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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.
| 151. |
Two point charges `100 mu C` and `5 mu C` are placed at points `A and B` respectively with `AB = 40 cm`. The work done by external froce in displacing the charge `5 mu C` from B to C, where `BC = 30 cm`, angle `ABC = (pi)/(2)` and `(1)/(4pi epsilon_(0)) = 9 xx 10^(9)Nm^(2)//C^(2)`A. `9 J`B. `(81)/(20) J`C. `(9)/(25) J`D. `-(9)/(4) J` |
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Answer» Correct Answer - D |
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| 152. |
Work done in moving a positive charge on an equipotential surafce isA. finite, Positive but not zeroB. finite, negative but not zeroC. zeroD. infinite |
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Answer» Correct Answer - C Work done `= (DeltaV) Q` For an equipotential surface. |
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| 153. |
If a unit positive charge is taken from one point to another over an equipotential surface ,thenA. work is done on the chageB. work is done by the chargeC. work done is constantD. no work is done |
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Answer» Correct Answer - D On the equipotential surafce, electric field is normal to the charged surface (where potential exists) so that no work will be done. |
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| 154. |
Figure shows some equipotential lines distributed in space. A charged object is moved from point `A` to point `5`. A. The work done in Fig. (i) is the greatestB. The work done in Fig. (ii) is leastC. The work done is the same in Fig. (i), Fig. (ii) and Fig. (iii)D. The work done in Fig. (iii) is greater than Fig. (ii) but equla to than in |
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Answer» Correct Answer - C The work done by a electrostatic force is given by `W_(12) = q(V_(2) - V_(1))`. Here initial and final potentials are same in all three case ans same charge is moved, so work done is same in all three cases. |
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| 155. |
Figure shown two equipotential lies `x, y` plane for an electric field. The scales are market. The `x`-component `E_(x)` and y-component `E_(y)` of the field in the space between these equipotential lines are respectively A. `+100 V//m, -200 V//m`B. `-100 V//m, +200 V//m`C. `+200 V//m, +100 V//m`D. `-200 V//m, -100 V//m` |
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Answer» Correct Answer - B `E_(x) = - (-2V)/(-2 xx 10^(-2) m) = - 100 Vm^(-1)` `E_(y) = - (-2V)/(1 xx 10^(-2)m) = 200 Vm^(-1)` |
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| 156. |
The work done in carrying a charge of `5 mu C` form a point `A` to a point `B` in an electric field is `10mJ`. The potential difference `(V_(B) - V_(A))` is thenA. `+2 kV`B. `-2 kV`C. `+200 V`D. `-200 V` |
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Answer» Correct Answer - A Work done `W = Q(V_(B) - V_(A))` `rArr (V_(B) - V_(A)) = (W)/(Q) = (10 xx 10^(-3))/(5 xx 10^(-6))J//C = 2kV` |
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| 157. |
There are two equipotential surafce as shown in figure. The distance between them is `r`. The charged of `-q` coulomb is taken from the surface A to B, the resultant work done will be A. `W = (1)/(4pi epsilon_(0)) (q)/(r )`B. `W = (1)/(4pi epsilon_(0)) (q)/(r^(2))`C. `W = -(1)/(4pi epsilon_(0)) (q)/(r^(2))`D. `W =` zero |
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Answer» Correct Answer - D The work done is given by `= q (V_(2) - V_(1)) = 0` |
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| 158. |
`A 40 muF` capacitor in a defibrillator is charged to `3000 V`. The energy stored in the capacitor is sent through the patient during a pulse of duration `2 ms`. The power delivered to the patient isA. `45 kW`B. `180 kW`C. `90 kW`D. `360 kW` |
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Answer» Correct Answer - C A capacitor is a device that stores energy in the electric field between a pair of conductors on which equal but opposite electric charges have been placed. The energy stored in a capacitor `= (1)/(2)CV^(2)` Given, `C = 40muF = 40 xx 10^(-6)F, V = 3000 V` `:. E = (1)/(2) xx 40 xx 10^(-6) xx (3000)^(2) = 180 J` Also `1W = 1 J//s :. 2ms = 2 xx 10^(-3) s` Hence, power `= (180 J)/(2 xx 10^(-3) s)` `= 90 xx 10^(3) W = 90 kW` |
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| 159. |
