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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.
| 1. |
An electric dipole is placed in a uniform electric field `vec(E )` of magnitude `40N//C`. Graph shows the magnitude of the square on the dipole versus the angle `theta` between the field `vec(E )` and the dipole moment `vec(p)`. The magnitude of dipole moment `vec(p)` is equal to: A. `1.25xx10^(-28)C-m`B. `2.0xx10^(-25)C-m`C. `2.5xx10^(-28)C-m`D. `5.0xx10^(-28)C-m` |
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Answer» Correct Answer - a `vec(tau)=vec(p)xxvec(E )` `tau_(max)=pEimplies 50xx10^(-28)= p.40` `implies p=1.25xx10^(-28)C-m` |
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| 2. |
Assertion: If bob of a simple pendulum is kept in a horizontal electric field, its period of oscillation will remain same. Reason: If bob is charged and kept in horizontal electric field, then the time period will be decreased.A. if both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of the Assertion.C. if Assertion is true, but the Reason is false.D. if both Assertion and Reason are false |
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Answer» Correct Answer - b When the bob is placed in an electric field, the bob time period of simple pendulum having charged bob is placed in a horizontal electric field then the period will be decreased because there will be a increase the restoring force |
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| 3. |
Assertion: If there exists coulombic attracation between two bodies both of them may not be charged. Reason: In coulombic attraction two bodies are oppositely charged.A. if both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of the Assertion.C. if Assertion is true, but the Reason is false.D. if both Assertion and Reason are false |
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Answer» Correct Answer - c Coulombic attraction exists even when one body is charged and the other is uncharged. |
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| 4. |
Two charged particles of charge `+2q` and `+q` have masses `m` and `2m` respectively. Then are kept in uniform electric field allowed to move for the same time. Find the ratio of their kinetic energies.A. `1:8`B. `16:1`C. `2:1`D. `3:1` |
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Answer» (b) Let `E=` uniform electric field Let `m_(1)=m, m_(2)=2m` Force on particles, `F_(1)=ma_(1)=2qE` `implies F_(2)=2ma_(2)=qE` `a_(1)=(2qE)/m` and `a_(2)=(qE)/(2m)` Final velocities of the particle `v_(1)=u_(1)=a_(1)t` `implies v_(1)=(2qE)/m timpliesv_(2)=(qE)/(2m)t` Let `K=` kinetic energy `K_(1)=1/2m ((qE)/(2m))^(2)t^(2)implies(K_(1))/(K_(2))=16/1` |
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| 5. |
Charge `+2Q` and `-Q` are placed as shown in figure. The point at which electric field intensit is zero will be A. somewhere between `-Q` and `+2Q`B. somewhere on the left of `-Q`C. somewhere on the right of `+2Q`D. somewhere on the right bisector of line joining `-Q` and `+2Q` |
| Answer» Correct Answer - (b) | |
| 6. |
Suppose we have large number of identical particles, very small in size. Any of them at `10 cm` separation repel with a force of `3xx10^(-10) N`. If one of them is at `10 cm` from a group of `n` others, how strongly do you expect it to be repelled?A. `3xx10^(-10) N`B. `3nxx10^(-10)N`C. `(3xx10^(-10))/(n)N`D. Can not be calculated |
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Answer» Correct Answer - b The group of `n` small charged particles must behaves as a single point charge, so that if can have a separation of `10 cm` from the one particle in question. Obviously force on this particle due to the group of `n` particles is `n` times he force due a single particle. Hence, force dut to group of `n` particles, `F= nxx3xx10^(-10)N` |
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| 7. |
Two ideantical point charges are placed at a separation of `d`. `P` is a point on the line joining the charges, at a distance `x` from any one charge. The field at `P` is `E, E` is plotted against `x` for value of `x` from close to zero to slightly less then `d`. Which of the following represents the resulting curveA. B. C. D. |
