Uncategorized - NEET Physics Chapterwise MCQs & PYQs

NEET Uncategorized MCQs & PYQs

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Question 251:

easy

43. A long solenoid of diameter $0.1 m$ has $2 \times 10^4$ turns per metre. At the centre of solenoid, a coil of 100 turns and radius $0.01 m$ is placed with its axis coinciding with the solenoid axis. The current in the solenoid reduces at a constant rate to $0 A$ from $4 A$ in $0.05 s$. If the resistance of the coil is $10 \pi^2 \Omega$, the total charge flowing through the coil during this time is: (2017-Delhi)

Total charge flowing is $q = \frac{\Delta \Phi}{R} = \frac{N_{coil} (\Delta B) A_{coil}}{R}$.
Change in magnetic field is $\Delta B = \mu_0 n \Delta I = (4\pi \times 10^{-7}) \times (2 \times 10^4) \times 4 = 32\pi \times 10^{-3} T$.
$q = \frac{100 \times (32\pi \times 10^{-3}) \times [\pi (0.01)^2]}{10 \pi^2} = \frac{100 \times 32\pi^2 \times 10^{-7}}{10 \pi^2} = 32 \times 10^{-6} C = 32 \mu C$.

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Question 252:

easy

44. Two coils of self-inductances $2 mH$ and $8 mH$ are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coils is: (2006)

For complete coupling between two coils, the coupling coefficient is $k = 1$.
The mutual inductance is given by $M = k \sqrt{L_1 L_2}$.
$M = 1 \times \sqrt{2 mH \times 8 mH} = \sqrt{16} = 4 mH$.

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Question 253:

easy

45. Two coil have a mutual inductance $0.005 H$. The current changes in first coil according to equation $I = I_0 \sin \omega t$ where $I_0 = 2 A$ and $\omega = 100pi rad/sec$. The maximum value of emf in second coil is: (1998)

Induced emf in the second coil is $e = -M \frac{dI}{dt} = -M \frac{d}{dt}(I_0 \sin \omega t) = -M I_0 \omega \cos \omega t$.
The maximum (peak) value of induced emf is $e_{max} = M I_0 \omega$.
$e_{max} = 0.005 \times 2 \times 100\pi = 0.01 \times 100\pi = \pi V$.

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Question 254:

easy

32. A long solenoid has 1000 turns. When a current of $4 A$ flows through it, the magnetic flux linked with each turn of the solenoid is $4 \times 10^{-3} Wb$. The self inductance of the solenoid is: (2016 – I)

Total magnetic flux linked with the solenoid is $N \Phi = L I$.
Given $N = 1000$, $\Phi = 4 \times 10^{-3} Wb$, and $I = 4 A$.
$L = \frac{N \Phi}{I} = \frac{1000 \times 4 \times 10^{-3}}{4} = 1 H$.

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Question 255:

easy

33. A long solenoid has 500 turns. When a current of 2 ampere is passed through it, the resulting magnetic flux linked with each turn of the solenoid is $4 \times 10^{-3} Tm^2$. The self inductance of the solenoid is: (2008)

Total flux linked is given by $N \Phi = L I$.
Here, $N = 500$, $\Phi = 4 \times 10^{-3} Wb$ (or $T m^2$), and $I = 2 A$.
$L = \frac{N \Phi}{I} = \frac{500 \times 4 \times 10^{-3}}{2} = 1.0 henry$.

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Question 256:

easy

34. If N is the number of turns in a coil, the value of self inductance varies as (1993)

Self-inductance of a coil/solenoid is given by $L = \frac{\mu_0 N^2 A}{l}$.
This relation clearly shows that self-inductance is directly proportional to the square of the number of turns.
Therefore, $L \propto N^2$.

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Question 257:

easy

35. What is the self-inductance of a coil which produces $5 V$ when the current changes from 3 ampere to 2 ampere in one millisecond? (1993)

Induced emf is given by $e = L \left|\frac{\Delta I}{\Delta t}\right|$.
Here $e = 5 V$, $|\Delta I| = |2 - 3| = 1 A$, and $\Delta t = 1 ms = 10^{-3} s$.
$L = \frac{e}{|\Delta I / \Delta t|} = \frac{5}{1 / 10^{-3}} = 5 \times 10^{-3} H = 5 mili-henry$.

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Question 258:

easy

36. If the number of turns per unit length of a coil of solenoid is doubled, the self-inductance of the solenoid will: (1991)

Self-inductance of a solenoid is $L = \mu_0 n^2 A l$, where $n$ is the number of turns per unit length.
Thus, self-inductance is directly proportional to the square of turns per unit length ($L \propto n^2$).
When $n$ is doubled, $L' = (2n)^2 = 4L$, so it becomes four times.

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Question 259:

easy

37. The current in self inductance $L = 40 mH$ is to be increased uniformly from 1 amp to 11 amp in 4 milliseconds. The e.m.f. induced in inductor during process is (1990)

Induced e.m.f. is given by $e = L \frac{\Delta I}{\Delta t}$.
Given $L = 40 mH = 40 \times 10^{-3} H$, $\Delta I = 11 - 1 = 10 A$, and $\Delta t = 4 ms = 4 \times 10^{-3} s$.
$e = 40 \times 10^{-3} \times \frac{10}{4 \times 10^{-3}} = 100 volt$.

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Question 260:

easy

38. The magnetic potential energy stored in a certain inductor is $25 mJ$, when the current in the inductor is $60 mA$. This inductor is of inductance: (2018)

The magnetic energy stored is $U = \frac{1}{2} L I^2$.
Given $U = 25 mJ = 25 \times 10^{-3} J$ and $I = 60 mA = 60 \times 10^{-3} A$.
$L = \frac{2U}{I^2} = \frac{2 \times 25 \times 10^{-3}}{(60 \times 10^{-3})^2} = \frac{50 \times 10^{-3}}{3600 \times 10^{-6}} = \frac{50000}{3600} \approx 13.89 H$.

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