Tamil Nadu Board 12th Standard Physics - Unit 1: Book Back Answers and Solutions
This post covers the book back answers and solutions for Unit 1 from the Tamil Nadu State Board 12th Standard Physics textbook. These detailed answers have been carefully prepared by our expert teachers at KalviTips.com.
We have explained each answer in a simple, easy-to-understand format, highlighting important points step by step under the relevant subtopics. Students are advised to read and memorize these subtopics thoroughly. Once you understand the main concepts, you’ll be able to connect other related points with real-life examples and confidently present them in your tests and exams.
By going through this material, you’ll gain a strong understanding of Unit 1 along with the corresponding book back questions and answers (PDF format).
Question Types Covered:
- 1 Mark Questions: Choose the correct answer, Fill in the blanks, Identify the correct statement, Match the following
- 2 Mark Questions: Answer briefly
- 3, 4, and 5 Mark Questions: Answer in detail
All answers are presented in a clear and student-friendly manner, focusing on key points to help you score full marks.
All the best, Class 12 students! Prepare well and aim for top scores. Thank you!
I. Multiple choice questions.
(b) B1 and B2
(a) point charge
(b) uniformly charged infinite line
(c) uniformly charged infinite plane
(d) uniformly charged spherical shell
Answer Key:
(c) uniformly charged infinite plane
(a) 4 mC
(b) 8 mC
(c) 5 mC
(d) 7 mC
Answer Key:
(b) 8 mC
Hint:
T = PE sin θ
T = (q × 21)E sin30°
q = (q × 10-2) ×
(b) A < B = C < D
(c) C < A = B < D
(d) D > C > B > A
Hint:
flux depends on charge
(a) less than before
(b) same as before
(c) more than before
(d) zero
Answer Key:
(c) more than before
Hint:
(b) 2 = 4 < 3 < 1
(c) 2 = 3 < 1 < 4
(d) 3 < 1 < 2 < 4
Answer Key:
(a) 1 = 4 < 2 < 3
Hint:
(a) 10 V
(b) -20 V
(c) +20 V
(d) -10 V
Answer Key:
(c) +20 V
Hint:
E =
dv = E.dx = l0x = 10 × 2
dv = 20 V
In a spherical shell, the electric field inside is zero. But the electric potential is constant.
(a) 8.80 × 10-17 J
(b) -8.80 × 10-17 J
(c) 4.40 × 10-17 J
(d) 5.80 × 10-17 J
Answer Key:
(a) 8.80 × 10-17 J
Hint:
WA→B = (VA – VB)q
=(7+4)ne
= 11 × 50 × 1.6 × 10-19
=8.8 × 10-17
(a) Q remains the same, C is doubled
(b) Q is doubled, C doubled
(c) C remains the same, Q doubled
(d) Both Q and C remain the same
Answer Key:
(c) C remains the same, Q doubled
Hint:
(a) Capacitance
(b) Charge
(c) Voltage
(d) Energy density
Answer Key:
(d) Energy density
14. Three capacitors are connected in a triangle as shown in the figure. The equivalent capacitance between points A and C is
(a) 1 μF
(b) 2 μF
(c) 3 μF
(d)
Answer Key:
(b) 2 μF
Hint:
(a) 3 × 10-2 C
(b) 4 × 10-2 C
(c) 1 × 10-2 C
(d) 2 × 10-2 C
Answer Key:
(a) 3 × 10-2 C
Hint:
Total charge Q = q1 + q2 = 4 × 10-2 C
charge on bigger sphere,
q2 = Q(
= 4 × 10-2 ×
q2 = 3 × 10-2 C
II. Short answer questions.
1. What is meant by quantisation of charge?Answer key:
q=ne
n is any integer (0 , + 1, + 2, + 3…)
This is called quantisation of charge.
Answer Key:
Coulomb’s law states that, Electrostatic force is directly proportional to the product of the magnitude of the point charges and inversely proportional to the square of the distance between two point charges.
S.No
|
Coulomb
Force
|
Gravitational
Force
|
1
|
It
may be attractive or repulsive.
|
It
is always attractive in nature.
|
2
|
It
depends upon medium.
|
It
does not depend upon the medium.
|
3
|
It
is always greater in magnitude.
K=9x109
Nm2 C-2
|
It
is lesser than coulomb force.
G=6.67x10-11
Nm2 Kg2.
|
Answer Key:
When a number of charges are interacting the total force of a given charge is the vector sum of the individual forces exerted on the given charge by all the other charges.
5. Define electric field.
Answer Key:
The electric field at the point P at a distance r from the point charge q is the force experienced by a unit charge .
SI unit: NC-1
Answer key:
Electric field vectors are visualized by the concept of electric field lines.
They form a set of continuous lines which represent the electric field in some region of space visually.
Answer Key:
If some charge is placed in the intersection point , then it has to move in two different directions at the same time which is physically impossible. Hence , electric field lines do not intersect.
Answer Key:
Electric diapole:
Two equal and opposite charges separated by a small distance constitute an electric dipole.
Magnitude of the electric dipole:
Magnitude of the electric dipole moment is equal to the product of the magnitude of one of the charges and the distance between them.
Answer Key:
The electric dipole moment for a collection of n point charges is given by where
Answer Key:
Electric potential at a point P is equal to the work done force to bring unit positive charge with constant velocity from infinity to the point P in the region of the external electric field.
Answer Key:
An equipotential surface is a surface on which all the points are at the same electric potential.
