Careers360 Logo
NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields

Access premium articles, webinars, resources to make the best decisions for career, course, exams, scholarships, study abroad and much more with

Plan, Prepare & Make the Best Career Choices

NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields

Edited By Vishal kumar | Updated on Sep 08, 2023 02:29 PM IST | #CBSE Class 12th

NCERT Solutions for Class 12 Physics Chapter 1 – Access and Download Free PDF

NCERT Solutions For Class 12 Physics Chapter 1 Electric Charges and Fields: Welcome to the updated NCERT book solutions for the very first chapter of the Class 12 syllabus. On this Careers360 page, you will find comprehensive electric charges and fields class 12 NCERT solutions covering questions from 1.1 to 1.34. Questions 1.1 to 1.23 belong to the exercise section, while questions 1.24 to 1.34 are part of the additional exercise. These class 12 physics ch1 NCERT solutions, crafted by subject experts, are presented in a simple and detailed manner. You can download the PDF for free and use them at your convenience.

Problems in Physics Class 12 Chapter 1 are important from the CBSE class 12 exam point of view. Try solving all the questions of NCERT Class 12 Physics Chapter 1 and if any doubt arises then check the CBSE NCERT solutions for Class 12 Physics Chapter 1 Electric Charges and Fields. Electrostatics is the study of electric charges at rest. The NCERT Class 12th Physics Electric Charges and Fields deal with the charging of a body, properties of charge, Columbus law, electric field, electric flux, Gauss law and application of Gauss law. Two main laws discussed in Physics Class 12 Chapter 1 are Gauss law and Columbus law. NCERT Solutions for Class 12 physics chapter 1 Electric Charges and Fields explains problems related to these topics.

Apply to Aakash iACST Scholarship Test 2024

Applications for Admissions are open.

The derivations in Physics Class 12 NCERT Chapter 1 Electric Charges and Fields of NCERT are very important in CBSE board exam point of view. As the second chapter of NCERT solutions for Class 12 is the continuation of the NCERT Class 12 Physics chapter 1 pdf, it is important to remember all points studied in the chapter. These class 12 electric field and charges ncert solutions will help you in a better understanding of the concepts you have studied. While studying NCERT Solutions for Class 12 Physics Chapter 1 formulas can be compared with the chapter Gravitation of Physics NCERT solutions class 11.

ALLEN JEE Exam Prep

Start your JEE preparation with ALLEN

Aakash iACST Scholarship Test 2024

Get up to 90% scholarship on NEET, JEE & Foundation courses

Free download class 12 physics chapter 1 exercise solutions PDF for CBSE exam.

NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields

Download PDF


NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields: Exercise Question Answer

Q 1.1 What is the force between two small charged spheres having charges of 2 \times 10^{-7} C and 3 \times 10^{-7}C placed 30 cm apart in air?

Answer:

Given,

q_{1} = 2 \times 10^{-7} C

q_{2} = 3 \times 10^{-7}C

r = 30 cm = 0.3 m

We know,

Force between two charged particles, q_{1} and q_{2} separated by a distance r.

F = \frac{1}{4\pi \epsilon _{0}} \frac{q_{1}q_{2}}{r^2}

= \frac{1}{4\pi \epsilon _{0}} \frac{2\times10^{-7} \times 3 \times 10^{-7}}{(30\times10^{-2}\ m)^2\ }

= (9\times10^9\ N)\times \frac{6\times10^{-14+4}}{900\ m^2\ } = 6\times10^{-3} N

Since the charges are of the same nature, the force is repulsive.

Q 1.2 (a) The electrostatic force on a small sphere of charge 0.4 \mu C due to another small sphere of charge -0.8 \mu C in air is -0.2N . (a) What is the distance between the two spheres?

Answer:

Given,

q_{1} = 0.4 \mu C

q_{2} = -0.8 \mu C

F = -0.2N (Attractive)

We know,

Force between two charged particles, q_{1} and q_{2} separated by a distance r.

F = \frac{1}{4\pi \epsilon _{0}} \frac{q_{1}q_{2}}{r^2}

\\ -0.2 = 9\times10^9 \times \frac{(0.4\times10^{-6})(-0.8\times10^{-6})}{r^2} \\ \implies r^2 = 9\times10^9\times0.32\times10^{-12}/0.2 \\ \implies r^2 = 9\times0.16\times10^{-2} \\ \implies r = 1.2\times10^{-1} m = 0.12 m =12 cm

Therefore, the distance between the two charged spheres is 12 cm.

Q 1.2(b) The electrostatic force on a small sphere of charge 0.4 \mu C due to another small sphere of charge -0.8 \mu C in air is 0.2N .(b)What is the force on the second sphere due to the first?

Answer:

Using Newton's third law, the force exerted by the spheres on each other will be equal in magnitude.

Therefore, the force on the second sphere due to the first = 0.2 N (This force will be attractive since charges are of opposite sign.)

Q 1.3 Check that the ratio \frac{ke^{2}}{Gm_{e}m_{p}} is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

Answer:

Electrostatic force

F=\frac{KQ^2}{r^2}

So the dimension of

[Ke^2]=[Fr^2] ..................(1)

The gravitational force between two bodies of mass M and m is

F=\frac{GMm}{r^2}

so dimension of

[Gm_em_p]=[Fr^2] .............(2)

Therefore from (1) and (2)

[\frac{ke^{2}}{Gm_{e}m_{p}} ] is dimensionless

or

Here,

K = 1/4\pi \epsilon _{0} , where 1515131900509914 is the permittivity of space. [1/ \epsilon _{0}] =[C/V.m] = [Nm^2 C^{-2}]

e = Electric charge ([e] = [C])

G = Gravitational constant. ([G]= [Nm^2kg^{-2}] )

m_{e} and m_{p} are mass of electron and proton ([ m_{e} ] = [ m_{p} ] = [Kg])

Substituting these units, we get

[\frac{ke^{2}}{Gm_{e}m_{p}} ] = [\frac{C^2 \times Nm^2C^{-2}}{Nm^2kg^{-2}\times kg\times kg} ] = M^{0}L^{0}T^{0}

Hence, this ratio is dimensionless.

Putting the value of the constants

\frac{ke^{2}}{Gm_{e}m_{p}} = \frac{9\times10^9\times(1.6\times10^{-19})^{2}}{6.67\times10^{11}\times9.1\times10^{-31}\times1.67\times10^{-27}}

=2.3 \times 10^{40}

The given ratio is the ratio of electric force \frac{ke^{2}}{R^2} to the gravitational force between an electron and a proton \frac{Gm_{e}m_{p}}{R^2} considering the distance between them is constant!

Q 1.4 (a) Explain the meaning of the statement ‘electric charge of a body is quantised’.

Answer:

The given statement "electric charge of a body is quantised" implies that charge on a body can take only integral values. In other words, only the integral number of electrons can be transferred from one body to another and not in fractions.

Therefore, a charged body can only have an integral multiple of the electric charge of an electron.

Q 1.4 (b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?

Answer:

On a macroscopic level, the amount of charge transferred is very large as compared to the charge of a single electron. Therefore, we tend to ignore the quantisation of electric charge in these cases and considered to be continuous in nature.