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Answer» Correct Answer - d At mid point, `E=0` Before mid point, `E` is positive. This is maximum near the charge and decreases towards mid point. After mid point, `E` is negative, The curve crosses `x`-axes at `x`-axis at `x= d//2`. From centre to end, `E` decreases. The Varitation is shown by curved. |
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| 8. |
Assertion: A line of force has sudden breaks. Reason: An electrostatic line of force is a continuous curve.A. if both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of the Assertion.C. if Assertion is true, but the Reason is false.D. if both Assertion and Reason are false |
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Answer» Correct Answer - a The tangent at a point an the electric line of force tells the direction of electric field charge from point to point. So, line of force are curved lines. Further they are continous curves and cannot have sudden breaks otherwise it will indicate the absence of electric field at the break point. |
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| 9. |
A point charge `Q` is placed at the centre of a circular wire of radius `R` having charge `q`. Find the force of electrostatic interaction between point charge and the wire. A. `(qQ)/(4piepsilon_(0)R^(2))`B. zeroC. `(q^(2))/(4pi epsilon_(0)R`D. none of these |
| Answer» Correct Answer - (b) | |
| 10. |
A charged particle of mass `m` and charge `q` is released from rest in an electric field of constant magnitude `E`. The kinetic energy of the particle after time `t` isA. `(2E^(2)t^(2))/(mq)`B. `(E^(2)q^(2)t^(2))/(2m)`C. `(Eq^(2)m)/(2t^(2))`D. `(Eqm)/(2t)` |
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Answer» Correct Answer - b `F=qE` `a= (f)/(m)= (qE)/(m)` `v= at= (qEt)/(m)` Kinetic energy `= 1/2m[(qEt)/(m)]^(2)=(E^(2)q^(2)t^(2))/(2m)` |
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| 11. |
An electron is projected with velocity `10^(7)m//s` at an angle `theta(=30^(@))` with horizontal, in a region of uniform electrilc fieldof `500N//C` vertically upwards. Find the maximum distance covered by an electron in vertical direction above its initial levelA. `14.2mm`B. `15 mm`C. `12.6mm`D. `14.2mm` |
| Answer» Correct Answer - (a) | |
| 12. |
If a charged particle is projected on a rough horizontal surface with speed `v_(0)`. Find the value of dynamic coefficient of friction, if the kinetic energy of system is constant. A. `(qE)/(mg)`B. `(qE)/m`C. `q/g`D. none of these |
| Answer» Correct Answer - (a) | |
| 13. |
A point charge is projected along the axis of circular ring of charge `Q` and radius `10sqrt(2)cm`. The distance of the point charge from centre of ring, where acceleration of charged particle is maximum, will beA. `10cm`B. `20cm`C. at infinityD. none of the above |
| Answer» Correct Answer - (a) | |
| 14. |
Two charges `+4e` and `+e` are at a distance `x` apart. At what distance,a charge `q` must be placed from charge `+e` so that is in equilibriumA. `x//2`B. `2x//3`C. `x//3`D. `x//4` |
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Answer» Correct Answer - D For equilibrium of `q` `|F_(1)|=|F_(2)|` `(BMS_OBJ_XII_C01_E01_066_S01.png" width="80%"> Which gives `x_(2)= (x)/(sqrt(Q_(1)/(Q_(2)))+1)=(x)/(sqrt(9e)/(e)+1)=(x)/(4)` |
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| 15. |
A charge of `1muC` is divided into two parts such that their charges are in the ratio of `2:3`. These two charges are kept at a distance `1m` apart is vacuum. Then, the electric force between them (in `N`) isA. `0.216`B. `0.00216`C. `0.0216`D. `2.16` |
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Answer» (b) Ratio of charges `=2:3` `:. q_(1)=2/5xx5xx1 mu C` and `q_(2)=35xx1 muC` Electrostatic force between the two charges `F=1/(4pi epsilon_(0))(q_(1)q_(2))/(r^(2))` `=(9xx10^(9)xx2xx10^(-6)xx3xx10^(-6))/(5xx5xx(1)^(2))` `=2.16xx10^(-3)N` `~~0.00216N` |
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| 16. |
An electric line of forces in the x-y plane is given by the equation `x^2+y^2=1`. A particle with unit poistive charge, initially at rest at the point `x=1`, `y=0` in the x-y plane, will move along the circular line of force.A. not move at allB. will move along straight lineC. will move along the circular line of forceD. informtion is insufficient to draw any conclusion |
| Answer» (c) Charge will move along the circular line of force because `x^(2)+y^(2)=1` is the equation of circle in `xy`- plane. | |
| 17. |
