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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
Assertion : When bodies are charged through friction, there is a transfer of electric charge from one body to another, but no creation or destruction of charge. Reason : This follows from conservation of electric charges.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - A Conservation of electric charges states that the total charge of an isolate system remains uncharged with time. |
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| 52. |
Assertion : When we rub a glass rod with silk, the rod gets positively charged and the silk gets negatively charged. Reason : On rubbing, electrons from silk cloth move to the glass rod.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - C When we rub a glass rod with silk cloth, electrons from the glass rod are transferred to the silk colth. Thus the rod gets positively charged and the silk gets negatively charged. |
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| 53. |
What happens to the weight of a body when it is charged positively? |
| Answer» When a body is positively charged it must loose some electrons. Hence, weight of the body will decrease. | |
| 54. |
A conducting sphere is negatively charged. Which of the following statements is true?A. The charge is uniformly distributed throughout the entire volume.B. The charge is located at the center of the sphere.C. The charge is located at the bottom of the sphere because of gravity.D. The charge is uniformly distributed on the surface of the sphere. |
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Answer» Correct Answer - D When a conducting sphere get charged, the charge is uniformly distributed on the surface of the sphere. |
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| 55. |
The number of electrons present in -1 C of charge isA. `6xx10^(18)`B. `1.6xx10^(19)`C. `6xx10^(19)`D. `1.6xx10^(18)` |
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Answer» Correct Answer - A By quantisation of charge, q = ne `or" "n=q/e=(-1C)/(-(1.6xx10^(-19)))approx6xx10^(18)` electrons |
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| 56. |
If `10^(9)` electrons move out of a body to another body every second, then the time required to get a total charge of 1 C on the other body isA. 250 yearsB. 100 yearsC. 198 yearsD. 150 years |
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Answer» Correct Answer - C The charge given out in one second `=1.6xx10^(-19)Cxx10^(9)=1.6xx10^(-10)C` Time required to accumulate a charge of 1 C `=1/(1.6xx10^(-10))=6.25xx10^(9)s=198` years |
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| 57. |
A cup contains 250 g of water. Find the total positive charge present in the cup of water.A. `1.34xx10^(19) C`B. `1.34xx10^(7) C`C. `2.43xx10^(19) C`D. `2.43xx10^(7) C` |
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Answer» Correct Answer - B Mass of water = 250 g, Molecular mass of water = 18 g Number of molecules in 18 g of water = `6.02xx10^(23)` Number of molecules in one cup of water `=250/18xx6.02xx10^(23)` Each molecule of water contains two hydrogen atoms and one oxygen atom, i.e., 10 electrons and 10 protons. `therefore" "` Total positive charge present in one cup of water `=250/8xx6.02xx10^(23)xx10xx1.6xx10^(-19)C=1.34xx10^(7)C` |
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| 58. |
If `10^(9)` electrons move out of a body to another body evergy second, how much time is required to get a total charge of 1 C on the other body? |
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Answer» In one second `10^(9)` electrons move out of the body. Therefore the charge given out in one second is `1.6xx10^(-19)xx10^(9)C=1.6xx10^(-10) C`. The time required to accumulate a charge of 1 C can then be astimated to be `1 C div (1.6xx10^(-10) C//s)=6.25xx10^(9)s=6.25xx10^(9) div (365xx24xx3600)` years = 198 years. Thus to collect a charge of one coulomb, from a body from which `10^(9)` electrons move out every second, we will need approximately 200 years. One coulomb is, therefore, a very large unit for many practical purposes. It is, however, also important to know what is roughly the number of electrons contained in a piece of one cubic centimetre of a material. A cubic piece of copper of side 1 cm contains about `2.5xx10^(24)` electrons. |
