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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.
| 1301. |
Assertion : In the following circuit e.m.f. is `2 V` internal resistance of the cell is `1 Omega` and `R = 1 Omega` the reading of the voltmeter is `1 V`. Reason : `V = E - ir`, where `E = 2 V, i= 1 A` and `R = 1 Omega`A. Statement-1 is True,Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True,Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1C. Statement -1 is True, Statement-2 is false.D. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - A Here `E=2V,I=2/2=1` amp and `r=1 Omega` `:. V=E=ir=2-(1)(1)=1` Volt |
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| 1302. |
Assertion : When the length of a conductor is doubled, its resistance will also get doulbed. Reason : Resistance is direclty proportional to the length of a conductor.A. If both assertion and reason are ture and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason both are false. |
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Answer» Correct Answer - A (a) `R = rho (l)/(A)` Where `l = 2 l, A = A//2` Because volume `V_(1) = V_(2))`. |
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| 1303. |
What are super-conductors? Write their two applications. |
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Answer» As the temperature of certain metals and alloys decreases, their resistance also decreases. When the temperature reaches a certain critical value called critical temperature, the resistance of material completely disappears, i.e., it becomes zero. Then the material behaves as a super-conductor. Thus super-conductor are those material conductors whose resistance disappear at critical temperature. The critacal temperautre is different materials. Super conductors are used (i) in power transmission (ii) to produce very high speed computers. |
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| 1304. |
Statement- 1 : The resistance of super-conductor is zero Statement- 2 : The super-conductors are used for the transmission of electric powerA. Statement - 1: is true ,Statement - 2: is true ,Statement - 2: is correct explanation of statement - 1,B. Statement - 1: is true ,Statement - 2: is true ,Statement - 2: is not a correct explanation of statement - 1,C. statement - 1 is correct and statement - 2 is falseD. statement - 1 is false and statement - 2 is true |
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Answer» Correct Answer - b Here both the statement-1 and statement-2 both are correct but the reason is not the correct explanation of assertion |
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| 1305. |
Assertion : When the length of a conductor is doubled, its resistance will also get doulbed. Reason : Resistance is direclty proportional to the length of a conductor.A. Statement-1 is True,Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True,Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1C. Statement -1 is True, Statement-2 is false.D. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - A `R=rho l/A` Where `l=2l,A=A//2` Because volume `V_(1)=V_(2)`. |
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| 1306. |
Assertion : When current through a bulb is increased by `2%` power increases by `4%` Reason : Current passing through the bulb is `prop (1)/(Resistance)`.A. Statement-1 is True,Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True,Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1C. Statement -1 is True, Statement-2 is false.D. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - B `P=I^(2)R` So `100%xx((dP)/P)=2((dI)/I)100%` |
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| 1307. |
Statement-1: A conductor carrying electric current becomes electricity charged. Statement-2: A conductor carrying electric current contains same number of positive and negative charges and thus conductor is electrically neutral.A. Statement-1 is True,Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True,Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1C. Statement -1 is True, Statement-2 is false.D. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - D The electrons are in motion which constitute electric current in a conductor but no of positive and hegative charges are same. |
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| 1308. |
Calculate the value of current and indicate its direction Also calculate the potential difference between the points (i) `B & A` (ii) `A & C` |