A point `Q` lies on the perpendicular bisector of an electrical dipole of dipole moment `p`, If the distance of `Q` from the dipole is `r` (much larger than the size of the dipole), then electric field at `Q` is proportional toA. `p^(-1)` and `r^(-2)`B. `p` and `r^(-2)`C. `P^(2)` and `r^(-3)`D. `p` and `r^(-3)` |
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Answer» Correct Answer - d `E_(equat o rial)= (kp)/(r^(3))i.e., E prop P` and `E prop r^(-3)` |
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| 18. |
Four point `+ve` charges of same magnitude`(Q)` are placed at four corners of a rigid square frame as shown in figure. The plane of the frame is perpendicular to `z`-axis. If a `-ve` point charge is placed at a distance `z` away from the above frame `(z lt lt L)` then A. `-ve` charge oscillates along the z axis.B. It moves away from the frameC. It moves slowly towards the frame and stays in the plane of the frameD. It passes through the frame only once. |
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Answer» Correct Answer - a The negative charge osscillates, the resultant force acts as a restoring force and proportional to displacement. When it reaches the plane `XY`, the resultant force is zero and the mass moves down due to inertia. Thus oscillation is set. |
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| 19. |
Four point `+ve` charges of same magnitude`(Q)` are placed at four corners of a rigid square frame as shown in figure. The plane of the frame is perpendicular to `z`-axis. If a `-ve` point charge is placed at a distance `z` away from the above frame `(z lt lt L)` then A. `-ve` charge oscillates along the z axis.B. It moves away from the frameC. It moves slowly towards the frame and stays in the plane of the frameD. It passes through the frame only once. |
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Answer» Correct Answer - a The negative charge oscillates, the resultant force acts as a restoring force and proportional to displacement. When it reaches the plane `xy`, the resultant force is zero and the mass moves down due inertia. Thus oscillation is set. |
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| 20. |
A hemisphere is uniformaly charged positively. The electric field at a point on a diameter away from the centre is directedA. perpendicular to the diameterB. parallel to the diameterC. at an angle tilted towards the diameterD. at an angle tilted away from the diameter |
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Answer» Correct Answer - a When the point is situated at a point on diameter away from the centre of hemisphere charged uniformly positively, the electric field is perpendicular to the diameter. The component of electric intensity parallel to the diameter cancel out. |
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| 21. |
Figure, shown above, shows three situations involving a charged particle and a uniformly charged spherical shell. The charges and radii of the shells are indicated in the figure. If `F_(1),F_(2)` and `F_(3)` are the magnitudes of the force on the particle due to the shell in situations (I),(II) and (III) then A. `F_(3) gt F_(2) gt F_(1)`B. `F_(2) gt F_(2)=F_(3)`C. `F_(3)=F_(2) gt F_(1)`D. `F_(1) gt F_(2) gt F_(3)` |
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Answer» Correct Answer - c `F_(1)` is zero as charge lies inside the shell (Electric field inside the shell due to its own charge is zero). `F_(2)=F_(3)` as charge over the surface of he shell behaves like charge placed at the centre of the shell. |
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| 22. |
An electric dipole is kept on the axis of a uniformly charged ring at distance d from the centre of the ring. The direction of the dipole moment is along the axis. The dipole moment is `p`, charge of the ring is `Q` & radius of the ring is `R`. The force on the dipole isA. `(pQ)/(3piepsilon_(0)sqrt(3)R^(2))`B. `(4pQ)/(3piepsilon_(0)sqrt(3)R^(2))`C. `(pQ)/(3piepsilon_(0)R^(2))`D. zero |
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Answer» Correct Answer - d Electric field due to ring and dipole moment are parallel to each other. |
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| 23. |
The following diagram shows the electric field lines between two opposite charges. The positive charge is indicated by the black circle, the negative charge by the white circle. An electron string from rest at the indicated position `(X)`, and accelerated to high speed by the electric field will follow most likely which trajectory?A. B. C. D. |
