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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.
| 2401. |
One mole of a monoatomic gas at 300K is mixed with two moles of diatomic gas (degree of freedom = 5) at 600K. The temperature of the mixture will beA. 456 KB. 531 KC. 495 KD. 501 K |
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Answer» Correct Answer - B |
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| 2402. |
The temperature coefficient of copper is `0.004^(@)C^(-1)`. Find the resistance of a 5 m long copper wire of diameter 0.2 mm at `100^(@)C`, if the resistivity of copper at `0^(@)C` is `1.7xx10^(-8) Omega` m. |
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Answer» Correct Answer - `3.8 Omega` |
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| 2403. |
Two coils connected in series have resistance of `600K Omega` and `300 Omega` at `20^@C` and temperature coefficient of` 0.001` and `0.004 (.^@C)^-1` respectively. Find resistance of the combination at a temperature of `50^@C`. What is the effective temperature coefficient of combination?A. `1/1000"degree"^(-1)`B. `1/250"degree"^(-1)`C. `1/500"degree"^(-1)`D. `3/1000"degree"^(-1)` |
| Answer» Correct Answer - B::C | |
| 2404. |
Two bulbs 60 W and 100 W designed for voltage 220 V are connected in series across 220 V source. The net power dissipated isA. `37.5` WB. 75WC. 80WD. 40 W |
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Answer» Correct Answer - A |
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| 2405. |
Two electic bulbs marked `25 W -220 V` and `100 W - 220 V` are connected in series to a 440 V supply. Which of the bulbs will fuse?A. 2.5 W bulb will fuseB. 100 W bulb will fuseC. both will fuseD. both will not fuse |
| Answer» Correct Answer - A | |
| 2406. |
Two electic bulbs marked `25 W -220 V` and `100 W - 220 V` are connected in series to a 440 V supply. Which of the bulbs will fuse?A. bothB. 100 WC. 25 WD. Neither |
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Answer» Correct Answer - C `P=Vi` `R=(V^(2))/(P)` |
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| 2407. |
Two electic bulbs marked `25 W -220 V` and `100 W - 220 V` are connected in series to a 440 V supply. Which of the bulbs will fuse?A. `25 W` bulbB. `100 W` bulbC. both bulbsD. none |
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Answer» Correct Answer - A (a) The resistance of `25 W` bulb will be higher than `100 W` bulb. So when connected in series across `440 V`, potential difference across `25 W` bulb will be more than `220 V`. Hence, this bulb is likely to fuse. |
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| 2408. |
A factory is served by a 220 V supply line. In a circuit protected by a fuse marked 10 A, the maximum number of 100 W lamps in parallel that can be turned on, isA. 11B. 22C. 33D. 66 |
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Answer» Correct Answer - B |
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| 2409. |
Two bulbs consume same energy when operated at `200 V and 300 V` , respectively . When these bulbs are connected in series across a dc source of `500 V`, thenA. ratio of potential differences across them is `3//2`B. ratio of potential differences across them is `9//4`C. ratio of powers consumed across them is `4//9`D. ratio of powers consumed across them is `2//3` |
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Answer» Correct Answer - C |
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| 2410. |
Two bulbs consume same energy when operated at `200 V and 300 V` , respectively . When these bulbs are connected in series across a dc source of `500 V`, thenA. ratio of potential difference across them is `3//2`B. ratio of potential difference across them is `4//9`C. ratio of power consumed across them is `4//9`D. ratio of power consumed across them is `2//3` |
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Answer» Correct Answer - B::C |
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| 2411. |
A heater is designed to operate with a power of `1000W` in a `100 V` line. It is connected in combination with a resistance of `10 Omega` and a resistance `R`, to a `100 V` mains as shown in figure. What will be the value of `R` so that the heater operates with a power of `62.5W`? |
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Answer» From `P=V^2/R` Resistance of heater, `R=V^2/P=((100)^2)/1000=10Omega` From `P=i^2 R` Current required across heater of power of `62.5 W`, `i=sqrt(P/R)=sqrt(62.5/10)=2.5A` Main current in the circuit `I=100/(10+(10R)/(10+R))=(100(10+R))/(100+20R)=(10(10+R))/(10+2R)` this current will distribute in inverse ratio of resistance between heater and R `:. i=(R/(10+R))I` or `2.5=(R/(10+R))[(10(10+R))/(10+2R)]=(10R)/(10+2R)` Solving this equation we get `R=5Omega` |
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| 2412. |
A 1 kQ heater is meant to operate at 200 V. (a) what is its resistance ? (b) How much power will it consume if the line voltage drops to 100 V? (c) how many units of electrical energy will it consume in a month (of 30 days) if it operates 10 hr daily at the specified voltage? |
