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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.
| 501. |
If each of the resistance in the network in figure `R`, the equivalent resistance between terminals `A` and `B` is A. `5R`B. `3R`C. RD. `R//2` |
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Answer» Correct Answer - C |
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| 502. |
In the circuit shown here, the readings of the ammeter and voltmeter are A. 6 A, 60 VB. `0.6` A, 6 VC. 6 A, 6 VD. 6/11 A, 6/11 V |
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Answer» Correct Answer - D |
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| 503. |
The equivalent resistance of the following diagram A and B is A. `(2)/(3)Omega`B. `9 Omega`C. `6 Omega`D. None of these |
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Answer» Correct Answer - D |
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| 504. |
Find the equivalent resistance across `AB` : A. 1`Omega`B. 2`Omega`C. 3`Omega`D. 4`Omega` |
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Answer» Correct Answer - A |
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| 505. |
A resistor circuit is constructed such that 12 resistor are arranged to form a cube as shown in fig. Each resistor has a resistance of `2Omega.` The potential difference of 30V is applied across two of the opposing points as shown In question 20, the potential difference between the points C and G is |
| Answer» Correct Answer - A | |
| 506. |
In the circuit shown in fignre-3.341, find the steady state charge on capacitor `C_(1)` A. `2 muC`B. `3 muC`C. `4 muC`D. zero |
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Answer» Correct Answer - A,C |
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| 507. |
Find the time constant for the given RC circuits in correct order (in `mus`) `R_(1)=1Omega,R_(2)=2Omega,C_(1)=4muF,C_(2)=2muF`A. `18,4,(8)/(9)`B. `18,(8)/(9),4`C. `4,18,(8)/(9)`D. `4,(8)/(9),18` |
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Answer» Correct Answer - B `tau=CR` `tau_(1)=(C_(1)+C_(2))(R_(1)+R_(2))=18mus` `tau_(2)=((C_(1)C_(2))/(C_(1)+C_(2)))(R_(1)R_(2))/(R_(1)+R_(2))=(8)/(6)xx(2)/(3)=(8)/(9)mus` `tau_(3)=(C_(1)+C_(2))((R_(1)R_(2))/(R_(1)+R_(2)))=(6)((2)/(3))=4mus` |
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| 508. |
What is equivalent time constant of RC circuit shown in fignre-3.340 ? A. 1.5 RCB. 3RCC. 2RCD. `(RC)/(2)` |
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Answer» Correct Answer - A,D |
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| 509. |
If the length of the filament of a heater is reduced by `10%`, the power of the heater willA. increaases by about `9%`B. increases by about `11%`C. increases by about `19%`D. decreases by about `10%` |
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Answer» Correct Answer - B `P=(V^(2))/Rimplies(P_(2))/(P_(1))=(R_(1))/(R_(2))=(l_(1))/(l_(2))=l/(0.9l)=10/9` `P_(2)=1.11 P` Increased about `11%` |
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| 510. |
N batteries each of emf E and internal resistance r are first connected in series and then in parallel to an external resistance R. if current through R is same in both cases then.A. `R=r^(2)//N`B. `R-Nr//2`C. `R=Nr`D. `R=r` |
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Answer» Correct Answer - 4 In serie s curren t `=(NE)/(Nr+R)` In parallel, current `=(E)/((r//N)+R)=(NE)/(r+NR)` since the currents in the two cases are equal, `(NE)/(Nr+R)=(NE)/(r+NR)` or `Nr+R=r+NR` or `Nr-r=NR-R` or `r(N-1)=R(N-1)` or r=R `therefore R=r`. |
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| 511. |
Two metallic s pheres of radii `R_(1)` and `R_(2)` are connected b y a thin wire. If `+q_(1)~ and `+q_(2)` are the charges on the two s pheres thenA. `q_(1)// q_(2)=R_(1)^(2)//R_(2)^(2)`B. `q_(1)//q_(2)=R_(1)//R_(2)`C. `q_(1)//q _(2)=R_(1)^(3)//R_(2) ^(3)`D. `q_(1)//q_(2)=(R_(1)^(2)-R_(2)^(2))//(R_(1)^(2)+R_(2)^(2))` |
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Answer» Correct Answer - 2 When the two metallic spheres are connected by a wire, they are at the saem potential for a spherical conductor capacitiy `C=4piepsilon_(0)R`. ltbr. `therefore` Potential `V=("charge")/("capacity")` `therefore V_(1)=V_(2)` or `(q_(1))/(4pi epsilon_(0)R_(1))=(q_(2))/(4piepsilon_(0)R_(2))` `therefore (q_(1))/(q_( 2))=(R_(1))/(R_(2))` |
