KIITEE Mock Test - 1 - JEE MCQ

Average value of the current, 

As .
Thus, 

Impedence of the circuit,

Peak value of the current,

Hence, r.m.s. value of the current is

Magnetic moment m = AI = r²I, where r is the radius of the circular loop.
Now, the circumference of the circle = length of the wire, i.e.
2πr = L
or r² = 
Therefore, m = πr²I = 
Hence the correct choice is (d)

PE = -2 KE

As the current is leading the voltage, then we have

if ω is the angular speed acquired by the coil when a charge Q is passed through it for very short time Δt, then 

Lowest priority atom is always away from the viewer. Priority is seen on the basis of atomic number; and if atomic numbers are same, then on the basis of second or third bond. In the above figure, priority order group or atom is: -Br > -Cl > -CH₃ > H, and rotation is clockwise. So, it shows R-configuration.

Before diving into the answer, let's briefly review the Aufbau principle and the quantum numbers. The Aufbau principle states that electrons are added to orbitals in increasing order of energy. In the standard periodic table, each orbital can accommodate two electrons with opposite spins. This is determined by the Pauli Exclusion Principle, which states that no two electrons in an atom can have the same set of quantum numbers. The four quantum numbers are:

1. Principal quantum number (n): represents the energy level and size of the orbital.
2. Azimuthal quantum number (l): represents the shape of the orbital (s, p, d, f).
3. Magnetic quantum number (m_l): represents the orientation of the orbital in space.
4. Spin quantum number (m_s): represents the intrinsic angular momentum or spin of the electron (+1/2 or -1/2).

Now let's analyze each statement:

1. In the new periodic table, Zr would continue to belong to the 5th period.

If each orbital accommodates three electrons instead of two, the number of electrons in each energy level would change, altering the periodic table structure. In the standard periodic table, Zr belongs to the 5th period with electron configuration [Kr] 5s2 4d2. However, in this new arrangement, the electron configuration would change, and Zr might not belong to the 5th period anymore.

2. Spin quantum number would overflow.

Since the spin quantum number has only two possible values (+1/2 and -1/2), it cannot accommodate three electrons in the same orbital without violating the Pauli Exclusion Principle. Hence, the spin quantum number would overflow.

3. The number of s-electrons would equal the number of p-electrons.

In the standard periodic table, the number of s-electrons and p-electrons differ, with s orbitals holding 2 electrons and p orbitals holding 6 electrons. In the new arrangement, both s and p orbitals would hold 3 electrons each, making the number of s-electrons equal to the number of p-electrons. However, this arrangement would violate the Aufbau principle as well as the Pauli Exclusion Principle.

In conclusion, all the statements are incorrect if we assume that each orbital can accommodate three electrons instead of two, ignoring the Aufbau principle.

A salt is precipitated only when the product of ionic concentrations is more than its solubility product.

So, AB will be precipitated only when the concentration of [B^-] is more than 10^-5 M.

From Quantum Mechanics viewpoint, the magnetic moment is dependent on both spin and orbital angular momentum contributions. The spin-only formula used is given by

And this can be modified to include the orbital angular momentum as follows.

Therefore, the value of the magnetic moment can differ because of the contribution of the total orbital motion of the electron to the magnetic moment. Thus, this is the correct answer.

According to Freundlich, the adsorption isotherms and adsorption isobars are plotted when temperature and pressure are constant respectively.
The extent of adsorption is directly proportional to the pressure but inversely proportional to the temperature.
Therefore, option (c) is incorrect and the correct relation is .

V.P does not depend on the amount of water.
The volume occupied by water molecules in vapour phase is 1 × 10^-4 dm³ which is approximately equal to 1 x 10^-3 m³(volume of the flask)
Ideal Gas Law -
PV = nRT
where
P - Pressure
V - Volume
n - No. of Moles
R - Gas Constant
T - Temperature

PV = nRT

Carbylamine test is given by aliphatic as well as aromatic primary amines, i.e. amines having --NH₂ group.

Hence, dimethyl amine, being a secondary amine, does not undergo carbylamine reaction.

We know, k = Ae-^Ea/RT
Hence, k1 = Ae^-150/RT …(1)
k₂ = Ae^-90/RT …(2)
From (1) and (2), we get

Taking log on both sides, we get

Given:
C : H : : 10.5 : 1
Their total weight = 11.5
For molecular weight of hydrocarbon in gas phase,
PV = (w/M) RT
1 x 1 = 2.8/M × 0.0821 × 400
M = 92 g (mol)^-1
Now, 11.5 g hydrocarbon has 1 g H.
Therefore, 92 g hydrocarbon has 92/11.5 = 8 g of H and 92 - 8 = 84 g carbon
g atoms of C = 84/12 = 7
g atoms of H = 8/1 = 8
Formula of hydrocarbon = C₇H₈

Put x = 1
Then, 

l = lower boundary point of median class
c.f. = Cumulative frequency of the class preceding the median class
f = Frequency of the median class
n = total number of values or observations
h = class length of median class

 || m
So, slope of  = slope of m

x = 9
So, length = 12 units and width = 9 units
Area =  × w = 12 units × 9 units = 108 sq. units

For the case when the number n is not a positive integer the binomial theorem becomes, for -1 < x < 1,