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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.
| 401. |
In a Wheatstone bridge circuit, `P=5 Omega, Q=6 Omega, R=10 Omega` and `S= 5 Omega`. Find the additional resistance to be used in series with S, so that the bridge is balanced. |
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Answer» Let the bridge be balanced when additional resitance x is put in series with S. Then, `(S +x)= Q/P R` or `x=Q/P R - S = 6/5 xx 10 - 5= 7 Omega`. |
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| 402. |
A current of 1.0 mA is flowing through a potentiometer wire of length 4 m and of resistance `4 Omega`, find the potential gradient of potentiometer wire |
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Answer» Here, `I= 1.0 mA = 10^(-3) A, L = 4 m, R= 4 Omega` Potential drop across potentiometer wire, `V=IR= 10^(-3) xx 4 V` Potential gradient, `K=V/L =4 xx 10^(-3)/4 = 10^(-3)V//m` |
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| 403. |
A 10 C of charge flows through a wire in 5 minutes. The radius of the wire is 1 mm. It contains `5xx10^(22)` electrons per `"centimetre"^(3)`. Calculate the current and drift velocity. |
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Answer» Correct Answer - `0.033 A, 1.326xx10^(-6)ms^(-1)` |
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| 404. |
`1.6 mA` current is flowing in conducting wire then the number of electrons flowing per second isA. `10^(11)`B. `10^(16)`C. `10^(19)`D. `10^(15)` |
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Answer» Correct Answer - B |
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| 405. |
A 60 coulomb of charge flows through a wire in half minute. The radius of the wire is 1 mm. The wire contains `5xx10^(22)` electrons per cubic centimetre. Calculate the current and drift velocity. |
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Answer» Here, `q=60 C, t=30` seconds, `r=10^(-3)m, n=5 xx 10^(22)//c c ~= 5xx10^(28)m^(-3)` Current `I=q/t = 60/30=2A` Drift veloccity, `v_(d)=I/(n e A)=I/( n e pi r^(2))` `2/((5xx10^(28)) xx (1.6xx10^(-19)) xx3.142 xx(10^(-3))^(2))` `=7.95 xx 10^(-5) ms^(-1)` |
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| 406. |
A current of 1 mA is flowing through a copper wire. How many electrons will pass a given point in one second `[e = 1.6 xx 10^(19)`Coulomb]A. `6.25 xx 10^(19)`B. `6.25 xx 10^(15)`C. `6.25 xx 10^(31)`D. `6.25 xx10^(8)` |
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Answer» Correct Answer - B |
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| 407. |
If three resistors of resistance `2Omega, 4Omega` and `5 Omega` are connected in parallel then the total resistance of the combination will beA. `(20)/(19)Omega`B. `(19)/(20)Omega`C. `(10)/(20)Omega`D. `(29)/(10)Omega` |
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Answer» Correct Answer - A The resistance `2 Omega, 4 Omega` and `5 Omega` are connected in parallel combination. Therefore, resultant resistance is given by `(1)/(R )=(1)/(2)+(1)/(4)+(1)/(5)=(19)/(20) rArr R = (20)/(19) Omega` |
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| 408. |
If the resistivity of a potentiometer wire be `rho` and area of cross-section be A , then what will be potential gradient along the wireA. `(I rho)/(A)`B. `(I)/(Arho)`C. `(IA)/(rho)`D. `IA rho` |
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Answer» Correct Answer - A |
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| 409. |
A current of 0.01 mA passes through the potentiometer wire of a resistivity of `10^(9)Omega` cm and area of cross-section `10^(-1) cm^(-2)` . The potential gradient isA. `10^(9) V/m`B. `10^(11) V/m`C. `10^(10) V/m`D. `10^(8) V/m` |
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Answer» Correct Answer - D Potential gradient is given by `k=(V)/(l)=(lR)/(l) " " (because V = lR)` `= (l xx rho l// A)/(l) " " (because R = (rho l)/(A))` `therefore k = (0.01xx10^(-3)xx10^(9)xx10^(-2))/(10^(-2)xx10^(_4))=10^(8) V//m` |
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| 410. |
A battery of emf `E` and internal resistance `r` is connected to a resistor of restance `r_(1)` and Q Joules of heat is produced in a certain time `t`. When te same battery is connected to another resistor of resistance `r_(2)` the same quantity of heat is produced in the same time `t`. Then the value of `r` isA. `(r_(1)^(2))/(r_(2))`B. `(r_(2)^(2))/(r_(1))`C. `(1)/(2)(r_(1)+r_(2))`D. `sqrt(r_(1)r_(2))` |
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Answer» Correct Answer - D `Q=(V^(2))/(r_(1))=(E^(2)r_(1))/((r+r_(1))^(2))` `Q=(V^(2))/(r_(2))=(E^(2)r_(2))/((r+r_(2))^(2))` |