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| 512. |
The potential difference in open circuit for a cell is 2.2 votls . When a `4 ohm`resistor is connected between its two electrodes that potential difference becomes 2 volts. The internal resistance of cell will beA. `1 ohm`B. `0.2 ohm`C. `2.5 ohm`D. `0.4 ohm` |
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Answer» Correct Answer - D (d) The potential difference in open circuit for the cell is 2.2 volt, it mens the emf of the battery should be 2.2 volts. Now a `4 ohm` resistor is connected between its two electrodes, then the potential difference across the battery becomes 2 volts `V = epsilon - ir implies 2 = 2.2 - ir` `implies ir = 0.2 V` But `i= (epsilon)/(r + R) = (2.2)/(r + 4)` Putting the value of `i` in (i), we get `r =0.4 ohm`. |
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| 513. |
A Steady current flows in a metalic conductor of non uniform cross section. The quantity/quantities which remain constant along the length of the conductor is/areA. Current electric field and drift speedB. Drift speed onlyC. Current and drift speedD. Current only |
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Answer» Correct Answer - D (d) If `E` is the electric field, then current density `J = sigma E` Also we know that current density `J = (i)/(A)` Hence `j` is difference for different area of cross-section. When `j` is different, then `E` is also different. Thus `E` is not constant. The drift velocity `v_(d)` is given by `v_(d) = (j)/(n e)` only current `i` will be constant. |
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| 514. |
Current through the `5Omega` resistor is A. 2AB. 4AC. zeroD. 1A |
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Answer» Correct Answer - A |
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| 515. |
A cell which has an emf `1.5` V is connectedin series with an external resistance of `10Omega`. If the potential difference across the cell is `1.25` V, then the internal resistance of the cell is `("in" Omega)`A. 2B. `0.25`C. `1.5`D. `0.3` |
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Answer» Correct Answer - A |
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| 516. |
`62.5 XX 10^(18)` electorns per second are flowing through a wire of area of cross-section `0.1 m^(2)`, the value of current flowing will beA. `1 A`B. `0.1 A`C. `10 A`D. `0.11 A` |
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Answer» Correct Answer - C (c ) `i= (n e)/(t) = (62.5 xx 10^(18) xx 1.6 xx 10^(-19))/(1) = 10` ampere |
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| 517. |
`62.5 xx 10^(18)` electrons per second are flowing through a wire of area of cross-section `0.1m^(2)`, the value of current flowing wil beA. 1AB. `0.1A`C. `10A`D. `0.11A` |
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Answer» Correct Answer - C |
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| 518. |
We have two wire `A` and `B` of the same mass and the same material. The diameter of the wire `A` is half of that `B`. If the resistance of wire `A` is `24 ohm` them the resistance of wire `B` will beA. 12 ohmB. 3.0 ohmC. 1.5 ohmD. None of the above |
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Answer» Correct Answer - C |
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| 519. |
The electron with change `(q=1.6xx10^(-19)C)` moves in an orbit of radius `5xx10^(-11)`m with a speed of `2.2xx10^(6)ms^(-1)`, around an atom. The equivalent current isA. `1.12xx10^(-6)A`B. `1.12xx10^(-3)A`C. `1.12xx10^(-9)A`D. `1.12A` |
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Answer» Correct Answer - B |
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| 520. |
Ampere- hour is the unit ofA. quantity of chargeB. potentialC. energy currentD. |
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Answer» Correct Answer - A `l=q/t` `q=lxxt` |
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| 521. |
Effective resistance between A and B is A. `15 Omega`B. `5 Omega`C. `(5)/(2) Omega`D. `20 Omega` |
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Answer» Correct Answer - B |
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| 522. |