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Answer» Correct Answer - c Trajectory will be tangent to electric field |
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| 24. |
A soap bubble is given a negative charge, then its radiusA. DecreaseB. IncreaseC. Remain sameD. Nothing can be predicted as uniformation is insufficient |
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Answer» Correct Answer - b Due to mutual repulsion of charges distributed on the surface of bubble. |
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| 25. |
A charge `q` is placed at the some distance along the axis of a uniformly charged disc of surface cahrge density `sigma`. The flux due to the charge `q` through the disc is `phi`. The electric force on charge `q` exerted by the disc isA. `sigma phi`B. `(sigma phi)/(4pi)`C. `(sigma phi)/(2pi)`D. `(sigma phi)/(3 pi)` |
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Answer» Correct Answer - a We know that if a point charge subtends a half-angle `alpha` on the circular cross-section of a disc, then flux passing through the disc is : `phi=(q)/(2 epsilon_(0))(1- cos alpha)` Also, if a point charge `q` lies on the axis of a charged disc of charge density `sigma` and subtends a half-anlge `alpha`, then it experiences a force `F= (q sigma)/(2 epsilon_(0))(1-cos alpha)= sigma phi` |
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| 26. |
What is the amount of charge possessed by `1 kg` of electronsA. `6.25xx10^(10)C`B. `1.76xx10^(11)C`C. `1.76xx10^(10)C`D. `1.25xx10^(10)C` |
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Answer» Correct Answer - b `n=(M)/(m_(e))= (1kg)/(9.1xx10^(-31))= 1.1xx10^(3)` Thereforre, charge on `1 kg` of electrons, `q= n e= 1.1xx10^(30)xx1.6xx10^(-19)= 1.76xx10^(11)C` |
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| 27. |
Assertion: On going away from a point charge or a small electric dipole, electric field decreases at the same rate in both the cases Reason: Electric field is inversly proportional to square of distance from the charge or an electric dipole.A. if both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of the Assertion.C. if Assertion is true, but the Reason is false.D. if both Assertion and Reason are false |
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Answer» Correct Answer - d The rate of decrease of electric field is different in the two cases. In case of a point charge, it decreases as `1//r^(2)` but in the case of electric dipole it decreases move rapidly, as `E prop 1//r^(3)`. |
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| 28. |
A polythene piece rubbed with wool is found to have negative charge of `3.6xx10^(-7)C`. Calculate the number of electrons trandferred from wool to polythene. If an electron has a mass `9.1 xx 10^(-31) kg`, find the mass trandferred to polythene.A. `2.25xx10^(10) kg`B. `6.25xx10^(-18) kg`C. `2.05xx10^(-18) kg`D. `4.15xx10^(-18) kg` |
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Answer» Correct Answer - c `n=(q)/e=(3.6xx10^(-7)C)/(1.6xx10^(-19)C)= 2.25xx10^(12)` Therefore, mass trandferred from wool to polythene, `M=nxxm_(e)= 2.25xx10^(12)xx9.1xx10^(-31)` `=2.05xx10^(-18) kg` |
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| 29. |
A large solid sphere with uniformly distributed positive charge has a smooth narrow tunnel along its direction. A small particle with negative charge, initially at rest far from the sphere, approaches it along the line of the tunnel, reaches its surface with a speed `v`, and passes through the tunnel. Its speed at the centre of the sphere will be |
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Answer» Correct Answer - b Let `m` and `-q` be the mass and the charge of the particle respectively. Let `v_(0)=`speed of the particle at the centre of the sphere. Using conservation of mechanical energy `Delta K+Delta U=0` `((1)/(2)mv_(s)^(2)-0)+[-q(V_(s)-V_(oo))=0` `(1)/(2)mv_(s)^(2)=q[V_(s)-0]=qV_(S)` Similarly `(1)/(2)mv_(0)^(2)=q[V_(c )-0]=qV_(c )` We know `V_(c )=(3)/(2)k(Q)/(R)V_(s)=k(Q)/(R)` Hence `(1)/(2)mv_(s)^(2)=q.k(Q)/(R )`....(i) and `(1)/(2)mv_(0)^(2)=q(3)/(2)(Q)/(R)`....(ii) Dividing Eq. (ii) by (i), `(v_(0)^(2))/(v^(2)) = (3)/(2) =1.5` or `v_(0) = sqrt(1.5)v` |