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| 411. |
A 20 m long potentiometer wire has a resistance of 20 Ohm. It is connected in series with a battery of emf 3V and resistance of `10 Omega`. The internal resistance of cell is negligible. If the lenth can be read accurately up to 1 mm, the potentiometer can read voltage:A. up to maximum of 0.2 mVB. with an accuracy of 0.2 mVC. with an accuracy of 0.1 mVD. upto maximum of 2V |
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Answer» Correct Answer - C::D |
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| 412. |
In the circuit shown, both cells are ideal and of fixed emf, the resistance of resistance `R_(1)` has fixed resistance and the resistance of resistor `R_(2)` can be varied (but the value of `R_(2)` is not zero). Then: A. The electric power delivered to resistor of resistance `R_(1)` is independent of `R_(2)`B. Electric power delivered by `E_(1)` is independent of `R_(2)`C. Electric power delivered by `E_(1)` is dependent on `R_(2)`D. Electric power delivered to `R_(1)` is dependent on `R_(2)` |
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Answer» Correct Answer - A::B |
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| 413. |
Six identical cunducting rods are joined as shown in Fig. Points A and D are maintained at temperatures `200^@C` and `20^@C` respectively. The temperature of junction B will beA. `120^(@)C`B. `100^(@)C`C. `140^(@)C`D. `80^(@)C` |
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Answer» Correct Answer - C |
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| 414. |
Three conducting rods of same material and cross-section are showm in Fig .7(CF).22. Temperatures of A, D and C are maintained at `20^(@)C, 90^(@)C` and `0^(@)C` . The ratio of the lengths of BC and BD, if there is no flow in AB, is A. `2//7`B. `7//2`C. `9//2`D. `2//9` |
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Answer» Correct Answer - B |
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| 415. |
The cell has an emf of `2 V` and the internal resistance of this cell is `0.1 Omega`, it is connected to resistance of `3.9 Omega`, the voltage across the cell will be |
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Answer» Here, `epsilon = 2 V, r = 0.1 Omega, R = 3.9 Omega` Current `I = epsilon/R +r = 2/3.9 + 0.1=0.5A` pot. Diff. across the terminals of the cell, i.e., terminal pot. Diff. `V = epsilon - Ir =2 - 0.5 xx 0.1 = 1.95 V` |
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| 416. |
One mole of a monatomic gas is involved in the cyclic process as shown in the PV graph. Extension of the line segments 1-2 and 3-4 pass through the origin and the curves 4-1 and 2-3 are isotherms. Find the ratio `(V_(3))/(V_(2))` if `(V_(2))/(V_(1))=2`. |
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Answer» Correct Answer - B |
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| 417. |
In circuit shown below, the resistance are given in ohms and the battery is assumed ideal with emf equal to 3 volt The voltage across the resistance `R_(4)` is A. `0.4V`B. `0.6V`C. `1.2V`D. `1.5V` |
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Answer» Correct Answer - A |
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| 418. |
In the circuit shown, the resistance are given ohms and the battery is assumed ideal with end equal to 3.0 volts. The resistance that dissipates the maximum power isA. `R_(1)`B. `R_(2)`C. `R_(3)`D. `R_(4)` |
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Answer» Correct Answer - A |
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| 419. |
A microammeter has as resistance of `100 Omega` and full scale range of `50 muA`. It can be used a voltmeter or as ahigher range ammeter provided a resistance is added to it. Pick the correct range and resistance combinations 50 V range with `10 kOmega` resistance in series b.`10 V` range with `200 kOmega` resistance in series c. 5mA rangw with `1Omega` resistance in parallel `10 mA` range with `1Omega` resistance in parallelA. (i),(ii)B. (ii),(iii)C. (iii),(iv)D. all |
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Answer» Correct Answer - B `G = 100 Omega, i_g = 50 mu A` (i) `V = i_g (R + G)` `50 = 50 xx 10^-6 (R + 100) rArr R = (10^6 - 100)ohm` (ii) `(10 = 50 xx 10^-6)(R + 100) rArr R = (2 xx 10^5 - 100) Omega, O.K`. (iii) `(i-i_g) S = i_g G` `(5 xx 10^-3 - 50 xx 10^-6) S = 50 xx 10^-6 xx 100` `S = (1)/(1 - 0.01) = 1.01 Omega, O.K` (iv) `(10 xx 10^-3 - 50 xx 10^-6) S = 50 xx 10^-6 xx 100` `(10 - 50 xx 10^-3) S = 5` `S = (5)/(10 - 0.05) = 0.5 Omega`. |
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| 420. |