Q 1.5 When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.

Answer:

When a glass rod is rubbed with a silk cloth, opposite charges appear on both the rod and the cloth.

The phenomenon of charging bodies by rubbing them against each other is known as charging by friction. Here, electrons are transferred from one body to another giving both the bodies an equal but opposite charge. The number of electrons lost by one body (attains positive charge due to loss of negatively charged electrons) is equal to the number of electrons gained by the other body (attains negative charge). Therefore, the net charge of the system is zero. This is in accordance with the law of conservation of charge

Q 1.6 Four point charges q_{A}= 2 \mu C , q_{B}= -5 \mu C , q_{C}= 2 \mu C , and q_{D}= -5 \mu C are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?

Answer:

Charges at (A, C) and (B, D) are pairwise diametrically opposite and also equal.

Therefore, their force on a point charge at the centre of the square will be equal but opposite in directions.

Now, AC = BD = \sqrt{2}\times(10\times10^{-2} m) = \sqrt{2}\times0.1 m

\therefore AO = BO = CO = DO = r = Half of diagonal = \sqrt{2}\times0.05 m

Force on point charge at centre due to charges at A and C = F_{A} = -F_{C} = \frac{k(2\mu C)(1\mu C)}{(r)^2}

Similarly, force on point charge at centre due to charges at B and D = F_{B} = -F_{D} = \frac{k(-5\mu C)(1\mu C)}{(r)^2}

\therefore Net force on point charge = \\F_{A} + F_{B} + F_{C} + F_{D} \\

= -F_{B} +F_{B} - F_{D} + F_{D} = 0 .

Hence, the charge at the centre experiences no force.

Q 1.7 (a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?

Answer:

A positive point charge experiences a force in an electrostatic field. Since the charge will experience a continuous force and cannot jump from one point to another, the electric field lines must be continuous.

Q 1.7 (b) Explain why two field lines never cross each other at any point?

Answer:

A tangent drawn at any point on a field line gives the direction of force experienced by a unit positive charge due to the electric field on that point. If two lines intersect at a point, then the tangent drawn there will give two directions of force, which is not possible. Hence two field lines cannot cross each other at any point.

Q 1.8 (a) Two point charges q_{A}=3 \mu C and q_{B}=-3 \mu C are located 20 cm apart in vacuum. (a) What is the electric field at the midpoint O of the line AB joining the two charges?

Answer:

1643006219871

Given, AB = 20 cm

Since, O is the midpoint of the line AB.

AO = OB = 10 cm = 0.1m

The electric field at a point caused by charge q, is given as,

E = \frac{kq}{r^2}

Where, q is the charge, r is the distance between the charges and the point O

k = 9x10 9 N m 2 C -2

Now,

Due to charge at A, electric field at O will be E_{A} and in the direction AO.

E_{A} =\frac{ 9\times10^9 \times 3\times10^{-6}}{0.1^2}

Similarly the electric field at O due to charge at B, also in the direction AO

E_{B} =\frac{ 9\times10^9 \times (-3\times10^{-6})}{0.1^2}

Since, both the forces are acting in the same direction, we can add their magnitudes to get the net electric field at O:

E' = E_{A} + E_{B} = 2E (Since their magnitudes are same)

E' =2\times \frac{ 9\times10^9 \times 3\times10^{-6}}{0.1^2} = 5.4\times 10^4 N/C along the direction AO.

Q 1.8 (b) Two point charges q_{A}=3\mu C and q_{B}=-3\mu C are located 20 cm apart in vacuum. If a negative test charge of magnitude 1.5 \times 10^{-9}C is placed at this point, what is the force experienced by the test charge?

Answer:

Let Q = -1.5 \times10^{-9} C

The force experienced by Q when placed at O due to the charges at A and B will be:

F = Q \times E'

where 'E' is the net electric field at point O.

F=1.5 \times 10^{-9} C \times 5.4 \times 10^4 N/C = 8.1\times10^{-3 } N

Q being negatively charged will be attracted by positive charge at A and repelled by negative charge at B. Hence the direction of force experienced by it will be in the direction of OA.

Q 1.9 A system has two charges q_{A}=2.5\times 10^{-7}C and q_{B}=-2.5\times 10^{-7} C located at points A: (0, 0, –15 cm) and B: (0,0, +15 cm), respectively. What are the total charge and electric dipole moment of the system?

Answer:

1643006161832Given,

q_{A}=2.5\times 10^{-7}C and q_{B}=-2.5\times 10^{-7}

The total charge of the system = q_{A} + q_{B} = 0

\therefore The system is electrically neutral. (All dipole systems have net charge zero!)

Now, distance between the two charges, d = 15 + 15 = 30 cm = 0.3 m

We know, The electric dipole moment of the system, p = q_{A} x d = q_{B} x d (i.e, the magnitude of charge x distance between the two charges)

\therefore p = 2.5\times10^{-7} C \times 0.3 m = 7.5\times10^{-8} Cm

The direction of a dipole is towards the positive charge . Hence, in the positive z-direction.

Q 1.10 An electric dipole with dipole moment 4\times 10^{-9} Cm is aligned at 30^{\circ} with the direction of a uniform electric field of magnitude 5\times 10^{4}NC^{-1} . Calculate the magnitude of the torque acting on the dipole.

Answer:

1643006119102Given,

Electric dipole moment, p = 4\times 10^{-9} Cm

\Theta = 30^{0} \ \therefore sin\Theta = 0.5

E = 5\times 10^{4}NC^{-1}

We know, the torque acting on a dipole is given by:

\tau = p \times E

\implies \tau = p Esin\Theta = 4\times10^{-9} \times5\times10^{4}\times0.5 \ Nm

\implies \tau =10^{-4}Nm

Therefore, the magnitude of torque acting on the dipole is 10^{-4}Nm

Q 1.11 (a) A polythene piece rubbed with wool is found to have a negative charge of 3\times 10^{-7} . (a) Estimate the number of electrons transferred (from which to which?)

Answer:

Clearly, polyethene being negatively charged implies that it has an excess of electrons(which are negatively charged!). Therefore, electrons were transferred from wool to polyethene.

Given, charge attained by polyethene = -3\times 10^{-7} C

We know, Charge on 1 electron = -1.6\times10^{-19} C

Therefore, the number of electrons transferred to attain a charge of -3\times 10^{-7} =

\frac{-3\times 10^{-7}}{-1.6\times10^{-19} C} = 1.8\times10^{12} electrons.

Q 1.11 (b) A polythene piece rubbed with wool is found to have a negative charge of 3\times 10^{-7}C . Is there a transfer of mass from wool to polythene?

Answer:

The charge attained by polyethene (and also wool!) is solely due to the transfer of free electrons.

We know, Mass of an electron = 9.1\times10^{-31}\ kg

The total mass of electron transferred = number of electrons transferred x mass of an electron

= 9.1\times10^{-31} \times 1.8\times10^{12}\ kg = 16.4\times10^{-19} \ kg

Yes, there is a transfer of mass but negligible.

Q 1.12 (a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5\times 10^{-7}C ? The radii of A and B are negligible compared to the distance of separation.

