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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.
| 1351. |
In the construction of potentiometer, the length of wire used isA. equal to one metreB. less than one metreC. greater than 1m upto 10 mD. all of these |
| Answer» Correct Answer - C | |
| 1352. |
Potentiometer was first invented byA. AmpereB. NewtonC. PoggendorfD. Millikan |
| Answer» Correct Answer - C | |
| 1353. |
When null point is obtained in the potentiometer the current is drawn fromA. cell onlyB. main battery onlyC. both the cell and main batteryD. neither cell nor main battery |
| Answer» Correct Answer - B | |
| 1354. |
Which of the following statement is not correct in metre bridge experiment ?A. the resistance wire must be of uniform cross sectional area.B. the readings should be taken at the middle of wire.C. the readings should be taken by interchanging the positions known and unknown resistance.D. the jockey should be changed. |
| Answer» Correct Answer - D | |
| 1355. |
Metre bridge is used toA. determine unknown resistanceB. measure currentC. measure P.D.D. all of these |
| Answer» Correct Answer - A | |
| 1356. |
For the accurate measurement of resistance by metre bridge, the null point should be obtainedA. towards left endB. towards right endC. at the middle of wireD. all of these |
| Answer» Correct Answer - C | |
| 1357. |
Two samples `1` and `2` are initially kept in the same state. The sample `1` is expanded through an isothermal process where as sample `2` through an adiabatic process upto the same final volume. The final temperature in process `1` and `2` are `T_(1)` and `T_(2)` respectively, thenA. `T_(1)gtT_(2)`B. `T_(1)=T_(2)`C. `T_(1)ltT_(2)`D. The relation between `T_(1)` and `T_(2)` cannot be deduced |
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Answer» Correct Answer - A |
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| 1358. |
If `2%` of the main current is to be passed through the galvanometer of resistance `G`, the resistance of shunt required isA. `(G)/(49)`B. `(G)/(50)`C. `49G`D. `50G` |
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Answer» Correct Answer - A |
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| 1359. |
If only 2% of the main current is to be passed through a galvanometer of resistance G , then the resistance of shunt will beA. `(G)/(50)`B. `(G)/(49)`C. `50 G`D. `49G` |
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Answer» Correct Answer - B |
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| 1360. |
If `2%` of the main current is to be passed through the galvanometer of resistance `G`, the resistance of shunt required isA. `G/49`B. `G/50`C. `49 G`D. `50G` |
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Answer» Correct Answer - A `I_G/I_S=S/G` `:. S=(I_G/I_S)G` `=(2/98)G=G/49` |
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| 1361. |
N identical current sources each of emf E and internal resistance r are connected to form a closed loop as shown in figure. The potential difference between points A and B which divides the circuit into n and (N-n) units is A. `NE`B. `(N-n)E`C. `nE`D. zero |
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Answer» Correct Answer - D By symmetry, `V_A=V_B` or `V_(AB)=0` |
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| 1362. |
If the length of the filament of a heater is reduced by `10%`, the power of the heater willA. increase by about `9%`B. increase by about `11%`C. increase by about `19%`D. decrease by about `10%` |
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Answer» Correct Answer - B `P=V^2/R` or `P prop 1/R` `P_2/P_1=R_1/R_1=l_1/l_2` (as R `prop` 1) `:. P_(2) = ((l_(1))/(l_(2))) P_(1) = ((l)/(0.9 l)) P_(1) = 1.11 P_(1)` so, power will increase by `11%` |
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| 1363. |
If the length of the filament ofa heater is reduced by 10% the power of the heater will :A. Increse by about 9%B. Increase by about 11 %C. Increase by about I9%D. Decrease by about I 0% |
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Answer» Correct Answer - D |
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| 1364. |
Define mean free path of electron in a conductor. |
| Answer» The average distance transvered by an electron during successive collisionsin a conductor is called mean free path of electron in a conductor. | |
| 1365. |
The equivalent resistance and potential difference between A and B for the circuit is respectively A. `4 Omega, 8 V`B. `8 Omega, 4V`C. `2 Omega, 2V`D. `16 Omega, 8V` |
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Answer» Correct Answer - A |