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Answer» The resistance of an elecric appliance is given by `R=(V_(S)^(2))/(W) so, R=((200)^(2))/(1000)=40Omega` (b) The actual power consumed by an electric appliance is given by `P=((V_(A))/(V_(S)))^(2)xxW` so, `P=((100)/(200))^(2)xx1000=250W` (c) The total electrical energy consumed by an electric appliance in a specified time is given by `E=(sumW_(1)h_(1))/(1000)kWh` so, `E=(1000xx(10xx30))/(1000)=300kWh` |
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| 2413. |
A ` 1 k W` heater is meant to operate at `200 V`. (a) What is the resistance? (b) How much power will it consume if the line voltage drops to `100 V`? ( c ) How many units of electrical energy will it consume in a month `( of 30 days)` if it operates `10 h `daily at the specified voltage `( 200 V)`? |
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Answer» Here, power of electric heater, P = 1 kW = 1000 W , V = 220V (i) Resistance of heater, `R = V^(2)/P = (220)^(2)/1000 = 48.4 Omega` (ii) Current drawn by heater, `I= P/V = 1000/220 = 4.545 A` (iii) Rate of dissipation of energy, `H = P/J = 1000/4.2 = 238.1 cal.s^(-1)` (iv) New power of heater `= (V_(1)^(2))/R = (200)^(2)/48.4` `= 826.45 W` |
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| 2414. |
In the two electron circuit determine the readings of ideal ammeter `(A)` ideal voltmeter `(V)` |
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Answer» Correct Answer - (a)`7.5 V, 1.5A`(b) - 1.5V, 7.5A` In circuit (a) Here two cell the connected in opposition so .Effective emf of the circuit `epsilon = 9 - 6 = 3V`, total resistance of circuit `R = 1+ 1 = 2 Omega` current recorded by ideal ammeter `I =- (epsilon )/(R ) = (3)/(2) = 1.5 A` As the current `I` flow from position to negative terminal inside the cell of `6 V`, i.e , the cell `6V` is being charged so terminal potential difference of the cell is `V = epsilon + I r = 6 + 1.5 xx 1= 7.5 V` Reading of voltmeter `= 7.5 V` In circuit (b) Effective emf of the circuit `epsilon = 9 - 6 = 15V` total resistance of circuit `R = 1+ 1 = 2 Omega` current recorded by ideal ammeter `I = (epsilon )/(R ) = (15)/(2) = 7.5 A` As the current `I` flow from `-v` to terminal in `+v` terminal inside the cell of `6 V,` i.e , the cell `6V` is being charged so terminal potential difference of the cell is `V = epsilon - Ir = 6 7.5 xx 1= -1.5 V` Reading of voltmeter `= -1.5 V` Reading of ammeter `= 7.5 A` |
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| 2415. |
A microameter has a resistance of `100 omega` and a full scale range of ` 50 muA`. It can be used as a voltmeter or as a higher range ammeter provides a resistance is added to it . Pick the correct range and resistance combination(s)A. 50 V range with `10kOmega` resistance in seriesB. `10V` range with `200kOmega` resistance in seriesC. `5mA` range with `1Omega` resistance in parallelD. 10 mA range with `1Omega` resistance in parallel |
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Answer» Correct Answer - B::C To increases the range of ammeter a parallel resistance (called shunt) is requried which is given by `S=((i_(g))/(i-i_(g)))G` for option (c) `S=((50xx10^(-6))/(5xx10^(-3)-50xx10^(-6)))(100)=1Omega` To change it into voltmeter, a high resistance `R` is put in series where R is given by `R=(V)/(i_(g))-G` For option (b) `R=(10)/(50xx10^(-6))-100=200kOmega` |
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| 2416. |
Two identical fuses are rated at `10A`. If they are joinedA. in parallel, the combination acts as a fuse of rating `20 A`B. in parallel, the combination acts as a fuse of rating `5 A`C. in series, the combination acts as a fuse of rating `10A`.D. in series, the combination acts as a fuse of rating `20A`. |
| Answer» Correct Answer - A::C | |
| 2417. |
A conductor of length l and area of cross-section A has n number of electrons per unit volume of the conductor. The current flowing through conductor in time t is given byA. `I=(nAl e)/(t)`B. `I=n A l`C. `I=(e)/(nA l t)`D. `I= (n e )/(t)` |
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Answer» Correct Answer - A `I=(q)/(t)=(n A l e )/(t)` |
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| 2418. |
A stready current i is flowing through a conductor of uniform cross-section. Any segment of the conductor hasA. Zero chargeB. Only positive chargeC. Only negative chargeD. Charge proportional to current i |
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Answer» Correct Answer - A |
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| 2419. |
A stready current i is flowing through a conductor of uniform cross-section. Any segment of the conductor hasA. negative chargeB. positive chargeC. zero chargeD. none of these |
| Answer» Correct Answer - C | |
| 2420. |
A stready current i is flowing through a conductor of uniform cross-section. Any segment of the conductor hasA. zero chargeB. only positive chargeC. only negative chargeD. charge proportional to current `i` |