Answer:

Since the radii of the spheres A and B are negligible compared to the distance of separation, we consider them as a point object.

Given,

charge on each of the spheres = 6.5\times 10^{-7}C

and distance between them, r = 50 cm = 0.5 m

We know,

F = k\frac{q_{1}q_{2}}{r^2}

Therefore, the mutual force of electrostatic repulsion(since they have the same sign of charge)

F = 9\times10^9Nm^{2}C^{-2}\times\frac{(6.5\times10^{-7}\ C)^2}{(0.5\ m )^2} = 1.5\times10^{-2} N

Q 1.12 (b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

Answer:

We know, force between two charged particles separated by a distance r is:

F = k\frac{q_{1}q_{2}}{r^2} = k\frac{q^2}{r^2} (\because q_{1} = q_{2} = q)

Now if q\rightarrow 2q\ and\ r\rightarrow r/2

The new value of force:

F_{new} =k\frac{(2q)^2}{(r/2)^2} = 16k\frac{q^2}{r^2} = 16F

Therefore, the force increases 16 times!

F_{new} =16F = 16\times1.5\times10^{-2} N = 0.24\ N

Q 1.13 Suppose the spheres A and B in Exercise 1.12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?

Answer:

When two spheres of the same size are touched, on attaining the equipotential state, the total charge of the system is equally distributed on both of them.

Therefore, (i) When uncharged third sphere C is touched with A, charge left on A = 0.5\times 6.5\times 10^{-7}C

and charge attained by C = 0.5\times 6.5\times 10^{-7}C

(ii) Now, charge on B + charge on C = 6.5\times 10^{-7}C + 0.5\times 6.5\times 10^{-7}C = 1.5\times 6.5\times 10^{-7}C

When touched, charge left on B = 0.5\times1.5\times 6.5\times 10^{-7}C

Therefore q_{A}\rightarrow 0.5\times q_{A}\ and\ q_{B}\rightarrow 0.75\times q_{B}

Therefore, F' = 0.5\times0.75 \times F = 0.375 \times 1.5\times10^{-2} N = \boldsymbol{5.7\times10^{-3}\ N}

Q 1.14 Figure shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

1643006268513 Answer:

Charges 1 and 2 are repelled by the negatively charged plate of the system

Hence 1 and 2 are negatively charged .

Similarly, 3 being repelled by positive plate is positively charged.

(charge to the mass ratio: charge per unit mass)

Since 3 is deflected the most, it has the highest charge to mass ratio .

Q 1.15 (a) Consider a uniform electric field E=3\times 10^{3}\ \widehat{i}\ N/C . (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane?

Answer:

Given,

E=3\times 10^{3}\ \widehat{i}\ \frac{N}{C}

Area of the square = 0.01^2\ m^2

Since the square is parallel to the yz plane, therefore it's normal is in x-direction.(i.e \widehat{i} direction )

therefore, flux through this surface:

\phi = E.A

\implies \phi = (3\times10^3\ \widehat{i}).(0.01\ \widehat{i}) Nm^2/C = \boldsymbol{30 Nm^2/C}

Q 1.15 (b) Consider a uniform electric field E=3 \times 10^{3}\ \widehat{i}\ N/C .What is the flux through the same square if the normal to its plane makes a 60^{\circ} angle with the x-axis?

Answer:

Now, Since the normal of the square plane makes a 60^{\circ} angle with the x-axis

cos\Theta = cos(60^{0}) = 0.5

therefore, flux through this surface:

\phi = E.A = EAcos\Theta

\implies \phi = (3\times10^3)(0.01)(0.5) Nm^2/C = \boldsymbol{15\ Nm^2/C}

Q 1.16 What is the net flux of the uniform electric field of Exercise 1.15 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?

Answer:

The net flux of the uniform electric field through a cube oriented so that its faces are parallel to the coordinate planes is z ero .

This is because the number of lines entering the cube is the same as the number of lines leaving the cube.

Alternatively,

using Gauss’s law, we know that the flux of electric field through any closed surface S is 1/\epsilon _{0} times the total charge enclosed by S.

i.e. \phi = q/\epsilon _{0}

where, q = net charge enclosed and \epsilon _{0} = permittivity of free space (constant)

Since there is no charge enclosed in the cube, hence \phi = 0 .

Q 1.17 (a) Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 \times 10^{3}\frac{Nm^{2}}{C} . (a) What is the net charge inside the box?

Answer:

Using Gauss’s law, we know that the flux of electric field through any closed surface S is 1/\epsilon _{0} times the total charge enclosed by S.

i.e. \phi = q/\epsilon _{0}

where, q = net charge enclosed and \epsilon _{0} = permittivity of free space (constant)

Given, \phi = 8.0 \times10^3\ Nm^2/C

\therefore q = \phi \times\epsilon _{0} = (8.0\times10^3 \times8.85\times10^{-12})\ C

\implies q = 70 \times10^{-9} C = \boldsymbol{0.07 \mu C}

This is the net charge inside the box.

Q 1.17 (b) Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0\times 10^{3}\frac{Nm^{2}}{C} .If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or Why not?

Answer:

Using Gauss's law, we know that \phi = q/\epsilon _{0}

Since flux is zero, q = 0, but this q is the net charge enclosed by the surface.

Hence, we can conclusively say that net charge is zero, but we cannot conclude that there are no charges inside the box.

Q 1.18 A point charge +10 \mu C is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)

esc1

Answer:

Let us assume that the charge is at the centre of the cube with edge 10 cm.

Using Gauss's law, we know that the flux of electric field through any closed surface S is 1/\epsilon _{0} times the total charge enclosed by S.

i.e. \phi = q/\epsilon _{0}

where, q = net charge enclosed and \epsilon _{0} = permittivity of free space (constant)

Therefore, flux through the cube: \phi = (10\times10^{-6} C)/\epsilon _{0}

Due to symmetry, we can conclude that the flux through each side of the cube, \phi' , will be equal.

\therefore \phi' = \frac{\phi}{6} = \frac{10^{-5}}{6\times\epsilon _{0}} = \frac{10^{-5}}{6\times\ 8.85\times10^{-12}} = \boldsymbol{1.9 \times 10^5\ Nm^2C^{-1}}

Q 1.19 A point charge of 2.0\mu C is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?

Answer:

Given,

q = net charge inside the cube = 2.0\mu C

Using Gauss’s law, we know that the flux of electric field through any closed surface S is 1/\epsilon _{0} times the total charge enclosed by S.

i.e. \phi = q/\epsilon _{0}

where, q = net charge enclosed and \epsilon _{0} = permittivity of free space (constant)

\therefore \phi = 2.0\times10^{-6}/8.85\times10^{-12}\ Nm^2C^{-1} = \boldsymbol{2.2\times10^5\ Nm^2C^{-1}}

(Note: Using Gauss's formula, we see that the electric flux through the cube is independent of the position of the charge and dimension of the cube! )

Q 1.20 (a) A point charge causes an electric flux of -1.0\times 10^{3}\frac{Nm^{2}}{C} to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge. If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?

