KEAM Paper 1 Mock Test - 2 - 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]

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 the system of polarizers A and B, the intensity of the emergent light is I/2. So, A and B have same alignment of transmission axis.
Let's assume that C is introduced at θ angle w.r.t. A.
Then,
Output Intensity = I/2
So, we have

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

Or 4 × 10^-5 Wb/m²

As the Rocket is fired from earth (E) towards the sun (s) at point (P) the distance from center of the earth.

The Force acts on the rocket are: These two gravitational forces are in opposite directions.

As we know gravitational force between two objects which are at distance R from each other. 
G = universal gravitational constant, m₁m₂ = mass of the objects, R = Distance between them
By the gravitational Force
Force on Rocket due to Earth = Force on Rocket due to sun

By simplification:

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,

When a particle connected to a string revolves in a circular path around a center, the centripetal force is provided by the tension produced in the string. Thus, in the given case, the net force on the particle is the tension T, i,e F = T = mv²/1 Where F is the net force acting on the particle.

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 =  

Resolving power (R.P.) of the astronomical telescope is its ability to form separate images of two neighbouring astronomical objects like stars etc.

where D is diameter of objective lens and λ is wave length of light used.

The magnetic field in a plane of the electromagnetic wave is given by, B = 2 x 10^-7 sin (0.5 × 10³x + 1.5 × 10¹¹t)
Comparing this equation with the standard wave equation: B_y = B₀ sin[Kx + ωt]
From the above equation, K = 0.5 x 10³ = propagation constant

The wavelength range of microwaves is 10^-3 m to 0.3 m. The wavelength of this wave lies between 10^-3m to 0.3 m, so the equation represents microwaves.

An infinite wire, carrying current, is bent in the following way.
Two semi-infinite parts and a bent finite part. We have to find the induced magnetic field at an external point o, at a distance R. The corresponding figure is shown below.

Magnetic field induced by a finite current carrying conductor at a distance R is given by, 
Where, θ₁ and θ₂ are the angles made by the endpoints of the wire to the external point.
And using right hand thumb rule, we get direction of magnetic field,
For the 1st semi-infinite part, θ₁= 0 and θ₂ = 90^o = π/2
Substitute the above values in equation (i), we get 
For the bent portion θ₁ = 45^o and θ₂ = 135^o 
Therefore, magnetic field will be,

For the second semi-infinite part, 
θ₁ = π/2 θ₂ = 0
Therefore, magnetic field will be, 
Total magnetic field at point o is, B = B₁ + B₂ + B₃

Let P₁ and P₂ be the partial pressures of gases A and B, respectively. Under equal conditions of temperature and volume, P₁ = P₂
Let P be the total pressure exerted by the mixture of two gases A and B. Using Dalton's law of partial pressures, P = P₁ + P₂
⇒ P = 2P₁ ⇒ P/P₁ = 2/1 ⇒ P : P₁ = 2 : 1

Current I = 1.6 Amps, Time t = 1 minute or 60 seconds
Two electrons are needed to liberate one ion of Cu⁺⁺, i.e. the charge required is = 2 × 1.6 × 10^-19 C = 3.2 × 10^-19C
The total charge flow in 1 minute is Q = It = 1.6A × 60s = 96C
Thus, the number of ions liberated is n

The argument in the trigonometric function should be a dimensionless quantity.
The dimensions of time (t)  angular velocity (ω) and amplitude (A) are [t] = [T], [ω] = [T^-1], [A] = [L]
A. The dimension of the argument in the equation:

Thus, option 'A' is not wrong on dimensional grounds.
B. In the equation: x = A / T sin (t/A), the dimension of the argument in the sine function is [t/A] = |T|/L

Thus, option 'B' is wrong on dimensional grounds.
C. The dimension of the argument in equation:

Thus, option 'C' is not wrong on dimensional grounds.
D. The dimension of the argument in equation: x = A sin (ωt) is [ωt] =  [T^-1] [T] = dimensionless quantity
Thus, option 'D' is not wrong on dimensional ground. All the equations, except x = A/T sin (t/A) i.e. option 'B' have dimensionless argument.

The given figure is shown below:

The above arrangement is equivalent to two parallel plate capacitors connected in parallel with the same separation d and area A.
Now, the capacitance of each capacitor is 
The equivalent capacitance of C₁ and C₂ is C = C₁ + C₂ 

Spring constant, k = 5 × 10³ N/m
The work required to stretch the spring by the displacement x is W = 1/2 kx² 
When the spring is stretched by 5 cm, work done is W₁ = 1/2 × (5 × 10³N/m) × (5 × 10^-2m)²
W₁ = 6.25J
When the spring is again stretched further by 5 cm, the net displacement of the spring from its equilibrium position is
x' = 5 cm, +5 cm = 10 cm
Thus, the work required to stretch it further by an additional 5 cm is W₂ = 1/2 x (5 × 10³N/m) × (10 × 10-2m)² 
⇒ W₂ = 25J
So, the extra work required to stretch the spring by an additional 
5 cmis ΔW = W₂ - W₁ ⇒ ΔW = 25J - 6.25J ⇒ ΔW = 18.75J or 18.75 N - m

y₁ = 6 cos(100πt) ...... (i)
y₂ = 4 cos(102πt) ...... (ii)
Standard wave equation is given by y = A_o cos(ωt) ....... (iii)
From equation (i) and (iii), we get angular frequency ω₁ = 100π = 2πf₁ ⇒ f₁ = 50 Hz
Amplitude, A₀₁ = 6 unit From equation (ii) and (iii), we get ω₂ =102π = 2πf₂ ⇒ f₂ =51 Hz
Amplitude, A₀₂ = 4 unit
Beat frequency = difference in frequencies of the two given waves = f₂ - f₁ = 51 Hz - 50 Hz = 1 Hz, i.e 1 beat per sec
Resultant amplitude A of superposition of two waves of amplitudes A₀₁ and A₀₂ is A =
For maximum resultant amplitude, phase 

For minimum resultant amplitude, phase 

Two beams of red and violet colors are made to pass separately through a prism (angle of the prism is 60°).
We have to find the angle of refraction at the position of minimum deviation.
In the position of minimum deviation, light passes symmetrically through the prism irrespective of the light wavelength used.

In the position of minimum deviation, r + r = 2r = A[ as deviation δ = 0]
  for both the colour.

Radius of the planet R = 1/10 x (radius of Earth)
Depth of the well = R/5 Linear mass density of the wire, μ = 10^-3 kg m^-1
We have to find the force applies at the top of the wire by a person holding it in.
Acceleration due to gravity on the planet will be given by g = GM/R²

(i) Acceleration due to gravity of earth  
Dividing equation (i) by (ii), we get 
Now, we will consider a mass element dm of width dx at depth x below the planet.
Mass of this element will be dm = μdx
Also, acceleration due to gravity at depth x below the planet will be 
Force on the small segment of wire 
Total force on the wire is obtained by integrating the above equation.

The free-body diagram of the system is shown below:

[ac is centripetal acceleration]
Centripetal acceleration is in the horizontal direction as seen in the diagram. So, no acceleration is there in a vertical direction. So, the net force in a vertical direction is zero.
Thus

If L is the original length of the string, then the increase in length is given by ΔL = (T/A) (L/Y)
where, Y: Young's modulus of the wire, A: cross-sectional area of the wire