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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. |
If the potential of a capacitor having capacity of `6 muF` is increased from 10 V to 20 V,then increase in its energy will beA. `12 xx 10^(-6) J`B. `9 xx 10^(-4) J`C. `4 xx 10^(-6) J`D. `4 xx 10^(-9) J` |
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Answer» Correct Answer - B The increase in energy of the capacitor `DeltaU = (1)/(2) C(V_(2)^(2) - V_(1)^(2)) = (1)/(2) (6 xx 10^(-6)) (20^(2) - 10^(2))` `= 3 xx 10^(-6) xx 300 = 9 xx 10^(-4) J` |
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| 2. |
The electric potential V at any point x,y,z (all in metre) in space is given by `V=4x^2` volt. The electric field at the point `(1m, 0, 2m)` is ……………`V/m`.A. 8 along negative X-axisB. 8 along positive X-axisC. 16 along negative X-axisD. 16 along positive Z-axis |
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Answer» Correct Answer - A |
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| 3. |
Electric potential is given by `V = 6x - 8xy^(2) - 8y + 6yz - 4z^(2)` Then electric force acting on `2C` point charge placed on origin will beA. 2 NB. 6 NC. 8 ND. 20 N |
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Answer» Correct Answer - D |
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| 4. |
Variation of electrostatic potential along the x - direction is shown in (Fig. 3.142). The correct statement about electric field is .A. x - component at point `B` is maximumB. x - component at point `A` is toward positive x - axisC. x - component at point `C` is along negative x - axisD. x - component at point `C` is along positive x - axis |
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Answer» Correct Answer - d We know, `E = -(d V)/(d x)`. At `B, dV//d x = 0`, hence `E_x = 0` At `A, d V//d x` is positive, hence `E_x` is negative. At `C, d V//dx` is negative, hence `E_x` is positive. |
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| 5. |
The potentaial function of an electrostatic field is given by `V = 2 x^2`. Determine the electric field strength at the point `(2 m, 0, 3 m)`.A. `vec E = 4 hat i(NC^-1)`B. `vec E = -4 hat i(NC^-1)`C. `vec E = 8 hat i(NC^-1)`D. `vec E = -8 hat i(NC^-1)` |
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Answer» Correct Answer - d `E _x = -(d V)/(d x) = -4 x = -4 xx 2 = -8` `E_y = 0, E_z = 0` Hence. `vec E = -8 hat i NC^-1`. |
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| 6. |
A ball of mass `1g` and charge `10^(-8) C` moves from a point `A`. Where potential is `600` volt to the point `B` where potential is zero. Velocity of the ball at the point `B` is `20 cm//s`. The velocity of the ball at the point `A` will beA. `22.8 cm//s`B. `228cm//s`C. `16.8 cm//s`D. `168 cm//s` |
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Answer» Correct Answer - A By using `(1)/(2) m(v_(1)^(2) - v_(2)^(2)) = QV` `rArr(1)/(2) xx 10^(-3) {v_(1)^(2) - (0.2)^(2)} = 10^(-8) (600 - 0)` `rArr v_(1) = 22.8 cm//s` |
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| 7. |
Find the ratio of electric work done in bringing a charge `q` from `A` to `B(W_(AB))` and that from `B` to `C(W_(BC))` in a sphere of charge `Q` distributed uniformly throughout its volume. A. `1`B. `1.5`C. `0.75`D. None of these |
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Answer» Correct Answer - A Work done by electric field `W_(AB) = q(V_(A) - V_(B))` `= q[(3Q)/(8 pi epsilon_(0)R)-(Q)/(4pi epsilon_(0)R)]` `= (Qq)/(8pi epsilon_(0)R)` `W_(BC) = q(V_(B) - V_(C))` `= q[(Q)/(4 pi epsilon_(0)R)-(Q)/(8pi epsilon_(0)R)]` `= (Qq)/(8pi epsilon_(0)R)` `rArr (W_(AB))/(W_(BC)) = 1` |
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| 8. |
The work done in bringing a `20` coulomb charge from point `A` to point `B` for disatnce `0.2 m` is `2 J`. The potential difference between the two points will be (in volt)A. `0.2`B. `8`C. `0.1`D. `0.4` |
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Answer» Correct Answer - C By using `W = Q.DeltaV rArr DeltaV = (2)/(20) = 0.1` volt |
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| 9. |
