Careers360 Logo

Popular Searches

    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.

    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.

    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

    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

    \\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

    University of Essex, Colchester
     Wivenhoe Park Colchester CO4 3SQ
    University College London, London
     Gower Street, London, WC1E 6BT
    The University of Edinburgh, Edinburgh
     Old College, South Bridge, Edinburgh, Post Code EH8 9YL
    University of Nottingham, Nottingham
     University Park, Nottingham NG7 2RD
    Lancaster University, Lancaster
     Bailrigg, Lancaster LA1 4YW
    Bristol Baptist College, Bristol
     The Promenade, Clifton Down, Bristol BS8 3NJ
    MCAT Exam Pattern 2023 (Section-wise) - Check Here!
    4 minSep 14, 2023 14:09 PM IST
    LSAT Abroad Eligibility Criteria 2024
    5 minSep 14, 2023 13:09 PM IST
    What is SAT
    3 minSep 14, 2023 13:09 PM IST
    New SAT Format 2023 - What You Need to Know
    4 minSep 14, 2023 13:09 PM IST

    Questions related to CBSE Class 12th

    Have a question related to CBSE Class 12th ?

    Hello student,

    Hope you are in good health, so the answer to your question is as follows:-

    First of all please get registered to delhi university admissions and fill the application fees.

    Then try to calculate your best 4 according to the course .

    As you have not mentioned which category you belong to , I am assuming you to be a general category student. So ,suppose  if your best four came out to be  somewhere like 89 , so if I see the last years cutoff for every BSC course the last cutoff comes out  to be 89 , 87 where you can get bsc home science and bsc life science according to last year cutoff , but it always the chance of matter  every year , cutoff always surprise us so you have to look every cutoff and when you find that you are getting the cutoff for that particular course do visit immediately to that college with all the correct papers and do the admission right  away .

    For further details you can go through the following link:-

    https://university.careers360.com/articles/du-cutoff

    Hope this helps!

    Hello pallavi,
    Your percentage is decent for good college in du university for  your desired stream..

    Check this college predictor to check in which colleges the chances of your admission are high

    https://university.careers360.com/delhi-university-college-predictor

    For more query please replay in the comments section..
    Best of luck!!!
    Hello dear,
    Your percentage is decent for good college in du university for  your desired stream..

    Check this college predictor to check in which colleges the chances of your admission are high

    https://university.careers360.com/delhi-university-college-predictor

    For more query please replay in the comments section..
    Best of luck!!!

    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.

    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
    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
    Graphic Designer

    Within the graphic design and graphic arts industry, a graphic designer is a specialist who designs and builds images, graphic design, or visual effects to develop a piece of artwork. In career as graphic designer, individuals primarily generate the graphics for publishing houses and printed or electronic digital media like pamphlets and commercials. There are various options for industrial graphic design employment. Graphic design career includes providing numerous opportunities in the media industry.

    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
    GIS Expert

    GIS officer work on various GIS software to conduct a study and gather spatial and non-spatial information. GIS experts update the GIS data and maintain it. The databases include aerial or satellite imagery, latitudinal and longitudinal coordinates, and manually digitized images of maps. In a career as GIS expert, one is responsible for creating online and mobile maps.

    3 Jobs Available
    Database Architect

    If you are intrigued by the programming world and are interested in developing communications networks then a career as database architect may be a good option for you. Data architect roles and responsibilities include building design models for data communication networks. Wide Area Networks (WANs), local area networks (LANs), and intranets are included in the database networks. It is expected that database architects will have in-depth knowledge of a company's business to develop a network to fulfil the requirements of the organisation. Stay tuned as we look at the larger picture and give you more information on what is db architecture, why you should pursue database architecture, what to expect from such a degree and what your job opportunities will be after graduation. Here, we will be discussing how to become a data architect. Students can visit NIT Trichy, IIT Kharagpur, JMI New Delhi

    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
    Risk Management Specialist

    Individuals who opt for a career as a risk management specialist are professionals who are responsible for identifying risks involved in business that may include loss of assets, property, personnel or cash flow. Credit risk manager responsibilities are to identifies business opportunities and eliminates issues related to insurance or safety that may cause property litigation. A risk management specialist is responsible for increasing benefits.