A microammeter has as resistance of `100 Omega` and full scale range of `50 muA`. It can be used a voltmeter or as ahigher range ammeter provided a resistance is added to it. Pick the correct range and resistance combinations 50 V range with `10 kOmega` resistance in series b.`10 V` range with `200 kOmega` resistance in series c. 5mA rangw with `1Omega` resistance in parallel `10 mA` range with `1Omega` resistance in parallelA. `50 V` range with `10 k Omega` resistance in series.B. `10 V` range with `200 k Omega` resistance in series.C. `5 mA` range with `1 Omega` resistance in parallel.D. `10 mA` range with `1 kOmega` resistance in parallel. |
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Answer» Correct Answer - B::C `(B) R=V/(I_(g))-G=10/(50xx10^(-6))-100=2xx10^(5)` `Omega=200 K Omega` `(C)S=G/((I/(I_(g))-1))=(100Omega)/(((5xx10^(-3))/(50xx10^(6))-1))=1 Omega` |
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| 421. |
Consider the circuit shown in the figure. Both the circuits are taking same current from battery but current through `R` in the second circuit is `(1)/(10) th` of current through `R` in the first circuit. If `R` is `11 Omega`, the value of `R_(1)` A. `9.9 Omega`B. `11 Omega`C. `8.8 Omega`D. `7.7 Omega` |
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Answer» Correct Answer - A (a) In figure (b) through `R_(2) = i- (i)/(10) = (9 i)/(10)` Potential difference across `R_(2) =` Potential difference across `R implies R_(2) xx (9)/(10) I = R xx (1)/(10)`, i.e., `R_(2) = (R )/(9) = (11)/(9) Omega` `R_(eq) = (R_(2) xx R)/((R_(2) + R)) = ((11)/(9) xx (11)/(1))/((11)/(9) + (11)/(1)) = (11)/(10) Omega` Total circuit resistance `= (11)/(10) + R_(1) = R = 11` `implies R_(1) = 9.9 Omega` |
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| 422. |
The current flowing in a coil of resistance `90 Omega` is to be reduced by 90%. What value of resistance should be connected in parallel with itA. `9 Omega`B. `90 Omega`C. `1000 Omega`D. `10 Omega` |
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Answer» Correct Answer - D |
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| 423. |
Consider the circuit shown in the figure. Both the circuits are taking same current from battery but current through `R` in the second circuit is `(1)/(10) th` of current through `R` in the first circuit. If `R` is `11 Omega`, the value of `R_(1)` A. `9.9 Omega`B. `11 Omega`C. `8.8 Omega`D. `7.7 Omega` |
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Answer» Correct Answer - A (a) In Figure `(b)` through `R_(2) = i-(i)/(10) =(9i)/(10)` Potentail difference across `R_(2) =` Potentail difference across `R implies R_(2) xx (9)/(10) i= R xx (i)/(10) implies R_(2) = (R )/(9) = (11)/(9) Omega` `R_(eq) = (R_(2) xx R)/((R_(2) + R)) = ((11)/(9) xx (11)/(1))/((11)/(9) + (11)/(1)) = (11)/(10) Omega` Total circuit resistance `= (11)/(10) + R_(1) = R = 11` `implies R_(1) = 9.9 Omega` |
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| 424. |
The material of fuse wire should should haveA. A high specific resistance and high melting pointB. A low specific resistance and low melting pointC. A high specific resistance and low melting pointD. A low specific resistance and a high melting point |
| Answer» Correct Answer - C | |
| 425. |
In the figure shown the current flowing `2 R` is: A. from left to rightB. from right to leftC. no currentD. none of these |
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Answer» Correct Answer - B (b) Right end of `2 R` will be at higher potential than left end. So current in `2 R` will flow from right to left end. |
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| 426. |
What is the reading of voltmeter in the following figure ? A. 3VB. 2VC. 5VD. 4V |
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Answer» Correct Answer - D |
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| 427. |
For a cell of e.m.f. 2 V , a balance is obtained for 50 cm of the potentiometer wire. If the cell is shunted by a `2Omega` resistor and the balance is obtained across 40 cm of the wire, then the internal resistance of the cell isA. `0.25 Omega`B. `0.50 Omega`C. `0.80 Omega`D. `1.00 Omega` |
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Answer» Correct Answer - B |
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| 428. |
The e.m.f. of a standard cell balances across 150 cm length of a wire of potentiometer. When a resistance of `2Omega` is connected as a shunt with the cell, the balance point is obtained at 100cm . The internal resistance of the cell isA. `0.1 Omega`B. `1 Omega`C. `2 Omega`D. `0.5 Omega` |
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Answer» Correct Answer - B |
