1. ⇒ (MHT CET 2023 13th May Morning Shift )

Three coils of inductance $L_{1} = 2 H, L_{2} = 3 H$ and $L_{3} = 6 H$ are connected such that they are separated from each other. To obtain the effective inductance of 1 henry, out of the following combinations as shown in figure, the correct one is

A. S

B. P

C. R

D. Q

Correct Option is (D)

The combinations $P$ and $S$ are series combinations. Hence, the effective inductance cannot be $1 H$ .

This leaves combinations $R$ and $Q$ .

For combination $R$ ,

$L_{1} + L_{2} = 5 H$

$\frac{1}{L_{1} + L_{2}} + \frac{L}{L_{3}} = \frac{1}{5} + \frac{1}{6} = \frac{6 + 5}{30} = \frac{11}{30}$

$∴ L_{eff} = \frac{30}{11} = 2.72 H$

$∴$ The correct combination would be $Q$ .

$\begin{aligned} \frac{1}{L_{1}} + \frac{1}{L_{2}} + \frac{1}{L_{3}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} \\ ∴ L_{eff} = 1 H \end{aligned}$

2. ⇒ (MHT CET 2023 12th May Evening Shift )

SI units of self inductance is

A. $\frac{V - A}{S}$

B. $\frac{V}{A - S}$

C. $\frac{V - S}{A}$

D. $\frac{A}{V - S}$

Correct Option is (C)

Currently no explanation available

3. ⇒ (MHT CET 2023 12th May Evening Shift )

Inductance per unit length near the middle of a long solenoid is $(μ_{0} =$ permeability of free space, $n =$ number of turns per unit length, $d =$ the diameter of the solenoid)

A. $μ_{0} π {(\frac{nd}{2})}^{2}$

B. $4 μ_{0} π (\frac{nd}{2})$

C. $(\frac{μ_{0} π n d}{2})$

D. $\frac{4 μ_{0} π}{n^{2} d^{2}}$

Correct Option is (A)

The inductance long solenoid is

$\begin{aligned} L = \frac{μ_{0} N^{2} A}{l} \\ L = μ_{0} {(\frac{N}{l})}^{2} \times π \times \frac{d^{2}}{4} \end{aligned}$

$∴$ The inductance per unit length near the middle of a long solenoid is:

$\frac{L}{l} = μ_{0} π {(\frac{nd}{2})}^{2} \dots . (∵ \frac{N}{l} = n)$

4. ⇒ (MHT CET 2023 12th May Morning Shift )

Two inductors of $60 mH$ each are joined in parallel. The current passing through this combination is $2.2 A$ . The energy stored in this combination of inductors in joule is

A. 0.0333

B. 0.0667

C. 0.0726

D. 0.0984

Correct Option is (C)

$L_{1} = L_{2} = L = 60 mH$

When two inductors are connected in parallel, their equivalent inductance is given by,

$\begin{aligned} \frac{1}{L_{eq}} = \frac{1}{L_{1}} + \frac{1}{L_{2}} \\ ∴ L_{eq} = \frac{L}{2} = 30 mH \\ u_{B} = \frac{1}{2} L_{eq} I^{2} \\ ∴ u_{B} = \frac{1}{2} \times 30 \times 10^{- 3} \times 2.2 \times 2.2 \\ ∴ u_{B} = 0.0726 J \end{aligned}$

5. ⇒ (MHT CET 2023 12th May Morning Shift )

Two coils have a mutual inductance of $0.004 H$ . The current changes in the first coil according to equation $I = I_{0} \sin ω t$ , where $I_{0} = 10 A$ and $ω = 50 π rad s^{- 1}$ . The maximum value of e.m.f. in the second coil in volt is

A. $5 π$

B. $4 π$

C. $2.5 π$

D. $2 π$

Correct Option is (D)

$\begin{aligned} | e_{s} | = M \frac{d I_{p}}{d t} \\ | e_{s} | = M \frac{d}{d t} I_{0} \sin ω t \\ | e_{s} | = M I_{0} ω \cos ω t \\ ∴ & {| e_{s} |}_{max} = M I_{0} ω = 0.004 \times 10 \times 50 π \\ ∴ & {| e_{s} |}_{max} = (2 π) volt \end{aligned}$

⇒ Electromagnetic Induction