    4 Jobs Available
    Bank Probationary Officer (PO)

    A career as Bank Probationary Officer (PO) is seen as a promising career opportunity and a white-collar career. Each year aspirants take the Bank PO exam. This career provides plenty of career development and opportunities for a successful banking future. If you have more questions about a career as  Bank Probationary Officer (PO), what is probationary officer or how to become a Bank Probationary Officer (PO) then you can read the article and clear all your doubts. 

    3 Jobs Available
    Credit Manager

    Credit Management refers to the process of granting credit, setting the terms it’s granted on, recovering the credit when it’s due, and confirming compliance with the organization's credit policy, among other credit-related operations. Individuals who opt for a career as Credit Manager should have hands-on experience with accounting software, a solid understanding of lending procedures, excellent analytical skills with the ability to create and process financial spreadsheets, negotiation skills, and a bachelor’s or master’s degree in a field relevant to finance or accounting. Ultimately, Credit Management job is to help organizations minimize bad debts and increase revenues from the loan.

    3 Jobs Available
    Investment Banker

    An Investment Banking career involves the invention and generation of capital for other organizations, governments, and other entities. Individuals who opt for a career as Investment Bankers are the head of a team dedicated to raising capital by issuing bonds. Investment bankers are termed as the experts who have their fingers on the pulse of the current financial and investing climate. Students can pursue various Investment Banker courses, such as Banking and Insurance, and Economics to opt for an Investment Banking career path.

    3 Jobs Available
    Insurance Analyst

    In the career as an insurance analyst, one can monitor the choices the customers make about which insurance policy options best suit their requirements. They research and make recommendations that have a real impact on the financial well-being of a client down the road. Insurance companies are helping people prepare themselves for the long term. Insurance Analysts find the documents of the claim and perform a thorough investigation, like travelling to places where the incident has occurred, gathering evidence, and working with law enforcement officers.

    3 Jobs Available
    Finance Executive

    A career as a Finance Executive requires one to be responsible for monitoring an organisation's income, investments and expenses to create and evaluate financial reports. His or her role involves performing audits, invoices, and budget preparations. He or she manages accounting activities, bank reconciliations, and payable and receivable accounts.  

    3 Jobs Available
    Bank Branch Manager

    Bank Branch Managers work in a specific section of banking related to the invention and generation of capital for other organisations, governments, and other entities. Bank Branch Managers work for the organisations and underwrite new debts and equity securities for all type of companies, aid in the sale of securities, as well as help to facilitate mergers and acquisitions, reorganisations, and broker trades for both institutions and private investors.

    3 Jobs Available
    Treasurer

    Treasury analyst career path is often regarded as certified treasury specialist in some business situations, is a finance expert who specifically manages a company or organisation's long-term and short-term financial targets. Treasurer synonym could be a financial officer, which is one of the reputed positions in the corporate world. In a large company, the corporate treasury jobs hold power over the financial decision-making of the total investment and development strategy of the organisation.

    3 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
    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
    Carpenter

    Carpenters are typically construction workers. They stay involved in performing many types of construction activities. It includes cutting, fitting and assembling wood.  Carpenters may help in building constructions, bridges, big ships and boats. Here, in the article, we will discuss carpenter career path, carpenter salary, how to become a carpenter, carpenter job outlook.

    2 Jobs Available
    Welder

    An individual who opts for a career as a welder is a professional tradesman who is skilled in creating a fusion between two metal pieces to join it together with the use of a manual or fully automatic welding machine in their welder career path. It is joined by intense heat and gas released between the metal pieces through the welding machine to permanently fix it. 

    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
    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. 

    2 Jobs Available
    Field Surveyor

    Are you searching for a Field Surveyor Job Description? A Field Surveyor is a professional responsible for conducting field surveys for various places or geographical conditions. He or she collects the required data and information as per the instructions given by senior officials. 

    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

    A veterinary doctor is a medical professional with a degree in veterinary science. The veterinary science qualification is the minimum requirement to become a veterinary doctor. There are numerous veterinary science courses offered by various institutes. He or she is employed at zoos to ensure they are provided with good health facilities and medical care to improve their life expectancy.

