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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.
| 1201. |
ABCD is a uniform circular wire of resistance `16 Omega` and `AOC and BD` are two uniform wires forming diameters at right angles, each of resistance `2 Omega`. The two straight wires do not touch each other at O. A battery of emf `10 V` is placed in `AO` as shown. (a) Find the current through the battery (figure a) (b) If the straight wires are tied at O so as to form a junction, find the current through the battery (figure b) |
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Answer» Correct Answer - (a) `(5)/(3)A` (b) `(35)/(12) A` |
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| 1202. |
Two copper wires of the same length have got different diameters a. which wire has greater resistance? b. greater specific resistance? |
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Answer» a. For a given wire, `R=rho l/A`, i. e. `Rprop1/A` So the thinner wire will have greater resistance b.Specific resistance (`rho`) is a material property. It does not depend of `l` or `A`. So, both the wire will have same specific resistance. |
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| 1203. |
Electric field inside a copper wire of length 10 meter, resistance 2 ohm connected to a 10 volt battery is (in SI units)A. 1B. 0.5C. 10D. 5 |
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Answer» Correct Answer - 1 Electric field `=(P.D.)/("length ")=(10V)/(10m)=1V//m` |
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| 1204. |
Two cells `E_(1)` and `E_(2)` in the circuit shown in figure, have emfs of 5 V and 9 V and internal resistance of `0.3 Omega` and `1.2 Omega` respestivley. Calculate the value of current flowing through the resistance of `3 Omega`. |
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Answer» Net emf `= E_(2) - E_(1) = 9-5 = 4V`. Total resistance `= 0.3 +1.2 +4.5 + (6 xx 3)/(6+3)=8 Omega` Current through the circuit, `I = (4V)/(8 Omega) = 0.5 A` Current through `3 Omega` resistance `= (6 xx 0.5)/(6+3) =1/3A` |
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| 1205. |
Two cells `E_(1)` and `E_(2)` in the circuit shown in figure, have emfs of 5 V and 9 V and internal resistance of `0.3 Omega` and `1.2 Omega` respestivley. Calculate the value of current flowing through the resistance of `3 Omega`. |
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Answer» Correct Answer - ` (1)/(3) A` As the two cell are connected in oppostion Net emf `= E_(2) - E_(1) = 9 - 5 = 4V` Total resistance `= 0.3 + 1.2 + 4.5 + (6 xx 3)/(6 + 3) = 8 Omega` Current through circuit, `I = (4)/(8) = 0.5 A` Current through circuit `3 Omega` resistance `= (6 xx 0.5)/(6 + 3) = (1)/(3) A` |
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| 1206. |
`1kg` piece of copper is drawn into a wire `1mm` thick and another piece into a wire `2mm` thick. Compare the resistance of these wires. |
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Answer» Correct Answer - `16: 1` Let `d` be the meterial of copper wire Let `l_(1) and l_(2)` be length of copper wires of diameter `1mm and 2 mm` respectively. As mass = volume `xx` density `[(piD^(2)//4)xxl]d`, so `1 = (pi (10^(-3))^(2))/(4) l_(1) xx d = (pi( 2xx 10^(-3))^(2))/(4)l_(2) xx d` or `l_(1) = 4 l_(2)` `:. R_(1) = (rhol_(1))/(piD_(1)^(2)//4) = (rho4l_(2))/(pi(10^(-3))^(2)//4)` or `R_(2) = (rhol_(2))/(piD_(2)^(2)//4) = (rhol_(2))/(pi(2 xx 10^(-3))^(2)//4)` so `(R_(1))/(R_(2)) = 16 or R_(1) = 16 R_(2)` |
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| 1207. |
In a potentiometer the balancing with a cell is at length of 220cm. On shunting the cell with a resistance of `3Omega` balance length becomes 130cm. What is the internal resistance of this cell.A. `4.5Omega`B. `7.8Omega`C. `6.3Omega`D. `2.08Omega` |
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Answer» Correct Answer - D `"Here", l_(1)=220cm, l_(2)=130cm, R=3Omega` `r=((l_(1)-l_(2))/(l_(2)))R=((220-130)/(130))xx3=2.08Omega` |
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| 1208. |
3V poteniometer used for the determination of internal resistance of a 2.4V cell. The balanced point of the cell in open circuit is 75.8cm. When a resistor of `10.2Omega` is used in the external circuit of the cell the balance point shifts to 68.3cm length of the potentiometer wire. The internal resistance of the cell is A. `2.5Omega`B. `2.25Omega`C. `1.12Omega`D. `3.2Omega` |
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Answer» Correct Answer - C Internal resistance of cell `r=R((l_(1))/(l_(2))-1)=1.5xx(65)/(32)=3.05V` `l_(2)=68.3cm or r=10.2((75.8)/(68.3)-1)=1.12Omega` |
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| 1209. |