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| 59. |
Which among the curves shown in Fig. cannot possibly represent electrostatic field lines? |
| Answer» By the property of electric field lines that they are not in the form of closed loops. Here, the lines form closed loop. So, it does not represent the electric field lines. | |
| 60. |
Which among the curves shown in Fig. cannot possibly represent electrostatic field lines? |
| Answer» By the property of electric field lines, two electric field lines, two electric field lines never intersect each other. Here, two lines intersect. So it does not represent the electric field lines. | |
| 61. |
Which among the curves shown in Fig. cannot possibly represent electrostatic field lines? |
| Answer» According to the property of electrostatic field lines, they never start from negative charge, here some of the lines start from negative charge. So, it cannot represent the electrostatic field lines. | |
| 62. |
Which among the curves shown in Fig. cannot possibly represent electrostatic field lines? |
| Answer» As the property of electric field lines that they start outwards from positive charge. Hence, it represents the electrostatic field lines. | |
| 63. |
A positive charge `Q` is uniformly distributed along a circular ring of radius `R`.a small test charge `q` is placed at the centre of the ring .The A. if q gt 0, and is displaced away from the centre in the plane of the ring, it will be pushed back towards the centre.B. if q lt 0 and is displaced away from the centre in the plane of the ring, it will never return to the centre and will continue moving till it hits the ring.C. if q lt 0 it will perform SHM for small displacement along the axis.D. all of the above |
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Answer» Correct Answer - D At the centre of the ring, E = 0 when a positive charge q gt 0 is displaced away from the centre in the plane of the ring, say to the right, force of repulsion on q, due to charge on right half increases and due to charge on left half decreases. Therefore, charge q is pushed back towards the centre. So option (a) is correct. When charge q is negative (q lt 0), force is of attraction. Therefore, charge q displaced to the right continues moving to the right till it hits the ring. Along the axis of the ring, at a distance r from the centre, `E=(Qr)/(4piepsilon_(0)(r^(2)+a^(2))^(3//2))` If charge q is negative (q lt 0), it will perform SHM for small displacement along the axis. |
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| 64. |
If `oint_(s) E.ds = 0` Over a surface, thenA. the electric field inside the surface and on it is zero.B. the electric field inside the surface is necessarily uniform.C. all charges must necessarily be outside the surface.D. all of these. |
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Answer» Correct Answer - C As `oint vecE*vec(ds)=q/(epsilon_(0))`, Where q is charge enclosed by the surface. when `oint vecE*vec(ds)=0,q=0` i.e., net charge enclosed by the surface must be zero. Therefore, all other charges must be outside the surface. This is because charges outside the surface do not contribute to the electric flux. |
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| 65. |
Assertion. When charges are shared between any two bodies, no charge is really lost but some loss of energy does occur. Reason. Some energy disappears in the from of heat, sparking etc.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - A Charges are conserved by the law of conservatin of charges. Energy is also conserved, if we take in account of the loss of energy by heat, sparkling etc. |
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| 66. |
Two charges `+- 10 mu C` are placed `5*0 mm` apart. Determine the electric field at (a) point P on the axis of dipole `15 cm` away from its center on the side of the positive charge. As shown in Figure and at (b) a point Q. `15 cm` away form O on a line passing through O and a line passing through O and normal to the axis of the dipole as shown in Fig. |