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Answer» Net emf`=E_(2)-E_(1)`, total resistance `=R+r_(1)+r_(2)` Current in the circuit is `I=((E_(2)-E_(1))/(R+r_(2)+r_(1)))=(4.0-2.0)/ (5.0+1.0+2 .0)=0.25A` (from C to A via B) (i) Potential difference between the points B and `A= V_(B)-V_(A)=E _(1)+Ir_(1)=2+0.25xx1=2.25V` (ii) Potential difference between the points A and C =potential drop across `5Omega=5xx0.25=1.25V` |
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| 1309. |
If not current flows through the galvanometer the find the value of unknown resistance X. Assume that resistance per unit length of wire AB is 0.2 `Omega//cm`. then also calculate total current given by battery. |
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Answer» For balanced condition `(X)/( R_(AJ))=(2)/(R_(JB))` `R_(AJ)=0.02xx60=1.2Omega,R_(JB )=0.02xx4 0= 0.8O mega rArr(X)/(1.2)=(2)/(0.8)rArrX=3Omega` Current given by the battery `I=(4)/(R_(eq))` where `R_(eq)=((3+2)(1.2+0.8))/(3+2+1.2+0.8)=(10)/(7)Omega` so `I=(4xx7)/(10)=2.8A` |
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| 1310. |
`V-I` graphs for a material is shown in the figure. The graphs are drawn at two different temperatures.A. `T_(1)-T_(2)propcot2theta`B. `T_(1)-T_(2)propsin2theta`C. `T_(1)-T_(2)proptan2theta`D. `T_(1)-T_(2)propcos2theta` |
| Answer» Correct Answer - A | |
| 1311. |
A long constan wire is connected across the terminals of an ideal battery. If the wire is cut in to two equal pieces and one of them is now connected to the same battery, what will be the mobility of free electrons not in the wire compared to that in the first case?A. same as that of previous velueB. double that of previous valueC. half that of previous valueD. four times that of previous value |
| Answer» Correct Answer - A | |
| 1312. |
Assertion: Drift velocity of electrons is independent of time. Reason: Electrons are accelerated in the presence of electric field.A. If both assertion and reason are true and reason is the correct explantion of assertion.B. If both assertion and reason are true and reason is not the correct explantion of assertion.C. If assertion is false but reason is true.D. If assertion is true but reason is false. |
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Answer» Correct Answer - B Drift velocity is the average velocity of electrons in presence of electric field, which is independent of time. |
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| 1313. |
Assertion : Insulators do no allow flow of current through them. Reason: Insulators have no free charge carrierA. If both assertion and reason are ture and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason both are false. |
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Answer» Correct Answer - A (a) Since current arises due to continuous flow of charged praticles. There is no free charge in insulator. Hence no flow of charges is possible. Therefore current do not flow through insulators. |
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| 1314. |
Assertion : Insulators do no allow flow of current through them. Reason: Insulators have no free charge carrierA. If both assertion and reason are true and reason is the correct explantion of assertion.B. If both assertion and reason are true and reason is not the correct explantion of assertion.C.D. If assertion is true but reason is false. |
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Answer» Correct Answer - A Current arises due to continuous flow of charged particles. There is no free charge carriers in insulator hence no flow of charges are possible. Therefore current do not flow through insulators. |
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| 1315. |
Current is allowed to flow in a metallic wire at a constant potential difference. When the wire becomes hot, cold water is poured on half portion of the wire. By doing so, its other portion becomes still more hot. Explain its reason. |
| Answer» When cold water is poured on half portion of hot wire, it gets cooles there. Its temperature falls and hence its resistance decreases, due to it, the current in the wire increase. As heat produced, `H prop I^(2)`, therefore, the other portion of the wire becomes more hot. | |
| 1316. |
Assertion : Electromotive force is a force which helps the electrons to flow and produce current. Reason : Electromotive force is independent of the voltage across the cell.A. If both assertion and reason are true and reason is the correct explantion of assertion.B. If both assertion and reason are true and reason is not the correct explantion of assertion.C. If assertion is false but reason is true.D. If assertion is true but reason is false. |