Answer:

Given,

\phi = -1.0\times 10^{3}\frac{Nm^{2}}{C}

Using Gauss’s law, we know that the flux of electric field through any closed surface S is 1/\epsilon _{0} times the total charge enclosed by S.

i.e. \phi = q/\epsilon _{0}

where, q = net charge enclosed and \epsilon _{0} = permittivity of free space (constant)

Therefore, flux does not depend on the radius of the sphere but only on the net charge enclosed. Hence, the flux remains the same although the radius is doubled.

\phi' = -10^{3}\frac{Nm^{2}}{C}

Q 1.20 (b) A point charge causes an electric flux of -1.0\times 10^{3}\frac{Nm^{2}}{C} to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge. What is the value of the point charge?

Answer:

Given,

\phi = -1.0\times 10^{3}\frac{Nm^{2}}{C}

Using Gauss’s law, we know that the flux of electric field through any closed surface S is 1/\epsilon _{0} times the total charge enclosed by S.

i.e. \phi = q/\epsilon _{0}

where, q = net charge enclosed and \epsilon _{0} = permittivity of free space (constant)

\therefore q = -10^{3}Nm^2C^{-1} \times 8.85\times10^{-12}N^{-1}m^{-2}C^{2} = -8.8\times10^{-9}\ C \\ = \boldsymbol{-8.8\ nC}

Q 1.21 A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 \times 10^{3}\frac{N}{C} and points radially inward, what is the net charge on the sphere?

Answer:

We know, for determining the electric field at r>R for a conducting sphere, the sphere can be considered as a point charge located at its centre.

Also, electric field intensity at a point P, located at a distance r, due to net charge q is given by,

E = k\frac{q}{r^2}

Given, r = 20 cm = 0.2 m (From the centre, not from the surface!)

\\ \therefore 1.5\times10^3 = 9\times10^9\times \frac{q}{0.2^2} \\ \implies q = \frac{1.5\times0.04}{9}\times10^{-6} = 6.67\times10^{-9}\ C

Therefore, charge on the conducting sphere is - 6.67\ nC (since flux is inwards)

Q 1.22 (a) A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0\frac{\mu C}{m^{2}} . (a) Find the charge on the sphere.

Answer:

Given,

Surface charge density = 80.0\mu Cm^{-2}

Diameter of sphere = 2.4 m \therefore radius of sphere, r = 1.2 m

The charge on the sphere, Q= surface charge density x surface area of the sphere

= (80\times10^{-6})\times(4\pi r^2) = 320\times22/7\times(1.2)^2 = \boldsymbol{1.45\times10^{-3}\ C}

Q 1.23 An infinite line charge produces a field of 9\times 10^{4}\frac{N}{C} at a distance of 2 cm. Calculate the linear charge density.

Answer:

Given,

\lambda = 9\times 10^{4}\frac{N}{C}

d = 2 cm = 0.02 m

We know, For an infinite line charge having linear charge density \lambda , the electric field at a distance d is:

E = k\lambda / d

\therefore 9\times10^4 = 9\times10^{9}\lambda / 0.02



The linear charge density is 10 \mu C/cm .

Q 1.24 (a) Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17\times 10^{-22}\frac{C}{m^{2}} What is E: (a) in the outer region of the first plate

Answer:

We know, electric field, E, due to an infinite plate (length>>thickness) having surface charge density \sigma = \sigma/2\epsilon_{0} .

(To note: It's independent of distance from the plate!)

In the region outside the first plate,

since both plates have the same surface charge density(in magnitude only), their electric fields are same in magnitude in this region but opposite in direction.

(E due to positive plate away from it and E due to negative plate towards it!)

Hence, the electric field in the outer region of the first plate is zero.

Q 1.24 (b) Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17\times 10^{-22}\frac{C}{m^{2}} . What is E:(b) in the outer region of the second plate

Answer:

We know, electric field, E, due to an infinite plate (length>>thickness) having surface charge density \sigma = \sigma/2\epsilon_{0} .

(To note: It's independent of distance from the plate and same everywhere!)

In the region outside the second plate,

Since both plates have the same surface charge density(in magnitude only), their electric fields are same in magnitude in this region but opposite in direction.

(E due to positive plate away from it and E due to negative plate towards it!)

Hence, the electric field in the outer region of the second plate is zero.

Q 1.24 (c) Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17\times 10^{-22}\frac{C}{m^{2}} . What is E:(c) between the plates?

Answer:

We know, electric field, E, due to an infinite plate (length>>thickness) having surface charge density \sigma = \sigma/\epsilon_{0} .

(To note: It's independent of distance from the plate!)

Let A and B be the two plates such that:

\sigma_{A}= 17\times10^{-22} Cm^{-2} = \sigma

\sigma_{B}= -17\times10^{-22} Cm^{-2} = - \sigma

Therefore,

The electric field between the plates, E = E_{A} + E_{B} = \sigma_{A}/2\epsilon_{0} + (-\sigma_{B}/2\epsilon_{0})

= \sigma/2\epsilon_{0} = 17\times10^{-22}/8.85\times10^{-12} = 1.92\times10^{-10 } NC^{-1}

NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields: Additional Exercises Question

Q 1.25 An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 \times 10^{4}NC^{-1} (Millikan’s oil drop experiment). The density of the oil is 1.26 g cm -3 . Estimate the radius of the drop.

Answer:

The force due to the electric field is balancing the weight of the oil droplet.

weight of the oil drop = density x volume of the droplet x g = \rho \times \frac{4}{3}\pi r^3 \times g

Force due to the electric field = E x q

charge on the droplet, q = No. of excess electrons x charge of an electron = 12\times q_{e} = 12\times 1.6\times10^{-19} = 1.92\times10^{-18} C

Balancing forces:

\rho \times \frac{4}{3}\pi r^3 \times g = E\times q

Putting known and calculated values:

\\ r^3 =\frac{3}{4\pi} E\times q / \rho\times g = \frac{3}{4}\times\frac{7}{22}\times(2.55\times10^4)\times 1.92\times10^{-18} / (1.26\times10^3 \times10 ) \\ = 0.927\times10^{-18} m^3

r = 0.975\times10^{-6} m = 9.75\times10^{-4} mm

Q 1.26 Which among the curves shown in Figure cannot possibly represent electrostatic field lines?

esc12

1643005781117

Answer:

(a) Wrong, because field lines must be normal to a conductor.

(b) Wrong, because field lines can only start from a positive charge. It cannot start from a negative charge,

(c) Right;

(d) Wrong, because field lines cannot intersect each other,

(e) Wrong, because electrostatic field lines cannot form closed loops.

Q 1.27 In a certain region of space, electric field is along the z-direction throughout. The magnitude of electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of 10^{5}NC^{-1} per metre. What are the force and torque experienced by a system having a total dipole moment equal to 10^{-7} Cm in the negative z-direction ?

Answer:

Force on a charge F=qE

but here E is varying along the Z direction.

Force can be written as,

F=q\frac{dE}{dz}dz=P\frac{dE}{dz}=10^{-7}\times10^5=10^{-2}N

Torque experienced =0 since both dipole and electric field are in the Z direction. The angle between dipole and the electric field is 180 degrees

\tau = \mathbf{P}\times \mathbf{E}=PEsin\180=0

Q 1.28 (a) A conductor A with a cavity as shown in Figure a is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor.