The effective resistance of two resistors in parallel is ` (12)/(7) Omega`. If one of the resistors is disconnected the resistance becomes `4 Omega`. The resistance of the other resistor isA. `4 Omega`B. `3 Omega`C. `(12)/(7) Omega`D. `(7)/(12) Omega` |
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Answer» Correct Answer - B |
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| 523. |
The effective resistance of two resistors in parallel is ` (12)/(7) Omega`. If one of the resistors is disconnected the resistance becomes `4 Omega`. The resistance of the other resistor isA. `4Omega`B. `3Omega`C. `(12)/(7)Omega`D. `(7)/(12)Omega` |
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Answer» Correct Answer - B Given, `R_(1)= 4Omega` Using the relation for parallel combination of two resistors `(R_(1)R_(2))/(R_(1)+R_(2))=(12)/(7), R_(1)=4 Omega, R_(2) = ?` or `" " (4R_(2))/(4+R_(2))=(12)/(7)` or `28R_(2)=48+12 R_(2)` or `16 R_(2)=48 rArr R_(2) = 3 Omega` |
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| 524. |
In the network shown in the figure, each of the resistance is equal to `2 Omega`. The resistance between the points and `B` is A. `1 Omega`B. `4 Omega`C. `3 Omega`D. `2 Omega` |
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Answer» Correct Answer - D |
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| 525. |
A cell of constant emf first connected to a resistance `R_(1)` and then connected to resistance `R_(2)`. If power deliverd in both cases is same, then the internal resistance of the cell isA. `sqrt(R_(1)R_(2))`B. `sqrt((R_(1))/(R_(2)))`C. `(R_(1)-R_(2))/(2)`D. `(R_(1)+R_(2))/(2)` |
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Answer» Correct Answer - A Current given by cell `l= E/(R+r)` Power delivered in firstcase ltbegt `P_(1)=l^(2)R_(1)=((E )/(R_(1)+r))^(2)R_(1)` Power delivered in first case `P_(2)=l^(2)R_(2)=((E )/(R_(2)+r))^(2)R_(2)` Power delivered is same in the both cases. `((E )/(R_(1)+r))^(2)R_(1)=((E )/(R_(2)+r))^(2)R_(2)`ltbegt `((E )/(R_(1)+r))^(2)R_(1)=((E )/(R_(2)+r))^(2)R_(2)` `R_(1)(R_(2)^(2)+r^(2)+2R_(2)r)=R_(2)(R_(1)^(2)+r^(2)+2R_(1)r)` `R_(1)R_(2)^(2)+R_(1)r^(2)+2R_(1)R_(2)r=R_(2)R_(1)^(2)+R_(2)r^(2)+2R_(1)R_(1)r` `R_(1)R_(2)^(2)-R_(2)R_(1)^(2)=R_(2)R_(1)^(2)-R_(1)r^(2)` `R_(1)R_(2)(R_(2)-R_(1))=r^(2)(R^(2)-R_(1))` `r=sqrtR_(1)R_(2)` |
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| 526. |
Five resistances are connected as shown in the figure. The effective resistance between the points A and B is A. `(10)/(3) Omega`B. `(20)/(3) Omega`C. `15 Omega`D. `6 Omega` |
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Answer» Correct Answer - A |
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| 527. |
In the arrangement of resistances shown below, the effective resistance between point `A` and `B` is A. `20 Omega`B. `30 Omega`C. `90 Omega`D. `110 Omega` |
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Answer» Correct Answer - A |
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| 528. |
Find the equivalent resistance between the points a and b A. `2 Omega`B. `4 Omega`C. `8 Omega`D. `16 Omega` |
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Answer» Correct Answer - B |
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| 529. |
Calculate the equivalent resistance between points A and B in each of the following networks of resistors : |
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Answer» Correct Answer - `(a) 12 Omega (b) 40//3 Omega (c ) 2 Omega (d) 10//3 Omega (e ) 16 Omega (f) 5 Omega` |
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| 530. |
A resistance of `R Omega` draws current from a potentiometer. The potentiometer has a total resistance `R_(0) Omega`. A voltage V is supplied to the potentiometer. Derive an expression for the voltage fed into the circuit when the slide contact is in the middle of potentiometer. |
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Answer» When slide is in the middle of the potentiometer wire, only half of the resistance of potentiometer wire `( = R_(0)//2)` will be between the points A and B. Hence effective resistance `(R_(1))` between A and B is `1/R_(1) = 1/R + 1/(R_(0)//2) or R_(1) = (R_(0)R)/(R_(0) + 2R)` Total resistance between A and C `= R_(1) + R_(0)/2` Current through the potentiometer wire will be `I=V/(R_(1) + R_(0)//2) = 2V/(2 R_(1) +R_(0))` The voltage `V_(1)` taken from the potentiometer will be the product of current I and resistance `R_(1)` , i.e., `V_(1) = IR_(1) = ((2V)/(2R_(1) + R_(0))) xx R_(1)` `=(2V)/((2(R_(0)R)/(R_(0)+2R))+R_(0)) xx (R_(0)R)/(R_(0)+2R)` ` = ( 2 VR)/( 2R + R_(0) + 2R) = (2VR)/(R_(0) +4R)` |