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Answer» Correct Answer - D Electromotive force is not a force, it is the voltage difference between the two terminals of a source in open circuit. |
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| 1317. |
A galvanometer is to be converted into an ammeter or voltmeter. In which of the following cases the resistance of the device is greatest?A. An ammeterofrange I0AB. A voltmeter ofrange 5VC. An ammeter of range SAD. A voltmeter of range I 0V |
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Answer» Correct Answer - B |
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| 1318. |
A carbon filament has resistance of `120Omega` at `0^(@)C` what must be te resistance of a copper filament connected in series with resistance and combined resistance remained constant at all temperature `(alpha_("carbon")=(-5xx10^(-4))/(``^(@)C),alpha_("copper")=(4xx10^(-3))/(``^(C))` |
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Answer» If `R_(0), R_(t)` be the resistance of a wire at `0^@C` and `t^@C` then temperature coefficient of resistance is `alpha = (R_(t) - R_(0))/R_(0)t` or change in resistance `= R_(t) - R_(0)=R_(0) alpha t` Let R be the resistance of copper palced in series with carbon at `0^@C`. If the combination has same resistance at all temperature, then at temperature `t^@C`, increase in resistance of copper = decrease in resistance of carbon `:. alpha_(Cu) R t = alpha_(c) R_(0) t` or `0.004 xx R xx t = 0.0007 xx 100 xx t` or `R = (0.0007 xx 100)/0.004 = 17.5 Omega` |
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| 1319. |
A uniform wire is cut into four segments. Each segment is twice as long as the earlier segment. If the shortest segment has a resistance of `4 Omega`, find the resistance of the original wire. |
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Answer» Correct Answer - `60 Omega` |
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| 1320. |
A uniform wire is cut into four segments. Each segment is twice as long as the earlier segment. If the shortest segment has a resistance of `2 Omega`, find the resistance of original wire. |
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Answer» Let `l` be the length of shortest segment. Then lengths of four segments will be `l, 2 l, 4l, 8l`. Their corresponding resistance will be `R, 2 R, 4 R` and 8R. Given ` R =2 Omega`. Resistance of the original wire `1=R + 2 R +4R + 8R` `=15R = 15xx2 = 30 Omega` |
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| 1321. |
The V - I graphs for two resistors and their series combination are shown in Fig 5.23. Which one of these graphs represents the series combination of the two resistors? Given reason for your answer. |
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Answer» Let for the given voltage V, the values of currents be `I_(1), I_(2)` and `I_(3)` for resistors 1,2,3 . `:. R_(1)= V/I_(1), R_(2) = V/I_(2) and R_(3) = V/I_(3)` From graph, we note that `I_(3) gt I_(2) gt I_(1)` `:. R_(3) lt R_(2) lt R_(1)` Hence graph 1 reprsents the series combination of other two resistnace. |
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| 1322. |
A uniform wire is cut 10 segments increasing in length in equal steps. The resistance of shortest segment is R and the resistance of the other segments increases in steps of `4 Omega`. If the resistance of the longest segment is 2R, find the value of R and the resistance of original wire. |
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Answer» Resistance of first segment `= R Omega` Resistance of second segment = `R + 4 xx 1 Omega` Resistance of third segment `=R + 4 xx 2 Omega` Resistance of tenth segment `= R + 4 xx 9` `= R + 36 Omega` As per question, `R + 36 = 2R or R = 36 Omega` Resistance of the origin wire `R + (R +4) + (R + 8) +(R + 12) +...` `+ (R + 36)` `10R +4(1+2+3+...+9)` `=10 xx 36+4 xx 45 = 540 Omega` |
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| 1323. |
Which one is not the correct statementA. 1 volt `xx` 1 coulomb = 1 jouleB. 1 volt `xx` 1 ampere = 1 joule/secondC. 1 volt `xx` 1 watt = 1 H.P.D. Watt-hour can be expressed in eV |
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Answer» Correct Answer - C |
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| 1324. |