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Answer» (a) Field at P due to charge `+10muC` `=(10^(8)C)/(4pi(8.854xx10^(-12)C^(2)N^(-1)m^(-2)))xx(1)/((15-0.25)^(3)xx10^(-4)m^(3))` `=4.13xx10^(8)NC^(-1) "along BP"` Field at P due to charge `-10muC` `=(10^(8)C)/(4pi(8.854xx10^(-12)C^(2)N^(-1)m^(-2)))xx(1)/((15-0.25)^(3)xx10^(-4)m^(3))` The resultant electric field at P due to the two charges at A and B is `=2.7xx10^(5)NC^(-1)"along BP".` In this example, the ratio OP/OB is quite large (= 60). Thus, we canexpect to get approximately the same result as above by directly usingthe formula for electric field at a far-away point on the axis of a dipole.For a dipole consisting of charges ± q, 2a distance apart, the electricfield at a distance r from the centre on the axis of the dipole has amagnitude `E=(2)/(4piepsilon_(0)r^(3))" " (r//a gt gt 1)` where p = 2a q is the magnitude of the dipole moment. The direction of electric field on the dipole axis is always along thedirection of the dipole moment vector (i.e., from –q to q). `Here, p=10^(-5)Cxx5xx10^(-3)m=5xx10^(-8)Cm` Therefore `=(10^(8)C)/(4pi(8.854xx10^(-12)C^(2)N^(-1)m^(-2)))xx(1)/((15)^(3)xx10^(-4)m^(3))=2.6xx10^(5)NC^(-1)` along the dipole moment direction AB, which is close to the result obtained earlier. (b) Field at Q due to charge `+ 10 muC at B` `=(10^(5)C)/(4pi(8.854xx10^(-12)C^(2)N^(-1)m^(-2)))xx(1)/("["15^(2)-(0.25)^(3)"]"xx10^(-4)m^(2))` `=3.99xx10^(6)NC^(-1) "along BQ"` ltBRgt Field at Q due to charge `-10muC at A` `=(10^(8)C)/(4pi(8.854xx10^(-12)C^(2)N^(-1)m^(-2)))xx(1)/("["15^(2)-(0.25)^(3)"]"xx10^(-4)m^(3))` `=3.99xx10^(6)NC^(-1) "along QA".` Clearly, the components of these two forces with equal magnitudes cancel along the direction OQ but add up along the direction parallel to BA. Therefore, the resultant electric field at Q due to the two charges at A and B is `=2xx(0.25)/(sqrt(15^(2)+(0.25)^(2)))=3.99xx10^(6)NC^(-1) "along BA"` `1.33xx10^(5) NC^(-1) "along BA"`As in (a), we can expect to get approximately the same result by directly using the formula for dipole field at a point on the normal to the axis of the dipole: `E=(P)/(4piepsilon_(0)r^(3)) " " (r//a gt gt 1)` `=(5xx10^(8)Cm)/(4pi(8.854xx10^(-12)C^(2)N^(-1)m^(-2)))xx(1)/((15)^(3)"xx10^(-6)m^(3))` `=1.33xx10^(5)NC^(-1)` The direction of electric field in this case is opposite to the direction of the dipole moment vector. Again, the result agrees with that obtained before. |
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| 67. |
An electric field is uniform and in the positive x direction for positive x and uniform with the same magnitude but in the negative x direction for negative x. It is given that `E=200 hat(i) N//C` for `x gt 0` and `E=-200 hat(i) N//C` for `x lt 0`. A right circular cylinder of length 20 cm and radius 5 cm has its centre at the origin and its axis along the x-axis so that one face is at `x=+10` cm and the other is at `x=-10` cm What is the net outward flux through each flat face? |
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Answer» We can see from the figure that on the left face E and `Delta S` are parallel. Therefore, the outward flux is `phi_(L)=E. Delta S=-200 hat(i). Delta S` `=+200 Delta S`, since `hat(i).Delta S=-Delta S` `=+200xx pi (0.05)^(2)=+1.57 Nm^(2) C^(-1)` On the right face, E and `Delta S` are parallel and therefore `phi_(R)=E. Delta S=+1.57 Nm^(2) C^(-1)`. |
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| 68. |
An electric field is uniform and in the positive x direction for positive x and uniform with the same magnitude but in the negative x direction for negative x. It is given that `E=200 hat(i) N//C` for `x gt 0` and `E=-200 hat(i) N//C` for `x lt 0`. A right circular cylinder of length 20 cm and radius 5 cm has its centre at the origin and its axis along the x-axis so that one face is at `x=+10` cm and the other is at `x=-10` cm What is the net outward flux through the xylinder? |
| Answer» Net outward flux through the cylinder `phi=1.57+1.57+0=3.14 Nm^(2) C^(-1)`. | |
| 69. |
What is the force between two small charged spheres having charges of `2xx10^(-7)C and 3xx10^(-7) C` placed 30cm apart in air ? |