    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
    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
    Paediatrician

    A career as paediatrician has emerged as one of India's most popular career choices. By choosing a career as paediatrician, not only in India but also overseas, one can find lucrative work profiles as demand for talented and professional paediatricians is increasing day by day. If you are passionate about children and have the patience to evaluate and diagnose their issues, you may have a good career as paediatricians. Paediatricians take care of children's physical, mental and emotional health from infancy to adolescence.

    3 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
    Surgical Technologist

    When it comes to an operation theatre, there are several tasks that are to be carried out before as well as after the operation or surgery has taken place. Such tasks are not possible without surgical tech and surgical tech tools. A single surgeon cannot do it all alone. It’s like for a footballer he needs his team’s support to score a goal the same goes for a surgeon. It is here, when a surgical technologist comes into the picture. It is the job of a surgical technologist to prepare the operation theatre with all the required equipment before the surgery. Not only that, once an operation is done it is the job of the surgical technologist to clean all the equipment. One has to fulfil the minimum requirements of surgical tech qualifications. 

    Also Read: Career as Nurse

    3 Jobs Available
    Ophthalmic Medical Technician

    Ophthalmic technician careers are one of the booming careers option available in the field of healthcare. Being a part of this field as an ophthalmic medical technician can provide several career opportunities for an individual. With advancing technology the job of individuals who opt for a career as ophthalmic medical technicians have become of even more importance as he or she is required to assist the ophthalmologist in using different types of machinery. If you want to know more about the field and what are the several job opportunities, work environment, just about anything continues reading the article and all your questions shall be answered.

    3 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
    Talent Agent

    The career as a Talent Agent is filled with responsibilities. A Talent Agent is someone who is involved in the pre-production process of the film. It is a very busy job for a Talent Agent but as and when an individual gains experience and progresses in the career he or she can have people assisting him or her in work. Depending on one’s responsibilities, number of clients and experience he or she may also have to lead a team and work with juniors under him or her in a talent agency. In order to know more about the job of a talent agent continue reading the article.

    If you want to know more about talent agent meaning, how to become a Talent Agent, or Talent Agent job description then continue reading this article.

    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
    Talent Director

    Individuals who opt for a career as a talent director are professionals who work in the entertainment industry. He or she is responsible for finding out the right talent through auditions for films, theatre productions, or shows. A talented director possesses strong knowledge of computer software used in filmmaking, CGI and animation. A talent acquisition director keeps himself or herself updated on various technical aspects such as lighting, camera angles and shots. 

    2 Jobs Available
    Multimedia Animator

    Films like Baahubali, Kung Fu Panda, Ice Age and others are both a sensation among adults and children, and the multimedia animation industry's future looks promising. A multi media jobs could be described as the activity of giving life to a non-living object. Cartoons are the work of animation. Multimedia animation is an illusion developed with the still photographs. Multimedia animators work in a specific medium. Some concentrate on making video games or animated movies. Multi media artists produce visual effects for films and television shows. Multimedia career produce computer-generated images that contain representations of the movements of an actor and then animating them into three-dimensional objects. Multi media artists draw beautiful landscapes or backgrounds.

    2 Jobs Available
    Videographer

    Careers in videography are art that can be defined as a creative and interpretive process that culminates in the authorship of an original work of art rather than a simple recording of a simple event. It would be wrong to portrait it as a subcategory of photography, rather photography is one of the crafts used in videographer jobs in addition to technical skills like organization, management, interpretation, and image-manipulation techniques. Students pursue Visual Media, Film, Television, Digital Video Production to opt for a videographer career path. The visual impacts of a film are driven by the creative decisions taken in videography jobs. Individuals who opt for a career as a videographer are involved in the entire lifecycle of a film and production. 

    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
    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
    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
    News Anchor

    A career as news anchor requires to be working closely with reporters to collect information, broadcast newscasts and interview guests throughout the day. A news anchor job description is to track the latest affairs and present news stories in an insightful, meaningful and impartial manner to the public. A news anchor in India needs to be updated on the news of the day. He or she even works with the news director to pick stories to air, taking into consideration the interests of the viewer.

