COMEDK Mock Test - 5 - JEE MCQ

ΔQ_cycle = 40 - 60 = ΔW
⇒ ΔW = -20J = W_BC + W_CA
⇒ W_CA = -20J - W_BC
= -20 - (-50)
= 30 J

Purely inductive circuit:
θ = π/2
cosπ/2= 0
Average power = 0
Power dissipation becomes zero when there is only a capacitor or inducer in a circuit, even though there is still current flowing through it.

Y = 0 + 0 = 0

∴ m₁d + m₂ (- x) = 0

or, m₁d = m₂x

The AC circuit containing a resistor, inductor, and a capacitor is called an LCR circuit.

The meanings of inductor, capacitor and resistance are as follows:

• The device which stores magnetic energy in a magnetic field is called an inductor.
• The device that stores electrostatic energy in an electric field is called a capacitor.
• The properties of a conductor which opposes the flow of current are called resistance.

The capacitor and inductor are the storage devices that store the energy in it.

The inductor and capacitor can’t dissipate energy. As the resistance opposes the flow of electric current. So, the resistance of any circuit dissipates the power.

τ_c = 2RC

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.

We know that if there are two solenoids of equal length and one solenoid is placed coaxially inside the other solenoid then the mutual inductance of solenoid 1 with respect to solenoid 2 will be equal to the mutual inductance of solenoid 2 with respect to solenoid 1.
The mutual inductance of both the solenoids is given as,

Where, n₁ = number of turns per unit length of solenoid 1, n₂ = number of turns per unit length of solenoid 2, r₁ = radius of the inner solenoid, and l = length of both the solenoids
By equation (1) it is clear that the mutual inductance of both the solenoids does not depend on the current in the solenoids.
Therefore, when the current in both the solenoids is doubled, the mutual inductance of both the solenoids S₁ and S₂ will remain unchanged. 

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.

When the displacement of a SHM is:
y=a sin wt+ b cos wt

• Amplitude of the SHM will be:
  A=√a²+b²

Here, a = 3, b = 4
Amplitude, A= √(3²+4²) = 5 cm

Hence option B is correct.

Weight of the rod will produce the torque,
τ = mg x L / 2 = I α = mL² / 3 α (∵ I_rod = ML² / 3)
Hence, angular acceleration, α = 3g / 2L

The image provides a solution for finding the dimensions of a in the Van der Waals equation for 1 mole of a real gas:
The Van der Waals equation is:

Step-by-step solution:
To find the dimensions of a, we analyze the term a/V², which has the same dimensions as pressure P.
The dimensional formula of pressure P is ML⁻¹T⁻².
The dimensional formula of V² is L⁶, so:

Now, we calculate the dimensions of a:

Check the options:
The dimensions of a match those of PV²:

Thus, the correct answer is B.

⇒ (2n)eAv_L = ne(2A)v_R

⇒ v_L/v_R = 1

Length , l = 4,234 m
Breadth, b = 1.005 m
Thickness , t = 2.01 x 10^-2 m
Volume = l x b x t
⇒ V = 4.234 x 1.005 x 0.0201 = 0.0855289 = 0.0855 m³ (significant figure = 3)

Induced e.m.f. =

∴ Induced e.m.f =

= 50 μV

According to Boyle's law, for a given mass of a gas at a constant temperature, 

Now this means that the P−V  curve is a hyperbola.

Hence Option 1 is correct because it shows P−V curve hyperbola.

For log P vs log V 
curve, PV = k = constant.

Taking log on both sides,
log P+log V = log k
P = −log V + log k
Thus, in the log P vs log V curve, the slope is downwards and negative.

Hence Option B is correct.

Option C is correct because as we know

Option D is incorrect because as mentioned, the P−V curve should be a rectangular hyperbola but it is shown that the P−V curve is a straight line.

Therefore, option D is correct answer.

As the temperature increases, the equilibrium constant decreases and yield also decreases.

Since it is an exothermic reaction, energy is released, The number of moles of products produced is also less than the number of moles of reactants involved, thereby releasing excess energy.

Therefore, the correct order is as follows:

T1 > T2 > T3

All the given ions are isoelectronic as they contain same number of electrons. For isoelectronic species,

ionic radius 

Hence, the correct increasing order of ionic size is

Mg²⁺ < Na⁺ < F⁻ < O²⁻ < N³⁻

The electronic configurations of the given elements are -

Removal of two electrons from Eu gives it the configuration [Xe] 4f⁷ which is a stable half-filled configuration.

For the reaction: P + 2Q → R, when concentration of 'Q' is doubled, the half-life does not change, hence reaction is first-order with respect to 'Q'.
This is because for the first order reaction, rate constant is independent from initial concentration of the reactant.
When the concentration of 'P' is made four times, the rate of reaction becomes double. It means the reaction is half-order with respect to 'P'.
Hence, the rate law expression is:

Overall order = 1 + 1/2 = 3/2
Units of the rate constant = (mol L^-1)^{(1 - n)} s^-1
So, unit of rate constant is L^1/2 mol^-1/2 s^-1.