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| 429. |
What is the equivalent resistance between A and B in the net work of resistance .Find the correct through `12 Omega` |
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Answer» Here `4 Omega` and `12 Omega` are in parallel. Their combined resistance `(4 xx 12)/(4 + 12) = 3 Omega` Similarly `6 Omega` and `3 Omega` are in parallel third combination resistance `= (3 xx 6)/(5 + 6) = 2 Omega` Now the `3 Omega` resistance are in series so the equivalent resistance between `A` and `B = 3 + 2 = 5 Omega` current in the circuit `1 = (10V)/(5 Omega) = 2A` potential difference across `12 Omega and 2 Omega = 2 xx 3 = 6V` current though `12 Omega = (6)/(12) = (1)/(2)A` |
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| 430. |
The length of a wire of a potentiometer is 100 cm, and the e.m.f. of its standard cell is E volt. It is employed to measure the e.m.f. of a battery whose internal resistance is `0.5 Omega`. If the balance point is obtained at I = 30 cm from the positive end, the e.m.f. of the battery is . where i is the current in the potentiometer wire.A. `(30 E)/(100)`B. `(30 E)/(100.5)`C. `(30 E)/((100 - 0.5))`D. `(30 (E - 0.5 i))/(100)`, where `i` is the current in the potentiometer |
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Answer» Correct Answer - A (a) From the principle of potentiometer `V prop l` `implies (V)/(E) = (l)/(L)`, Where `V = ` emf of battery, `E = ` emf of standard cell, `L=` Length of potentiometer wire `V = (El)/(L) = (30 E)/(100)` |
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| 431. |
Resistance of 100 cm long potentiometer wire is `10 Omega`, it is connected to a battery (2 volt) and a resistance R in series. A source of 10 mV gives null point at 40 cm length, then external resistance R isA. `490 Omega`B. `790 Omega`C. `590 Omega`D. `990 Omega` |
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Answer» Correct Answer - B |
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| 432. |
Residence of 100 cm long potentiometer wire is `10 Omega`, it is connected to a battery (2 volt) and a resistance R in series. A source of 10 mV gives null point at 40 cm length, then external resistance R isA. `490 Omega`B. `790 Omega`C. `590 Omega`D. `790 Omega` |
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Answer» Correct Answer - B (b) `E = (e)/((R + R_h + r)) (R )/(L) xx l` `10 xx 10^(-3) = (2)/((10 + R + 0)) xx (10)/(1) xx 0.4 implies R = 790 Omega` |
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| 433. |
The point B in the circuit is earthed what will be the potential at point D? |
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Answer» Total resistance of current `= 3 + 10 + 15+ 20= 50Omega` main current in circuit `= (50)/(50) = 1 amp` resistance between `B` and `D = 10 + 12 + 15 = 25 Omega`potential difference across `B` and `D = 25 xx 1 = 25V` Here `B` is at highet potential `D` becomes is flowing from `B` to `D`when `B`is earthed potential of `b`zero therfore potential of`D` is `-25V` |
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| 434. |
A voltmeter connected in series with a resistance `R_1` to a circuit indicates a voltage `V_1=198V`. When a series resistor `R_2=2R_1` is used, the voltmeter indicates as voltage `V_2=180V`. If the resistance of the voltmeter is `R_V=900Omega`, then the applied voltage across `A` and `B` is A. `210V`B. `200V`C. `220V`D. `240V` |
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Answer» Correct Answer - C In series `PD` distribution in direct ratio of resistance In first case `198/(V_(AB)-198)=900/R_1` ………i In second case `180/(V_(AB)-180)=900/(2R_1)`……….ii Solving the two equations, we get `V_(AB)=220V` |
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| 435. |
What is immeterial for an electric fuse wire ?A. Its specific resistanceB. Its radiusC. Its lengthD. Current flowing through it |
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Answer» Correct Answer - C (c ) Length is immaterial for an electic fuse wire. |
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| 436. |
Two conductors are made of the same material and have the same length. Conductor `A` is a solid wire of diameter `1 mm`. Conductor `B` is a hollow tube of outer diameter `2 mm` and inner diameter `1mm`. Find the ratio of resistance `R_(A)` to `R_(B)`.A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - C |
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| 437. |
Two conductors are made of the same material and have the same length. Conductor `A` is a solid wire of diameter `1 mm`. Conductor `B` is a hollow tube of outer diameter `2 mm` and inner diameter `1mm`. Find the ratio of resistance `R_(A)` to `R_(B)`. |