The current density varies radical distance r as `J=ar^(2)`, in a cylindrical wire of radius R. The current passing through the wire between radical distance `R//3 and R//2` is,A. `(65piaR^(4))/(2592)`B. `(25piaR^(4))/(72)`C. `(65pia^(2)R^(3))/(2938)`D. `(81pia^(2)R^(4))/(144)` |
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Answer» Correct Answer - 1 Current density `J=ar^(2)` C urrent `I=int jdA` `I=int_(R//3)^( R//2) ar^(2)(2pi rdr)` [where dA ` 2pirdr`) `=2pia[(r^(4))/(4)]_(R//3)^(R//2)=(65piaR^(4))/(25892)` |
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| 1210. |
Two identical batteries each of emf `E= 2`volt and internal resistance `r=1` ohm are available `t`. produce heat in an external resistance by passing a current through it. What is the maximum power that can be developed across an external resistance `R` using these batteries? |
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Answer» Correct Answer - B As derived in the above question `P_(max)=E^2/(4r)` here, `E =` net emf `= 2+2=4V` and `r =` net internal resistance `=1+1=2Omega` `:. P_(max)=((4)^2)/((4)(2))=2W` |
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| 1211. |
A battery of internal resistance `2 Omega` is connected ot a variable resistor whose value can vary from `4 Omega` to `10 Omega`. The resistance is initially set at `4 Omega`. If the resistance is now increased thenA. power consumed by it will decreaseB. power consumed by it will increaseC. power consumed by it may increase or may decreaseD. power consumed will first increase then decrease |
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Answer» Correct Answer - A (a) `P = I^(2) R = ((V)/(R ))^(2) R = (epsilon^(2))/((R + r)^(2)) R` `epsilon` is constant and `(R + r)` increases rapidly then `P darr` |
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| 1212. |
Consider the circuit given here with the following parameters E.M.F of the cell `=12 V`. Internal resistance of the cell `=2 Omega`. Resistance `R = 4 Omega`. Which one of the following statements in trueA. Rate of energy loss in the source is = 8 WB. Rate of energy conversion in the source is 16 WC. Power output in is = 8 WD. Potential drop across R is `= 16 V` |
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Answer» Correct Answer - A |
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| 1213. |
A dry cell has an e.m.f. of `1.5 V` and an internal resistance of `0.05 Omega`. The maximum current obtainable from this cell for a very short time interval isA. 30AB. 300AC. 3AD. `0.3A` |
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Answer» Correct Answer - A |
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| 1214. |
Assertion: In the part of the circuit shown in figure, maximum power is produced across `R`. Reason : Power `P=V^2/R` A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - A::B In parallel, V=constant `:.` From the equation `P=V^2/Rimplies Pprop 1/R` |
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| 1215. |
Assertion : Current `I` is flowing through a cylindrical wire of non-uniform cross-section as shown. Section of wire near `A` will be more heated compared to the section near B. Reason : Current density near `A` is more.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - B `R=(rhol)/Aor R/l=` resistance per unit length `=rho/Aprop1/A` Near `A`, areas of cross section is less. Therefore, resistance per unit length will be more. Hence from the equation `H=i^2Rt`, heat generation near A will be more. Current density `J=i/A` or `Jprop1/A` (as i is same)` |
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| 1216. |
Assertion : In the circuit shown in figure ammeter and voltmeter are non-ideal. Why positions of ammeter and voltmeter are changed, reading of ammeter will increase while of voltmeter will decrease. Reason : Resistance of an ideal ammeter is zero while that of an ideal voltmeter is infinite.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - B Even if ammeter is non ideal its resistance should be small and net parallel resistance is less than the smallest individual resistance. `:. R_("net")lt` resistance of ammeter in changed situation. Hence, net resistance of the circuit will decrease. So, main current will increase. But maximum percentage of main current will pass through ammeter (in parallel combination) as its resistance is less. Hence, reading of ammeter wil increase. Initial voltmeter reading`= `emf of battery Final voltmeter reading=emf of battery -potential drop across shown resistancce. Hence, voltmeter reading will decrease. |
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| 1217. |
Assertion: In the circuit shown in figure after closing the switch `S` reading of ammeter will increase while that of voltmeter will decrease. Reason : Net resistance decreases as parallel combination of resistors is increased,A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - D Since net resistance decreases, therefore main current increases Hence, net potential difference across voltmeter also increases. |