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Answer» Repulsive force of magnitude `6 xx 10^(−3) N` Charge on the first sphere, `q_(1) = 2 xx 10^(−7) C` Charge on the second sphere, `q_(2) = 3 xx 10^(−7) C` Distance between the spheres, `r = 30 cm = 0.3 m` Electrostatic force between the spheres is given by the relation, `F=(q_(1)q_(2))/(4 piepsilon_(0) r^(2))` Where, ` epsilon_(0)` = Permittivity of free space `(1)/((4 piepsilon_(0)) = 9 xx 10^(9) Nm^(2) C^(-2)` `F = (9 xx 10^(9) xx 2 xx 10^(-7) xx 3 xx 10^(-7))/((0.3)^(2)) = 6 xx 10^(-3)N` Hence, force between the two small charged spheres is `6 xx 10^(−3)` N. The charges are of same nature. Hence, force between them will be repulsive. |
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| 70. |
The force between two small charged spheres having charges of `1xx10^(-7)C and 2xx10^(-7)C` placed 20 cm apart in air isA. `4.5xx10^(-2)N`B. `4.5xx10^(-3)N`C. `5.4xx10^(-2)N`D. `5.4xx10^(-3)N` |
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Answer» Correct Answer - B Here, `q_(1)=1xx10^(-7)C,q_(2)=2xx10^(-7)C,r=20cm=20xx10^(-2)m` `As" "F=(q_(1)q_(2))/(4piepsilon_(0)r^(2))=(9xx10^(9)xx1xx10^(-7)xx2xx10^(-7))/((20xx10^(-2))^(2))` `=4.5xx10^(-3)N` |
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| 71. |
Consider a region inside which there are various types of charges but the total charge is zero ,.At points outside the regionA. the electric field is necessarily zero.B. the electric field is due to the dipole moment of the charge distribution only.C. the dominant electric field is inversely proportional to `r^(3), for large r (distance from origin).D. the work done to move a charged particle along a closed path, away from the region will not be zero. |
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Answer» Correct Answer - C When there are various types of charges in a region, but the total charge is zero, the region can be supposed to contain a number of electric dipoles. Therefore, at point outside the region, the dominant electric field `prop1/r^(3)` for large r. |
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| 72. |
Write the expression for electric intensity due to a charged conducting spherical shell at points outside and inside the shell. |
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Answer» (a) Intensity of electric field ay any point inside a spherical shel is zero. (b) Intensity of electric field at anu point outside a uniformly charged spherical shell is `E=1/(4 pi epsi_(0)). q/r^(2)` |
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| 73. |
Five equal charges each of value q are placed at the corners of a regular pentagon of side a. The electric field at the centre of the pentagon is A. `q/(4piepsilon_(0)r^(2))`B. `(q^(2))/(4piepsilon_(0)r^(2))`C. `(2q)/(4piepsilon_(0)r^(2))`D. zero |
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Answer» Correct Answer - D The electric field at the centre of pentagon would be zero. |
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| 74. |
In question number 45, what will be the electric field at centre O, if the charge from one of the corners (say A) is removed?A. `q/(4piepsilon_(0)r^(2))` along OAB. `(2q)/(4piepsilon_(0)r^(2))` along OBC. `(q^(2))/(4piepsilon_(0)r^(2))` along OCD. `(2q)/(4piepsilon_(0)r^(2))` along OA |
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Answer» Correct Answer - A When a charge q from corner A is removed, electric field at O is `E_(1)=q/(4piepsilon_(0)r^(2)` along OA |
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| 75. |
Four point charges are placed at the corners of a square ABCD of side 10 cm, as shown in figure. The force on a charge of `1muC` placed at the centre of square is A. 7 NB. 8 NC. 2 ND. zero |
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Answer» Correct Answer - D Force of repulsion on `1muC` charge at O due to `3muC` charge, at A and C are equal and opposite. So they cancel each other. Similarly, force of attraction of `1muC` charge at O due to `-4muC` charges at B and D are also equal and opposite. So they also cancel each other. Hence the net force on the charge of `1muC` at O is zero. |
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