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| 1366. |
If you are provided three resistance `2 Omega, 3 Omega` and `6 Omega`. How will you connect them so as to obtain the equivalent resistance of `4 Omega`A. B. C. D. None of these |
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Answer» Correct Answer - C |
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| 1367. |
Length of a potentiometer wire is kept long and uniform to ahcieve:-A. uniform and more potential gradientB. non-uniform and more potential gradientC. uniform and less potential gradientD. non-u niform and less potential gradient. |
| Answer» Correct Answer - 3 | |
| 1368. |
The specific resistance of a conductor increases with:A. increase in temperatureB. increase in cross-section areaC. increase in cross-section and decrease in length.D. decrease in cross-section area. |
| Answer» Correct Answer - 1 | |
| 1369. |
In the circuit shown below `E_(1) = 4.0 V, R_(1) = 2 Omega, E_(2) = 6.0 V, R_(2) = 4 Omega` and `R_(3) = 2 Omega`. The current `I_(1)` is A. `1.6A`B. `1.8A`C. `1.25A`D. `1.0A` |
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Answer» Correct Answer - B |
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| 1370. |
In the given circuit, it is observed that the current I is independent of the value of the resistance `R_6`. Then the resistance values must satisfy .A. `R_(1) R_(2) R_(5) = R_(3) R_(4) R_(6)`B. `(1)/(R_(5)) + (1)/(R_(6)) = (1)/(R_(1) + R_(2)) + (1)/(R_(3) + R_(4))`C. `R_(1) R_(4) = R_(2) R_(3)`D. `R_(1)R_(3) = R_(2)R_(4) = R_(5)R_(6)` |
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Answer» Correct Answer - C (c ) As `I` is independent of `R_(6)` no current flows through `R_(6)` this requires that the junction of `R_(1)` and `R_(2)` is at the same potential as the junction of `R_(3)` and `R_(4)`. This must satisfy the condition `(R_(1))/(R_(2)) = (R_(3))/(R_(4))`, as in the Wheatstone brigde |
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| 1371. |
In the circuit shown here, `E_(1) = E_(2) = E_(3) = 2 V` and `R_(1) = R_(2) = 4 ohms`. The current flowing between point `A` and `B` through battery `E_(2)` is A. ZeroB. 2 amp from A to BC. 2 amp from B to AD. None of the above |
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Answer» Correct Answer - B |
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| 1372. |
The thermal energy developed in a current-carrying resistor is given by`U=i^(2)Rt`and also by`U=Vit.`should we say that U is proportional to`i^(2)`or to i? |
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Answer» Correct Answer - A |
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| 1373. |
An electric motor operates on a `50V` d.c. supply draw a current of `15 A`. If the motor yields a mechanical power of `150 W` estimate the power dissipated across its windings. Also find the efficiency of the motor ? |
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Answer» Correct Answer - `600 W,20%` Total power of moter `P = VI = 50 xx 15 = 750 W` mechhanical power of motor `P = 150 W` power dissipated across the winding of motor `= 750 - 150 = 600 W` Efficiency of moter `= (P)/(P) xx 100 = (150)/(150) xx 100 = 20%` |
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| 1374. |
In the circuit shown in fig. the cell has emf 10V and internal resistance `1 Omega` A. The current though the `3Omega`resistor is 1 A.B. The current though the `3Omega` resistor is 0.5 AC. The current through the `4Omega`resistor is 0.5 AD. The current through the `4Omega` resistor is 0.25 A |
| Answer» Correct Answer - A::D | |
| 1375. |
N cells each of e.m.f. E & internal resistance r are grouped into sets of K cells connected in series. The (N/K) sets ae connected in parallel to a load of resistance R, then,A. Maximum power is delivered to the load is `K=sqrt((NR)/r)`B. Maximum power is delivered to the load if `K=sqrt((r)/(NR))`C. Maximum power delivered to the load is `E^(2)/(4Nr)`D. Maximum power deliverd to the load is `(NE)^(2)/(4r)` |
| Answer» Correct Answer - A::D | |
| 1376. |
Which of the adjoining graphs represents ohmic resistanceA. B. C. D. |
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Answer» Correct Answer - A |
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| 1377. |
The magnitude and direction of the current in the circuit shown will be A. `(7)/(3)A` from a to b through eB. `(7)/(3)A` from b to a through eC. 1A from b to a through eD. 1A from a to b through e |
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Answer» Correct Answer - D |
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| 1378. |