1643005593691

Answer:

We know that the electric field inside a conductor is zero.

Using Gauss' law, if we draw any imaginary closed surface inside the solid, net charge must be zero.(Since E= 0 inside)

Hence there cannot be any charge inside the conductor and therefore, all charge must appear on the outer surface of the conductor.

Q 1.28 (b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q

1643005642848

Answer:

We know, electric field inside a conductor is zero.

Now, imagine a Gaussian surface just outside the cavity inside the conductor. Since, E=0 (Using Gauss' law), hence net charge must be zero inside the surface. Therefore, -q charge is induced on the inner side of the cavity(facing conductor B).

Now consider a Gaussian surface just outside the conductor A. The net electric field must be due to charge Q and q. Hence, q charge is induced on the outer surface of conductor A. Therefore, the net charge on the outer surface of A is Q + q.

Q 1.28 (c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.

1643005677851

Answer:

We know that the electric field inside a conductor is zero.

Therefore, a possible way to shield from the strong electrostatic fields in its environment is to enclose the instrument fully by a metallic surface.

Q 1.29 A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is \left (\frac{\sigma}{2 \epsilon_{0}} \right ) \widehat{n} , where \widehat{n} is the unit vector in the outward normal direction, and σ is the surface charge density near the hole.

Answer:

1643005712828

Let the surface area of the sphere be S.

And assume that the hole is filled. For a point B, just above the hole, considering a gaussian surface passing through B, we have

\oint E.dS = q/\epsilon_{0}

Now, since the electric field is always perpendicular to the surface of the conductor.

\\ \therefore \oint E.dS = E.S= q/\epsilon_{0} = (\sigma.S)/\epsilon_{0} \\ \implies E = \sigma/\epsilon_{0}

Using Superposition principle, E =E_{1} + E_{2} ,

where E_{1} is due to the hole and E_{2} is due to the rest of the conductor. (both pointing outwards, i.e away from the centre)

Again, for a point A, just below the hole, Electric field will be zero because of electrostatic shielding.

Using the superposition principle, this will be due to E_{1} pointing inwards(towards the centre) and due to E_{2} (Pointing away from the centre)

0 =E_{1}-E_{2} \implies E_{1}=E_{2}

Using this relation, we get:

E =E_{1} + E_{2} = 2E_{1} \implies E_{1} = E/2 = \sigma/2\epsilon_{0}

Since this is pointing outwards,

\boldsymbol{E_{1}} = \sigma/2\epsilon_{0}\ \boldsymbol{\widehat{n}} is the electric field in the hole.

(Trick: 1. Assume the hole to be filled.

2. Consider 2 points just above and below the hole.

3. Electric fields at these points will be due to the hole and rest of the conductor. Use superposition principle.)

Q 1.30 Obtain the formula for the electric field due to a long thin wire of uniform linear charge density E without using Gauss’s law.

[Hint: Use Coulomb’s law directly and evaluate the necessary integral.]

Answer:

Let AB be a long thin wire of uniform linear charge density λ.

Let us consider the electric field intensity due to AB at point P at a distance h from it as shown in the figure.


1515132007897481

The charge on a small length dx on the line AB is q which is given as q = λdx.

So, according to Coulomb’s law, the electric field at P due to this length dx is

dE'=\frac{1}{4\pi \epsilon}\frac{\lambda dx}{(PC)^2}


But 1515132009443638

1515132010256363

This electric field at P can be resolved into two components as dEcosθ and dEsinθ. When the entire length AB is considered, then the dEsinθ components add up to zero due to symmetry. Hence, there is only dEcosθ component.

So, the net electric field at P due to dx is

dE' = dE cosθ

1515132011019361 ………………….(1)

In ΔPOC,

1515132011793417

⇒ x = h tan θ

Differentiating both sides w.r.t. θ,

1515132012524776

⇒ dx = h sec 2 θ dθ …………………….(2)

Also, h 2 + x 2 = h 2 + h 2 tan 2 θ

⇒ h 2 + x 2 = h 2 (1+ tan 2 θ)

⇒ h 2 + x 2 = h 2 sec 2 θ ………….(3)

(Using the trigonometric identity, 1+ tan 2 θ = sec 2 θ)

Using equations (2) and (3) in equation (1),

1515132013293275

, 1515132014201895

The wire extends from 1515132014938349 to 1515132015707190 since it is very long.

Integrating both sides,

1515132016477995

1515132017214494

1515132017979544

151513201871543

1515132019451670

This is the net electric field due to a long wire with linear charge density λ at a distance h from it.

In the question linear charge density =E

therefor

E'=\frac{E}{2\pi \epsilon h}

Q 1.31 It is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so called ‘up’ quark (denoted by u) of charge + \left ( \frac{2}{3} \right ) e, and the ‘down’ quark (denoted by d) of charge \left (- \frac{1}{3} \right ) e, together with electrons build up ordinary matter. (Quarks of other types have also been found which give rise to different unusual varieties of matter.) Suggest a possible quark composition of a proton and neutron.

Answer:

Given, a proton and a neutron consist of three quarks each.

And, ‘up’ quark is of charge + \left ( \frac{2}{3} \right ) e, and the ‘down’ quark of charge \left (- \frac{1}{3} \right ) e

Let the number of 'up' quarks be n. Therefore, the number of 'down' quarks is (3-n).

\therefore The net charge = \\ n\times\frac{2}{3}e + (3-n)\frac{-1}{3}e = (n-1)e

Now, a proton has a charge +1e

\therefore (n-1)e = +1e \implies n =2

Proton will have 2 u and 1 d, i.e, uud

Similarly, the neutron has a charge 0

\therefore (n-1)e = 0 \implies n =1

Neutron will have 1 u and 2 d, i.e, udd

Q 1.32 (a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E = 0) of the configuration. Show that the equilibrium of the test charge is necessarily unstable.

Answer:

For equilibrium to be stable, there must be a restoring force and hence all the field lines should be directed inwards towards the null point. This implies that there is a net inward flux of the electric field. But this violates Gauss's law, which states that the flux of electric field through a surface not enclosing any charge must be zero. Hence, the equilibrium of the test charge cannot be stable.

Therefore, the equilibrium is necessarily unstable.

Q 1.32 (b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.

Answer:

Two charges of same magnitude and sign are placed at a certain distance apart. The mid-point of the line joining these charges will have E =0.

When a test charged is displaced along the line towards the charges, it experiences a restoring force(which is the condition for stable equilibrium). But if the test charge is displaced along the normal of the line, the net electrostatic force pushes it away from the starting point. Hence, the equilibrium is unstable.

Q 1.33 A particle of mass m and charge (–q) enters the region between the two charged plates initially moving along x-axis with speed vx .The length of plate is L and an uniform electric field E is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is \frac{qEL^{2}}{2mv_{x}^{2}} .

Answer:

Let s be the vertical deflection, t be the time taken by the particle to travel between the plates

\therefore s = ut + \frac{1}{2}at^2

Here , u =0 , since initially there was no vertical component of velocity.

The particle in the elecric field will experience a constant force (Since, Electric field is constant.)