Assertion : If a current is flowing through a conducting wire of non-uniform cross-section, drift speed and resistance both will increase at a section where cross-sectional area is less. Reason : Current density at such sections is more where cross-sectional area is less.A. If both Assertion and Reason are true and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are true and Reason is not correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true. |
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Answer» Correct Answer - B |
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| 1325. |
Every atom makes one free electron in copper. If 1.1 ampere current is flowing in the wire of copper having 1 mm diameter, then the drift velocity (approx.) will be (Density of copper `=9 xx 10^(3) kg m^(-3)` and atomic weight = 63)A. `0.3 mm//sec`B. `0.1 mm//sec`C. `0.2mm//sec`D. `0.2cm//sec` |
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Answer» Correct Answer - B |
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| 1326. |
Every atom makes one free electron in copper. If 1.1 ampere current is flowing in the wire of copper having 1 mm diameter, then the drift velocity (approx.) will be (Density of copper `=9 xx 10^(3) kg m^(-3)` and atomic weight = 63)A. `0.01 mm s^(-1)`B. `0.02mm s^(-1)`C. `0.2 mm s^(-1)`D. `0.1 mm s^(-1)` |
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Answer» Correct Answer - 4 No. of atom in 1 gm `Cu=(N_(A))/(63)` No. of atom/`cm^(3)=(N_(A))/( 63)xx9` `because` each atom contributes one electron So no. of electron `(n)//m^(3)=(N_(A))/(63)xx9xx10^(6)` Now I=`"ne"Av_(d)` Put values I,n,e,A and calculate for `v_(d)` |
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| 1327. |
A resistor of `5 Omega` is connected in series with a parallel combination of a number of resistors each of `5 Omega`. If the total resistance of the combination is `6 Omega` , how many resistor are in parallel? |
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Answer» Let n resistance each of `5 Omega` be connected in parallel. Their effective resistance is `1/R_(p) =1/5 + 1/5 +1/5 + .. N times =n/5 or R_(p)=5/n` As this parallel combination of resistance is connected in series with `5 Omega` resistance, the total resistance of the combination is `R=R_(p) + 5 = (5/n)+5 or 5/6 = 6-5=1` or `n=5` |
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| 1328. |
Find the net resistance between point A and B in the circuit |
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Answer» Correct Answer - `2.4 Omega` In closed circuit ECDE, arm ECD is parallel to arm ED so `R_(ED) = (6 xx (5+ 3))/(6 + (3+ 3)) = 3 Omega` The resistebnce `R_(ED)` is in series `3 Omega` resistance of arm `FD `and they are in parallel its `s Omega` resistance of arm `EF` so total resistance between A and B is `R_(AB) = (4 xx (3+3))/(4+ (3+3)) = (24)/(10)` `= 2.4 Omega` |
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| 1329. |
A parallel combination of three resistors take a currect of `7.5 A` from `30 V` supply. If the two resistor are `10 Omega` and `12 Omega`. find the third one. |
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Answer» Correct Answer - `R_(3) = 15 Omega` `R_(P) = 30//7.5 = 4 Omega` `(1)/(R_(P)) = (1)/(R_(1)) + (1)/(R_(2)) + (1)/(R_(3)) or (1)/(4) + (1)/(10) + (1)/(12) + (1)/(R_(3))` or `(1)/(R_(1)) = (1)/(4) - ((1)/(10) + (1)/(12)) = (4)/(60) = (1)/(15)` or `R_(3) = 15 Omega` |
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| 1330. |
Which of the following statement is only currect statement for a potentiometer?A. it cannot measure potential differenceB. it cannot measure the capacitance of a capacitorC. it cannot measure currentD. it cannot measure resistance |
| Answer» Correct Answer - b | |
| 1331. |
A meter-bridge is based on the principle of |
| Answer» Correct Answer - Wheatstone bridge | |
| 1332. |
A meter bridge cannot be used to measure …… resistances |
| Answer» Correct Answer - very high or very low | |
| 1333. |
A balance point is obtained in the potentiometer wire .If the fall of potential along the potentiometer wire Due to……is greater than the …..in the balanced |
| Answer» Correct Answer - dividing cell ; emf of the cell | |
| 1334. |
The fall of potential per unit length of wire is called…….. |
| Answer» Correct Answer - potential gradient | |
| 1335. |
A 25 watt, 220 volt bulb and a 100 watt, 220 volt bulb are connected in series across 440 volt lineA. only 100 watt bulb will fuseB. only 25 watt bulb will fuseC. none of the bulb will fuseD. both bulbs will fuse |