In the adjoining circuit, the battery `E_(1)` has an e.m.f of 12 volt and zero internal resistance while the battery E has an e.m.f of 2volt . If the galvanometer G reads zero, then the value of the resistance X in ohm is A. 10B. 100C. 500D. 200 |
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Answer» Correct Answer - B |
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| 1379. |
In the adjoining circuit, the battery `E_(1)` has an e.m.f me of 12 volt and zero internal resistance while the battery E has an e.m.f me of 2volt . If the galvanometer G reads zero, then the value of the resistance X in ohm is A. 250B. 100C. 50D. 200 |
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Answer» Correct Answer - B |
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| 1380. |
Equivalent resistance between A and B will be A. 2`Omega`B. 18`Omega`C. 6`Omega`D. `3.6Omega` |
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Answer» Correct Answer - D |
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| 1381. |
The potential drop across the `3 Omega` resistor is A. 1 VB. `1.5 V`C. 2 VD. 3 V |
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Answer» Correct Answer - A |
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| 1382. |
In a network as shown in the figure the potential difference across the resistance 2R is (the cell has an emf of E and has no internal resistance): A. 2EB. `(4E)/(7)`C. `(E)/(7)`D. E |
| Answer» Correct Answer - 2 | |
| 1383. |
Two cells , each of `emf E` and internal resistance `r`, are connected in parallel across a resistor `R`. The power delivered to the resistor is maximum if `R` is equal toA. R = r/2B. R = rC. R = 2rD. R = 0 |
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Answer» Correct Answer - A |
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| 1384. |
Two cells , each of `emf E` and internal resistance `r`, are connected in parallel across a resistor `R`. The power delivered to the resistor is maximum if `R` is equal toA. `R=r`B. `R=2r`C. `R=(3r)/(2)`D. `R=(r)/(2)` |
| Answer» Correct Answer - 4 | |
| 1385. |
An electron revolves `6 xx 10^(1)` times/sec in circular loop. The current in the loop isA. `0.96 mA`B. `0.06 mu A`C. `28.8 A`D. None of these |
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Answer» Correct Answer - A |
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| 1386. |
A rod of certain metal is `1.0 m` long and `0.6 cm` in diameter. Its resistance is `3.0 xx 10&(-3)` ohm. Another disc made of the same metal is `2.0 cm` in diameter and `1.0 mm` thick. What is the resistance between the round faces of the disc?A. `1.35 xx 10^(-8)` ohmB. `2.70 xx 10^(-7)` ohmC. `4.05 xx 10^(-6)` ohmD. `8.10 xx 10^(-5)` ohm |
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Answer» Correct Answer - B |
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| 1387. |
A rod of certain metal is `1.0 m` long and `0.6 cm` in diameter. Its resistance is `3.0 xx 10&(-3)` ohm. Another disc made of the same metal is `2.0 cm` in diameter and `1.0 mm` thick. What is the resistance between the round faces of the disc?A. `1.35 xx 10^(-8) ohm`B. `2.70 xx 10^(-7) ohm`C. `4.05 xx 10^(-6) ohm`D. `8.10 xx 10^(-5) ohm` |
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Answer» Correct Answer - B (b) Resistivity of the material of the rod `rho = (RA)/(1) = (3 xx 10^(-3) pi (0.3 xx 10^(-2))^(2))/(1) = 27 xx 10^(-9) pi Omega xx m` Resistance of disc `R = (("Thickness"))/(("Area of cross section"))` `= 27 xx 10^(-9) pi xx ((10^(-3)))/(pi xx (1 xx 10^(-2))^(2)) = 2.7 xx 10^(-7) Omega`. |
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| 1388. |
A cell of e.m.f. E is connected with an external resistance R , then p.d. across cell is V . The internal resistance of cell will beA. `((E-V)R)/(E)`B. `((E-V)R)/(V)`C. `((V-E)R)/(V)`D. `((V-E)R)/(E)` |
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Answer» Correct Answer - B |
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| 1389. |
Equal potentials are applied on an iron and copper wire of same length. In order to have the same current flow in the two wires, the ratio r (iron)/r(copper) of their radii must be (Given that specific resistance of iron `=1.0 xx 10^(-7) ohm-m` and specific resistance of copper `=1.7 xx 10^(-8) ohm-m`)A. About 1.2B. About 2.4C. About 3.6D. About 4.8 |
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Answer» Correct Answer - B |
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| 1390. |