Assertion: If the distance between parallel plates of a capacitor is halved and dielectric constant is made three times, then the capacitor becomes `6` times. Reason: Capacity of the capacitor does not depend upon the nature of the meterial.A. If both assertion and reason are true and reson is the 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 the assertion is true but the reason are false.D. If both the assertion and reason are false. |
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Answer» Correct Answer - B By the formula capacitance of a cpacitor `C_(1) = epsilon_(0) xx (KA)/(d) prop (K)/(d)` Hence, `(C_(1))/(C_(2)) = (K_(1))/(d_(1)) xx (d_(2))/(K_(2)) = (K_(1))/(K_(2)) xx (d//2)/(3K) = (1)/(6)` or `C_(2) = 6C_(1)` Again for capacity of a capacitor `C = (Q)/(V)` Therefore, capacity of a capacitor does not depend upon the nature of the material of the capacitor. |
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| 10. |
A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is ismilarly charged to a potential difference 2V. The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the poistive terminal of one is connected to the negative terminal of the other. The final energy of the configuration isA. ZeroB. `(23CV^(2))/(6)`C. `(3CV^(2))/(2)`D. `(9CV^(2))/(2)` |
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Answer» Correct Answer - C Total charge `= (2C)(2V) + (C ) (-V) = 3CV` `:.` Common potential `= (3CV)/(3C) = V` `:.` Energy `= (1)/(2)(3C)(V)^(2) = (3)/(2)CV^(2)` |
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| 11. |
Assertion: The force with which one plate of a parallel plate capacitor is attracted towards the other plate is equal to square of surface density per `epsilon` per unit area. Reason: The electric field due to one charged plate of the capacitor at the location of the other is equal to surface density per `epsilon`.A. If both assertion and reason are true and reson 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 assertion and reason both are false. |
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Answer» Correct Answer - D The electric field due to one charged plate at the location of the other is `E = sigma//2 epsilon_(0)` and the force per unit area is `F = sigma E = sigma^(2)//2 epsilon_(0)`. |
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| 12. |
A particle `A` of mass `m` and charge `Q` moves directly towards a fixed particle `B`, which has charge `Q`. The speed of `A` is a `v` when it is far away from `B`. The minimum separtion between the particles is not proportional toA. `Q^(2)`B. `(1)/(v^(2))`C. `(1)/(v)`D. `(1)/(m)` |
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Answer» Correct Answer - C `(1)/(2)mv^(2) = (1)/(4pi epsilon_(0)) (Q^(2))/(r_(min))` or `r_(min.) = (2Q^(3))/((4pi epsilon_(0))mv^(2))` Clearly `r_(min) prop Q^(2) prop (1)/(m) prop(1)/(v^(2))` |
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| 13. |
A proton moves from a large distance with a speed u m/s directly towards a free proton originally at rest. Find the distance of closest of closest approach for the two protons in terms of mass of proton m and its charge e. |
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Answer» Since the particle rest is free to move, when one particle approaches to other, due to electrostatic replusion, the other particle will also start moving. So the velocity of the first particle will decrease while that of the other particle will increase, and at the closest approach both will move with the same velocity. So, if v is the common velocity of each particle at the closest approach, as no external forces are acting on them (the system of particles), the linear momentum of the system of particles will be conserved. By conservation of momentum, we get `m u = mv + mv`, `i.e., v = (1)/(2) u` And by conservation of energy, we have `(1)/(2) m u^2 = (1)/(2) m v^2 + (1)/(2) m V^2 + (1)/(4 pi epsilon _0) (e^2)/( r)` So, `r = (e^2)/(pi epsilon_0 m u^2)` `(as v = u//2)`. |
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| 14. |
