# CBSE Board Exam 2020: Important MCQs (with Answers) for Class 12 Physics - Chapter 5-Magnetism and Matter; Also useful for JEE Main, UPSEE, WBJEE

Check important MCQs (with Answers) for Class 12 Physics Board Exam 2020 (for Chapter 5 - Magnetism and Matter). These MCQs are very useful for competitive exams like UPSEE 2020, JEE Main, WBJEE etc.

Created On: Jan 28, 2020 12:18 IST
Modified On: Jan 28, 2020 15:33 IST CBSE Class 12 Physics Board Exam Important MCQ Chapter 5 Magnetism and Matter

CBSE Board Exams 2020 for Class 12 are going to start from 15th February. Students of CBSE Class 12 having Physics as one of their subjects can check some important MCQs (with Answers) based on Class 12 Physics - Chapter 5, Magnetism & Matter. These questions are very important for the preparation of upcoming CBSE 12th Physics Board Exam 2020 & many other competitive exams like UPSEE  2020, JEE Main 2020, WBJEE 2020, etc. Here, you will also get links to access some very valuable articles for the preparation of the upcoming CBSE Board Exam 2020.

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Important MCQs (with Answers) for Class 12 Physics: Chapter 5 Magnetism and Matter:

1. Which of the following option is correct for a diamagnetic material?

Sol: (c)

2. What is the net magnetic moment of an atom of a diamagnetic material?

(a) Greater than 1

(b) Less than 1 but greater than zero

(c) Less than zero but greater than ‒1

(d) Zero

Sol: (d)

In diamagnetic materials all the electron are paired so there is no permanent net magnetic moment per atom.

3. What is the S.I. unit of susceptibility?

(a) A m‒1

(b) T A m‒1

(c) T m A‒1

(d) No units

Sol: (d)

Sol:  (b)

If magnet is rotated from an initial orientation θ1 to final orientation θ2 then stored potential energy will be equal to the work required

5. It is considered that the earth consists of a huge bar magnet of magnetic moment 8.0 × 1022 Am2. What is the magnitude of the earth’s magnetization? (Radius of earth = 6400 km)

(a) 100 Am‒1

(b) 80.2 Am

(c) 72.8 Am‒1

(d) 30 Am

Sol: (c)

Given, radius of earth = R = 6400 km = 6.4 × 106 m, magnetic moment = M = 8.0 × 1022 Am2.

Assuming earth to be sphere,

6. A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22o with the horizontal. The horizontal component of the earth’s magnetic field at the place is 0.35 G. The earth’s magnetic field at the place is: (Given cos22o = 0.97)

(a) 0.91 G

(b) 0.64 G

(c) 0.38 G

(d) 0.12 G

Sol: (c)

Given angle of dip = θ = 22o. If BH is the horizontal component of earth’s magnetic field, then the earth’s magnetic field is given by BE = BH /cos θ⇒BE = 0.35/0.97 = 0.38 G.

7. The figure given below is the hysteresis loop for two elements A and B. Identify the elements A and B.

(a) A is steel whereas B is soft iron

(b) B is steel whereas A is soft iron

(c) B is hard iron whereas A is soft iron

(d) Insufficient information is provided in the question

Sol: (b)

Hysteresis loop for soft iron is narrow and large whereas for steel it is wide and short.

Here B is steel and A is soft iron.

8. What is the value of angle of dip at the magnetic equator?

(a) 0o

(b) 90o

(c) 45o

(d) Nearly 30o

Sol: (a)

The value of angle of dip at magnetic equator is zero degree.

9. A magnetic needle lying parallel to the magnetic requires W units of work to turn it through 60o. Find the value of torque needed to maintain the needle in this position?

(a) W

(b) 2W

(c) W √3

(d) (W/2) √3

Sol: (c)

As, W = MB (1 ‒ cos 60o) = MB/2.

Now, torque = MB sin 60o = √3 MB/2 = W √3.

10. A magnet of magnetic moment M is rotated through 360o in a magnetic field of magnitude H. The net work done will be:

(a) ‒MH

(b) 2MH

(c) Zero

(d) MH

Sol: (c)

Work done, W = ‒ MH (cos θ2 ‒ cos θ1) = ‒ MH (cos360o ‒ cos 0o) = zero.

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