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Answer» Correct Answer - A |
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| 2421. |
In a straight conductor of uniform cross-section charege q is flowing for time t. Let s be the specific charge of an electron. The momentum of all the free electrons per unit length of the free electrons per unit length of the conductor, due to their drift velocity only isA. `(q)/(ts)`B. `((q)/(ts))^(2)`C. `sqrt((q)/(ts))`D. `qts` |
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Answer» Correct Answer - A `I=nAev_(d)` or `v_(d)=(I)/(nAe)=((q)/(t))/(nAe)` No. of free electrons per unit length of conductor `N=nAxx1` `therefore` Momentum of all the free electrons is `p=Nmv_(d)` |
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| 2422. |
Shown a conductor of length `l` having a circular cross section. The radius of cross section varies linearly form `a to b`. The resistivity of the material is`(rho)`.Assuming that `b-altltl`,find the resistance of the conductor. |
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Answer» Since radius of left end is a and that of right end is b, therefore increase in radius over length l is (b-a). Hence rate of increase of radius per unit length `=((b-a)/(l))` Increase in radius over length `x=((b-a)/(l))x` since radius at left end is a, radius at distance `x=r=a+((b-a)/(l))x` Area at this particualr section `A=pir^(2)=pi[a+((b-a)/(l))xx]^(2)` Hence curret density `J=(i)/(A)=(i)/(pir^(2))=(i)/(pi[a+(x(b-a))/(l)]^(2))` |
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| 2423. |
A straight conductor of uniform cross-section carries a current `I`. Let `s=` specific charge of an electron. The momentum of all the free electrons per unit length of the conductor, due to their drift velocity only, isA. i.sB. `sqrt(i)//(s)`C. i/sD. `(i/s)^(2)` |
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Answer» Correct Answer - C Current flowing in a straight conductor `i="nAev"_(d)` or `v_(d)=(i)/("nAe")` Total number of free electrons in the unit length of conductor, `N = nA xx l = nA xx 1` (length, l = 1m). Total linear momentum of all the free electrons per unit length `= N_(m)v_(d)=nAm xx (i)/("nAe")=(i0)/((e//m))=(i)/(s)` where, s is charge per unit mass (e/m). |
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| 2424. |
A straight conductor of uniform cross section carries a time varying current, which varies at the rate `di//dt = I`. If `s` is the specific charge that is carried by each charge carrier of the conductor and l is the length of the conductor, then the total force experienced by all the charge carrier per unit length of the conductor due to their drift velocities only isA. I sB. `(I)/(s)`C. `sqrt((I)/(s))`D. `((I)/(s))^2` |
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Answer» Correct Answer - B `I = n e A v_d, s = ( e)/(m)` Momentum of all free electrons//length `p = ((nAl)mv_d)/(l) = (n A ev_d m)/(e)` =`(n e Av_d)/(e//m) = (I)/(s)`. |
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| 2425. |
Calculate the number of free electrons passing through the filament of an electric lamp in one hour when the current through the filament is 0.32 A. |
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Answer» Correct Answer - `7.2xx10^(21)` |
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| 2426. |
`10^(20)` electrons, each having a charge of `1.6xx10^(-19) C, pass from a point A towards another point B in 0.1 s. What is the current in ampere ? What is its direction ? |
| Answer» Correct Answer - Given, `n | |
| 2427. |
One billion electrons pass from a point P towards another point Q in `10^(-3)` s. What is the current in ampere ? What is its direction ? |
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Answer» Correct Answer - `1.6xx10^(-7)` A, direction of current is form Q to P |
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| 2428. |
70 calories of heat required to raise the temperature of 2 moles of an ideal gas at constant pressure from `30^@Cto 35^@C.` The amount of heat required (in calories) to raise the temperature of the same gas through the same range `(30^@C to 35^@C)` at constant volume is:A. 30B. 50C. 70D. 90 |
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Answer» Correct Answer - B |
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| 2429. |
The value of the resistance `R` in figure is adjusted such that power dissipated in the `2Omega` resistor is maximum. Under this condition A. R = 0B. `R = 8 Omega`C. power dissipated in the `2 Omega` resistors is 72 WD. power dissipated in the `2 Omega` resistors is 8 W |
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Answer» Correct Answer - A::C |
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| 2430. |
The value of the resistance `R` in figure is adjusted such that power dissipated in the `2Omega` resistor is maximum. Under this condition A. `R=0`B. `R=8 Omega`C. power dissipated in the `2 Omega` resistor is `72 W`.D. power dissipated in the `2 Omega` resistor is `8 W`. |