F = ma = -qE (Using Newton's Second Law, F = ma)

\therefore a = -qE/m (-ve sign implies here in downward direction)

Again, t = Distance covered/ Speed = L/ v_{x}

(In x-direction, since there is no force, hence component of velocity in x-direction remains constant = v_{x} .

And, the distance covered in x-direction = length of the plate = L)

Putting these values in our deflection equation,

\\ s = ut + \frac{1}{2}at^2 = (0)(L/v_{x}) + \frac{1}{2}(-qE/m)(L/v_{x})^2 \\ \implies s =\frac{-qEL^2}{2mv_{x}^2}

(S is -ve, which implies it deflect in downwards direction.)

\therefore The vertical deflection of the particle at the far edge of the plate is \frac{qEL^{2}}{2mv_{x}^{2}} .

This motion is similar to the motion of a projectile in a gravitational field, which is also a constant force. The force acting on the particle in the gravitational field is mg whiles in this case, it is qE. The trajectory will be the same in both cases.

Q 1.34 Suppose that the particle in is an electron projected with velocity v_{x}=2.0 \times 10^{6}ms^{-1} . If E between the plates separated by 0.5 cm is 9.1 \times 10^{2}\frac{N}{C} , where will the electron strike the upper plate? (|e|= \left | e \right |=1.6 \times 10^{-19}, m_{e}=9.1\times 10^{-31}kg )

Answer:

\therefore The vertical deflection of the particle at the far edge of the plate is s=\frac{qEL^{2}}{2mv_{x}^{2}}

given s= 0.5cm=0.005cm

calculate for L from the above equation

L=\sqrt{\frac{2smv_x^2}{qE}}=\sqrt{\frac{2\times 0.005\times 9.1\times 10^{-31}(2\times10^6)^2 }{1.6\times 10^{-19}\times9.1\times10^2}}

L=1.6 cm


class 12 physics chapter 1 ncert solutions is very important for board exams, JEE, and NEET. It covers basic concepts like electric charges, Coulomb's law, and electric fields. Understanding these basics is essential for success in board exams, where questions test these fundamentals. In competitive exams like JEE and NEET, a strong grasp of class 12 physics chapter 1 exercise solutions is crucial for the physics section. Moreover, this chapter's concepts are interconnected with later chapters in Class 12, forming the groundwork for more advanced physics topics. So, mastering electric charges and fields class 12 is not only exam-important but also essential for building a strong physics foundation.

NCERT Solutions for Class 12 Physics- Chapter Wise

JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%
Download EBook

Electric charges and fields class 12 NCERT solutions: Important Formulas and Diagrams

In class12 physics ch1 ncert solutions, you'll find important formulas and diagrams that aid in understanding electrostatic concepts. These resources are valuable for exam preparation and building a strong foundation in electrostatics.

  • Coulomb's Law

JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

JEE Main Important Physics formulas

As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters

\\F\propto \frac{Q_{1}Q_{2}}{r^{2}}\\F=\frac{KQ_{1}Q_{2}}{r^{2}}

Where:

K is the proportionality, Constant and Q1 and Q2 are two-point charges.

1694162286856

1694162285301

The vector form of Coulomb's Law:

1694162339206

1694162340518

  • Electric Field

Electric Field Intensity(E):

1694162341409

1694162340878

Where:

F is the force experienced by qo. The SI unit of E is V/m.

Electric Field Due To A Point Charge:

1694162370131

1694162370799


Electric Flux :

1694162371444

>> Flux is a scalar quantity. The SI unit of flux is volt-meter or newton-meter.

Electric Dipole:

1694162422500

Electric Dipole Moment: 1694162421869

Where: 2l is the dipole length.

>> The electric dipole moment is a vector quantity and the S.I unit is coulomb-meter (C-m).

  • Gauss's Law

The total electric flux through a closed surface enclosing a charge q is equal to 1694162421097 times the total charge q enclosed by the surface.


1694162423056

Topics Covered in Class 12 NCERT Chapter Electric Charges and Fields:

The main topics covered in the ch 1 Physics class 12 are:

Charge and its properties

This NCERT Class 12 topic discuss the concept of charge with practical examples of charging through friction, conduction and induction. Also, the concepts of properties like conservation, additivity and quantisation of charge are discussed in the chapter electric charges and fields. Questions based on this are present in electric charges and fields Class 12 solutions.

Coulomb's law and its applications

Coulomb's law discusses the relation between the force exerted between two charges separated by a distance. For example, questions 1 and 12 of NCERT solutions for Class 12 chapter 1 uses this law to solve it.

  • Electric field: electric field and properties of electric field lines are discussed in this topic

  • Electric dipole: The concepts of dipole and field due to dipole at the axial, equatorial and general point and torque due to electric dipole are discussed.

  • Continuous charge distribution: Linear charge, surface charge and volume charge distributions are covered in this class 12 NCERT Physics chapter 1 topic

  • Electric flux, Gauss's law and its applications: The concepts of electric flux, the relation between electric charge and flux and application of Gauss law to different charge configurations are discussed in class 12 NCERT.

All the topics mentioned in Physics Class 12 chapter 1 are important and students are advised to go through all the concepts mentioned in the topics. Questions from all the above topics are covered in the electric charges and fields NCERT solutions.

Importance of NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields in Board Exams:

The physics paper for the CBSE board exam is for 70 marks. In 2019 CBSE board exam 9 marks questions were asked from class 12 chapter 1 physics that includes the first two chapters, and this makes solving NCERT class 12 physics chapter 1 quite important. Practicing problems from the NCERT class 12 Physics solutions chapter 1 help students to score well in the final exam. Students can also use NCERT class 12 Physics chapter 1 solutions pdf download button to read chapter 1 physics class 12 NCERT solutions offline.

Key Features of Class 12 Electric Field and Charges NCERT Solutions

  1. Comprehensive Coverage: This chapter covers fundamental concepts related to electric charges, Coulomb's law, electric fields, and their applications.

  2. Exercise and Additional Exercise Questions: It includes a wide range of exercises and additional exercise questions to practice and assess understanding.

  3. Detailed Solutions: Class12 physics ch1 ncert solutions provided in the NCERT textbook and supplementary materials offer step-by-step explanations for better comprehension.

  4. Foundation for Advanced Topics: The concepts introduced in this class 12 physics chapter 1 ncert solutions serve as the building blocks for more complex topics in electromagnetism and modern physics.

  5. Preparation for Exams: A strong grasp of this electric charges and fields class 12 is essential for success in board exams, JEE, and NEET.

  6. Interconnected Concepts: ncert solution class 12 physics chapter 1 establishes connections to later chapters in the Class 12 physics syllabus like the gravitation chapter, emphasizing its significance.

    Also Check NCERT Books and NCERT Syllabus here:

NCERT Exemplar Class 12 Solutions

Subject Wise Solutions

Excluded Content:

Certain portions of the chapter have been omitted or removed. These exclusions include:

  1. Activity involving paper strips and the creation of an electroscope from Section 1.2 on Electric Charge.
  2. The concept related to earthing in Section 1.3 on Conductors and Insulators.
  3. The topic of Charging by Induction from Section 1.4.
  4. Exercises numbered 1.13 and the range of exercises from 1.25 to 1.34.