    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 cost got reduced the viewership for these types of content has increased on a large scale. Therefore, the career as vlogger has a lot to offer. If you want to know more about the career as vlogger, how to become a vlogger, so on and so forth then continue reading the article. Students can visit Jamia Millia Islamia, Asian College of Journalism, Indian Institute of Mass Communication to pursue journalism degrees.

    3 Jobs Available
    Videographer

    Careers in videography are art that can be defined as a creative and interpretive process that culminates in the authorship of an original work of art rather than a simple recording of a simple event. It would be wrong to portrait it as a subcategory of photography, rather photography is one of the crafts used in videographer jobs in addition to technical skills like organization, management, interpretation, and image-manipulation techniques. Students pursue Visual Media, Film, Television, Digital Video Production to opt for a videographer career path. The visual impacts of a film are driven by the creative decisions taken in videography jobs. Individuals who opt for a career as a videographer are involved in the entire lifecycle of a film and production. 

    2 Jobs Available
    SEO Analyst

    An SEO Analyst is a web professional who is proficient in the implementation of SEO strategies to target more keywords to improve the reach of the content on search engines. He or she provides support to acquire the goals and success of the client’s campaigns. 

    2 Jobs Available
    Gemologist

    A career as a gemologist is as magnificent and sparkling as gemstones. A gemologist is a professional who has knowledge and understanding of gemology and he or she applies the same knowledge in his everyday work responsibilities. He or she grades gemstones using various equipment and determines its worth. His or her other work responsibilities involve settling gemstones in jewellery, polishing and examining it. 

    4 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

    Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers. 

    Resource Links for Online MBA 

    3 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. 

    2 Jobs Available
    Commercial Manager

    A Commercial Manager negotiates, advises and secures information about pricing for commercial contracts. He or she is responsible for developing financial plans in order to maximise the business's profitability.

    2 Jobs Available
    QA Manager

    Quality Assurance Manager Job Description: A QA Manager is an administrative professional responsible for overseeing the activity of the QA department and staff. It involves developing, implementing and maintaining a system that is qualified and reliable for testing to meet specifications of products of organisations as well as development processes. 

    2 Jobs Available
    Garment Technologist

    From design to manufacture, garment technologists oversee every stage of clothing production. Individuals are actively engaged in determining the perfect fabric and ensuring that production remains inside the budget. Garment Technologists operate very closely with the designing team, pattern cutters and consumers.

    2 Jobs Available
    Reliability Engineer

    Are you searching for a Reliability Engineer job description? A Reliability Engineer is responsible for ensuring long lasting and high quality products. He or she ensures that materials, manufacturing equipment, components and processes are error free. A Reliability Engineer role comes with the responsibility of minimising risks and effectiveness of processes and equipment. 

    2 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
    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
    ITSM Manager

    ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks. 

    3 Jobs Available
    Big Data Analytics Engineer

    Big Data Analytics Engineer Job Description: A Big Data Analytics Engineer is responsible for collecting data from various sources. He or she has to sort the organised and chaotic data to find out patterns. The role of Big Data Engineer involves converting messy information into useful data that is clean, accurate and actionable. 

    2 Jobs Available
    Test Analyst

    Test Analyst Job Description: A Test Analyst is responsible for ensuring functionality of computer software and hardware equipment, or other products depending on the industry before setting them into the market. His or her role involves designing, developing and administering a series of tests and evaluating them. The role demands to identify potential issues with the product. 

    2 Jobs Available
    Cloud Solution Developer

    A Cloud Solutions Developer is basically a Software Engineer with specialisation in cloud computing. He or she possesses a solid understanding of cloud systems including their operations, deployment with security and efficiency with no little downtime. 

    2 Jobs Available
    CRM Technology Consultant

    A Customer Relationship Management Technology Consultant or CRM Technology Consultant is responsible for monitoring and providing strategy for performance improvement with logged calls, performance metrics and revenue metrics. His or her role involves accessing data for team meetings, goal setting analytics as well as reporting to executives.  

    2 Jobs Available
    QA Manager

    Quality Assurance Manager Job Description: A QA Manager is an administrative professional responsible for overseeing the activity of the QA department and staff. It involves developing, implementing and maintaining a system that is qualified and reliable for testing to meet specifications of products of organisations as well as development processes. 

    2 Jobs Available
    Back to top