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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.
| 301. |
Assertion : If three identical bulbs are connected in series as shown in figure then on closing the switchs. Bulb `C` short circuited and hence illumination of bulbs `A` and `B` decreases. Reason : Voltage on `A` and `B` decreases 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 - D (d) When switch `S` is closed, bulb `C` is short circuited, so voltage `V` distributes only in two parts i.e., voltage on Bulb `A` and `B` increases as compared previously. Hence illumination of Bulb `A` and `B` increases. |
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| 302. |
An electric kettle has two heating coils. When one of them is switched on water in it boils in 6 minutes and when other is switched on water boils in 4 minutes. In what time will the water boil if both coil are switched on simultaneouslyA. 1.6 minB. 2.8 minC. 2.4 minD. 3 min |
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Answer» Correct Answer - C `t_(S)=t_(1)+t_(2),t_(P)=(t_(1)t_(2))/(t_(1)+t_(2))` |
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| 303. |
What is the most proable cause of super- conductivity? |
| Answer» The cause of super-conductivity is that, the free electrons in super-conductor are no longer independent but become mutually dependent and coherent when critical temperature is reached. The ionic vibrations which could deflect the free electrons in metals are unable to deflect this coherent or cooperative cloud of electrons in super-conductors. That is why, there is no resistance offered by the super-conductors to the flow of electrons. | |
| 304. |
Three equal resistors connected in series across a source of emf together dissipate 10W of power. What would be the power dissipated if te same resistors are connected in parallel across the same source of emf?A. 10 WB. 30 WC. 90 WD. `(10//3) W` |
| Answer» Correct Answer - C | |
| 305. |
n resistors each of resistance R first combine to give maximum effective resistgance and then combine to give minimum. The ratio of the maximum resistance isA. nB. `n^(2)`C. `n^(2)-1`D. `n^(3)` |
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Answer» Correct Answer - B To get maximum equivalent resistance all resistances must be connected in series `therefore (R_("eq"))_("max")=R+R+R...n"times"=nR` To get minimum equivalent resistance all resistances myst be connected in parallel. `therefore (1)/(R_("eq"))_("min")=(1)/(R)+(1)/(R)+....n"time", (1)/(R_("eq"))_("min")=(n)/(R)` `(R_("eq"))_("min")=(R)/(n)therefore (R_("eq"))_("max")/(R_("eq"))_("min")=(nR)/(R//n)=n^(2)` |
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| 306. |
A thermodynamic system undergoes cyclic process `ABCDA` as shown in figure. The work done by the system is A. `p_(0)V_(0)`B. `2p_(0)V_(0)`C. `(p_(0)V_(0))/(2)`D. zero |
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Answer» Correct Answer - D |
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| 307. |
When a wire drawn until its radius decreases by `3%` then percentage of increase in resistance is-A. `10%`B. `9%`C. `6%`D. `12%` |
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Answer» Correct Answer - D `Rprop(1)/(r^(4))` |
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| 308. |
Figure shows three resistor configurations `R_(1),R_(2)` and `R_(3)` connected to `3V` battery. If the power dissipated by the configurations `R_(1),R_(2)` and `R_(3)` is `P_(1),P_(2)` and `P_(3)` respectively, then A. `P_(1)gtP_(2)gtP_(3)`B. `P_(1)gtP_(3)gtP_(2)`C. `P_(2)gtP_(1)gtP_(3)`D. `P_(3)gtP_(2)gtP_(1)` |
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Answer» Correct Answer - C `P_(1)=(V^(2))/(R_(1))impliesP_(1)=((3)^(2))/1impliesP_(1)=9 W` `P_(2)=(V^(2))/(R_(2))impliesP_(2)=((3)^(2))/((1//2))impliesP_(2)=18 W` `P_(3)=(V^(2))/(R_(3))impliesP_(3)=((3)^(2))/2impliesP_(3)=9/2 W` So `P_(2)gtP_(1)gtP_(3)` |
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| 309. |
Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in temperature. If at room temperature, `100W, 60W and 40W` bulbs have filament resistances `R_(100), R_(60) and R_(40)`, respectively, the relation between these resistances isA. `1/(R_(100))=1/(R_(40))+1/(R_(60))`B. `R_(100)=R_(40)+R_(60)`C. `R_(100)gtR_(60)gtR_(40)`D. `1/(R_(100))gt1/(R_(60))gt1/(R_(40))` |
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Answer» Correct Answer - D `P=(V^(2))/Rimplies1/R=P/(V^(2))` `1/(R_(100))gt1/(R_(60))gt1/(R_(40))` |
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| 310. |
Statement-1 : In a Meter Bridge experiment, null point for an unknown resistance is measured. Now, the unknown resistance is put inside an enclosure maintained at a higher temperature. The null point can be obtained at the same point as before by decreasing the value of the standard resistance. Statement-2 : Resistance of metal increases with increase in temperature.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 Since unknown resistance increases so keep balance point same standard resistance has to be increases. |
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| 311. |
Statement-1 : In a Meter Bridge experiment, null point for an unknown resistance is measured. Now, the unknown resistance is put inside an enclosure maintained at a higher temperature. The null point can be obtained at the same point as before by decreasing the value of the standard resistance. Statement-2 : Resistance of metal increases with increase in temperature.A. STATEMENT-1 Is True, STATEMENT-2 is True, STATEMENT-2 is a correct explaintion for STATEMENT-1B. STATEMENT-1 is True, STATEMENT-2 is True, STATEMENT-2 is a correct explanation for STATEMENT-1C. STATEMENT-1 is True, STATEMENT-2 is FalseD. STATEMETNT-1 is False, STATEMETN-2 is True. |
| Answer» Correct Answer - D | |
| 312. |
In a potentiometer of ten wires each of `1m`, the balance point is obtained on the sixth wire. To shift the balance point to eighth wire, we shouldA. increase resistance in the primary circuitB. decrease resistance in the primary circuitC. decrease resistance in series with the cell whose emf.has to be measured.D. increase resistance in series with the cell whose emf. Has to be measured. |
| Answer» Correct Answer - A | |
| 313. |
In given figure, the potentiometer wire AB has a resistance of `5 Omega` and length 10 m . The balancing length AM for the emf of 0.4 V is A. `0.4m`B. `4m`C. `0.8m`D. `8m` |
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Answer» Correct Answer - D |
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| 314. |
A voltmeter essentially consists ofA. A high resistance, in series with a galvanometerB. A low resistance, in series with a galvanometerC. A high resistance in parallel with a galvanometerD. A low resistance in parallel with a galvanometer |
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Answer» Correct Answer - A |
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| 315. |
A potentiometer consists of a wire of length 4 m and resistance `10Omega`. It is connected to a cell of emf 2V.The potential gradient of the wire isA. `0.5 V//m`B. `10 V//m`C. `2V//m`D. `5V//m` |
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Answer» Correct Answer - A |
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| 316. |
The length of a potentiometer wire is 600 cm and it carries a current of `40 mA`. For a cell of emf `2V` and internal resistance `10Omega`, the null point is found to be at `500 cm`. On connecting a voltmeter acros the cell, the balancing length is decreased by `10 cm` The resistance of the voltmeter isA. `500Omega`B. `290Omega`C. `490Omega`D. `20Omega` |
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Answer» Correct Answer - C `r=R(l_1/l_2-1)` `:. 10=R(500/490-1)` Solvng this equation we get `R=490 Omega` Further `r = R(E/V-1)` or `10=490(2/V-1)` Solvinng we get `V=1.96V` |
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| 317. |
figure-3.332 shows a potentiometer arrangement with` R_(AB)= IOOmega` and rheostat of variable resistance x. For x = 0 null deflection point is found at 20cm from A. For unknown value of x null deflection point was at 30cm from A, then the value ofx is: A. `10Omega`B. `5Omega`C. `2Omega`D. `1Omega` |