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| 1218. |
Assertion:In the part of a circuit shown in figure, given that `V_(b) gt V_(a)`. The current should flow from b to a. Reason: Direction of current inside a battry is always from negative terminal to positive terminal.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - C If current flows from `a` to `bm`, then equation will become `V_a-ir-E=V_b` or `V_a-V_b=E+ir` so, `V_a-V_b` is always positive. Hence `V_a` is always greater than `V_b`. |
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| 1219. |
Out of five resistance of `ROmega` each 3 are connected in parallel and are joined to the rest 2 in series. Find the resultant resistance.A. `(3//7)ROmega`B. `(7//3)ROmega`C. `(7//8)ROmega`D. `(8//7)ROmega` |
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Answer» Correct Answer - B |
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| 1220. |
Two batteries A and B each of e.m.f. 2 V are connected in series to an external resistance R = 1 ohm . If the internal resistance of battery A is 1.9 ohms and that of B is 0.9 ohm , what is the potential difference between the terminals of battery A A. 2 VB. `3.8` VC. zeroD. `4.8` V |
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Answer» Correct Answer - C |
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| 1221. |
Two batteries A and B each of e.m.f. 2 V are connected in series to an external resistance R = 1 ohm . If the internal resistance of battery A is 1.9 ohms and that of B is 0.9 ohm , what is the potential difference between the terminals of battery A A. 2VB. 3.8VC. ZeroD. None of the above |
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Answer» Correct Answer - C |
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| 1222. |
A battery of e.m.f. E and internal resistance r is connected to a variable resistor R as shown here. Which one of the following is true A. Potential difference across the terminal of the battery is maximum when `R= r`B. Power delivered to the resistor is maximum when `R = r`C. Current in the circuit is maximum when `R= r`D. Current in the circuit is maximum when `R gt gt t` |
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Answer» Correct Answer - B |
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| 1223. |
A cell of emf `E` and internal resistance `r` connected in the secondary gets balanced against length `l` of potentiometer wire. If a resistance `R` is connected in parallel with the cell, then the new balancing length for the cell will beA. `((R)/(R-r))`B. `((R-r)/(R))`C. `((R)/(r))`D. `((R)/(R+r))l` |
| Answer» Correct Answer - D | |
| 1224. |
When a resistor of `11 Omega` is connected in series with an electric cell, the current following in it is `0.5 A`. Instead, when a resistor of `5 Omega` is connected to the same electric cell in series, the current increases by `0.4 A` The internal resistance of the cell isA. `1.5 Omega`B. `2 Omega`C. `2.5 Omega`D. `3.5Omega` |
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Answer» Correct Answer - C |
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| 1225. |
Sensitivity of potentiometer can be increased byA. decreasing the length of potentiometer wireB. increasing potential gradient on its wireC. increasing emf of battery in the primary circuitD. decreasing the potential gradient on its wire |
| Answer» Correct Answer - D | |
| 1226. |
The direction of current in a cell isA. `(-)` ve pole to `(+)` ve pole during deschargingB. `(+)` ve pole to `(-)` ve pole during dischargingC. Always `(-)` ve pole to `(+)` ve poleD. always flows `(+)` ve pole to `(-)` ve pole |
| Answer» Correct Answer - A | |
| 1227. |
Back emf of a cell is due toA. Electrolytic polarizationB. Peltier effectC. Magnetic effect of currentD. Internal resistance |
| Answer» Correct Answer - A | |
| 1228. |
How much work is required to carry a `6muC` charge from the negative terminal to the positive terminal of a 9 V battery?A. `54xx10^(-3)J`B. `54xx10^(-6)J`C. `54xx10^(-9)J`D. `54xx10^(-12)J` |
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Answer» Correct Answer - B |
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| 1229. |
The internal resistance of a cell is the resistance ofA. Electrodes of the cellB. Vessel of the cellC. Electrolyte used in the cellD. Material used in the cell |
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Answer» Correct Answer - C |
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| 1230. |
Consider four circuits shown in the figure below. In which circuit power dissipated is greatest (Neglect the internal resistance of the power supply)?A. B. C. D. |
