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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.
| 1. |
If a shunt `1//10` of the coil resistance is applied to a moving coil galvanometer, its sensitivity becomesA. `10` foldB. `11` foldC. `(1)/(10)` foldD. `(1)/(11)` fold |
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Answer» Correct Answer - D `(I_(g))/(I) = (S)/(S + G) = ((G //10))/((G//10) + G) = (1)/(11)` Initially, `alpha_(1) = theta//I_(g)` (i) Finally, after the shunt is used, `alpha_(f) = theta//I`(ii) `:. (alpha_(f))/(alpha_(I)) = (theta//I)/(theta//I_(g)) = (I_(g))/(I) = (1)/(11)` So current sensitivity becomes `1//11 - fold`. |
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| 2. |
In Fig.`6.59`, when an ideal voltmetre is connected across `4000 Omega` resistance, it reads `30 V`. If the voltmeter is connected across `3000 Omega` resistance, it will read A. `20 V`B. `22.5 V`C. `35 V`D. `40 V` |
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Answer» Correct Answer - B Let `I` be the current in the circuit, then `I xx 4000 = 30 V`. If voltmeter is put across `3000 Omega` resistance, then reading of voltmeter is `I xx 3000 = (30)/(4000) xx 3000 = 22.5 V` |
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| 3. |
In the above question, the reading of ammeter is `200//x mA`. What is the value of `x`? |
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Answer» Correct Answer - `(3)` `(4 - V_(1))/(100) = (V_(1) - 1)/(100) + (V_(1) - 0)/(100)` or `V_(1) = (5)/(3) V` Current in ammeter (II)`= (V_(1) - 1)/(R ) = ((5//3) - 1)/(200) = (2)/(300) A = (20)/(3) mA` |
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| 4. |
In a potentiometer experiment the balancing with a cell is at length 240 cm. On shunting the cell with a resistance of `2Omega`, the balancing length becomes 120 cm.The internal resistance of the cell isA. `2 Omega`B. `4 Omega`C. `0.5`D. `1 Omega` |
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Answer» Correct Answer - A `r = (l_(1) - l_(2))/(l_(2)) R = (240 - 120)/(120) xx 2 Omega = 2 Omega` |
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| 5. |
A galvanometer has a resistance of `3663Omega`. A shunt `S` is connected across it such that (`1//34`) of the total current passes through the galvanometer. Then the value of the shunt isA. `3663 Omega`B. `111 Omega`C. `107.7 Omega`D. `3555.3 Omega` |
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Answer» Correct Answer - B `(I_(g))/(I) = (S)/(S + G)` or `(1)/(34) = (S)/(S + G)` `:. S = (G//33) = (3663//33) = 107.7 Omega` |
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| 6. |
An ammetre is obtained by shunting a `30 Omega` galvanometer with a `30 Omega` resistance. What additional shunt should be connected across it to double the range ?A. `15 Omega`B. `10 Omega`C. `5 Omega`D. none of these |
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Answer» Correct Answer - A Total range is doubled. i.e., `4l_(g)` Now shunt required is `S = (I_(g))/(4l_(g) - I_(g)) xx G = 10 Omega` `(1)/(30) + (1)/(x) = (1)/(10)` or `x = 15 Omega` |
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| 7. |
A milliammeter of range `10 mA` has a coli of resistance `1 Omega`. To use it as an ammeter of range `1 A`, the required shunt must have a resistance ofA. `(1)/(101) Omega`B. `(1)/(100) Omega`C. `(1)/(99) Omega`D. `(1)/(9) Omega` |
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Answer» Correct Answer - C `S = (i_(g) G)/((I - i_(g))) = (0.01 xx1)/(1 - 0.01) = (1)/(99) Omega` |
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| 8. |
What shunt resistance is required to make the `1.00 mA`, `20 Omega` Galvanometer into an ammeter with a range of 0 to `50.0A`? |
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Answer» We have `I_(2) = 1.00mA=1.00xx 10^(-3) A` and `R_(2) = 2.0.0 Omega`, and the ammetre should be able to handle maximum current , `I = 50.0 xx 10^(-3) A`. Solving equation `I_(g) r_(g) = (I - I_(g)) R_(S)` for `R_(S)`, we get `R_(S) = (I_(g)R_(g))/(I - I_(g)) = ((1.00 xx 10^(-3) A)(20.0 Omega))/(50.0 xx 10^(-3) A- 1.00 xx 10^(-3) A) = 0.408 Omega` |
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| 9. |
The ammeter has range 1 ampere without shunt. The range can be varied by using different shunt resistance. The graph between shunt resistance and range will have the nature A. PB. QC. RD. S |
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Answer» Correct Answer - B |
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| 10. |
The graph which represents the relation between the total resistance `R` of a multi range moving coil voltmeter and its full scale deflection A. (i)B. (ii)C. (iii)D. (iv) |
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Answer» Correct Answer - D |
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| 11. |
In `Q.36`, the external resistance that must be connected is series with the main circuit so that the total current in the main circuit remains unaltered even when the galvanometer is shunted isA. `3663 Omega`B. `111 Omega`C. `107.7 Omega`D. `3555.3 Omega` |
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Answer» Correct Answer - D Compensation exernal resistance `G - (SG)/(S + G) = 3663 - 107.7 = 3555.3 Omega` |
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| 12. |