Answer Key:
1. The work done to move a charge q between any two points A and B, W = q (VB – VA).
2. If the points A and B lie on same equipotential surface Work done is zero because VA = VB
3. The electric field is always normal to an equipotential surface.
Answer Key:
Electric potential energy is defined as the work done in bringing the various charges to their respective positions from infinitely large mutual separation.
Answer Key:
The number of electric field lines crossing a given area kept normal to the electric field lines is called electric flux.
Answer Key:
The energy stored per unit volume of space is defined electrostatic energy density.
U – electrostatic potential energy
E – electric field
Answer Key:
The phenomenon of protecting a region of space from any external electric field is called as electrostatic shielding .
Consider a cavity inside the conductor whatever the charges at the surfaces and whatever the electrical disturbances outside, the electric field inside cavity is zero.
Ex: Faraday Cage
18. What is Polarisation?
Answer Key:
Polarisation is defined as the total dipole moment per unit volume of the dielectric.
χe= electric susceptibility
Answer Key:
Answer Key:
The capacitance C of a capacitor is defined as the ratio of the magnitude of charge on either of the conductor plates to the potential difference existing between them.
Answer Key:
1. The total charge of the charged conductor near the sharp edge reduces.
2. Leakage of charges from the sharp points to the charged conductor.
3. Corona discharge also known as "action of points"
III. Long answer questions.
1. Discuss the basic properties of electric charges.1. Electric Charge :
- Objects in the universe are made up of atoms which made up of protons , neutrons, and electrons.
- These particles have mass , an inherent property of particles.
- Electric charge is another intrinsic and fundamental property of particles.
- The SI unit of charge is Coulomb.
- The total electric charge in the universe is constant. Charge can neither be created nor be destroyed.
- In any physical process net change in charge will always be zero.
- Objects are electrically neutral before rubbing process happen.
- After rubbing simply transfers charges from one object to the others.
1. When a glass rod rubbed with silk cloth then negative charge transferred from glass to silk.
2. As a result , glass rod is positively charged and silk cloth is negatively charged.
- The charge q on any object is equal to an integral multiple of fundamental unit of charge e. This is called quantisation of electric charge. Charge of electron = - 1.6 X 10-19 C .
- q=ne
- n is any integer (0 , + 1, + 2, + 3…)
- This is called quantisation of charge.
Answer Key:
S.No
|
Coulomb Force
|
Gravitational Force
|
1
|
It may be attractive or repulsive. |
It is always attractive in nature.
|
2
|
It depends upon medium. |
It does not depend upon the medium.
|
3
|
It is always greater in magnitude.
K=9x109 Nm2 C-2
|
It is lesser than coulomb force.
G=6.67x10-11 Nm2
Kg2.
|
Answer Key:
Electric Field :
Electric field at a point P, at a distance r from the point charge q is the force experienced by a unit point charge.
Formula:
Important aspects of Electric field:
1. Force experienced by the test charge q0 is
2. q is positive , electric field points away from the source .
3. q is negative , electric field points towards the source .
4. Electric field is independent of test charge q0 and depends only on source charge q.
5. Electric field is a vector quantity which has unique direction and magnitude .
6. As distance increases electric field decreases in magnitude.
7. Test charge q0 is small , not modify the electric field of source charge.
8. Electric field equation is only valid for point charges.
9. There are two kinds of the electric field. Uniform and non - uniform electric field.
10. Uniform electric field : Have same direction and constant magnitude at all points.
11. Non-uniform electric field : Different directions and different magnitudes or both at different points in space.
Answer Key:
Diagram:
Explanation:
Consider an electric dipole placed on the x-axis . A point C is located at a distance of r from the midpoint O of the dipole on the axial line. Electric dipole moment vector
i) Electric field due to a electric dipole on its axial line:
1. Let a point C at a distance r from the midpoint O of the dipole on equatorial plane .
2. The point C is equi distant from +q and – q , magnitude of the electric fields is same.
3.
4. One component parallel to dipole axis and other perpendicular to it.
5. Perpendicular components |
6. Magnitude of total electric field is sum of the parallel component of
Diagram:
Answer Key:
Diagram:
Torque on dipole due to uniform electric field:
1. Consider an electric dipole moment
2. Charge +q experience a force:q
3. Charge -q experience a force:-q
4. Total force acts on the dipole is zero. But two forces constitute a couple.
5.The dipole experience a torque tends to rotate the dipole.
Derivation:
1.
1.Magnitude of torque is τ=pE sin θ and is maximum when θ=90°.
2.If
Answer Key:
Diagram:
Explanation:
Consider a positive charge q kept fixed at a origin . Let P be a point at distance r from charge q.
Derivation:
1. Electric potential at the point P:
3.
4. The infinitesimal displacement vector
5.
6.
Charge q
|
Potential
|
V Value
|
Distance
|
Positive
|
V=
|
Decreases
|
Increases
|
Neagative
|
V=
|
Increases
|
Increases
|
Answer Key:
Diagram:
Explanation
Consider two equal and opposite charges separated by a small distance 2a.
The point P is located at a distance r from the midpoint of the dipole.
Let θ be the angle between the line OP and dipole axis AB.
Let r1 be the distance of point P from +q. Let r2 be the distance of point P from -q.
1. Potential at P due to charge +q:
2. Potential at P due to charge -q:
3. Total Potential:
IV. Exercises.
Physics
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