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Answer» Correct Answer - B `R=(V^(2))/(P),V=iR` |
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| 1336. |
There are 5 tube-lights each of 40 W in a house. These are used on an average for 5 hours per day. In addition, there is an immersion heater of 1500 W used on an average for 1 hour per day. The nmber of units of electricity are consumed in a month isA. 25 unitsB. 50 unitsC. 75 unitsD. 100 units |
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Answer» Correct Answer - C `P=(E)/(t),1K.W.H=1` unit |
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| 1337. |
An electric immersion heater of `1.08 kW` is immersed in water . After it has reaches a temperature of `100^(@)C` , how much time will be required to produce `100 g` of steam?A. 50 sB. 420 sC. 105 sD. 105 s |
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Answer» Correct Answer - D |
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| 1338. |
In the given circuit, with steady current, the potential drop across the capacitor must be A. VB. `V//2`C. `V//3`D. `2V//3` |
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Answer» Correct Answer - C |
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| 1339. |
In potentiometer experiment, a cell is balanced by length 120cm. When a cell is shunted by resistance of `5Omega`, the balancing length is 80 cm. The internal resistance of cell isA. `2.5Omega`B. `3Omega`C. `4Omega`D. `5Omega` |
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Answer» Correct Answer - A `l_(1)=120cm, R_(1)=5Omega, l_(2)=80 cm, r_(1)=?` `r_(1)=R_(2)((l_(1)-l_(2))/(l_(2)))` `=((120-80)/(80))=(5xx40)/(80)=(200)/(80)=2.5 Omega`. |
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| 1340. |
A battery `epsilon_(1)` of 4 V and a variable resistance Rh are connected in series with the wire AB of the potentiometer. The length of the wire of the potentiometer of 1 meter. When a cell `epsilon_(2)` of emf 1.5 V is connected between points A and C, no currents flows through `epsilon_(2)` Length AC=60 cm. (i) Find the potetial difference between the ends A and B of the potentiometer. (ii) Would the method work, if the battery `epsilon_(1)` is replaced by a cell of emf of 1 V ? |
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Answer» (i) Let V be the pot. Diff. between the ends A and B of the potentiometer wire. Then `V/100 = epsilon_(2)/60` or `V = epsilon_(2) xx 100/60 = 1.5 xx 100/60 = 2.5V` (ii) If battery `epsilon_(1)` is replaced by a cell of emf 1 V, then method would not work. As `epsilon_(1) lt epsilon_(2)`, the balance point cannot be obtained on the potentiometer wire. |
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| 1341. |
A potentiometer wire of length 1 m is connected to a driver cell of emf 3 V. When a cell of emf 1.5V is used in the secondary circuit, the balance points is found to be 60 cm. On replacing this cell with a cell of inknown emf, the balance points shifts to 80 cm. (i) Calculate unknown emf of the cell. (ii) Explain with reason, whether the circuit works, if the drivere cell is replaced with a cell of emf 1 V. (iii) Does the high resistance R, used in the secondary circuit affect the balance point? Justify your answer. |
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Answer» `epsilon_(1)/epsilon_(2) = l_(1)/l_(2)` or `epsilon_(2) = epsilon_(1) xx l_(2)/l_(1) = 1.5 xx 80/60 =2.0V` (ii) When the driver cell of emf 2 V is replaced by a cell emf 1 V, the circuit will not work because the fall of potential on the entire potentiometer wire will become less than the emf of cell 1.5 V to be balanced on the wire. Due to it, balance point cannot be obtained on potentiometer wire. (iii) When the high resistance R is used in the secondary circuit, the position of balance point on the potentiometer wire will not be affected bacause no current flows through the cell at the balance point. |
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| 1342. |
A potentiometer wire of length 1.0 m has a resistance of 15 ohm. It is connected to a 5 V source in series with a resistance of `5 Omega`. Determine the emf of the primary cell which has a balance point at 60 cm. |
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Answer» L = 1m = 100cm r = 15Ω \(\epsilon\) = 5V R = 5Ω \(\epsilon' = \frac{\epsilon r}{R + r} \times \frac rL\) \(\epsilon' = \frac{5\times 15}{(5 + 15)} \times \frac{60}{100}\) \(\epsilon' = \frac{4500}{2000}\) \(\epsilon'= 2.25V\) |