When a charge of `3` coulombs is placed in a uniform electric field, it experiences a force of `3000` netwon. Within this field, potential difference between two points separated by a distance of `1 cm` isA. `10` voltsB. `90` voltsC. `1000` voltsD. `3000` volts |
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Answer» Correct Answer - A `V = Ed = (3000)/(3) xx 10^(-2) = 10V` |
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| 160. |
A particle `A` has chrage `+q` and a particle `B` has charge `+4q` with each of them having the same mass `m`. When allowed to fall from rest through the same electric potential difference, the ratio of their speed `(v_(A))/(v_(B))` will becomeA. `1:2`B. `2:1`C. `1:4`D. `4:1` |
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Answer» Correct Answer - A `(1)/(2)mv_(A)^(2) = qV` `(1)/(2)mv_(B)^(2) = 4qV` Dividing, `(v_(A)^(2))/(v_(B)^(2)) = (qV)/(4qV)` or `(v_(A)^(2))/(v_(B)^(2)) = (1)/(4)` or `(v_(A))/(v_(B)) = (1)/(2)` |
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| 161. |
A particle `A` has chrage `+q` and a particle `B` has charge `+4q` with each of them having the same mass `m`. When allowed to fall from rest through the same electric potential difference, the ratio of their speed `(v_(A))/(v_(B))` will becomeA. `2 : 1`B. `1 : 2`C. `1 : 4`D. `4 : 1` |
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Answer» Correct Answer - B Using `v = sqrt((2QV)/(m))` `rArr v prop sqrt(Q) rArr (v_(A))/(v_(B)) = sqrt((Q_(A))/(Q_(B))) = sqrt((q)/(4q)) = (1)/(2)` |
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| 162. |
A condenser of capacitance `10 muF` has been charged to `100V`. It is now connected to another uncharged condenser in parallel. The common potential becomes `40 V`. The capacitance of another condenser isA. `15 muF`B. `5 muF`C. `10 muF`D. `16.6 muF` |
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Answer» Correct Answer - A By using `V = (C_(1)V_(1) + C_(2)V_(2))/(C_(1) + C_(2))` `rArr 40 = (10 xx 100 + C_(2) xx 0)/(10 + C_(2)) rArr C_(2) = 15 muF` |
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| 163. |
Two condenser of capacities `1 muF` and `2 muF` are connected in series and the system is charged to `10` volts. Then the `P.D.` on `1 muF` capacitor (in volts) will beA. `40`B. `60`C. `80`D. `120` |
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Answer» Correct Answer - C Charges developed are same so `C_(1)V_(1) = C_(2)V_(2) rArr (V_(1))/(V_(2)) = 2` `V_(1) + V_(2) = 120 rArr V_(1) = 80` volts |
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| 164. |
A condenser of capacity `50 mu F` is charged to `10` volts. Its energy is equal toA. `2.5 xx 10^(-3)J`B. `2.5 xx 10^(-4)J`C. `5 xx 10^(-2)J`D. `1.2 xx 10^(8)J` |
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Answer» Correct Answer - A `U = (1)/(2) CV^(2) = (1)/(2) xx 50 xx 10^(-6) xx (10)^(2) = 2.5 xx 10^(-3) J` |
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| 165. |
The distance between the plates of a parallel plate condenser is `4mm` and potential difference is `60` volts. If the distance between the plates is increased to `12mm`, thenA. The potential difference of the condenser will become `180` voltsB. The `P.D.` will become `20` voltsC. The `P.D.` will remain unchangedD. The charge on condenser will reduce to one third |
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Answer» Correct Answer - A For capacitor `(V_(1))/(V_(2)) = (d_(1))/(d_(2)) rArr V_(2) = (V_(1) xx d_(2))/(d_(1)) = (60 xx 12)/(4) = 180 V` |
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| 166. |
The capacity of a condenser is `4 xx 10^(6)` farad and its potential is `100` volts. The energy released on discharging it fully will beA. `0.02 J`B. `0.04 J`C. `0.025 J`D. `0.05 J` |
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Answer» Correct Answer - A `U = (1)/(2) CV^(2) = (1)/(2) xx 4 xx 10^(-6) xx (100)^(2) = 0.02 J` |
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| 167. |
Find out the following a. `V_A - V_B` b. `V_B - V_C` c. `V_C - V_A` d. `V_D - V_C` e. `V_A - V_D` f. Arrange the order of potential for points (A, B,C and D). . |
