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TN 12th Computer Applications மின்னணு செலுத்தல் முறைகள் Sample Question Papers Study Material - QB365 Set A

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TN 12th Computer Applications மின் - வணிகம் Sample Question Papers Study Material - QB365 Set A

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12th Standard English Medium Physics Subject Electromagnetic Induction and Alternating Current Book Back 1 Mark Questions with Solution Part - I

Published on: 01/06/2021

QB365 provides detailed and simple solution for every Book back Questions in class 12 Physics Subject. It will helps to get more idea about question pattern in every book back questions with solution.

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Std: 12th Standard - Physics

chapter: - Book back Questions

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Questions:

Multiple Choice Question

5 x 1 = 5

In an electrical circuit, R, L, C, and AC voltage source are all connected in series. When L is removed from the circuit, the phase difference between the voltage and current in the circuit is \(\frac{\pi}{3}\). Instead, if C is removed from the circuit, the phase difference is again \(\frac{\pi}{3}\). The power factor of the circuit is

(a)

1/2

(b)

1/\(\sqrt2\)

(c)

(d)

\(\sqrt3\)/2

In a transformer, the number of turns in the primary and the secondary are 410 and 1230 respectively. If the current in primary is 6A, then that in the secondary coil is

(a)

2 A

(b)

18 A

(c)

12 A

(d)

1 A

The current i flowing in a coil varies with time as shown in the figure. The variation of induced emf with time would be

(a)

(b)

(c)

(d)

The flux linked with a coil at any instant t is given by \(\Phi\)_B = 10t²− 50t + 250. The induced emf at t = 3s is

(a)

−190 V

(b)

−10 V

(c)

10 V

(d)

190 V

An electron moves on a straight line path XY as shown in the figure. The coil abcd is adjacent to the path of the electron. What will be the direction of current, if any, induced in the coil?

(a)

The current will reverse its direction as the electron goes past the coil

(b)

No current will be induced

(c)

abcd

(d)

adcb

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Answers Key:

Multiple Choice Question

5 x 1 = 5

\(\Phi = \frac{\pi}{3}-\frac{\pi}{3}=0 \)
Power factor = cosФ = cos 0 = 1

(a)
2 A

Solution
e = -L\(\frac{dl}{dt}\)

\(\phi_B =10 t^2-50 t+250 \)
\(e =\frac{-d \phi_B}{d t} \)
\(=\frac{-d}{d t}\left(10 t^2-50 t+250\right) \)
=-(20 t - 50)
=-20 t + 50
When, t = 3 s, e =-20(3) + 50 = -60 + 50
e = -10V

The direction of conventional current is always opposite to the direction of flow of electrons.