An electron is fired directly towards the center of a large metal plate that has excess negative charge with surface charge density `=2.0xx10^(-6) C//m^(2)`. If the initial kinetic energy of electron is 100 eV and if it is to stop due to repulsion just as it reaches the plate, how far from the plate must it be fired ?A. 0.44 mmB. 0.20 mmC. 1 mmD. 0.30 mm |
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Answer» Correct Answer - a Electric field due to large metal plate is `sigma//epsilon_0`. Since the electron has energy of `100 e V`, so it can cross a potential difference of `100 V` only. So `100 = (sigma)/epsilon_0 d` (solve to get d.). |
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| 15. |
The electric potential varies in space according to the relation `V = 3 x + 4 y`. A particle of mass `0.1 kg` starts from rest from point `(2,3.2)` under the influence of this field. The charge on the particle is `+1 mu C`. Assume `V` and (x,y) are in `S I` units. The component of electric field in the y - direction `(E_y)` is.A. `3 Vm^-1`B. `-4 Vm^-1`C. `5 Vm^-1`D. `8 Vm^-1` |
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Answer» Correct Answer - B `E_x = - (delta V)/(delta x) = -3 Vm^-1, E_y = (delta V)/(delta y) = -4 Vm^-1` `a_x = (q E_x)/(m) = (1 xx 10^-6 xx 3)/(0.1) = -3 xx 10^-5 ms^-2` `a_y = (q E_y)/(m) = (1 xx 10^-6 xx 4)/(0.1) = -4 xx 10^-5 ms^-2` Time taken to cross the x - axis Using `s = ut + (1)/(2) at^2` we get `3.2 = (1)/(2) xx 4 xx 10^-5 xx t^2` `t = 400 s` `v_x = a_x t = -3 xx 10^-5 xx 400` =`12 xx 10^-3 ms^-1` `v_y = a_y t = -4 xx 10^-5 xx 400` =`16 xx 10^-3 ms^-1` `v = sqrt(v_x^2 + v_y^2) = 20 xx 10^-3 ms^-1`. |
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| 16. |
The electric potential varies in space according to the relation `V = 3 x + 4 y`. A particle of mass `0.1 kg` starts from rest from point `(2,3.2)` under the influence of this field. The charge on the particle is `+1 mu C`. Assume `V` and (x,y) are in `S I` units. The component of electric field in the x - direction `(E_x)` is.A. `-3 Vm^-1`B. `4 Vm^-1`C. `5 Vm^-1`D. `8 Vm^-1` |
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Answer» Correct Answer - A `E_x = - (delta V)/(delta x) = -3 Vm^-1, E_y = (delta V)/(delta y) = -4 Vm^-1` `a_x = (q E_x)/(m) = (1 xx 10^-6 xx 3)/(0.1) = -3 xx 10^-5 ms^-2` `a_y = (q E_y)/(m) = (1 xx 10^-6 xx 4)/(0.1) = -4 xx 10^-5 ms^-2` Time taken to cross the x - axis Using `s = ut + (1)/(2) at^2` we get `3.2 = (1)/(2) xx 4 xx 10^-5 xx t^2` `t = 400 s` `v_x = a_x t = -3 xx 10^-5 xx 400` =`12 xx 10^-3 ms^-1` `v_y = a_y t = -4 xx 10^-5 xx 400` =`16 xx 10^-3 ms^-1` `v = sqrt(v_x^2 + v_y^2) = 20 xx 10^-3 ms^-1`. |
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| 17. |
The electric potential varies in space according to the relation `V = 3 x + 4 y`. A particle of mass `0.1 kg` starts from rest from point `(2,3.2)` under the influence of this field. The charge on the particle is `+1 mu C`. Assume `V` and (x,y) are in `S I` units. The velocity of the particle when ot crosses the x - axis is.A. `20 xx 10^-3 ms^-1`B. `40 xx 10^-3 ms^-1`C. `30 xx 10^-3 ms^-1`D. `50 xx 10^-3 ms^-1` |
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Answer» Correct Answer - A `E_x = - (delta V)/(delta x) = -3 Vm^-1, E_y = (delta V)/(delta y) = -4 Vm^-1` `a_x = (q E_x)/(m) = (1 xx 10^-6 xx 3)/(0.1) = -3 xx 10^-5 ms^-2` `a_y = (q E_y)/(m) = (1 xx 10^-6 xx 4)/(0.1) = -4 xx 10^-5 ms^-2` Time taken to cross the x - axis Using `s = ut + (1)/(2) at^2` we get `3.2 = (1)/(2) xx 4 xx 10^-5 xx t^2` `t = 400 s` `v_x = a_x t = -3 xx 10^-5 xx 400` =`12 xx 10^-3 ms^-1` `v_y = a_y t = -4 xx 10^-5 xx 400` =`16 xx 10^-3 ms^-1` `v = sqrt(v_x^2 + v_y^2) = 20 xx 10^-3 ms^-1`. |
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| 18. |
The potential at a point, due to a positive charge of `100 mu C` at a distance of `9 m`, isA. `10^(4) V`B. `10^(5) V`C. `10^(6) V`D. `10^(7) V` |
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Answer» Correct Answer - B By using `V = 9 xx 10^(6) xx (Q)/(r ) = 9 xx 10^(9) xx (100 xx 10^(-6))/(9) = 10^(5)V` |
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| 19. |