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| 531. |
The current drawn from the battery shown in the figure is ltBrgt A. `(V)/(R)`B. `(V)/(2R)`C. `(2V)/(R)`D. `(3V)/(2R)` |
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Answer» Correct Answer - B |
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| 532. |
A resistance R is to be measured using a meter bridge. Student chooses the standared resistance S to be `100 Omega`. He finds the null point at `l_(1)=2.9cm`. He is told to attempt to improve the accuracy. Which of the folllowing is a useful way?A. He should measure `l_(1)` more accuratelyB. He should change S to `1000Omega` and repeat the experimentC. He should change S to 3`Omega` and repeat the experimentD. He should given up hope of a more accurate measurement with a meter bridge |
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Answer» Correct Answer - C |
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| 533. |
Potentiometer is an ideal voltmeter as voltmeter draws some current through the circuit while potentiometer needs no current to work. Potentiometer works on the principle of e.m.f comparison. In working conditions, a constant current flows through out the wire of potentiometer using standard cell of e.m.f. `e_(1)`. The wire of potentiometer is made of uniform material and cross sectional area and it has uniform resistance per unit length. the potential gradient, depends upon the current in the wire. A potentiometer with a cell of e.m.f `2 V` and internal resistance `0.4 Omega` is used across the wire `AB`. A standard cadmium cell of e.m.f. `1.02 V` gives a balance point at `66.3 cm` length of wire. The standard cell is the balance point found similarly turns out to be `82.3 cm` length of the wire. The length of potentiometer wire `AB` is `1m`.If the resistance is connected across the cell `e`, the balancing length will be: |
| Answer» Correct Answer - C | |
| 534. |
In the circuit shown in the figure, the power dissipated in the circuit is `P_(0)` if an ideal cell is connected across A and B. Same power is dissipated in the circuit if the same cell is connected across C and D. When the cell was connected across A and D or across B and C, the power dissipated in the circuit is found to be `3P_(0)`. Calculate the power dissipated in the circuit if the cell is connected across A and C. |
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Answer» Correct Answer - `((15+6sqrt(7))/(16+6sqrt(7)))P_(0)` |
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| 535. |
When 10 cells in series are connected to the ends of a resistance of `59 Omega`, the current is found to be 0.25 A, but when the same cells after being connected in parallel are joined to the ends of a `0.05 Omega `, the current is 25 A. Calculate the internal resistance and emf of each cell. |
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Answer» Correct Answer - `0.1 Omega, 1.5 V` |
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| 536. |
In Figure `E_1=12V` and `E_2=8V` (a) What is the direction of the current in the resistor?(b) Which battery is doing positive work?(c) Which point, `A` or `B`, is at the higher potential? |
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Answer» Correct Answer - A::B::C a. `E_1gtE_2` therefore, net curren is anti clockwise or from `B` to` A`. b.current through `E_1` is normal. Hence,it is doing the positive work. c. Current flows from `B` to `A` `:. V_BgtV_A` |
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| 537. |
An ideal monoatomic gas of one mole undergoes a cyclic process ADBCA as shown in figure. X and Y are two isothermal curves at temperature `211^(@)C` and `127^(@)C` respectively. If the work done by the gas in cyclic process is `lambdaR`, then find the value of `lambda`. (R is the gas constant) |
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Answer» Correct Answer - D |
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| 538. |