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Answer» Correct Answer - A::C Power will be maximum in `2 Omega` when current through it is maximum. It happens when `R=0` `I_(max)=6A` `P_(2Omega)=(6)^(2)+2=72` watt |
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| 2431. |
Mark out the correct optionsA. An ammeter should have small resistanceB. An ammeter should have large resistanceC. An voltmeter should have small resistanceD. An voltmeter should have large resistance |
| Answer» Correct Answer - A::D | |
| 2432. |
A galvanometer is to be converted into an ammeter or voltmeter. In which of the following cases the resistance of the device is largest?A. an ammeter of range `10A`B. a voltmeter of range `5V`C. an ammeter of range `5A`D. a voltmeter of range `10 V` |
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Answer» Correct Answer - D For conerversion in voltmeter `R=(V/(I_(g))-1)G` For conversion in ammeter `S=G/((I/(I_(g))-1))` |
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| 2433. |
A galvanometer is to be converted into an ammeter or voltmeter. In which of the following cases the resistance of the device is largest?A. an ammeter of range `10A`B. a voltmeter of range `5V`C. an ammeter of range `5A`D. a voltmeter of range `10V` |
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Answer» Correct Answer - D For making voltmeter of higher range, moere resistance is required. |
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| 2434. |
Assertion : Current is passed through a metallic wire, heating it red. Half of its portion is cooled by cold water jacket, then rest of the half portion become more hot. Reason: Resistances decreases due to decrease in temperature and so current through wire increases.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 trueD. If assertion is true but reason is false. |
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Answer» Correct Answer - A When half the portion of the wire is cooled, its resistance decreases due to decrease in temperature. As a result of this total resistance of circuit decrease, so current through each portion of wire increases. That is, rest of the half portion becomes more hot. |
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| 2435. |
Assertion : In metre bridge experiment, a high resistance is always connected in series with a galvanometer. Reason : As resistance increases, current through the circuit increases,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.D. If assertion is true but reason is false. |
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Answer» Correct Answer - C The resistance of the galvanometer is fixed. In meter bridge experiments, to protect the galvanometer from a high current, high resistance is connected to the galvanometer. |
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| 2436. |
Circuit diagram of metre bridge is shown in figure. The null point is found at a distance of 40cm from A. it now a resistance of `12Omega` is connected in parallel with S, the null point occurs at 64cm. |
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Answer» Fro m the first balance point `(R)/(S)=(40)/(100-40)=(40)/(60)` If `12Omega` is connected in parallel with S then `S_(eq)=(12S)/(12+S)` so from new balance conditions `(R)/(S_(eq))=(64)/(36)` `rArr (R(S+12))/(12S)=(64)/(36)rArr((40)/(36))((S+12)/(12))=(64)/(36)rArrS+12=32rArrS=20Omega` So `R=((2)/(3))(20)=(40)/(3)Omega` |
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| 2437. |
Assertion : In metre bridge experiment, a high resistance is always connected in series with a galvanometer. Reason : As resistance increases, current through the circuit increases,A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is falseD. If the assertion and reason both are false |
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Answer» Correct Answer - C |
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| 2438. |
Assertion: Electric field inside a current carrying wire is zero. Reason: Net charge on the current carrying wire is non zero.A. If both Assertion `&` Reason are True `&` the Reason is a correct explanantion of the AssertionB. If both Assertion `&` Reason True but Reason is not a correct explanation of the AssertionC. If Assertion is True but the Reason is FalseD. If both Assertion `&` Reason are False. |
| Answer» Correct Answer - 4 | |
| 2439. |
Given the resistances of `1 Omega, 2 Omega, 3 Omega` how will be combine them to get an equivalent resistance of (i) (11/3) `Omega` (ii) (11/5) `Omega` (iii) `6 Omega` (iv) (6//11) `Omega` |