In the circuit `P != R, ` the reading of the galvanometer is same with switch S open or closed. Then A. `I_R=I_G`B. `I_P=I_G`C. `I_Q=I_G`D. `I_Q=I_R` |
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Answer» Correct Answer - A |
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| 13. |
A galvenometer has a resistance of `30 Omega`, and a current of `2 mA` is needed for a full- scale deflection. What is the resistance and how is it to be connected to conver the galvanometer of `0.2 V` range? |
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Answer» Given `G = 30 Omega` and `I_(g) = 2 mA` (i) To convert the galvanometer into an ammeter of range `0.3 A`, `(I - I_(G)) S = I_(G)` or `(0.3 - 0.002)S = 0.002 xx 30` or `S = (0.002 xx 30)/(0.298) = 0.2013 Omega` (ii) To convert the galvanometer into a voltmeter of range `0.2 V`, `V = I_(G)(R + G)` or `0.2 = 2 xx 10^(-3) (30 + R)` or `R = 70 Omega` |
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| 14. |
Consider a circular as shows in Fig. `6.5` (a). We want to measure the current `i` flowing in the circuit. For this we connect an ammeter of resistance `R_(A)` as shows in Fig. `6.5`(b). Find the percentage error in the current. |
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Answer» Actual current is `I = E//R`. Let the current measured by the ammeter be `i_(i)`, then `i_(i) = E//(R + R_(A))`, clearly `L_(i)` is less than `i`. Percentage error is `(I - i_(i))/(i) xx 100 = ((E//R - E//(R + R_(A)))/(E//R)) xx 100 ((R_(A))/(R + R_(A))) xx 100` |
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| 15. |
Whether the switch K is open or closed, the reading of galvanometer is the same. If I denotes the current then : A. `I_(R_4)` = `I_G`B. `I_(R_5) = I_G`C. `I_(R_3) = I_G`D. `I_(R_4) = I_(R_3)` |
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Answer» Correct Answer - C |
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| 16. |
The measurement of voltmeter in the following circuit is A. 2.4 VB. 3.4 VC. 4.0 VD. 6.0 V |
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Answer» Correct Answer - D |
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| 17. |
In the circuit shows in Fig. 6.63, the cell is ideal with emf `9 V`. If the resistance of the coil of galvanometer is `1 Omega`, then A. no current flows in the galvanometerB. charge flowing through `8 mu F` is `40 mu C`C. potential difference across `10 mu F` is `5 V`D. potential difference across `10 mu F` is `4 V` |
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Answer» Correct Answer - A::B::D Since `R_(1)//R_(2) = C_(2)//C_(1)`, the wheatstone bridge is balanced. Hence, `V_(C) = V_(D)`. No current passes through the galvanometer. Hence, option (a) is correct. Potential difference across `R_(1) =` potential difference across `C_(1) = 4V` Potential difference across `R_(2) =` potential difference across `C_(2) = 5 V` Potential difference across `5 Omega` is the potential difference across `8 mu F` capacitor. So, charge across `8 mu F` capacitor is `Q = CV = 8 mu F xx 5V = 40 mu C`. hence, option (b) is correct and so is option(d). |
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| 18. |
In the circuit shown, the galvanometer shows zero current. The value of resistance R is : A. `1 Omega`B. `2 Omega`C. `4 Omega`D. `9 Omega` |
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Answer» Correct Answer - A |
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| 19. |
A potentiometer is connected across `A `and `B` and a balance is obtained at `64.0 cm`. When the potentiometer lead at `B` is moved to `C`, a balance is found at `8.0 cm`. If the potentiometer is now connected across `B` and `C`, a balanced will be found at A. `8.0 cm`B. `56.0 cm`C. `64.0 cm`D. `72.0 cm` |
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Answer» Correct Answer - B `E_(1) prop 64` `E_(1) - E_(2) prop 8` `E_(2) prop l` `:. 64 - l = 8` or `l = 64 - 8 = 56 cm` |
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| 20. |
How will the reading in the ammeter `A` of Fig. 6.39 be affected if another identical bulb `Q` is connected in parallel to `P` as shows. The voltage in the mains is maintained at a constant value. A. The reading will be reduced to one-half.B. the reading will not be affected.C. The reading wil be double of the previous one.D. The reading will be increased fourfold. |
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Answer» Correct Answer - C Resistance is halved. Current is doubled. |
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| 21. |
Tha scale of a galvanometer is divided into `150` equal divisions. The galvanometer be designed to read (i) `6` A per division and (ii) `1 V` per division? |
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Answer» Full-scale voltage `= 150` divisions`//2` divisions per `m V` `= 75 m V` Full-scale current`= 150` divisions`//10` divisions per `m A` `= 15 mA` So resistance of galvanometer is `G = ("Full-scale voltage")/("Full-scale current"1) = (75 xx 10^(-3))/(15 xx 10^(-3)) 5 Omega` (i) Range of ammeter is `I = 150 xx 6 = 900 A`. So, `S = (I_(G)G)/(I - I_(G)) = (15 xx 10^(-3) xx 5)/((900 - 15 xx 10^(-3))) = 8.3 xx 10^(-5) Omega` (ii) Range of voltmeter is `V = 150 xx 1 = 150 V`. So, `R = (V)/(I_(G)) - G = (150)/(15 xx 10^(-3)) - 5 = 9995 Omega`. |