As shown in the figure, charges `+q and -q` are placed at the vertices `B` and `C` of an isoscles triangle. The potential at the vertex `A` is A. `(1)/(4pi epsilon_(0)).(2q)/(sqrt(a^(2) + b^(2))`B. ZeroC. `(1)/(4pi epsilon_(0)).(q)/(sqrt(a^(2) + b^(2))`D. `(1)/(4pi epsilon_(0)).((-q))/(sqrt(a^(2) + b^(2))` |
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Answer» Correct Answer - B Potential at `A =` Potential due to `(+q)` charge `+`Potential due to `(-q)` charge `= (1)/(4pi epsilon_(0)). (q)/(sqrt(a^(2) + b^(2))) + (1)/(4pi epsilon_(0)) ((-q))/(sqrt(a^(2) + b^(2))) = 0` |
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| 20. |
The electric field in a region is given by `vecE = (A/x^3) vecI.` Write a suitable SI unit for A. Write an experssion for. the potential in the region assuming the potential at. infinity to be zero. |
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Answer» As `E = A // x^3`, potential in the region is `V = - underset(oo) overset(x)(int) vec(E) . vecd x = - underset(oo) overset(x)int (A/(x^3) i). (dx i)` =`-A underset(oo) overset(x)int x^-3 d x = -A [ (x^-2)/(-2)]_oo^x = A/(2 x^2)`. |
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| 21. |
`A, B` and `C` are three points in a unifrom electric field. The electric potential is A. Maximum at `A`B. Maximum at `B`C. Maximum at `C`D. Same at all three points `A, B` and `C` |
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Answer» Correct Answer - B The electric field will be maximum at `B`, as the direction of electric field is along decreasing potential `V_(B) gt V_(C) gt V_(A)`. |
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| 22. |
Figure shown three points. `X, Y` and `Z` forming an equilaternal triangle of side `s` in an unifrom electric field of strength `E`. A unit positive test charge is moved from `X` to `Y`, from `Y` to `Z`, and from `Z` back to `X`. Which one of the following correctly gives the work done against electrical forces in moving the charge along the various parts o fthis path ?A. `{:("X to Y"," Y to Z"," Z to X",),(+Es,+"Es "cos 60^(@).,-"Es "cos 60^(@).,):}`B. `{:("X to Y"," Y to Z"," Z to X",),(0,-"Es "cos 60^(@).,+"Es "cos 60^(@).,):}`C. `{:("X to Y"," Y to Z"," Z to X",),(+Es,+"Es "cos 60^(@).,-"Es "cos 60^(@).,):}`D. `{:("X to Y"," Y to Z"," Z to X",),(0,-"Es "sin 60^(@).,+"Es "sin 60^(@).,):}` |
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Answer» Correct Answer - D From `E = (dV)/(dx)`, work done in moving a unit positive test charge in an unifrom electric field of strength `E` is given by work done `V = int dV = - int Edx` `= - E int dx = - Ed` where `d` is the distancetravelled parallel to the direction of `E`. From `X` to `Y, d = 0`. Therefore work done `= 0` From `Y` to `Z, d = +s sin 60^(@)` Therefore work done is `Es sin 60^(@)` From `Z` to `X, d = - x sin 60^(@)` Therefore work done is `- Es sin 60^(@)` |
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| 23. |
Assertion: Work done in moving a charge between any two points in a unifrom electric field is independent of the path followed by the charge, between these points. Reason: Electrostatic forces are non-conservative.A. If both assertion and reason are true and reson is the 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 the assertion is true but the reason are false.D. If both the assertion and reason are false. |
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Answer» Correct Answer - C Electroststic forces are conservative, so work done in moving a charge in unform electric field does not depend on path followed. Hence, assertion is true but reason is false. |
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| 24. |
The electric potential at a point in free space due to a charge `Q` coulomb is `Q xx 10^(11)` volts. The electric field at that point isA. `4 pi epsilon_(0)Q xx 10^(22) V//m`B. `12 pi epsilon_(0)Q xx 10^(20) V//m`C. `4 pi epsilon_(0)Q xx 10^(20) V//m`D. `12 pi epsilon_(0)Q xx 10^(22) V//m` |
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Answer» Correct Answer - A The electric potential due to charge `Q` is `V = (1)/(4pi epsilon_(0)). (Q)/(r )` (i) where `r` is the distance of observation point from the charge. At the same point, electric field is `E = (1)/(4pi epsilon_(0)).(Q)/(r )` (ii) Combining Eqs. (i) and (ii), we have `E = (4pi epsilon_(0)V^(2))/(Q) = (4pi epsilon_(0) xx (Q xx 10^(11))^(2))/(Q)` `= 4pi epsilon_(0)Q xx 10^(22) V//m` |