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Answer» (i) The potential difference between two points in electric field is `|Delta V_(AB)| = - vec E . vec(AB) = E d cos theta = E. d` =`E d = 20 xx 2 xx 10^(-2) = 0.4` So `V_A - V_B = 0.4 V` Negative because in the direction of electric field, potential always decreases. (ii) `|Delta V_(BC) | = E d = 20 xx 2 xx 10^(-2) = 0.4` d is the distance between the equipotential passing through B and C, so `V_B - V_C = 0.4 V` (iii) `|Delta V_(CA) | = E d = 20 xx 4 xx 10^(-2) = 0.8` So `V_C - V_A = -0.8 V` Negative because in the direction of electric field, potential always decreases. (iv) `|Delta V_(DC) | = E d = 20 xx 0 = 0` So `V_D - V_C = 0` Because the effective distance between D and C is zero. Also potential difference between C and D is zero as both the points lie on same equipotential. (v) `|Delta V_(A D)| = Ed = 20 xx 4 xx 10^(-2) = 0.8` Here d is the distance between equipotential pasing through A and D. So `V_A - V_D = 0.8 V` Because in the direction of electric field, potential always decreases. (vi) The order of potential is `V_A gt V_B gt V_C = V_D`. |
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| 168. |
An infinite number of concentric rings carry a charge `Q` each alternately positive and negative. Their radii are `1, 2, 4, 8..` meters in geometric progression as shown in the figure. The potential at the centre of the rings will be A. zeroB. `(Q)/(12pi epsilon_(0))`C. `(Q)/(8pi epsilon_(0))`D. `(Q)/(6pi epsilon_(0))` |
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Answer» Correct Answer - D Potetnial at centre : `V = (kQ)/(1) - (kQ)/(2) + (kQ)/(4)`…. `kQ[1-(1)/(2)+(1)/(4)…] = (1)/(4pi epsilon_(0)) Q ((1)/(1+1//2)) = (Q)/(4pi epsilon_(0))` |
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| 169. |
A parallel plate capacitor is charged and the charging battery is then disconnected. If the plates of the capacitor are moved farther apart by means of insulating handles:A. the charge on the capacitor increasesB. the voltage across the plates decreasesC. the capacitance increasesD. the electrostatic energy stored in the capacitor increases |
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Answer» Correct Answer - D When the battery is diconnected, the charge will remain same in any case. Capacitance of a parallel plate capacitor is given by `C = (epsilon_(0)A)/(d)` When `d` is increased, capacitance will decreases and because the charge remains the same, so according to `q = CV` the voltage will increase, Hence the electrostatics energy stored in the capacitor will increase. |
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| 170. |
Two charges `q_(1)` and `q_(2)` are placed `30 cm` apart, as shown in the figure. A third charge `q_(3)` is moved along the arc of a circle of radius `40 cm` from `C` to `D`. The change in the potential energy o fthe system is `(q_(3))/(4pi epsilon_(0))k`., where `k` is A. ` 8 q_(2)`B. `8 q_(1)`C. `6 q_(2)`D. `6 q_(1)` |
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Answer» Correct Answer - A |
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| 171. |
The value of electric potential at any point due to any electric dipole isA. `k. (vecp xx vecr)/(r^(2))`B. `k. (vecp xx vecr)/(r^(3))`C. `k. (vecp .vecr)/(r^(2))`D. `k. (vecp .vecr)/(r^(3))` |
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Answer» Correct Answer - D Potential due to dipole in general position is given by `V = (k.p cos theta)/(r^(2)) rArr V = (k.p cos theta)/(r^(3)) = (k.(vecp.vecr))/(r^(3))` |
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| 172. |
For the circuit shown in figure, which of the following statements is true ? A. With `S_(1)` closed, `V_(1) = 15 V, V_(2) = 20V`B. With `S_(3)` closed, `V_(1) = V_(2) = 25V`C. With `S_(1)` and `S_(2)` closed `V_(1) = V_(2) = 0`D. With `S_(1)` and `S_(3)` closed `V_(1) = 30 V` and `V_(2) = 20 V` |
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Answer» Correct Answer - D When `S_(1)` is closed `V_(1) = 30V, V_(2) = 20 V` When `S_(3)` is closed `V_(1) = 30 V, V_(2) = 20V` When `S_(1)` and `S_(2)` are closed `V_(1) = 30V, V_(2) = 20 V` When `S_(1)` and `S_(3)` are closed `V_(1) = 30V, V_(2) = 20V` When `S_(1), S_(2)` and `S_(3)` are closed `V_(1) + V_(2) = 0V, V_(1) = 0, V_(2) = 0V` |
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| 173. |
The electric field lines are closer together near object `A` than they are near object `B`. We can conclude that.A. the potential near `A` is greater than the potential near `B`.B. the potential near `A` is less than the potential near `B`.C. the potential near `A` is equal to the potential near `B`.D. nothing about the relative potentials near `A and B`. |