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Answer» Correct Answer - A |
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| 1231. |
How much work in required to carry a `6 mu C` charge from the negative terminal to the positive terminal of a 9V batteryA. `54 xx 10^(-3)J`B. `54 xx 10^(-6)J`C. `54 xx 10^(-9)J`D. `54 xx 10^(-12)J` |
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Answer» Correct Answer - B |
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| 1232. |
Which depolarizers are used to neutralizes hydrogen layer in cellsA. Potassium dichromiteB. Manganese dioxideC. 1 or 2D. hydrogen peroxide |
| Answer» Correct Answer - C | |
| 1233. |
The electric power transferred by a cell to an external resistance is maximum when the external resistance is equal to (r internal resistance)A. `(r)/(2)`B. `2r`C. `r`D. `r^(2)` |
| Answer» Correct Answer - C | |
| 1234. |
The terminal Pd of a cell is equal to its emf ifA. external resistance is infinityB. internal resistance is zeroC. both 1 and 2D. internal resistance is `5Omega` |
| Answer» Correct Answer - C | |
| 1235. |
To supply maximum current cells should be arrange inA. seriesB. parallelC. mixed groupingD. depends on the internal and external resistance |
| Answer» Correct Answer - D | |
| 1236. |
When cells are arranged in seriesA. the current capacity decreasesB. the current capacity increasesC. the emf increasesD. the emf decreases |
| Answer» Correct Answer - C | |
| 1237. |
Internal resistance of a cell depends onA. concentration of electrolyteB. distance between the electrodesC. area of electrodeD. all the above |
| Answer» Correct Answer - D | |
| 1238. |
In a circuit two or more cells of the same emf are connected in parallel in orderA. increases the pd across resistance in the circuitB. decreases pd across a resistance in the circuitC. facilitate drawing more current from the battery systemD. Change the emf across the system of batteries |
| Answer» Correct Answer - C | |
| 1239. |
The value of internal resistance of ideal cell isA. zeroB. infinityC. `1Omega`D. `2Omega` |
| Answer» Correct Answer - A | |
| 1240. |
In a potentiometer the balance length with standard cadmium cell is 509 cm. The emf of a cell which when connected I the place of the standard cell gave a balance length of 750 cm is (emf of standard cell is 1.018 V)A. 1.5 VB. 0.5 VC. 1.08 VD. 1.2 V |
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Answer» Correct Answer - A `(E_(1))/(E_(2))=(l_(1))/(l_(2))3` |
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| 1241. |
Then 6 identical calls of no internal resistance are connected in series in the second arycircuit of a potentiometer, the balancing length is `l` if two of them are wrongly connected to balacing length becomesA. `(l)/(4)`B. `(l)/(3)`C. `l`D. `(2l)/(3)` |
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Answer» Correct Answer - B `NEalphal_(1),(N-2m)Ealphall_(2)` |
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| 1242. |
A steady current flows through the conductor of variable cross-sectional area. Now select the correct options.A. The value of current is same through the cross sections 1,2, and 3B. The current density is the maximum at the cross sections 3C. The drift velocity is greater at the cross-section 1D. The electric field is maximum at the cross section 3. |
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Answer» Correct Answer - A::B::D `I=nAv_(d)e` `v_(d)=(I)/(nAe)` We know for a conductor during the flow of steady current, `E=(v)/(e)=(Ri)/(l)` `E=(eli)/(Axxl)=(ei)/(A)` `therefore` electric field is maximum at cross-section (3) |
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| 1243. |
Select the correct statement(s)A. A current carrying conductor is electrically neutral.B. In a current carrying conductor, it is possible, all the free electrons can have same drft velocity.C. In a current carrying conductor, the speed of the all the free electrons need not be same.D. The electric field at any point inside the current carring conductor is non-zero, The electic field just outside the current carrying conductor is non-zero. |
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Answer» Correct Answer - A::B::C::D based on drift velocity |
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| 1244. |
Consider two identical galvanometers and two identical resistors with resistance `R`. If the internal resistance of the galvanometers `R_(c) lt R//2`, which of the following statement(s) about any one of the galvanometers is (are) true?A. The maximum voltage range is obtained when all the components are connected in seriesB. The maximum voltage range is obtained when the two resistors and one galvanometer are connected in series and the second galvanometer is connected in parallel to the first galvanometerC. The maximum current range is obtained whhen all the components are connected in parallelD. The maximum current range is obtained when the two galvanometers are connected in series and the combination is connected in parallel with both the resistors. |
