 0) & C₂(9, 12) and radii R₁ = 1 & R₂ = 2, respectively.

Thus, the given circles are external to each other.

 

∴ The Shortest distance between given circles are AB = C₁C₂ − (R₁ + R₂).
= 15 − (1 + 2)
= 12

Let, replace , as it can change the value of the term, but the number of terms won't be affected.
Now, 
Which contains (2n+1) terms
Hence,  (x + 1 + 1/x)ⁿ  contains (2n+1) terms
∴ 2n + 1 = 301
⇒ n = 150
Which is greater than 149.

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.

Let (x, y) is the set of points equidistant from point (2, 3) and the line 3x + 4y − 2 = 0.
So, comparing 
with SP² = e² × PM², here e = 1
So the given equation represents a parabola.

Let D = (x, y, z)

Thus,

(AD)² + (BD)² + (CD)² = (x - 1)² + (y - 1)² + z² + (x - 5)² + (y - 1)² + z + (x - 2)² + (y - 7)² + z²

Expanding and simplifying:

= 3x² - 18x + 3y² - 18y + 84 + 3z²

= 3(x - 3)² + 3(y - 3)² + 30 + 3z²

Thus, M = 31 = 3 + 1 = 4

If x = α is the common root of both equations, then
α⁴ + 6α² + 25 = 0
⇒ α⁴ = -25 - 6α²
Using this value of α⁴ in the second equation:
3(-25 - 6α²) + 4α² + 28α + 5 = 0
⇒ -75 - 18α² + 4α² + 28α + 5 = 0
⇒ -14α² + 28α - 70 = 0
⇒ 2α² - 4α + 10 = 0
⇒ α² - 2α + 5 = 0
On comparing with x² + bx + c, we get
b = -2, c = 5
So, P(x) = x² - 2x + 5
⇒ P(1) = (1)² - 2(1) + 5 = 4