A fuse `F_(1)` is connected across a source of variable voltage and the voltage is increased gradually. The fuse blows out just when the reading of the voltmeter and ammeter reaches `1.0 V and 1.0 A` respectively (see figure (i)). The experiment is repeated with another fuse `F_(2)` and the reading of the voltmeter and ammeter when it blows out is `2.4 V and 1.2 A` respectively. (a) The two fuses are connected in parallel as shown in figure (ii). Voltage is increased gradually. find the reading of the ammeter when any one of the fuses blows out. (b) The two fuses are connected in series as shown in figure (iii). find the reading of the voltmeter at the point one of the fuses blows out. |
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Answer» Correct Answer - (a) 1.5 A (b) 3.0 V |
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| 1391. |
In the circuit shown in the figure. find `I_(1) and I_(2)`. |
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Answer» Correct Answer - `I_(1)=I_(2)=0` |
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| 1392. |
Potential difference across the terminal of a non ideal battery isA. zero when it is short circuitedB. less than its emf when current flows from negative terminal to positive terinal inside the batteryC. zero when no current is drawn from the batteryD. greater than its emf when current flows from positive terminal to negative terminal inside the battery |
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Answer» Correct Answer - A::B::D |
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| 1393. |
Potential difference across the terminal of a non ideal battery isA. zero when it is short circultedB. less than its emf when current flows from negative terminal to positive terminal inside the batteryC. zero when no current is drwon from the batteryD. greater then its emf when current flows from positive terminal to negative inside the battery. |
| Answer» Correct Answer - A::B::D | |
| 1394. |
The magnitude and direction of the current in the circuit shown will be A. `(7)/(3)A` from a to eB. `(7)/(3)A` from b to eC. 1A from b to eD. 1A from a to e |
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Answer» Correct Answer - D |
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| 1395. |
A battery of emf E and internal resistance r is connected across a resistance R. Resistance R can be adjusted to any value greater then or equla to zero. A graph is plotted between the current passing through the resistance (I) and potential (V) across it. Select the correct alternatives. A. Internal resistance of the battery is `5 Omega`B. EMF of the battery is 10VC. Maximum curent which can be taken from the battery is 2AD. EMF of the battery is 5V |
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Answer» Correct Answer - A::B::C |
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| 1396. |
The current in the resistance R will be zero if A. `E_(1)r_(1)=E_(2)r_(2)`B. `(E_(1))/(r_(1))=(E_(2))/(r_(2))`C. `(E_(1)+E_(2))r_(1)=E_(1)r_(1)`D. `(E_(1)-E_(2))r_(1)=E_(2)r_(1)` |
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Answer» Correct Answer - B |
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| 1397. |
A battery of emf E and internal resistance r is connected across a resistance R. Resistance R can be adjusted to any value greater than or equal to zero. A graph is plotted between the current passing through the resistance (I) and potential difference (V) across it. Select the correct alternative(s). A. Internal resistance of the battery is `5 Omega`B. emf of the battery is 10 VC. maximum crrent which can be taken form the battery is 2 AD. V-I graph can never be a stright line as shown in figure. |
| Answer» Correct Answer - A::B::C | |
| 1398. |
In the network shown calculate current through the cell `(I_1)` and the current `I_(2)` through the `2 R` resistance. |
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Answer» Correct Answer - `I_(1)=(2V)/(R) ; I_(2)=0` |
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| 1399. |
In the circuit shown in figure A. Power supplied by the battery is 200 wattB. current flowing in the circuit is 5 AC. potential difference across `4 Omega` resistance is equal to the potential difference across `6 Omega` resistanceD. current in wire AB is zero |
| Answer» Correct Answer - A::C | |
| 1400. |
In the network shown in fig. , points A, B, and C are at potentials of 70 V, 0, and 10V, respectively. A. Point D is at a potential of 40 VB. The currents in the sections AD, DB, DC are in the ratio 3:2:1C. The currents in the sections AD, DB, DC are in the ratio 1:2:3D. The network draws a total power of 200 W. |
| Answer» Correct Answer - A::B::D | |