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Answer» Correct Answer - A::C For maximum voltage range across a galvanometer all the elements must be connected in series. For maximum current range through a galvanometer, all the elements should be connected in parallel. |
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| 1245. |
What amount of heat will be generated in a coil of resistance `R` due to a charge q passing through it if the current in the coil a. decreases down to zero uniformly during a time interval `t_0`? b. decrases down to zero having its value every `t_0` seconds?A. `(q^(2)ln2)/(3Deltat)R`B. `(q^(2)ln2)/(4Deltat)R`C. `(q^(2)ln2)/(Deltat)R`D. `(q^(2)ln2)/(2Deltat)R` |
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Answer» Correct Answer - D Obviously the current through thhe coil is given by `i=i_(0)((1)/(2))^((t)/(Deltat))` Then charge `q=int_(0)^(infty)idt=int_(0)^(infty)i_(0)2^((-t)/(Deltat))dt=(i_(0)Deltat)/(1n2)` So, `i_(0)=(qln2)/(Deltat)` And hence heat generated in the circuit in the time interval `t[0,infty]` `H=int_(0)^(infty)i^(2)Rdt=int_(0)^(infty)[(q1n2)/(Deltat)2^((-t)/(Deltat))]^(2)Rdt=(q^(2)1n2)/(2Deltat)R` |
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| 1246. |
What amount of heat will be generated in a coil of resistance `R` due to a charge q passing through it if the current in the coil a. decreases down to zero uniformly during a time interval `t_0`? b. decrases down to zero having its value every `t_0` seconds?A. `(2q^(2)R)/(3Deltat)`B. `(3q^(2)R)/(2Deltat)`C. `(4q^(2)R)/(3Deltat)`D. `(3q^(2)R)/(4Deltat)` |
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Answer» Correct Answer - C As current I is function of time and at `t=0` and `Deltat`, it equal `i_(0)` and zero respectively it may be represented as, `i=i_(0)(1-(t)/(Deltat))` Thus `q=int_(0)^(Deltat)idt=int_(0)^(Deltat)i_(0)(1-(t)/(Deltat))dt=(i_(0)Deltat)/(2)` so, `i_(0)=(2q)/(Deltat)` The heat generated `H=int_(0)^(Deltat)i^(2)Rdt=int_(0)^(Deltat)[(2q)/(Deltat)(1-(t)/(Deltat))]^(2)Rdt=(4q^(2)R)/(3Deltat)` |
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| 1247. |
A cell of e.m.f. 2V and negligible internal resistance is connected to resistor `R_(1)` and `R_(2)` as shown in the figure. The resistance of the Voltmeter `R_(1)` and `R_(2)` are `80Omega,40Omega` and `80Omega` respectively. The reading of the voltmeter is:- A. 1.78VB. 1.60VC. 0.80VD. 1.33V |
| Answer» Correct Answer - 3 | |
| 1248. |
In the figure shown, each resistor is of `20 Omega` and the cell has emf 10 volt with negligible internal resistance. Then rate of joule heating in the circuit is (in watts) A. `(20)/(3)`B. `(40)/(3)`C. `(10)/(3)`D. zero |
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Answer» Correct Answer - C |
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| 1249. |
In order to heat a liquid an electric heating coil is connected is to a cell of emf `E = 12 V` and internal resistance `r = 1 Omega`. There are three options for selecting the resistance `(R)` of the heating coil. `R` can be chosen as `1 Omega, 2 Omega or 4 Omega`. The cell has a rating of `2000 mAh` (milli Ampere hour) and it is to be used to heat the liquid till it expires. [The cell maintains constant emf till it lasts] (a) Which value of R will you chose so that maximum heat `(H_0)` is transferred to the liquid before the cell expires? Calculate `H_(0)`. (b) Which value of `R` will chose so that heat is transferred to the liquid at fastest possible rate? What percentage of `H_(0)` (as obtained in (a)) is transferred to the liquid in this case by the time the cell expires ? |
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Answer» Correct Answer - (a) `4 Omega,69.12 kJ` (b) `1 Omega,62.5 %` |
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| 1250. |
In a wheat stone bridge experiment to determine the unknown resistance `R_(4)`, the values of `R_(1) and R_(2)` were taken to be `10 Omega and 1 Omega` respectively. It was found that the galvanometer will show exact zero deflection when value of `R_(3)` is taken as `643 Omega lt R_(3) lt 644 Omega`. Now `R_(1)` is changed to `100 Omega` (`R_(2)` remains unchanged). (a) If you have been asked to obtain a balanced bridge, which values of `R_(3)` will you try with? (b) If balance is obtained for `6432 Omega lt R_(3) klt 6433 Omega` write the measured value of unknown resistance `R_(4)`. |
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Answer» Correct Answer - (a) We should try with `6430 Omega lt R_(3) lt 6440 Omega` (b) `R_(4) = 64.325 Omega` |
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