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Answer» Correct Answer - C |
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| 318. |
Assertion: Potentiometer measures correct value of emf of a cell. Reason: No current flows through cell at null point of potentiometer.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 | |
| 319. |
Figure shows a potentiometer arrangemeter with `R_(AB)=10 Omega` and rheostat of variable resistance d. For `x=0` null deflection point is found at `20 cm` from `A`. For unknown value of `x` null deflection point was at `30 cm` from `A`, then the value of x is A. `10Omega`B. `5Omega`C. `2Omega`D. `1Omega` |
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Answer» Correct Answer - B `E_0=V_(AC)=(i)_(AC)(R)_(AC)` `=(E/10)(10/1xx0.2)` `=E/5` In second case, `E_0=((E)/(10+x)) ((10)/1xx0.3)`……….ii Solving eqn i and ii we get `x=5Omega` |
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| 320. |
In the given potentiometer arrangemeter the null point A. can be obtained for any value of `V`B. can be obtained only if `V lt V_0`C. can be obtained only if `V gt V_0`D. can never be obtained |
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Answer» Correct Answer - D `V` and `V_0` are oppositely connected. |
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| 321. |
Resistance in the two gaps of a meter bridge are `10 ohm` and `30 ohm` respectively. If the resistances are interchanged he balance point shifts byA. `33.3 cm`B. `66.67 cm`C. `25 cm`D. `50 cm` |
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Answer» Correct Answer - D |
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| 322. |
Resistance in the two gaps of a meter bridge are `10 ohm` and `30 ohm` respectively. If the resistances are interchanged he balance point shifts byA. 33.3cmB. 66.67cmC. 25cmD. 50cm |
| Answer» Correct Answer - 4 | |
| 323. |
Assertion: At any junction of a network algebraic sum of various currents is zero Reason: At steady state there is no accumulation of charge at the junction.A. Both (A) and (R) are true and (R) is the correct explanantion of A.B. Both (A) and (R) are true but (R) is not the correct explanation of A.C. (A) is true but (R) is falseD. (A) is false but (R) is true. |
| Answer» Correct Answer - A | |
| 324. |
In the circuit shown if figure, `XY` is potentiometer wire `100 cm` long. The circuit is connected as shown. With switches `S_(2)` and `S_(3)` open, a balance point is found at `Z`. After switched `S_(1)` has remaind closed for some time, it is found that contact at `Z` must be moved towards `Y` to maintain a balance. which of the following in most likely reason for this ? A. The cell `V_(1)` is running downB. The cell `V_(2)` is running downC. The wire `XZ` is getting warm and its resistance is increasingD. The resistor `R_(1)` is getting warm and increasing in value. |
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Answer» Correct Answer - A (a) If `V_(1)` decrease, then potential differential across `xz` will decrease. So to have same potential difference i.e., `V_(2)` across `xz`, length `xz` has to be increased or `z` should be moved towares `y`. (b) is correct, because for this `z` should have been moved toward `x`. (c ) Will be incorrect because `xz` can get warm up if current through it increase which is not possible. (d) is incorrect, because which is no current in `R_(1)`. |
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| 325. |
In the petentiometer arrangemeter shown, the driving cell `A` has emf `epsilon` and internal resistance `r`. The emf of the cell `B` is `(epsilon)/(2)` and internal resistance `2r`. The petentiometer wire `CD` is `100 cm` long. If balance is obtained with length `CJ = l`, then A. `l = 50 cm`B. `l gt 50 cm`C. `l lt 50 cm`D. Balance cannot be obtained |
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Answer» Correct Answer - B (b) As potential gradient `k` in the potentiometer wire will be less than `(epsilon//100)` `(v//cm)` because of presence of `r`. So for the battery of emf `(epsilon//2)` the balance point `l` will be greater than `50 cm` as shown. `l = (epsilon//2)/(k(lt epsilon//100)) implies l gt 50 cm` |
