JEE Main Mock Test - 4 - JEE MCQ

Given: N = 2N₁ and B = 2B₁
The magnetic flux through the electric generator when the magnetic field is B, current flowing is A and the number of turns is N
φ = N B A cos θ ....(1)
The magnetic flux through the electric generator when the magnetic field and number of turns of the coil of an electric generator is doubled
φ₁ = N₁ B₁ A cosθ   ...(2)
⇒ φ₁ = (2N)(2B) A cos θ = 4 N B A cos θ = 4φ [∵φ = NBA cos θ]

f = Cm^xk³. Writing dimensions on both sides:

Comparing dimensions on both sides, we have

Aliter. Remembering that frequency of oscillation of loaded spring is

which gives 

Three forces are acting on a particle of mass m initially in equilibrium. If the first two forces R₁ and R₂ perpendicular to each other, R₃ has to balance the resultant of the other two forces. So,

Now, the force R₃ is removed
So, the particle moves with the resultant force R₁ and R₂
Now, the acceleration is
R₃ = ma
a = R₃/m
Hence, the acceleration of the particle is R₃/m.

By work - energy theorem
Total work done = change in kinetic energy
v₂ = 6 × 2 + 4 = 16 m/s =
v₀ 0 + 4 = 4 m/s

Mean position of the particle is mg/k distance below the unstretched position of spring. Therefore, amplitude of oscillation is A = mg/k

For the wavelength of the first member of Lyman series:

For the wavelength of the first member of Paschen series:

1.5 is the nearest option.

For deflection of θ/2, current also reduces to I/2 with shunt resistance S.
Hence,

Or 4 × 10^-5 Wb/m²

The cylinder is completely insulated from its surroundings. As a result, no heat is exchanged between the system (cylinder) and its surroundings. Thus, the process is adiabatic.
As we know that, the value of the ratio of specific heat for standard gas = 1.40.

For an adiabatic process, we have: 
The final volume is compressed to half of its initial volume.

The internal resistance of the cell = r
Balance point of the cell in open circuit, I₁ = 76.3cm
An external resistance (R.) is connected to the circuit with R = 9.5Ω
New balance point of the circuit, I₂ = 64.8cm
Current flowing through the circuit = 1
Using the relation connecting resistance and emf is,

Given: Wavelength of sodium line (λ₁) = 589nm Wavelength at which sodium line is observed (λ₂) = 589.6nm
Change in wavelengths is given by Δλ = λ₂ - λ₁ 
Substituting the values in above equation we get, Δλ 589.6 - 589 = 0.6nm
Velocity in terms of wavelength is given by,

Substituting the values in above equation we get, 

As per the given criteria, Refractive index of air, μ₁ = 1, Refractive index of glass, μ₂ = 1.5, Radius of curvature, R = 20 cm, Object distance, u = -100 cm
We know that,

Here, m = 0.01 kg, l = 0.3 mm = 0.3 × 10^-3 m, g = 10 ms^-2 V= 0.085 ms^-1, A = 0.1 m²
The metal block moves to the right because of the tension in the string. The tension T is equal in magnitude to the weight of the suspended mass m. 
Thus, the shear force F is: F = T = mg = 0.010 kg x 9.8 ms^-2 = 9.8 x 10^-2 N
Shear stress on the fluid =  

1. It is not a tribasic acid.
Option (a) is incorrect for H₃BO₃

2. Aqueous solution, instead of donating the −OH group it breaks the H−OH bond of water and accepts the −OH group from it leaving in the solution hence it is mono basic acid.

On doubling the [OCl⁻] keeping the [I⁻] and [OH⁻] constant, the rate of formation of [OI⁻] is doubled, hence the order of reaction with respect to [OCl⁻] is one.

On doubling the [I⁻] keeping the [OCl⁻] and [OH⁻] constant, the rate of formation of [OI⁻] is doubled, hence the order of reaction with respect to I−is one.

On changing the [OH⁻] to a half-value, the rate of formation of [OI⁻] is doubled. Hence, the order of reaction with respect to [OH⁻] is −1.

Hence the rate law is r = k[OCl⁻] [I⁻] [OH⁻]⁻¹ = k[OCl⁻] [I⁻]/[OH⁻]

The correct order of relative basic strength will be

In NH₃, the lone pair is completely available for donation, hence is most basic whereas NH₂ − NH₂ and NH₂ − OH can be considered as derivatives of NH₃. Here, H is replaced by −NH₂ forming NH₂ − NH₂ and −OH forming NH₂ − OH being highly electron withdrawing due to −O will decrease the election density most on −N.
∴ Least basic elect;on withdrawing −I-effect of
−NH₂ is less as compared to - OH.
Hence correct order of basic strength is
NH₃ > NH₂ − NH₂ > NH₂ − OH

Detol is a mixture of chloroxylenol and terpeneol.

The standard deviation of a set remains unchanged if each data is increased or decreased by a constant.
However, it changes similarly when data is multiplied or divided by a constant.
∴ The SD for the new data set will be = −2 × 3.5 = −7

Given:

The given number is 'GEOGRAPHY'

Calculation:

The word 'GEOGRAPHY' has 9 letters. It has the vowels E, O, A in it, and these 3 vowels must always come together. Hence these 3 vowels can be grouped and considered as a single letter. That is, GGRPHY(EOA).

Let 7 letters in this word but in these 7 letters, 'G' occurs 2 times, but the rest of the letters are different.

Now,

The number of ways to arrange these letters = 7!/2!

⇒ 7 × 6 × 5 × 4 × 3 = 2520

In the 3 vowels(EOA), all vowels are different

The number of ways to arrange these vowels = 3!

⇒ 3 × 2 × 1 = 6

Now, 

The required number of ways = 2520 × 6 

⇒ 15120

∴ The required number of ways is 15120.

Given:

Equation₁ = ax + 9y = 1

Equation₂ = 9y - x - 1 = 0 

Concept used:

If linear equations are a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0. Here, the equations have an infinite number of solutions, if

a₁/a₂ = b₁/b₂ = c₁/c₂

Calculation:

We have equations,

ax + 9y = 1 

⇒ ax + 9y - 1 = 0

and, 9y - x - 1 = 0

⇒ x - 9y + 1 = 0

Here, a₁ = a, b₁ = 9, c₁ = -1

and, a₂ = 1, b₂ = -9, c₂ = 1

As we know that 

a1/a2 = b1/b2 = c1/c₂

⇒ a/1 = 9/-9 = -1/1

⇒ a = -1 = -1

⇒ a = -1

∴ The value of a is -1.

(e^y) = xe^x + e^xc
At y(0) = 0,
c = 1
y = x + ln(x + 1)
y(1) = 1 + ln2

Hence, P(X = 2) = 28/256 = 7/64

cos²x + (2cos²x − 1)² + (4cos³ x − 3 cos x)² = 1
⇒16 cos⁶x − 20 cos⁴x + 6cos²x = 0 ⇒ 2cos²x(2cos²x − 1)(4cos²x − 3) = 0
⇒ cos²x = 0 ; cos²x = 1/2 = cos² π/4 ; cos²x = 3/4 = cos² π/6
⇒ x = nπ ± π/2 ; x = mπ ± π/4 ; x = kn ± π/6, n, m, k ∈ Z

Given: (x² + 1)² + x² = 109
Let x² + 1 = t
t² + t - 1 = 109
 (t - 10)(t + 11) = 0
 t = 10 = x² + 1 = a₂₂