These exclusions from class 12 electric charges and fields ncert solutions should be noted by students, and they should refer to their course materials for a precise understanding of what is covered in their syllabus.

Frequently Asked Question (FAQs)

1. What is the weightage of the chapter electric fields and charges for CBSE board exam

From the Class 12 Physics Chapter 1 NCERT solutions electric charges and fields 3 to 5 marks questions can be expected for CBSE bord exam. Questions based on the NCERT exercise questions can be expected for the exam. So practising the NCERT solutions for Class 12 Physics chapter 1 is important. The questions may be theory based, numerical or derivations.

2. Where can I find complete solutions of NCERT class 12 Physics

The detailed NCERT Class 12 Physics solutions chapter 1 is provided by careers 360. All the exercise and additional exercise questions of NCERT chapter Electric charges and Fields are solved with necessary explanations.

3. Whether the unit electrostatics is helpful in higher studies?

Yes, it is helpful in engineering streams and for research in the science field. The concepts studied in electrostatics will be useful in electrical and electronics-related branches.

4. What are the important topics of physics NCERT class 12 chapter 1

The two important laws discussed in NCERT solutions for Class 12 Physics Chapter 1  are important. That is the Coulombs law and the Gauss law. The derivations using Gauss laws and numerical related to both laws are important.

5. What is the weightage of electric charges and fields in JEE Mains

In JEE mains around 3.3% of questions are asked from the Physics class 12 chapter 1 questions and answers. That is 1 to 2 questions can be expected. 

6. What is the weightge of NCERT chapter electric charges and fields for neet exam

For NEET exam 4% questions can be expected from the NCERT physics chapter 1 Electric Charges and Fields. 

7. What are the main topics to be covered for NCERT Class 12 Physis Chapter 1 solutions
  • Electric Charge
  • Conductors And Insulators
  • Charging By Induction
  • Basic Properties Of Electric Charge
  • Coulomb’s Law
  • Electric Field
  • Electric Field Lines
  • Electric Flux
  • Electric Dipole 
  • Continuous Charge Distribution
  • Gauss’s Law And Applications

8. Can I prepare for the CBSE Exam with the NCERT Solutions for Class 12 Physics Chapter 1?

The experts at Careers360 created the ncert solution for class 12 physics chapter 1 to assist students in acing the board test with confidence. To boost pupils' confidence, the key principles are presented in the most organised manner possible. NCERT Solutions address every little nuance to aid students in their board exam preparation.

Articles

Explore Top Universities Across Globe

Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.


As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.


Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.


Believe in Yourself! You can make anything happen


All the very best.

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects  and  we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

If you'll do hard work then by hard work of 6 months you can achieve your goal but you have to start studying for it dont waste your time its a very important year so please dont waste it otherwise you'll regret.

Yes, you can take admission in class 12th privately there are many colleges in which you can give 12th privately.

View All

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

Bio Medical Engineer

The field of biomedical engineering opens up a universe of expert chances. An Individual in the biomedical engineering career path work in the field of engineering as well as medicine, in order to find out solutions to common problems of the two fields. The biomedical engineering job opportunities are to collaborate with doctors and researchers to develop medical systems, equipment, or devices that can solve clinical problems. Here we will be discussing jobs after biomedical engineering, how to get a job in biomedical engineering, biomedical engineering scope, and salary. 

4 Jobs Available
Data Administrator

Database professionals use software to store and organise data such as financial information, and customer shipping records. Individuals who opt for a career as data administrators ensure that data is available for users and secured from unauthorised sales. DB administrators may work in various types of industries. It may involve computer systems design, service firms, insurance companies, banks and hospitals.

4 Jobs Available
Ethical Hacker

A career as ethical hacker involves various challenges and provides lucrative opportunities in the digital era where every giant business and startup owns its cyberspace on the world wide web. Individuals in the ethical hacker career path try to find the vulnerabilities in the cyber system to get its authority. If he or she succeeds in it then he or she gets its illegal authority. Individuals in the ethical hacker career path then steal information or delete the file that could affect the business, functioning, or services of the organization.

3 Jobs Available
Data Analyst

The invention of the database has given fresh breath to the people involved in the data analytics career path. Analysis refers to splitting up a whole into its individual components for individual analysis. Data analysis is a method through which raw data are processed and transformed into information that would be beneficial for user strategic thinking.

Data are collected and examined to respond to questions, evaluate hypotheses or contradict theories. It is a tool for analyzing, transforming, modeling, and arranging data with useful knowledge, to assist in decision-making and methods, encompassing various strategies, and is used in different fields of business, research, and social science.

3 Jobs Available
Geothermal Engineer

Individuals who opt for a career as geothermal engineers are the professionals involved in the processing of geothermal energy. The responsibilities of geothermal engineers may vary depending on the workplace location. Those who work in fields design facilities to process and distribute geothermal energy. They oversee the functioning of machinery used in the field.

3 Jobs Available
Remote Sensing Technician

Individuals who opt for a career as a remote sensing technician possess unique personalities. Remote sensing analysts seem to be rational human beings, they are strong, independent, persistent, sincere, realistic and resourceful. Some of them are analytical as well, which means they are intelligent, introspective and inquisitive. 

Remote sensing scientists use remote sensing technology to support scientists in fields such as community planning, flight planning or the management of natural resources. Analysing data collected from aircraft, satellites or ground-based platforms using statistical analysis software, image analysis software or Geographic Information Systems (GIS) is a significant part of their work. Do you want to learn how to become remote sensing technician? There's no need to be concerned; we've devised a simple remote sensing technician career path for you. Scroll through the pages and read.

3 Jobs Available
Geotechnical engineer

The role of geotechnical engineer starts with reviewing the projects needed to define the required material properties. The work responsibilities are followed by a site investigation of rock, soil, fault distribution and bedrock properties on and below an area of interest. The investigation is aimed to improve the ground engineering design and determine their engineering properties that include how they will interact with, on or in a proposed construction. 

The role of geotechnical engineer in mining includes designing and determining the type of foundations, earthworks, and or pavement subgrades required for the intended man-made structures to be made. Geotechnical engineering jobs are involved in earthen and concrete dam construction projects, working under a range of normal and extreme loading conditions. 

3 Jobs Available
Cartographer

How fascinating it is to represent the whole world on just a piece of paper or a sphere. With the help of maps, we are able to represent the real world on a much smaller scale. Individuals who opt for a career as a cartographer are those who make maps. But, cartography is not just limited to maps, it is about a mixture of art, science, and technology. As a cartographer, not only you will create maps but use various geodetic surveys and remote sensing systems to measure, analyse, and create different maps for political, cultural or educational purposes.

3 Jobs Available
Budget Analyst

Budget analysis, in a nutshell, entails thoroughly analyzing the details of a financial budget. The budget analysis aims to better understand and manage revenue. Budget analysts assist in the achievement of financial targets, the preservation of profitability, and the pursuit of long-term growth for a business. Budget analysts generally have a bachelor's degree in accounting, finance, economics, or a closely related field. Knowledge of Financial Management is of prime importance in this career.