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| 326. |
In a potentiometer experiment it is found that no current passes through the galvanometer when the terminals of the cell are connected across `0.52 m` of the potentiometer wire. If the cell is shunted by a resistance of `5Omega` balance is obtained when the cell connected across` 0.4m` of the wire. Find the internal resistance of the cell.A. `0.5 Omega`B. `1 Omega`C. `1.5 Omega`D. `2 Omega` |
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Answer» Correct Answer - C (c ) `r = [(l_(1))/(l_(2)) - 1] = 5 [(0.52)/(0.4) - 1] = 5 [1.3 - 1] = 1.5 Omega` |
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| 327. |
It is observed in a potentiometer experiment that no current passes through the galvanometer, when the terminals of the potentiometer, wire. On shunting the cell by a `2Omega` resistance, the balancing length is reduced to half. The internal resistance of the cell is:-A. `4Omega`B. `2Omega`C. `9Omega`D. `18Omega` |
| Answer» Correct Answer - 2 | |
| 328. |
In a potentiometer experiment it is found that no current passes through the galvanometer when the terminals of the cell are connected across `0.52 m` of the potentiometer wire. If the cell is shunted by a resistance of `5Omega` balance is obtained when the cell connected across` 0.4m` of the wire. Find the internal resistance of the cell.A. `2Omega`B. `2.5 Omega`C. `1 Omega`D. `1.5 Omega` |
| Answer» Correct Answer - D | |
| 329. |
A potentiometer wire is 5 m long and a potential difference of 6V is maintained between its ends. Find the emf of a cell which balance against a length of 180 cm of the potentiometer wire. |
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Answer» Length of potentiometer wire L = 5 m Potentiall difference V = 6 volt Potential gradient `phi = (V)/(L) = (6)/(5) = 1.2 V //m` Balancing length l = 180 cm = 1.80 cm Emf of the cell `E = phi l` `= 1.2 xx 1.8 = 2.16 V`. |
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| 330. |
12 wires of different resistances are connected as shown. Find the equivalent resistance A and F.A. `(60)/(29)`B. `(50)/(29)`C. `(40)/(29)`D. `(20)/(29)` |
| Answer» Correct Answer - A | |
| 331. |
Twelve identical resistances arranged on all edges of a cube. The resistors are all the same. Then find the equivalent resistance between the edges A and B as shown in figure. A. between the diagonal of cube is `(5 R)/(6)`B. between the diagonal of face of cube is `(3 R)/(4)`C. between the ends of a side is `(7 R)/(12)`D. All options are correct |
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Answer» Correct Answer - D See solved example. |
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| 332. |
Six resistors are connected so as to form the edges of a tetrahedron ABCD, the resistances of opposite pairs being equal. (Note that the resistors AB and CD do not touch each other) find the equivalent resistance between A and C.A. `(r_(3))/(2)[(2r_(1)r_(2)+r_(3)(r_(1)+r_(2)))/((r_(1)+r_(3))(r_(2)+r_(3)))]`B. `(r_(2))/(2)[(2r_(1)r_(2)+r_(3)(r_(1)+r_(2)))/((r_(1)+r_(3))(r_(2)+r_(3)))]`C. `(r_(1))/(2)[(2r_(1)r_(2)+r_(3)(r_(1)+r_(2)))/((r_(1)+r_(3))(r_(2)+r_(3)))]`D. `(r_(3))/(2)[(2r_(1)r_(3)+r_(2)(r_(1)+r_(2)))/((r_(1)+r_(3))(r_(2)+r_(3)))]` |
| Answer» Correct Answer - A | |
| 333. |
A steady voltage drop is maintained across a potentiometer wire which is 5m long. A Daniel cell is balanced by a length of 3m of wire. If length of potentiometer wire is increased to 6 m, find new length of wire across which Daniel cell will be balanced.A. 3.6mB. 6.6mC. 6.3 mD. 3.3m |
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Answer» Correct Answer - A `(E_(1))/(l_(1))=I rho = (V)/(L)` `:.(V)/(E)=(L)/(l_(1))=I rho = ` constant `:.(l_(1.2))/(l_(1.1))=(L_(2))/(L_(1)) " " :.l_(1.2)=3.6m` |
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| 334. |