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Answer» For a solid wire of resistance `R_(A^(,)) l_(1) = l, p_(1) = p, D_(1) = 1 mm, A_(1) = piD_(1)^(2)//4 = pi(1)^(2)//4mm^(2)` `R_(A) = (p_(1)l_(1))/A_(1) = (pl)/(pi(1)^(2)//4) = (4pl)/pi` For hollow tube of resistance `R_(B), l_(2) = l, p_(2) = p, A_(2) = pi(D_(2)^(2)-D_(1)^(2))//4 = pi(2^(2)-1^(2))//4 = 3 pi//4 mm^(2)` `R_(B) = (p_(2)l_(2))/A_(2) = (pl)/((3pi//4)) = (4pl)/(3pi) :. R_(A)//R_(B) = 3` |
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| 438. |
The resistance of an ideal ammeter isA. InfiniteB. Very highC. SmallD. Zero |
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Answer» Correct Answer - D |
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| 439. |
Find potential of J with:- A. 40VB. 60VC. 20VD. 30V |
| Answer» Correct Answer - 3 | |
| 440. |
The electric resistance of a certain wire of iron is R . If its length and radius are both doubled, thenA. the resistance will be halved and the specific resistance will remain unchanged.B. the resistance will be halved and the specific. Resistancw will be doubledC. the resistance and the specific resistance, will both remain unchangedD. the resistance will be doubled and the specific resistance will be halved. |
| Answer» Correct Answer - 1 | |
| 441. |
A galvanometer of `25 Omega` resistance can read a maximum current of 6 mA . It can be used as a voltmeter to measure a maximum of 6 V by connecting a resistance to the galvanometer. Identify the correct choice in the given answersA. `1025 Omega` in seriesB. `1025 Omega` in parallelC. `975 Omega` in seriesD. `975 Omega` in parallel |
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Answer» Correct Answer - C |
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| 442. |
A galvanometer can be used as a voltmeter by connectingA. high resistance in seriesB. low resistance in seriesC. high resistance in parallelD. low resistance in parallel |
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Answer» Correct Answer - A (a) To covert a galvanometer inot a voltmeter, a high value resistance is to be connected in series with it. |
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| 443. |
A galvanometer can be used as a voltmeter by connectingA. a high resistance in series with its coilB. a low resistance in parallel with its coilC. a low resistance in series with its coilD. a high resistance in parallel with its coil. |
| Answer» Correct Answer - 1 | |
| 444. |
The device used for measuring potential difference is known as:A. potential differenceB. currentC. internal resistanceD. All of these |
| Answer» Correct Answer - 4 | |
| 445. |
A battery is charged at a potential fo 15 V for 8 h when the current folwing is 10A. The battery on discharge supplies a current of 5A fo 15h . The mean terminal voltage during discharge is 14V. The watt-hour efficiency of the battery isA. 0.8B. 0.9C. 0.875D. 0.825 |
| Answer» Correct Answer - 3 | |
| 446. |
A battery is charged at a potentail of `15 V` for 8 hours when the current flowing is `10 A` for 15 hours. Then mean terminal voltage during discharge is `14 V`. The Watt-hour efficiency of the battery isA. `82.5 %`B. `80%`C. `90%`D. `87.5%` |
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Answer» Correct Answer - D (d) Watt hour efficiency `= ("Discharging energy")/("Charging energy")` `= (14 xx 5 xx 15)/(15 xx 8 xx 10) = 0.875 = 87.5%` |
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| 447. |
In the given current distribution what is the value of I A. 3AB. 8AC. 2AD. 5A |
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Answer» Correct Answer - C |
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| 448. |
What is the equivalent resistance between A and B in the figure below if `R = 3 Omega` A. `9 Omega`B. `12 Omega`C. `15 Omega`D. None of these |
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Answer» Correct Answer - D |
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| 449. |
A battery is charged at a potential of 15 V for 8 hours when the current flowing is 10 A . The battery on discharge supplies a current of 5 A for 15 hours . The mean erminal voltage during discharge is 14 V. The "Watt-hour" efficiency of the battery isA. `82.5%`B. `80%`C. `90%`D. `87.5%` |
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Answer» Correct Answer - D |
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| 450. |
What is the equivalent resistance between A and B A. `(2)/(3)R`B. `(3)/(2)R`C. `(R)/(2)`D. `2R` |
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Answer» Correct Answer - C |
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