4 Jobs Available
Data Analyst

The invention of the database has given fresh breath to the people involved in the data analytics career path. Analysis refers to splitting up a whole into its individual components for individual analysis. Data analysis is a method through which raw data are processed and transformed into information that would be beneficial for user strategic thinking.

Data are collected and examined to respond to questions, evaluate hypotheses or contradict theories. It is a tool for analyzing, transforming, modeling, and arranging data with useful knowledge, to assist in decision-making and methods, encompassing various strategies, and is used in different fields of business, research, and social science.

3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Underwriter

An underwriter is a person who assesses and evaluates the risk of insurance in his or her field like mortgage, loan, health policy, investment, and so on and so forth. The underwriter career path does involve risks as analysing the risks means finding out if there is a way for the insurance underwriter jobs to recover the money from its clients. If the risk turns out to be too much for the company then in the future it is an underwriter who will be held accountable for it. Therefore, one must carry out his or her job with a lot of attention and diligence.

3 Jobs Available
Finance Executive
3 Jobs Available
Operations Manager

Individuals in the operations manager jobs are responsible for ensuring the efficiency of each department to acquire its optimal goal. They plan the use of resources and distribution of materials. The operations manager's job description includes managing budgets, negotiating contracts, and performing administrative tasks.

3 Jobs Available
Investment Director

An investment director is a person who helps corporations and individuals manage their finances. They can help them develop a strategy to achieve their goals, including paying off debts and investing in the future. In addition, he or she can help individuals make informed decisions.

2 Jobs Available
Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
Transportation Planner

A career as Transportation Planner requires technical application of science and technology in engineering, particularly the concepts, equipment and technologies involved in the production of products and services. In fields like land use, infrastructure review, ecological standards and street design, he or she considers issues of health, environment and performance. A Transportation Planner assigns resources for implementing and designing programmes. He or she is responsible for assessing needs, preparing plans and forecasts and compliance with regulations.

3 Jobs Available
Plumber

An expert in plumbing is aware of building regulations and safety standards and works to make sure these standards are upheld. Testing pipes for leakage using air pressure and other gauges, and also the ability to construct new pipe systems by cutting, fitting, measuring and threading pipes are some of the other more involved aspects of plumbing. Individuals in the plumber career path are self-employed or work for a small business employing less than ten people, though some might find working for larger entities or the government more desirable.

2 Jobs Available
Construction Manager

Individuals who opt for a career as construction managers have a senior-level management role offered in construction firms. Responsibilities in the construction management career path are assigning tasks to workers, inspecting their work, and coordinating with other professionals including architects, subcontractors, and building services engineers.

2 Jobs Available
Urban Planner

Urban Planning careers revolve around the idea of developing a plan to use the land optimally, without affecting the environment. Urban planning jobs are offered to those candidates who are skilled in making the right use of land to distribute the growing population, to create various communities. 

Urban planning careers come with the opportunity to make changes to the existing cities and towns. They identify various community needs and make short and long-term plans accordingly.

2 Jobs Available
Highway Engineer

Highway Engineer Job Description: A Highway Engineer is a civil engineer who specialises in planning and building thousands of miles of roads that support connectivity and allow transportation across the country. He or she ensures that traffic management schemes are effectively planned concerning economic sustainability and successful implementation.

2 Jobs Available
Environmental Engineer

Individuals who opt for a career as an environmental engineer are construction professionals who utilise the skills and knowledge of biology, soil science, chemistry and the concept of engineering to design and develop projects that serve as solutions to various environmental problems. 

2 Jobs Available
Naval Architect

A Naval Architect is a professional who designs, produces and repairs safe and sea-worthy surfaces or underwater structures. A Naval Architect stays involved in creating and designing ships, ferries, submarines and yachts with implementation of various principles such as gravity, ideal hull form, buoyancy and stability. 

2 Jobs Available
Orthotist and Prosthetist

Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

6 Jobs Available
Veterinary Doctor
5 Jobs Available
Pathologist

A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

5 Jobs Available
Speech Therapist
4 Jobs Available
Gynaecologist

Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth. 

4 Jobs Available
Oncologist

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

3 Jobs Available
Audiologist

The audiologist career involves audiology professionals who are responsible to treat hearing loss and proactively preventing the relevant damage. Individuals who opt for a career as an audiologist use various testing strategies with the aim to determine if someone has a normal sensitivity to sounds or not. After the identification of hearing loss, a hearing doctor is required to determine which sections of the hearing are affected, to what extent they are affected, and where the wound causing the hearing loss is found. As soon as the hearing loss is identified, the patients are provided with recommendations for interventions and rehabilitation such as hearing aids, cochlear implants, and appropriate medical referrals. While audiology is a branch of science that studies and researches hearing, balance, and related disorders.

3 Jobs Available
Hospital Administrator

The hospital Administrator is in charge of organising and supervising the daily operations of medical services and facilities. This organising includes managing of organisation’s staff and its members in service, budgets, service reports, departmental reporting and taking reminders of patient care and services.

2 Jobs Available
Actor

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs. 

4 Jobs Available
Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available
Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages.

Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available
Radio Jockey

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
Choreographer

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

2 Jobs Available
Videographer
2 Jobs Available
Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications. 

2 Jobs Available
Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available
Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook. 

5 Jobs Available
Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. 

Ever since internet costs got reduced the viewership for these types of content has increased on a large scale. Therefore, a career as a vlogger has a lot to offer. If you want to know more about the Vlogger eligibility, roles and responsibilities then continue reading the article. 

3 Jobs Available
Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
Linguist

Linguistic meaning is related to language or Linguistics which is the study of languages. A career as a linguistic meaning, a profession that is based on the scientific study of language, and it's a very broad field with many specialities. Famous linguists work in academia, researching and teaching different areas of language, such as phonetics (sounds), syntax (word order) and semantics (meaning). 

Other researchers focus on specialities like computational linguistics, which seeks to better match human and computer language capacities, or applied linguistics, which is concerned with improving language education. Still, others work as language experts for the government, advertising companies, dictionary publishers and various other private enterprises. Some might work from home as freelance linguists. Philologist, phonologist, and dialectician are some of Linguist synonym. Linguists can study French, German, Italian

2 Jobs Available
Public Relation Executive
2 Jobs Available
Travel Journalist

The career of a travel journalist is full of passion, excitement and responsibility. Journalism as a career could be challenging at times, but if you're someone who has been genuinely enthusiastic about all this, then it is the best decision for you. Travel journalism jobs are all about insightful, artfully written, informative narratives designed to cover the travel industry. Travel Journalist is someone who explores, gathers and presents information as a news article.

2 Jobs Available
Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
QA Manager
4 Jobs Available
Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Production Manager
3 Jobs Available
Merchandiser
2 Jobs Available
QA Lead

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans. 

2 Jobs Available
Metallurgical Engineer

A metallurgical engineer is a professional who studies and produces materials that bring power to our world. He or she extracts metals from ores and rocks and transforms them into alloys, high-purity metals and other materials used in developing infrastructure, transportation and healthcare equipment. 

2 Jobs Available
Azure Administrator

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems. 

4 Jobs Available
AWS Solution Architect

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party. 

4 Jobs Available
QA Manager
4 Jobs Available
Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
ITSM Manager
3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

3 Jobs Available
Back to top