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NCERT Exemplar Solutions for Class 12 Physics - Chapter 6: Electromagnetic Induction (MCQ I)

Sep 5, 2017 08:49 IST
NCERT Exemplar Solutions for Class 12 Physics - Chapter 6
NCERT Exemplar Solutions for Class 12 Physics - Chapter 6

NCERT Exemplar Solutions for CBSE Class 12 Physics Chapter 6 – Electromagnetic Induction are available here. With this article, you will get solutions from question number 6.1 to question number 6.6. These questions are basically multiple choice questions with single correct answer (MCQ I). These questions can be asked in CBSE Class 12 Physics board exams and other competitive exams like JEE Main, NEET, WBJEE etc.

NCERT Exemplar Solutions for CBSE Class 12 Physics, Chapter 6 (from question number 6.1 to 6.6) are given below:

Question 6.1:

NCERT Exemplar Solutions for Class 12 Physics - Chapter 6: Electromagnetic Induction (Question 6.1)

Solution 6.1: (c)

NCERT Exemplar Solutions for Class 12 Physics - Chapter 6: Electromagnetic Induction (Solution 6.1)

NCERT Exemplar Textbook: CBSE Class 12 Physics – All Chapters

Question 6.2:

NCERT Exemplar Solutions for Class 12 Physics - Chapter 6: Electromagnetic Induction (Question 6.2)

Solution 6.2: (b)

NCERT Exemplar Solutions for Class 12 Physics - Chapter 6: Electromagnetic Induction (Solution 6.2)

Question 6.3:

A cylindrical bar magnet is rotated about its axis. A wire is connected from the axis and is made to touch the cylindrical surface through a contact. Then,

(a) a direct current flows in the ammeter A

(b) no current flows through the ammeter A

(c) an alternating sinusoidal current flows through the ammeter A with a time period T = 2π/ω

(d) a time varying non-sinusoidal current flows through the ammeter A

NCERT Exemplar Solutions for Class 12 Physics - Chapter 6: Electromagnetic Induction (Question 6.3 - Diagram)

Solution 6.3: (b)

As cylindrical bar magnet is rotated about its axis, therefore, no change in flux linked with the circuit takes place and no emf induces. Hence, no current flows through the ammeter A.

Question 6.4:

There are two coils A and B as shown in figure. A current starts flowing in B as shown, when A is moved towards B and stops when A stops moving. The current in A is counter clockwise. B is kept stationary when A moves. We can infer that

(a) there is a constant current in the clockwise direction in A

(b) there is a varying current in A

(c) there is no current in A

(d) there is a constant current in the counter clockwise direction in A

NCERT Exemplar Solutions for Class 12 Physics - Chapter 6: Electromagnetic Induction (Question 6.4 - Diagram)

Solution 6.4: (d)

Here, induced emf in B is due to the variation of magnetic flux associated with it.

If the current in A would be variable, there must be an induced emf (current) in B even if the A stops moving.

If the current in A is constant and A stops moving the current in B becomes zero.

Question 6.5:

Same as problem 4 except the coil A is made to rotate about a vertical axis (figure). No current flows in B if A is at rest. The current in coil A, when the current in B (at t = 0) is counter-clockwise and the coil A is as shown at this instant, t = 0, is

(a) constant current clockwise

(b) varying current clockwise

(c) varying current counter clockwise

(d) constant current counter clockwise

NCERT Exemplar Solutions for Class 12 Physics - Chapter 6: Electromagnetic Induction (Question 6.5 - Diagram)

Solution 6.5: (a)

The anticlockwise flow of the current in B is equivalent to North Pole of magnet. If the coil A is above to it then the magnetic field lines are emanating upward to coil A. If coil A start rotating at t = 0, the current in A is constant along clockwise direction according to the Lenz's law.

Question 6.6:

The self inductance L of a solenoid of length I and area of cross-section A, with a fixed number of turns N increases as

(a) l and A increase

(b) l decreases and A increases

(c) l increases and A decreases

(d) both land A decrease

Solution 6.6: (b)

The self-inductance of a long air filled solenoid is given by, L = m n2 A l, where, n = N/l.

Here, A is cross-sectional area and l is length, n is number of turns (N) per unit length.

Clearly, L increases when l decrease and A increases.

NCERT Solutions for CBSE Class 12 Physics: All Chapters

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