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| 25. |
Three is an electric field E in the x - direction. If the work done by the electric field in moving a charge of `0.2 C` through a distance of `2 m` along a line making an angle `60^(@)` with the x- axis is `4 J`, then what is the value of E ?A. `sqrt(3)N//C`B. `4 N//C`C. `5 N//C`D. None of these |
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Answer» Correct Answer - D |
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| 26. |
Three is an electric field E in the x - direction. If the work done by the electric field in moving a charge of `0.2 C` through a distance of `2 m` along a line making an angle `60^(@)` with the x- axis is `4 J`, then what is the value of E ?A. `sqrt(3) NC^-1`B. `NC^-1`C. `NC^-1`D. `20 NC^-1` |
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Answer» Correct Answer - d `F = qE` `W = q E xx 2 cos 60^(@)` or `4 = 0.2 E xx 2 xx (1)/(2)` or `E = 20 NC^-1`. |
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| 27. |
Mark the correct statement :.A. An electron and a proton when released from rest in uniform electric field experience the same force and the same acceleration.B. Two equipotential surfaces may intersect.C. A solid conducting sphere holds more charge than a hollow conducting sphere of the same radius.D. No work is done in taking a positive charge from one point to another inside a negatively charged metallic sphere. |
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Answer» Correct Answer - d Option (a) is wrong because force will be the same, but acceleration will be different beacause masses of electrons and protons are different. Option (b) is wrong because at a point there can be only one potential. Option ( c) is wrong because charge always lies on the outer surface of a conductor. Option (d) is correct because the whole conductor will be an equopotential body. |
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| 28. |
Mark the correct statement :.A. If `E` is zero at a certain point, then `V` should be zero at that point.B. If `E` is not zero at a certain point, then `V` should not be zero at that point.C. If `V` is zero at a certain point, then `E` should be zero at that point.D. If `V` is zero at a certain point, then `E` may or may not be zero. |
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Answer» Correct Answer - d `V` is a scalar quantity, and `E` is a vector quantity. |
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| 29. |
Consider two conducting spheres of radii `R_(1) and R_(2)` with `R_(1) gt R_(2)`. If the two are at same potential, the larger sphere has more charge than the smaller sphere, then charge density of the smaller sphere is more than that of the larger one. |
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Answer» Correct Answer - 1 |
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| 30. |
Two spherical conductors of radii `R_1 and R_2` are separated by a distance much larger than the radius of eighter sphere. The spheres are connected by a conducting wire as shown in (Fig. 3.128). If the charges on the spheres in equilibrium are `q_1 and q_2`, respectively, what is the ratio of the field strength at the surfaces of the spheres ? .A. `R_2//R_1`B. `R_2^2//R_1^2`C. `R_1//R_2`D. `R_1^2//R_2^2` |
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Answer» Correct Answer - a Their potential will be same, i.e., `V_1 = V_2` or `(kq_1)/R_(1) = (kq_2)/R_(2)` or `q_(1)/q_(2) = R_(1)/R_(1)` or `(E_1)/(E_2) = (kq_1//R_1^2)/(kq_2//R_2^2) = q_(1)/q_(2)(R_(2)/R_(1))^2` =`R_(1)/R_(1) (R_(2)/R_(1))^2 = R_(2)/R_(1)`. |
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| 31. |
Two metal spheres `("radii" r_1, r_2 "with" r_1 lt r_2)` are very far apart but are connected by a thin wire. If their combined charges is Q, then what is their common potential ?A. `k Q//(r_1 + r_2)`B. `kQ//(r_1 - r_2)`C. `-k Q//(r_1 + r_2)`D. `-kQ//r_1 r_2` |