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| 30. |
A uniformly charged non conducting disc with surface charge density `10nC//m^(2)` having radius `R=3 cm`. Then find the value of electric field intensity at a point on the perpendicular bisector at a distance of `r=2cm` A. `348.6N//C`B. `305.6N//C`C. `251.2N//C`D. `116.8N//C` |
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Answer» Correct Answer - c `E=(sigma2pi)/(4piepsilon_(0))[1-(x)/(sqrt(R^(2))+x^(2))]` `E=9xx10^(9)xx10xx10^(-9)xx6.28[1-(2)/(sqrt(4)+9)]` `E= 90xx6.28[1-(2)/(sqrt(13))]` `E= 251.2N//C` |
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| 31. |
Assertion: When a body acquires positive charge, its mass decreases Reason: A body acquires positive charge when it loses electrons.A. if both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of the Assertion.C. if Assertion is true, but the Reason is false.D. if both Assertion and Reason are false |
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Answer» Correct Answer - a If a body acquires a positive charge, than it meand that it lost few electrons. In this way its mass decreases. |
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| 32. |
A copper atom consists of copper nucleus surrounded by `29` electrons. The atomic weight of copper is `63.5mol e^(-1)`. Let us now take two pieces of copper each weighing `10g`. Let us trandfer one elcetron from one piece to another for every `100` atoms in that piece. What will be the Coulomb force between the two pieces after the trandfer of electrons, if they are `1cm` apart? Avogadro number`= 6xx10^(23) mol e^(-1)`, charge on an electron `=-1.6xx10^(-19)C`.A. `1.12xx10^(18)N`B. `4.24xx10^(18)N`C. `2.06xx10^(18)N`D. `5.16xx10^(18)N` |
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Answer» Correct Answer - c Here, Avogadro number, `N= 6xx10^(23) mol e^(-1)` and Atomic weight of copper, `A= 63.5 g mol e^(-1)` Therefore, number of atoms in `10g` of copper. `n=(N)/(A)xx10=(6xx10^(23)xx10)/(63.5)= 9.45xx10^(22)` Since one electron per `100` atoms is trandferred from copper piece `A` to `B` the charge on copper piece `A`, `q_(A)= +(n)/(100)xxe= +151.2C` Obviously, charge on copper piece `B`, `q_(B)= -151.2C` Force between the two copper pieces, `F= (1)/(4piepsilon_(0)).(q_(A)xxq_(B))/(r^(2))` Setting `r= 1 cm= 0.01 m`, we get `F= 2.06 xx10^(8)N` (attractive) |
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| 33. |
Conduction electrons are almost uniformly distributed within a conducting plate. When placed in an electric field `vec(E )`, the electric field within the plateA. Is zeroB. Depends upon `E`C. Depends upon `vec(E )`D. Depends upon the atomic number of the conducting element |
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Answer» Correct Answer - a Electric field inside a conductor is zero. |
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| 34. |
A solid metallic sphere has a charge `+3Q`. Concentric with this sphere is a conducting spherical shell having charge `-Q`. The radius of the sphere is `a` and that of the spherical shell is `b(bgta)` What is the electric field at a distance `R(a lt R lt b)` from the centreA. `(Q)/(2pi epsilon_(0)R)`B. `(3Q)/(2pi epsilon_(0)R)`C. `(3Q)/(4pi epsilon_(0)R^(2))`D. `(4Q)/(4pi epsilon_(0)R^(2))` |
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Answer» Correct Answer - c Electric field at a distance `R` is only due to sphere because electrice field due to shell inside it is always zero. Hence electric field `=(1)/(4piepsilon_(0)).(3Q)/(R^(2))` |
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| 35. |
When a body is earth connected, electrons from the earth flow into the body. This means the body is `….`A. UnchangedB. Charge positivelyC. Charge negativelyD. an insulator |
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Answer» Correct Answer - b When a positively charged body connected to earth, electrons flows from earth to body and body becomes neutral. |
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| 36. |
In a region with uniform electric field, the number of lines of force per unit area is `E`. If a spherical metallic conductor is placed in this region, the number of lines of force per unit area inside the conductor will beA. `E`B. more than `E`C. less than `E`D. zero |
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Answer» Correct Answer - d No field inside the metallic conductor. |