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Answer» Correct Answer - a Potential decreases in the direction of electric field. So it epends on whether the lines of forces are from A to B or from B to A. |
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| 174. |
We want to produce a proton with kinetic energy `4.3 xx 106-15 J`. Through what difference of potential should we accelerate the proton to obtain this kinetic energy assuming that it starts from rest and no other forces are present ? |
| Answer» `e V = 4.3 xx 10^(-15)` or `1.6 xx 10^(-19) V = 4.3 xx 10^(-15) V = 26875 V` | |
| 175. |
A positive point charge `+Q` is fixed in space .A negative point charge `-q` of mass `m` revolves around a fixed charge in elliptical orbits .The fixed charge `+Q `is at one focus of the ellipse.The only force acting on negative charge is the electrostatic force due to positive charge is the electrostatic force due to positive charge.Then which of the following statement is true A. Linear momentum of negative point charge is conservedB. Angular momentum of negative point charge about fixed positive charge is conservedC. Total kinetic energy of negative point charge is conservedD. The sum of electrosatatic potential energy and kinetic energy of system of both point charges is conserved. |
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Answer» Correct Answer - B::D |
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| 176. |
An electron moves from the positive to the negative terminal of a battery (9 V). How much potential energy did it gain or lose ? |
| Answer» It will loose the energy of `e V = 1.6 xx 10^(-19) xx 9 = 14.4 xx 10^-19 J`. | |
| 177. |
The electric potential at a point on the axis of an electric dipole depends on the distance `r` of the point from the dipole asA. `prop (1)/(r )`B. `prop (1)/(r ^(2))`C. `prop r`D. `prop (1)/(r^(3))` |
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Answer» Correct Answer - B Electric potential due to dipole in its general position is given by `V = (k.p cos theta)/(r^(2)) rArr prop (1)/(r^(2))` |
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| 178. |
Assertion: Conductors having equal positive charge and volume, must also have same potential. Reason: Potential depends only on charge and volume of conductor.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 Electric potential of a charged conductor depends not only on the amunt of charge and volume but also on the sphere of the conductor. Hence if their shapes are different, they may have different eelctric potential. |
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| 179. |
Can there be a potential difference between two adjacent conductors that carry same amount of positive charge ? |
| Answer» Yes, if shape and size of the conductors are different. | |
| 180. |
Equipotentials at a great distance from a collection of charges whose total sum is not zero are approximatelyA. spheresB. planesC. paraboloidsD. ellipoids |
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Answer» Correct Answer - A |
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| 181. |
An electric dipole of dipole moment `vecp` is oriented parallel to a unifrom electric field `vecE`, as shown It is rotated to one of four orientations shown below. Rank the final orientations according to the change in the potential energy of the dipole-field system, most negative to most positive. A. (i), (ii), (iv), (iii)B. (iv), (iii), (ii), (i)C. (i), (ii), (iii), (iv)D. (iii), (ii) and (iv) tie, then (i) |
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Answer» Correct Answer - C `U = -PE cos theta` `rArr` As `theta` increases, `U` increase |
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| 182. |
The electirc potential at a point `(x, y, z)` is given by `V = -x^(2)y - xz^(3) + 4` The electric field `vecE` at that point isA. `vecE = hati (2xy +z^(3)) + hatj x^(2) + hatk 3xz^(2)`B. `vecE = hati 2xy + hatj (x^(2) + y^(2)) + hatk (3xz - y^(2))`C. `vecE = hati z^(3) + hatj xyz + hatk z^(2)`D. `vecE = hati (2xy - z^(3)) + hatj xy^(2) + hatk 3z^(2) x` |
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Answer» Correct Answer - A Electric field at a point is equal to the negative gradient of the electrostatic potential at that point. Potential gradient relates with electric field according to the following relation `E = (-dV)/(dr)` `vecE = - (delV)/(delx) = [-(delV)/(delx)hati - (delV)/(dely) hatj - (delV)/(delx)hatk]` `= [hati (2xy + z^(3))+ hatj x^(2) + hatk 3xz^(2)]` |