Two metal wires of identical dimesnios are connected in series. If `sigma_(1)` and `sigma_(2)` are the conducties of the metal wires respectively, the effective conductivity of the combination isA. `(2sigma_(1)sigma_(2))/(sigma_(1)+sigma_(2))`B. `(sigma_(1)+sigma_(2))/(2sigma_(1)sigma_(2))`C. `(sigma_(1)+sigma_(2))/(sigma_(1)sigma_(2))`D. `(sigma_(1)sigma_(2))/(sigma_(1)+sigma_(2))` |
| Answer» Correct Answer - A | |
| 335. |
A potentiometer wire is `100 cm` long hand a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obatined at `50 cm` and `10 cm` from the positive end of the wire in the two cases. The ratio of emfs is:A. `3 : 2`B. `5 : 1`C. `5 : 4`D. `3 : 4` |
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Answer» Correct Answer - A (a) `(E_(1) + E_(2))/(E_(1) + E_(2)) = (50)/(10)` `implies (2E_(1))/(2E_(2)) = (50 + 10)/(50 - 10) implies (E_(1))/(E_(2)) = (3)/(2)` |
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| 336. |
A circuit contains an ammeter, a battery of `30 V` and a resistance `40.8 ohm` all connected in series. If the ammeter has a coil of resistance `480 ohm` and a shunt of `20 ohm`, the reading in the ammeter will beA. 0.5AB. 0.25AC. 2AD. 1A |
| Answer» Correct Answer - A | |
| 337. |
In the circuit shown in figure, the emfs of batteries are `E_1`, and `E_2` which have internal resistances `R_1`, and `R_2`. At what value of the resistance `R` will the thermal power generated in it be the highest? What it is? |
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Answer» Correct Answer - A::B::D The two batteries are in parallel.. Thermal power generated in `R` will be miximum when, total internal resitance = total external resitance `or R=(R_1R_2)/(R_1+R_2)` `E_(eq)=((E_1/R_1+E_2/R_2))/((1/R_1+1/R_2))` `=((E_1R_2+E_2R_1)/(R_1+R_2))` `R_("net")=(2R_1R_2)/(R_1+R_2)` `:. i=E_(eq)/R_("net")=(E_1R-2+E_2R_1)/(2R_1R_2)` Maximum power through R `P_(max)=i^2R=((E_1R_2+E_2R_1)^2)/(4R_1R_2(R_1+R_2))` |
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| 338. |
The potential gradient of potentiometer is `0.2 "volt"//m`. A current of 0.1 amp is flowing throgh a coil of 2 ohm resistance. The balancing length in meters for the potential difference at the ends of this coil will beA. 2B. 1C. 0.2D. 0.1 |
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Answer» Correct Answer - B `0.1xx2=0.2l rArr l=1m` |
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| 339. |
A potentiometer wire of Length `L` and a resistance `r` are connected in series with a battery of e.m.f. `E_(0)` and a resistance `r_(1)`. An unknown e.m.f. `E` is balanced at a length `l` of the potentiometer wire. The e.m.f. `E` will be given by :A. `(LE_(0)r)/((r +r_(1)l))`B. `(E_(0)r)/(lr_(1)l)`C. `(E_(0)r)/((r + r_(1)l)).(1)/(L)`D. `(E_(0)l)/(L)` |
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Answer» Correct Answer - c Current through the potentionmeter wire is `1 = (E_(0))/(r+r_(1))` Potential difference across potentiometer wire `V = Ir = (E_(0)r)/(r+r_(1))` Potential gradient across potentiometer wire `K =(V)/(L)=((E_(0)r)/(r+r_(1)))(1)/(L)` Now `E = Kl = ((E_(0)r)/(r+r_(1)))(l)/(L)` |
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| 340. |
The potentiometer wire 10m long and 20 ohm resistance is connected to a 3 volt emf battery and a 10ohm resistance. The value of potential gradient in volt/m of the wire will beA. `0.02`B. `0.3`C. `0.2`D. `1.3` |
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Answer» Correct Answer - C |
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| 341. |
Find the value of the unknown resistance `X`, in the following circuit if no current flows through the arm AO from the battery of `6V` and negligible internal resistance. Also find the Current flowing from the battery. |
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Answer» Correct Answer - `1A` When no current flow through the arm `AO` the wheatstone bridge in balanced .Then `(2)/(X) = (3)/(6) or X = 4 Omega ` Total resistance of circuit `R = ((2 + 4) xx (3 + 6))/((2 +n 4) + (3+ 6)) + 2.4 = 6 Omega` Current is circuit `I = (epsilon)/(R ) = (6)/(6) = 1A` |
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| 342. |