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Answer» Correct Answer - a `V_C = V_1 = V_2` `(kq_1)/r_(1) = (kq_2)/r_(1)` `q_(1)/q_(2) = r_(1)/r_(1)` …(i) `q_1 + q_2 = Q` …(ii) From Eqs. (i) and (ii), we get `q_1 = (Q r)/(r_1 + r_2)` Putting `V_(1)`, we get `V_(C )=(kQ)/(r_(1)-r_(2))` |
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| 32. |
An electron having charge `e` and mass `m` starts from the lower plate of two metallic plates separated by a distance `d`. If the potential difference between the plates is `V`, the time taken by the electron to reach the upper plate is given by .A. `sqrt((2 m d^2)/(e V))`B. `sqrt(( m d^2)/(e V))`C. `sqrt(( m d^2)/(2 e V))`D. `(2 m d^2)/( e V)` |
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Answer» Correct Answer - a `E = V/d, F = eE = e V// d` `a = (F)/(m) = (e V)/(m d)` `d = (1)/(2) at^2` or `t = sqrt((2d)/A)` or `t = sqrt((2 d m d)/(e V)) = sqrt((2 md^2)/(e V))`. |
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| 33. |
As shown in (Fig. 3.90), two large parallel vertical conducting plates separated by distance D are charge so that their potential are `+V_0` and `- V_0`. A small conducting ball of mass m and radius r `("where" R lt lt d)` is hung midway between the plates. The thread of length L supporting the ball is a conducting wire connected to ground, so the potential of the ball is fixed at `V = 0`. The ball hangs straight down in stable equilibrium when `V_0` is sufficiently small. Show that the equilibrium of the ball is unstable if `V_0` exceeds the critical value `[k_e d^2 mg//(4 R L)]^(1//2)`. |
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Answer» The plates create uniform electric field to the right in the picture, with magnitude `[V_0 - (-V_0)]// d = 2V_0 //d`. Assume the ball swings a small distance x to the right. It moves to a place where the voltage created by the plates is lower by `- E x = -(2 V_0)/d x` Its ground connection maintains it at `V = 0` by allowing charge q to flow from ground onto the ball, where `- (2V_0 x)/d + (k_e q)/R = 0` `q = (2 V_0 x R)/(k_e d)` Then the ball feels electric force `F = q E = (4 V_0^2 x R)/(k_e d^2)` to the right. for equilibrium, this must be balanced by the horizontal component of string tension according to `T cos theta = mg` `T sin theta = (4 V_0^2 x R)/(k_e d^2) tan theta = (4 V_0^2 x R)/(k_e d^2 mg) = (x)/(L)` for small x Then `V_0 = ((k_e d^2 m g)/(4 R L))^(1//2)` If `V_0` is less than this value, the only equilibrium position of the ball is hanging straight down. If `V_0` exceeds this value, the ball will saving over to one plate or the other. |
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| 34. |
The energy of a charged capacitor is given by the expression `(q =` charge on the conductor and `C =` its capacityA. `(q^(2))/(2C)`B. `(q^(2))/(C )`C. `2qC`D. `(q)/(2C^(2))` |
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Answer» Correct Answer - A `q = CV` and `U = (1)/(2)CV^(2) = (q^(2))/(2C)` |
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| 35. |
Asseration: A point charge `q` is placed at centre of spherical cavity inside a spherical conductor as shown. Another point charge `Q` is placed outside the conductor as shown. Now as the point charge `Q` is pushed away from conductor, the potential difference `(V_(A) - V_(B))` between two points `A` and `B` within the cavity of sphere remains constant. Reason: The electric field due to charges on outer surface of conductor and outside the conductor is zero at all points inside teh conductor.A. If both assertion and reason are true and reson is the 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 the assertion is true but the reason are false.D. If both the assertion and reason are false. |
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Answer» Correct Answer - A The electric field inside the cavity depends only on point charge `q`. Hence `V_(A) - V_(B)` remains constant even if point charge `Q` is shifted. Here reason is correct explanation of assertion. |
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| 36. |