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Answer» It is to be noted that (a) the effective resistance of parallel combination of resistors is less than the individual resistance and (b) the effective resistance of series combination of resistors is more than individual resistance. Case (i) Parallel combination of `1 Omega` and `2 Omega` is connected is series with `3 Omega` Effective resistance of `1 Omega` and `2 Omega` in parallel will be given by `RP = (1 xx 2)/(1 + 2) = (2)/(3) Omega` `:.` Equivalent resistance of `(2)/(3) Omega` and `3 Omega` in series `= (2)/(3) + 3 = (11)/(3) Omega` , Case (ii) : Parallel combination of `2 Omega` and `3 Omega` is connected in series with `1 Omega` Equivalent resistance of `2 Omega` and `3 Omega` in parallel `= (2 xx 3)/(2 + 3) = (6)/(5) Omega` Equivalent resistance of `(6)/(5) Omega` and `1 Omega` in series `= (6)/(5) + 1 = (11)/(5) Omega` Case (iii) : All the resistance are to be connected in series show `:.` Equivalent resistance `= 1 + 2 + 3 = 6 Omega` (Case (iv) : All the resistance are to be connected in parallel `:.` Equivalent resistance (R ) is given by `(1)/(R ) = (1)/(1) + (1)/(2) + (1)/(3)` `= (6 + 3 + 2)/(6) = (11)/(6)` or `r = (6)/(11) Omega` |
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| 2440. |
A dry cell delivering 2 A h a termin al voltage is 1.41V. What is the internal resistance of the cell if its open- circuit voltage is 1.59V? |
| Answer» From V=E-ir we have `r=(E-V)/(1)=(1.59-1.41)/(2)=0.09Omega` | |
| 2441. |
Assertion : Voltameter measures current more accurately than ammeter. Reason : Relative error will be small if measured from voltameter.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is falseD. If the assertion and reason both are false |
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Answer» Correct Answer - A |
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| 2442. |
Assertion : The resistance of super-conductor is zero. Reason : The super-conductors are used for the transmission of electric power.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is falseD. If the assertion and reason both are false |
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Answer» Correct Answer - B |
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| 2443. |
Assertion: A voltmeter must b e connected in parallel in a circuit and it should have a high resistance. Reason: The voltmeter in the circuit must not affect the P.D. it is to measure.A. If both Assertion `&` Reason are True `&` the Reason is a correct explanantion of the AssertionB. If both Assertion `&` Reason True but Reason is not a correct explanation of the AssertionC. If Assertion is True but the Reason is FalseD. If both Assertion `&` Reason are False. |
| Answer» Correct Answer - 1 | |
| 2444. |
Statement I: Since all the current coming to our house returns to powerhouse ( as current travels in a closed loop), there is no need to pay the electricity bill. Statement II: The electricity bill is paid for the power used , not for the current used.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 Power used `=i^(2)R` hence Power is consumed not the current. |
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| 2445. |
A and B are two square plates of same metal and same thickness but length of B is twice that of A . Ratio of resistances of A and B is A. `4:1`B. `1:4`C. `1:1`D. `1:2` |
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Answer» Correct Answer - C |
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| 2446. |
Assertion : Electric field outside the conducting wire which carreis a constant is zero. Reason : Net charge on conducting wire is zero.A. If both Assertion `&` Reason are True `&` the Reason is a correct explanantion of the AssertionB. If both Assertion `&` Reason True but Reason is not a correct explanation of the AssertionC. If Assertion is True but the Reason is FalseD. If both Assertion `&` Reason are False. |
| Answer» Correct Answer - 1 | |
| 2447. |
Assertion : Electric field outside the conducting wire which carreis a constant is zero. Reason : Net charge on conducting wire is zero.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is falseD. If the assertion and reason both are false |
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Answer» Correct Answer - A |
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| 2448. |
Assertion : Electric field outside the conducting wire which carreis a constant is zero. Reason : Net charge on conducting wire is zero.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 When current flow through a conductor it always remains uncharged, hence no electric field is produced outside it. |
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| 2449. |
Following figures show four situations in which positive and negative charges move horizontaly through a region and give the rate at which each charge moves. Rank the situations according to the effective current through the region greatest first. A. `i = ii = iii = iv`B. `i gt ii gt iii gt iv`C. `i = ii =iii gt iv`D. `i = ii = iii lt iv` |
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Answer» Correct Answer - C |
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| 2450. |
Assertion : Electric field outside the conducting wire which carreis a constant is zero. Reason : Net charge on conducting wire is zero.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) When current flows through a conductor is always remains unchanged, Hence no electric field is produced outside it. |
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