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| 37. |
An electric dipole is placed at an angle of `30^(@)` with an electric field intensity `2xx10^(5)N//C`. It experiences a torque equal to `4Nm`. The charge on the dipole, if the dipole is length is `2 cm`, isA. `5 mC`B. `7muC`C. `8mC`D. `2 mC` |
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Answer» Correct Answer - d `tau=PE sin theta` `tau=ql E sin theta` `4= qxx2xx10^(-2)xx2xx10^(5) sin 30^(@)` `implies q= 2mC` |
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| 38. |
For a dipole `q= 2xx10^(-6)C` and `d= 0.01m`. Calculate the maximum torque for this dipole if `E= 5xx10^(5)N//C`A. `1xx10^(-3)Nm^(-1)`B. `10xx10^(-3)Nm^(-1)`C. `10xx10^(-3)Nm`D. `1xx10^(2)Nm^(2)` |
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Answer» Correct Answer - c `tau_(max)=pE= q(2l)E= 2xx10^(-6)xx0.01xx5xx10^(5)=10xx10^(-3)N-m` |
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| 39. |
A neutral water molecule `(H_(2)O)` in its vapour state has an electric dipole moment of magnitudes `6.4xx10^(-30)C-m`. How far apart are the molecules centres of positive and negative chargeA. `4m`B. `4 mm`C. `4 mu m`D. `4 p m` |
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Answer» Correct Answer - d There are `10` electrons and `10` protons in a neutral water molecule. So its dipole moment is `p=q (2l)= 10e(2l)` Hence length of the dipole i.e., distance between centres of positive and negative charge is `2l=(p)/(10e)=(6.4xx10^(-20))/(10xx1.6xx10^(-19))= 4xx10^(-12)m= 4p m` |
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| 40. |
Two charges `+3.2xx10^(-19)` and `-3.2xx10^(-19)C` placed at `2.4A` apart from an electric dipole. It is placed in a uniform electric field of intensity `4xx10^(5) "volt"//m`. The electric dipole moment isA. `15.36 xx 10^(-29) "coulomb" xx m`B. `15.36xx10^(-19) "coulomb" xx m`C. `7.68 xx 10^(-29) "coulomb" xx m`D. `7.68xx10^(-19) "coulomb" xx m` |
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Answer» Correct Answer - c Dipole moment `p=q(2l)` `= 3.2xx10^(-19)xx(2.4xx10^(-10))= 7.68xx10^(-29)C-m` |
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| 41. |
Assetrion: Electric lines of force never cross each other. Reason: Electric field at a point superimpose to give one resultant electric fieldA. If both assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both assertion and reason are true but Reason is not the correct explanation of the Assertion.C. If Assertion is true, but the Reason is false.D. If both assertion and reason are false |
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Answer» Correct Answer - a Electric field lines of force can never intersect each other, because if they do so then at the point of intersection two tangent can be drawn which would mean two directions of force at that point which is impossible. If there are `n` point charges `q_(1),q_(2).....q_(n)` then each of them will produce the same intensity at any point which it would have produced in the absence of other point charges, Hence, intensity of `vec(E )` is resultant of `vec(E )_(1),vec(E )_(2).....vec(E )_(n)` as `vec(E )_(1)+vec(E)_(2)+.....+vec(E )_(n)` |
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| 42. |
The velocity of an electron at point `A_(1)` is `V_(0)` where cross sectional areas is `A`. The velocity of electron just the end of contraction at point `B`, where cross sectional area is `2A` is `V_(1)`. Find the correct option: A. `V_(1) lt V_(0)`B. `V_(1)=V_(0)`C. `V_(1) gt V_(0)`D. `V_(1)=(V_(0))/(2)` |
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Answer» Correct Answer - c As the electrons comes near the throat, positive charges are induced on the throat. The induced charges attract the electron and velocity increases. So `V_(1)gtV_(0)`. |
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| 43. |
Consider the four field patterns shown. Assuming there are no charge in the regions shown, which of the patterns represent a possible electrostatic field?A. B. C. D. |
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Answer» Correct Answer - b Pattern (a) can be eliminated because field lines cannot simultaneously originate from and converge at a single point, (c ) can be eliminated because there are no charges in the region, and so there are no sources of field lines, (d) can be eliminated beacuse electrostatic fields lines do not close on themselves. |