If a charged spherical conductor of radius `10 cm` has potential `V` at a point distant `5 cm` from its centre, then the potential at a point distant `15 cm` from the centre will beA. `(1)/(3) V`B. `(2)/(3) V`C. `(3)/(2) V`D. `3 V` |
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Answer» Correct Answer - B Potential inside the sphere will be same as that on its surface i.e, `V = V_(surface) = (q)/(10)` stat volt `V_(out) = (q)/(15)` stat volt `:. (V_(out))/(V) = (2)/(3) rArr V_(out) = (2)/(3)V` |
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| 37. |
Inside a hollow charged spherical conductor, the potentialA. is constantB. varies directly as the distance from the centreC. varies inversely as the disatnce from the centreD. varies inversely as the square of the distance from the centres |
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Answer» Correct Answer - A Inside the hollow sphere at any point the potential is constant. |
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| 38. |
A hollow metal sphere of radius `5 cm` is charged so that the potential on its surface is `10 V`. The potential at the centre of the sphere isA. `0 V`B. `10 V`C. same as at point `5 cm` away from the surfaceD. same as at point `25 cm` away from the surface |
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Answer» Correct Answer - B Since potential inside the hollow sphere is same as that on the surface. |
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| 39. |
A hollow conducting sphere of radius `R` has a charge `(+Q)` on its surface. What is the electric potential within the sphere at a distance `r = (R )/(3)` from its centre ?A. ZeroB. `(1)/(4pi epsilon_(0))(Q)/(r )`C. `(1)/(4pi epsilon_(0))(Q)/(R )`D. `(1)/(4pi epsilon_(0))(Q)/(r^(2)) ` |
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Answer» Correct Answer - C Inside a conducting body, potential is same everywhere and equals to the potential of its surface |
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| 40. |
The potential at a distance `R//2` from the centre of a conducting sphere of radius `R` will beA. `0`B. `(Q)/(8pi epsilon_(0)R)`C. `(Q)/(4pi epsilon_(0)R)`D. `(Q)/(2pi epsilon_(0)R)` |
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Answer» Correct Answer - C Potential inside the conducting sphere is same as that of surface i.e., `(Q)/(4pi epsilon_(0)R)` |
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| 41. |
The respective radii of the two spheres of a spherical condenser are `12 cm` and `9 cm`. The dielectirc constant of the medium between them is `6`. The capacity of the condenser will beA. `240 pF`B. `240 mF`C. `240 F`D. None of the above |
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Answer» Correct Answer - A `C = 4pi epsilon_(0)K [(ab)/(b - a)] = (1)/(9 xx 10^(9)).6 [(12 xx 9 xx 10^(-4))/(3 xx 10^(-2))]` `= 24 xx 10^(-11) = 240 pF` |
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| 42. |
The electric potential inside a conducting sphereA. increases from cente to surfaceB. decreases from cente to surfaceC. remians constant from centre to surfaceD. is zero at every point inside |
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Answer» Correct Answer - C Electric potential inside a conductor is constant and it is equal to that on the surface of conductor. |
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| 43. |
A conducting sphere of radius `R` is given a charge `Q`. The electric potential and the electric field at the centre of the sphere respectively areA. Zero and `(Q)/(4pi epsilon_(0)R^(2))`B. `(Q)/(4pi epsilon_(0)R^(2))` and zeroC. `(Q)/(4pi epsilon_(0)R)` and `(Q)/(4pi epsilon_(0)R^(2))`D. Both are zero |
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Answer» Correct Answer - B Electric field at centre of conducting sphere is zero And electric potential is constant inside the sphere and equal to electric potential at the surafce of conducting sphere is equal to `V = (Q)/(4pi in_(0)R)` |
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| 44. |
