JEE Main Test 145

Let the radius of the initial circular coil be r

Thus length of the coil L = 2 (2\pi r)

Now the coil is rewound so as to have 4 turns such that the length of the coil is same.

\therefore L = 4 (2\pi r') = 2 (2\pi r) \implies r' = \dfrac{r}{2}

Magnetic field at the centre of the coil B_c = \dfrac{\mu_o}{4\pi} \dfrac{i}{r} (2\pi n) where n is the number of turns of the coil.

For 1st case: n = 2 \implies B = \dfrac{\mu_o}{4\pi} \dfrac{i}{r} (2\pi \times 2) = \dfrac{\mu_o i}{r} .........(1)

For 2nd case: n' = 4 \quad r' = \dfrac{r}{2}

\implies B' = \dfrac{\mu_o}{4\pi} \dfrac{i}{\frac{r}{2}} (2\pi \times 4) = 4\dfrac{\mu_o i}{r}

From (1), we get B' = 4B

The Diode is forward biased, hence current will flow in the circle and since Diode is Ideal forward bias voltage is zero.

Concept:

Hybridization:

→ The process of intermixing the orbitals of slightly different energies so as to redistribute their energies, resulting in the formation of a new set of orbitals of equivalent energies and shape.

For example, when one 2s and three 2p-orbitals of carbon hybridize, there is the formation of four new sp³ hybrid orbitals.

Types of hybridisation : sp , sp² , sp³ , sp³d , sp³d²  and sp³d³ .

Explanation:

→ PCl₅ : P configuration is 1s² 2s² 2p⁶ 3s² 3p³ . it is in the ground state. 

Excited configuration is 1s² 2s² 2p⁶ 3s¹ 3p³ 3d¹ here 3s¹ 3p³ 3d¹ are 5 half filled orbitals having almost the same energy will intermix and form new 5 hybrid orbitals of hybridisation sp³d. hence geometry is Trigonal bipyramidal.

→ SF₆  : S ground state configuration is 1s² 2s² 2p⁶ 3s² 3p⁴ .

Excited configuration is 1s² 2s² 2p⁶ 3s¹ 3p³ 3d² here 3s¹ 3p³ 3d² are 6  half filled orbitals having almost same energy will intermix and form new 6 hybrid orbitals of hybridisation sp³d². hence geometry is Octahedral.

→ BrF₅ : Br ground state configuration is [Ar] 3d¹⁰ 4s² 4p⁵ .

Excited state configuration is  [Ar] 3d¹⁰ 4s² 4p³ 4d² .

Here 1 full filled and 5 half filled orbitals (total 6) having almost the same energy will hybridise with each other. This intermixing will form  sp3d² hybridization. 1 sp³d² orbital will occupy the lone pair and the other 5 will be occupied by F.

So here 5 bond pairs and one lone pair hence its shape is square pyramidal.

→ BF₃ : B  ground state configuration is 1s² 2s² 2p¹ .

Excited configuration is 1s² 2s¹ 2p² . Here 2s¹ 2p² are 3 half-filled orbitals having almost the same energy that will intermix and form new 3 hybrid orbitals of hybridization sp².

Hence geometry is Trigonal planar.

Hence answer is Option 2

CONCEPT:

Dalton's Atomic Theory

• Dalton's atomic theory proposed several postulates regarding the nature of atoms and matter.
• Some of these postulates have been found to be incorrect based on modern scientific evidence.
• The key postulates under consideration are:
  • Atoms of different elements differ in mass.
  • Matter consists of divisible atoms.
  • Compounds are formed when atoms of different elements combine in a fixed ratio.
  • All the atoms of a given element have different properties including mass.
  • Chemical reactions involve reorganization of atoms.

EXPLANATION:

• Incorrect Postulates:
• (B). Matter consists of divisible atoms.
  • This postulate is incorrect because atoms are divisible into subatomic particles (protons, neutrons, and electrons).
• (D). All the atoms of a given element have different properties including mass.
  • This postulate is incorrect because atoms of a given element have the same number of protons and are usually identical in mass. However, the existence of isotopes (atoms of the same element with different masses) contradicts this statement.

CONCLUSION:

The correct answer is  (B), (D) only

Concept:

Any relation R is set to be reflexive, if (a, a) ∈ R for all a ∈ A that is, every element of A is R-related to itself

Any relation R is symmetric only if (b, a) ∈ R is true when (a,b) ∈ R. In a symmetric relation, if a=b is true then b=a is also true.

For transitive relation, if (x, y) ∈ R, (y, z) ∈ R, then (x, z) ∈ R.

If a relation is reflexive, symmetric, and transitive at the same time, it is known as an equivalence relation.

Calculation:

Given:

A relation R over a class of n × n real matrices A and B as "ARB iff there exists a non-singular matrix P such that PAP⁻¹ = B"

Now for reflexive:

(B, B) ∈ R ⇒ B = PBP^-1 which must be true for P = I, 

So R is reflexive

For symmetry

As (B, A) ∈ R for matrix P

B = PAP^-1 ⇒ P^-1B = P^-1PAP^-1

P^-1BP = IAP^-1P = IAI -----(Since P^-1P = I)

P^-1BP = A ⇒ A = P^-1BP

(A, B) ∈ R for matrix P^-1

So R is symmetric

For transitivity

B = PAP^-1 and A = PCP^-1

B = P (PCP^-1) P^-1

B = P²C (P^-1)² ⇒ B = P²C (P²)^-1

(B, C) ∈ R for matrix P²

So R is transitive

If a relation is reflexive, symmetric, and transitive at the same time, it is known as an equivalence relation.

So R is an equivalence relation.

So It is evident that the value of f(1) Lies in between (6,9)