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| 44. |
Three point charges are placed at the corner of an equilateral triangle. Assuming only electrostatic forces are acting.A. the system can never be in equilibriumB. the system will be equilibrium if the charges rotate about the centre of the triangleC. the system will be in equilibrium if the charges have different magnitudes and different signsD. the system will be in equilibrium if the charges have the same magnitude but different signs. |
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Answer» Correct Answer - a No system can be equilibrium under the influence of electrostatic force only. |
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| 45. |
Two identical simple pendulums, `A` and `B` are suspended from the same point. The bobs are given positive charges, with `A` having more charge than `B` making angles `theta_(1)` and `theta_(2)` with the vertical respectively. Which of the following is correct?A. `theta_(1) gt theta_(2)`B. `theta_(1) lt theta_(2)`C. `theta_(1)= theta_(2)`D. The tension in `A` is greater than that in `B` |
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Answer» Correct Answer - c Coulomb force between unequal charge are equal. |
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| 46. |
Consider the charge configuration and a spherical Gaussian surface as shown in the figure. When calculating the flux of the electric field over the spherical surface, the electric field will be due to. A. `q_(2)`B. Only the positive chargesC. All the chargesD. `+q_(1)` and `-q_(1)` |
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Answer» Correct Answer - c The electric field is due to all charges present whether inside or outside the given surface. |
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| 47. |
If the electric flux entering and leaving an enclosed surface respectively is `phi_1` and `phi_2`, the electric charge inside the surface will beA. `(phi_(1)+phi_(2))epsilon_(0)`B. `(phi_(2)-phi_(1))epsilon_(0)`C. `(phi_(1)+phi_(2))//epsilon_(0)`D. `(phi_(2)-phi_(1))//epsilon_(0)` |
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Answer» Correct Answer - b `varphi_(n et) = (1)/(epsilon_(0))xxQ_(enc)implies Q_(enc)=(varphi_(2)-varphi_(1))epsilon_(0)` |
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| 48. |
An electric dipole is put in north-south direction in sphere filled with water. Which statement is correctA. Electric flux is coming towards sphereB. Electric flux is coming out of sphereC. Electric flux entering into sphere and leaving the sphere are sameD. Water does not permit electric flux to enter into sphere |
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Answer» Correct Answer - c In electric dipole, the flux coming out from positive charges is equal to he flux coming in at negative charge i.e., total charge on sphere `=0`. From Gauss law, total flux passing through the sphere `=0` |
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| 49. |
Assertion: A charge particle is free to move along electric field. It does not move along an electric line of force? Reason: Its initial position decides that it will move along the line of force or nor.A. if both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of the Assertion.C. if Assertion is true, but the Reason is false.D. If Assertion is false but Reason is true. |
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Answer» Correct Answer - d The charged particle may or may not move along an electric line of force. If the charged particle was initially at rest, it will move along an electric line of force. In case the charged particle has some initial velocity making certain angle with a line of force, then its resultant path will not be along the line of force. |
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| 50. |
What is the electric flux linked with closed surface? A. `10^(11)N-m^(2)//C`B. `10^(12)N-m^(2)//C`C. `10^(10)N-m^(2)//C`D. `8.86xx10^(13)N-m^(2)//C` |
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Answer» Correct Answer - b Electric flux, `phi= (q)/(epsilon_(0))` Where `q=` total charg e enclosed by closed surface `phi= (1.25+7+1-0.4)/(epsilon_(0))` `=(8.85C)/(8.85xx10^(-12)C^(2)N^(-1)m^(-2))` `10^(12)N-m^(2)//C` |
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