An ideal monoatomic gas undergoes a process in which its internal energy U and density `rho` vary as `Urho`= constant. The ratio of change in internal energy and the work done by the gas is |
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Answer» Correct Answer - C |
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| 539. |
Two idential container joined by a small pipe initially contain the same gas at pressure `p_(0)` and absolute temperature `T_(0)`. One container is now maintained at the same temperature while the other is heated to `2T_(0)`. The common pressure of the gas |
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Answer» Correct Answer - D |
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| 540. |
Two moles of a diatomic ideal gas is taken through `pT=` constant. Its temperature is increased from T to 2T. Find the work done by the system? |
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Answer» Correct Answer - D |
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| 541. |
In the circuit shown in figure the current flowing through 25 V cell is A. 7.2AB. 10AC. 12AD. 14.2A |
| Answer» Correct Answer - C | |
| 542. |
In a circuit a cell with internal resistance r is connected to an external resistance R. The condition for the maximum current that drawn from the cell isA. `R=r`B. `Rltr`C. `Rgtr`D. `R=0` |
| Answer» Correct Answer - D | |
| 543. |
If an ammeter is connected in parallel to a circuit, it is likely to be damaged due to excessA. CurrentB. VoltageC. ResistanceD. All of these |
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Answer» Correct Answer - A |
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| 544. |
How many electrons pass through a lamp in one minute, if the current is 300 m A? |
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Answer» Here, `I= 300 mA = 300 xx 10^(-3)A` , `t=60 s, n=?` `I= ("ne")/(t)` or `n= (It)/(e) = ((300 xx 10^(-3))xx60)/(1.6 xx 10^(-19))` `1.125 xx 10^(20)` |
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| 545. |
An electron gun emits `2.0xx10^(16)` electrons per second. What is the associated current ? |
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Answer» Correct Answer - `3.2xx10^(-3)`A |
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| 546. |
If `10^(6)` electrons pass from a point towards another point B in a conductor in one microsecond. Find the magnitude and direction of current. Give charge of an electron is `1.6 xx 10^(-19)` C. |
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Answer» Here, `n=10^(6) , t=10^(-6)s`, `e =1.6 xx 10^(-19)C` Now, current `I= ("ne")/(t) = (10^(6) xx (1.6xx10^(-19)))/10^(-6)` `=1.6 xx 10^(-7)A` The direction of current is from B to A |
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| 547. |
In an atom an electron revolves around the nucleus in a circular orbit at the rate of `6xx10^(15)` revolutions per second. Calculate the equivalent current in milliampere. Take value of electromic charge `= 1.6xx10^(-19)C`. |
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Answer» Here, `v=6 xx 10^(15) s^(-1)`, `e=1.6 xx10^(-19)C` Equivalent current I=ev `=(1.6 xx 10^(-19))xx6xx10^(15)=9.6xx10^(-4)A` `=9.6 xx10^(-4) xx 10^(3)mA` `=0.96 mA` |
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| 548. |
In a hydrogen atom, the electron makes about `0.6xx10^(16)` revolutions per second around the nucleus. Determine the average current at any point on the orbit of the electron. |
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Answer» Correct Answer - 0.96 mA |
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| 549. |
If `2 xx 10^(20)` electrons pass through a loop in one minute, What is the current in milliampere ? |
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Answer» Correct Answer - `533 mA` Here , `n = 2 xx 10^(20), t = 60 s` `I = (n e)/(t) = ((2 xx 10^(20))xx (1.6xx 10^(-19)))/(60)` `= 0.533 A = 533 mA` |
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| 550. |
When there is an electric current through a conducting wire along its length, then an electric field must existA. Outside the wire but normal to itB. Outside the wire but parallel to itC. Inside the wire but parallel to itD. Inside the wire but normal to it |
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Answer» Correct Answer - C |
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