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| 183. |
The diagram below show regions of equipotential: A positive chrages is moved from `A` to `B` in each diagram. A. In all the four cases the work done is the sameB. Minimum work is required to move `q` is figure `(a)`C. Maximum work is required to move `q` in figure `(b)`D. Maximum work is required to move `q` is figure `(c )` |
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Answer» Correct Answer - A `W = q DeltaV` as `DeltaV` is same in all conditions, work will be same. |
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| 184. |
Four chrages `2C, -3C, -4C` and `5C` respectively are placed at all the corners of a square. Which of the following statements is true for the point of intersection of the diagonals ?A. Electric fields is zero but electric potential is non-zeroB. Electric fields is non-zero but electric potential is zeroC. Both electirc field and electirc potential is zeroD. Neither electric fields nor electric potential is zero. |
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Answer» Correct Answer - B `V = 9 xx 10^(9) [(2)/(r ) -(3)/(r ) - (4)/(r ) - (5)/(r )] = 0` Electric fields due to all the charges shell not cancel out because of different direction. |
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| 185. |
Two opposite and equal chrages `4 xx 10^(-8)` coulomb when placed `2 xx 10^(-2) cm` away, from a dipole. If dipole is placed in an ecternal electric field `4 xx 10^(8)` newton//coulomb, the value of maximum torque and the work done in rotating it through `180^(@)` will beA. `64 xx 10^(-4) Nm` and `64 xx 10^(-4) J`B. `32 xx 10^(-4) Nm` and `32 xx 10^(-4) J`C. `64 xx 10^(-4) Nm` and `32 xx 10^(-4) J`D. `32 xx 10^(-4) Nm` and `64 xx 10^(-4) J` |
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Answer» Correct Answer - D Dipole moment `p = 4 xx 10^(-8) xx 2 xx 10^(-4) = 8 xx 10^(-12) m` Maximum torque `= pE = 8 xx 10^(-12) x 4 xx 10^(8)` `= 32 xx 10^(-4) Nm` Work done in rotating through `180^(@) = 2pE` `= 2 xx 32 xx 10^(-4) = 64 xx 10^(-4) J` |
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| 186. |
(Figure 3.51) shows two concentric conducting shells of radii `r_(1)` and `r_(2)` carrying uniformaly distributed charges `q_(1)` and `q_(2)`, respectively. Find an expression for the poptential of each the potential of each shell. . |
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Answer» The potential of each sphere consists of two points : one due to its own charge, and the other due to the charge on the other sphere. Using the principle of superposition, we have `V_1 = V_(r_(1), "surface") + V_(r_(2), "inside")` and `V_2 = V_(r_(1), "outside") + V_(r_(2), "surface")` Hence, `V_1 = (1)/(4 pi epsilon _0) q_(1)/r_(1) + (1)/(4 pi epsilon_0) q_(2)/r_(2)` and `V_2 = (1)/(4 pi epsilon_0)q_(1)/r_(2)+(1)/(4 pi epsilon_0)q_(2)/r_(2)`. |
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| 187. |
The electric potential at a point distant 0.1m from the middle of a short electric dipole on a line inclined at an angle of `60^(@)` with the dipole axis is 900V. Calculate the dipole moment. |
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Answer» Correct Answer - 600 Cm |
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| 188. |
Three capacitors of capacitance `3 muF, 10 muF` and `15 muF` are connected in series to a voltage source of `100 V`. The charge on `15 muF` isA. `50 muC`B. `100 muC`C. `200 muC`D. `280 muC` |
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Answer» Correct Answer - C `(1)/(C_(eq)) = (1)/(3) + (1)/(10) + (1)/(15) rArr C_(eq) = 2muF` Charge on each capcitor `Q = C_(eq) xx V rArr 2 xx 100 = 200 muC` |
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| 189. |
A uniform electric field of `30NC^(-1)` exists along the X-axis calculate the potential difference `V_(B)-V_(A)` between the points A (4m,2m) and B(10m,5m). |
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Answer» Correct Answer - `-180`V |
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| 190. |
A uniform electric field of `20NC^(-1)` exists in the vertically downward direction. Determine the increase in the electric potential as one goes up through a height of 50 cm. |
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Answer» Correct Answer - 10V |