Assertion: Electrons move away from a low potential to light potential region. Reason: Because electrons have negative chargesA. If both assertion and reason are true and reson is the 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 the assertion is true but the reason are false.D. If both the assertion and reason are false. |
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Answer» Correct Answer - A Electron has negative charge, in electric field negative charge moves from lower potential to higher potential. |
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| 45. |
Potential at a point x-distance from the centre inside the conducting sphere of radius `R` and charged with charge `Q` isA. `(Q)/(R )`B. `(Q)/(x)`C. `(Q)/(x^(2))`D. `xQ` |
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Answer» Correct Answer - A Potential at any point inside the charged spherical conductor equals to the potential at the surafce of the conductor i.e. `Q//R` |
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| 46. |
Between the plates of a parallel plate condenser, a plate of thickness `t_(1)` and dielectric constant `k_(1)` is placed. In the rest of the space, there is another plate of thickness `t_(2)` and dielectric constant `k_(2)`. The potential difference across the condenser will beA. `(Q)/(A epsilon_(0))((t_(1))/(k_(1))+(t_(2))/(k_(2)))`B. `(epsilon_(0)Q)/(A)((t_(1))/(k_(1))+(t_(2))/(k_(2)))`C. `(Q)/(A epsilon_(0))((k_(1))/(t_(1))+(k_(2))/(t_(2)))`D. `(epsilon_(0)Q)/(A) (k_(1)t_(1) + k_(2)t_(2))` |
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Answer» Correct Answer - A Potential difference across the condenser `V = V_(1) + V_(2) = E_(1)t_(1) + E_(2)t_(2) = (sigma)/(K_(1)epsilon_(0)) + (sigma)/(K_(2)epsilon_(0)) t_(2)` `rArr V = (sigma)/(epsilon_(0)) ((t_(1))/(K_(1))+(t_(2))/(K_(2))) = (Q)/(A epsilon_(0)) ((t_(1))/(K_(1))+(t_(2))/(K_(2)))` |
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| 47. |
The area of the plates of a parallel plate condenser is `A` and the distance between the plates is `10 mm`. There are two dielectric sheets in it, one of dieectirc constant `10` and thickness `6 mm` an dthe other of dielectric constant `5` and thickness `4 mm`. the capacity of the condenser isA. `(12)/(35)epsilon_(0)A`B. `(2)/(3)epsilon_(0)A`C. `(5000)/(7)epsilon_(0)A`D. `1500 epsilon_(0)A` |
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Answer» Correct Answer - C `C = (epsilon_(0)A)/(((t_(1))/(k_(1))+(t_(2))/(k_(2)))) = (epsilon_(0)A)/((6 xx 10^(-3))/(10) + (4 xx 10^(-3))/(5)) = (5000)/(7) epsilon_(0) A` |
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| 48. |
In a parallel plate condenser, the radius of each circula plate is `12 cm` ans the distance between the plates is `5mm`. There is a glass slab of `3 mm` thick and of radius `12 cm` with dielectric constant `6` between its plates. The capacity of the condenser will beA. `144 xx 10^(-9) F`B. `40pF`C. `160 pF`D. `1.44 muF` |
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Answer» Correct Answer - C `C = (epsilon_(0)A)/(d-t + (t)/(K)) = (1)/(4 pi xx 9 xx 10^(9)).(pi(0.12)^(2))/((2+(1)/(2))10^(-3))` `= (2 xx 144 xx 10^(-10))/(36 xx 5) = 160 pF` |
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| 49. |
As shown in the figure, a very thin sheet of aluminium is placed in between the plates of the condenser. Then the capcaity A. will increasesB. will decreaseC. remains unchangedD. may increase or decrease |
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Answer» Correct Answer - C Since aluminium is a metal, therefore field inside this will be zero. Hence it would not affect the field in between the two plates, so capacity `= (q)/(V) = (q)/(Ed)` remains unchanged. |
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| 50. |
Electric potential at any point is `V = - 5 x + 3y + sqrt(15) z`, then the magnitude of the electric field isA. `3 sqrt(2)`B. `4 sqrt(2)`C. `5sqrt(2)`D. `7` |
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Answer» Correct Answer - D `E_(x) = - (dV)/(dx) = - (-5) = 5, E_(y) = - (dV)/(dy) = - 3` and `E_(z) = - (dV)/(dz) = - sqrt(15)` `E_("net") = sqrt(E_(x)^(2) + E_(y)^(2) + E_(z)^(2)) =sqrt((5)^(2) + (-3)^(2) + (-sqrt(15)^(2))) =7` |
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