The given figure represents an arrangement of potentiometer for the calculation of internal resistance (`r`) of the unknown battery (`E`). The balance length is `70.0 cm` with the key opened and `60.0 cm` with the key closed. `R` is `132.40Omega`. The internal resistance (r) of the uknown cell will be A. `22.1Omega`B. `113.5Omega`C. `154.5Omega`D. `10Omega` |
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Answer» Correct Answer - A `r=R(l_1/l_2-1)=132.4(70/60 - 1)=22.1Omega` |
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| 343. |
In a potentiometer of one metre length, an unknown emf voltage source is balanced at `60 cm` length of potentiometer wire, while a `3` volt battery is balanced at `45 cm` length. The the emf of the unknown voltage source is.A. 3 VB. 2.25 VC. 4 VD. 4.5 V |
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Answer» Correct Answer - C `(E_1)/(l_1) = (E_2)/(l_2)` `(E_1)/(60) = (3)/(45) rArr E_1 = 4 V`. |
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| 344. |
A `6V` battery of negligible internal resistance is connected across a uniform wire of length `1m`. The positive terminal of another battery of emf `4V` and internal resistance `1Omega` is joined to the point `A` as shown in figure. The ammeter shows zero deflection when the jockey touches the wire at the point `C`. The `AC` is equal to A. `2//3 m`B. `1//3 m`C. `3//5 m`D. `1//2m` |
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Answer» Correct Answer - A `x=6/1 v//m implies 6xxl_(AC)=4` `l_(AC)=2/3m` |
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| 345. |
Three identical cells each of emf 2 V and unknown internal resistance are connected in parallel. This combination is connected to a `5 Omega` resistor. If the terminal voltage across the cells is 1.5 V, what is the internal resistance of each cell ? |
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Answer» Correct Answer - `5 Omega` |
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| 346. |
In the curcuit shown in figure `R_1=R_2=R_3=10Omega`. Find the currents through `R_1` and `R_2` |
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Answer» Correct Answer - A `V_(R_1)=0` `:. i_(R_1)=0` `V_(R_2)=V_(R_3)=10V` `:. I_(R_2)=i_(R_3)=10/10=1A` |
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| 347. |
In the circuit shown in figure a `12 V` battery with unknown internal resistance `r` is connected to another battery with unknown emf E and internal resistance `1Omega ` and to a resistance of `3Omega` carrying a current of `2A`. The current through the rechargeable battery is `1 A` in the direction shown. Figure the unknown current `i` internal resistance r and the emf E. |
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Answer» Correct Answer - A::B::C Applying loop law equation in upper loop, we have `E+12-ir-1=0`……….i Applying loop law equation in lower, we have where `i=1+2=3A` `E+t6-1=0`……………ii Solving these two equations we get `E=-5V` and `r=2Omega` |
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| 348. |
A galvanometer has resistance of `7Omega` and given a full scale deflection for a current of `1.0A`. How will you convert it into a voltmeter of range 10VA. `3Omega` in seriesB. `3 Omega` in aprallelC. `17 Omega` in seriesD. `30 Omega` in series |
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Answer» Correct Answer - A |
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| 349. |
A galvanometer of `100 Omega` resistance gives full scale deflection when 10 mA of current is passed. To convert it into 10 A range ammeter, the resistance of the shunt required will beA. `-10 Omega`B. `1 Omega`C. `0.1 Omega`D. `0.01 Omega` |
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Answer» Correct Answer - C |
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| 350. |
The full scale deflection current of a galvanometer of resistance `1Omega` is `5 mA`. How will you convert it into a voltmeter of range of `V`? |
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Answer» Correct Answer - A `V=i_g(G+R)` `:. R=V/i_g-G=` series resistance connected with galvonometer `=(5/(5xx10^-3))-1=999Omega` |
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