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| 191. |
As shown in figure a dust particle with mass m = 5.0 × 10–9 kg and charge q0 = 2.0 nC starts from rest at point a and moves in a straight line to point b . What is its speed v at point b? A. `26 ms^-1`B. `34 ms^-1`C. `46 ms^-1`D. `14 ms^-1` |
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Answer» Correct Answer - c Apply conservation of mechanical energy between points `a and b` `(KE + PE)_a = (KE + PE)_b` `0 + (k(3 xx 10^-9)q_(0))/(0.01) - (k(3 xx 10^-9)q_0)/(0.02)` =`(1)/(2) mv^2 + (k( 3 xx 10^-9)q_0)/(0.02) - (k(3 xx 10^-9)q_0)/(0.01)` Put the values and get `v = 12 sqrt(15) = 46 ms^-1`. |
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| 192. |
About an electric field which of the following statements are not true ?A. If `E=0,V` must be zeroB. If `V=0,E` must be zeroC. If `E ne 0,V` cannot be zeroD. If `V ne 0,E` cannot be zero |
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Answer» Correct Answer - A::B::C::D |
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| 193. |
Two metal spheres, one fo radius R and the other of radius 2R, both have same surface charge density s. They are brought in contact and seprated. What will be new surface charge densitites on them ? |
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Answer» Correct Answer - `sigma_(1)=(5)/(3)sigma and sigma_(2)=(5)/(6)sigma` |
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| 194. |
A uniform electric field exists in `x-y` plane. The potential of points `A (-2m, 2m), B(-2m, 2m)` and `C(2m, 4m)` are `4V, 16 V` and 1`2 V` respectively. The electric field isA. `(4 hat(i)+5 hat(j))(V)/(m)`B. `(3 hat(i)+4 hat(j))(V)/(m)`C. `-(3hat(i)+4hat(j))(V)/(m)`D. `(3hat(i)-4 hat(j))(V)/(m)` |
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Answer» Correct Answer - D |
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| 195. |
In the previous question, if `V` is the electric potential of the first sphere, what would be the electric potential of the second sphere ?A. 2 VB. `V//2`C. `V//4`D. V |
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Answer» Correct Answer - a `V = E R`. If `R` is doubled, `V` also gets doubled. |
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| 196. |
There exists a uniform electric filed in the space as shown. Four points `A, B, C` and `D` are marked which are equildistant from the origin. If and `V_(D)` are their potentials respectively, then A. `V_(B)gtV_(A)gtV_(C)gtV_(D)`B. `V_(A)gtV_(B)gtV_(D)gtV_(C)`C. `V_(A)=V_(A)gtV_(C)=V_(D)`D. `V_(B)gtV_(C)gtV_(A)gtV_(D)` |
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Answer» Correct Answer - B |
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| 197. |
The potential field of an electric field `vec(E)=(y hat(i)+x hat(j))` isA. `V=-xy+`constantB. `V=-(x+y)+` constantC. `V=-(x^(2)+y^(2))+` constantD. V = constant |
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Answer» Correct Answer - A |
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| 198. |
Two identical thin ring, each of radius R meters, are coaxially placed a distance R metres apart. If `Q_1` coulomb, and `Q_2` coulomb, are repectively the charges uniformly spread on the two rings, the work done in moving a charge q from the centre of one ring to that of the other isA. ZeroB. `(q(Q_(1)-Q_(2))(sqrt(2)-1))/(sqrt(2).4piepsilon_(0)R)`C. `(2sqrt(2)(Q_(1)+Q_(2)))/(4 pi epsilon_(0)R)`D. `(q(Q_(1)+Q_(2))(sqrt(2)+1))/(sqrt(2).4 pi epsilon_(0)R)` |
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Answer» Correct Answer - B |
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| 199. |
The distance between `H^(+)` and `CI^(-)` ions in `HCI` molecule is `1.28 Å`. What will be the potential due to this dipole at a distance of `12 Å` on the axis of dipole ?A. `0.13 V`B. `1.3 V`C. `13 V`D. `130 V` |
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Answer» Correct Answer - A `V = 9 xx 10^(9).(p)/(r^(2))` `= 9 xx 10^(9) xx ((1.6 xx 10^(-19)) xx 1.28 xx 10^(-10))/((12 xx 10^(-10))^(2)) = 1.13 V` |
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| 200. |
Assertion: Surface of asymmetrical conductor can be treated as equipotential surface. Reason: Charges can easily flow in a conductor.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 - A Potential is constant on the surface of a sphere so it behaves as an equipotential suarface. |
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