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| 1343. |
The circuit shows in Fig . `6.33` shows the use of potentiometer to measure the internal resistance of a cell. (a) When the key is open, how does the balance point change, if the driver cell decreases ? (b) When the key is closed, how does the balance point change, if `R` is increased, keeping the current from the driver cell constant ? |
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Answer» (i) When the key K is open and current from the driver cell increases, the potential gradient across the potentiometer wire increases. Due to it, the balance point shifts towards the end A of potentiometer wire. (ii) When the key k is closed and R is decreased, keeping the current from the driver cell constant, the terminal potential difference of the cell decrease. Due to it, the balance point shifts towards the end A of potentiometer wire. |
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| 1344. |
A potentiometer wires has a length L and a resistance `R_(0)`. It is connected to a battery and a parallel resistance combination of R and S as shown in fogure. Find an exprssion for the potential gradient of the potentiometer wire. |
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Answer» Total resistance of potentiometer circuit `R_(0) + (RS)/(R+S)` Current in the circuit, `I = epsilon/(R_(0) + RS //(R +S))` Potential differnce across the wire AB `V = IR_(0) = epsilon/(R_(0) + RS//(R +S)) xx R_(0)` Potential gradient, `K = V/L = (epsilon R_(0))/(L (R_(0) +( RS)/(R+S)))` The emf of the test cell must be less than the total potential drop (V) across the wire AB. The positive terminal of test cell must be connected to end A of the wire. |
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| 1345. |
Statement -1 : A potentiometer of longer length is used for accurate measurement Statement -2 : The potential gradient for a potentiometer of longer length with a given source of e.m.f becomes smallA. Statement - 1: is true ,Statement - 2: is true ,Statement - 2: is correct explanation of statement - 1,B. Statement - 1: is true ,Statement - 2: is true ,Statement - 2: is not a correct explanation of statement - 1,C. statement - 1 is correct and statement - 2 is falseD. statement - 1 is false and statement - 2 is true |
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Answer» Correct Answer - a Here both the statement - 1 and statement - 2 are correct |
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| 1346. |
Assertion : A potentiometer of longer length is used fof accrurate measurement. Reason : The potential gradient for a potentiometer of longer length with a given source of e.m.f. becomes small.A. If both assertion and reason are ture and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason both are false. |
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Answer» Correct Answer - A (a) Sensitivity `prop (1)/(Potential) prop` (Length of wire) |
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| 1347. |
The variation of potential difference `V` with length `l` in case of two potentiometers `X` and `Y` is as shows in Fig. `6.21`. Which of these two will you perfer for comparing the emfs of the two cells and why ? |
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Answer» A potentiometer is said to be sensitive if fall of potential per unit length, i.e., potential gradient `(dV//dl)` is small. For given length, l, `V_(y) lt V_(x)` `:. ((dV)/(dl))_(y) lt ((dV)/(dl))_(x)` Therefore, potentiometer Y will be preferref for comparing emfs of the two cells |
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| 1348. |
Why do we prefer potentiometer with a longer bridge wire? |
| Answer» A potentiometer with longer bridge wire will have a small potential gradient. Due to it, it becomes more sensitive. Hence it is preferred. | |
| 1349. |
How can you make a potentiometer of given wire length more sensitive using a resistance box? |
| Answer» This can be achieved by using more resistance from the resistance box in the potentiometer wire circuit. Which in turn will decrease the current in the potentiometer wire and comsequently, the potential gradient of wire will decrease. As a result of it , the sensitivity of the potentiometer will increase. | |
| 1350. |
A potentiometer is more sensitive whenA. its wire is of small lengthB. its wire is of large lengthC. applied P.D. is largeD. potential gradient along the wire is very low |
| Answer» Correct Answer - D | |