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| 22. |
A cell of itnernal resitance `1.5 Omega` and of e.m.f. `1.5` volt balances `500 cm` on a potentiomter wire. If a wirr of 15 `Omega` is connected between the balance point and the cell, then tha balance point will shiftA. To zeroB. By 500 cmC. By 750 cmD. None of the above |
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Answer» Correct Answer - D |
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| 23. |
In the simple potentionmeter circuit, where the length `AB` of the potentiometer wire is `1 m`, the resistors `X` and `Y` have values of `5 Omega` and `2 Omega`, respectively. When `X` is shunted by a wire, the balance point is found to be `0.625 m` from `A`. What is the resistance of the shunt? |
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Answer» Let `R` be the resistance of the shunted wire. The effective resistance of `R` and `5 Omega` in parallet is `5 R//(5 + R)`. At balance point, `(5R//(5 + R))/(2) = (0.625)/(1 - 0.625) = (0.625)/(0.0375) =(5)/(3)` On solving, we get `R = 10 Omega`. |
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| 24. |
A `100 V` voltmeter of internal resistance `20 k Omega` in series with a high resistance `R` is connected to a `110 V` line. The voltmeter reads `5 V`, the value of `R` isA. `210 kOmega`B. `315 kOmega`C. `420 kOmega`D. `440 kOmega` |
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Answer» Correct Answer - C |
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| 25. |
In the circuit shown, a meter bridge is in its balanced state. The meter bridge wire has a resistance `.1 ohm//cm`. The value of unknown resistance `X` and the current drawn from the battery of negligible resistance is A. `6 Omega, 5 A`B. `10 Omega, 0.1 A`C. `4 Omega, 1.0 A`D. `12 Omega, 0.5 A` |
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Answer» Correct Answer - C |
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| 26. |
Statement I: The wire of a potentioment should be of uniform area of cross section. Statement II: It satisfies the requirement of the principle of a potentiometer.A. Statement `1` is true, Statement `2`is True , Statement `2` is correct explanation for Statement `1`.B. Statement `1` is True, Statement `2` is True, Statement `2` is NOT a correct explantion for Statement `1`.C. Statement `1` is True, Statement `2` is False.D. Statement `1` is False, Satement `2` is True. |
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Answer» Correct Answer - A Principle of a potentiomete states that drop of potential across any segament of the potentiometer wire is directly across any segment of the potentiometer wire is directly proportional to its length. This can be satisfied if wire of the potentiometer has a uniform area of cross section. |
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| 27. |
In the above questions, if the polarity of `E_(2)` is reversed, thenA. current in both ammeters will flow in same directionB. current in both ammeters will flow in opposite directionsC. current in both ammeters can be same if `R_(1) gt R_(2)`D. current in both can be same if `R_(1) lt R_(2)` |
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Answer» Correct Answer - B::C If polarity of `E_(2)` is reversed, then `I_(2) = (E_(2))/(R_(2))` and `I_(1) = (E_(1) + E_(2))/(R_(1))` for `I_(1) = I_(2), R_(1) gt R_(2)` |
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| 28. |
For measurement of potential difference, potentiometer is perferred in comparison to voltmeter becauseA. Statement `1` is true, Statement `2`is True , Statement `2` is correct explanation for Statement `1`.B. Statement `1` is True, Statement `2` is True, Statement `2` is NOT a correct explantion for Statement `1`.C. Statement `1` is True, Statement `2` is False.D. Statement `1` is False, Satement `2` is True. |
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Answer» Correct Answer - A In a balanced condition, the potentiomaeter does not draw any current, anf hence does not disturb the circuit. |
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| 29. |
Two resistances are connected in the two gaps of a meter bridge. The balance point is `20 cm` from the zero end. When a resistance `15 Omega` is connected in series with the smaller of two resistance, the null point+ shifts to `40 cm`. The smaller of the two resistance has the value.A. 3B. 6C. 9D. 12 |
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Answer» Correct Answer - C |
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| 30. |
In the circuit shows in Fig. `6.20`, a meter bridge is in its balance state. The meter bridge wire has a resistance of `1 Omega cm^(-1)`. Calculate the value of the unknow resistance `X` and the battery of negligible internal resistance. |
| Answer» For balanced wheatstone bridge, `(X)/(40) = (3)/(60)` or `X = 2 Omega` | |
| 31. |
A potentiometer is an ideal voltmeter since a voltmeter draws some current through the circuit while potentiometer needs no current to work. A potentiometer works on the principle of emf comparison. In working condition, a constant currant flows throughout the wire of a 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 emf `2 V` and internal resistance `0.4 Omega` is used across the wire `AB`. A standard cadmium cell of emf `1.02 V` gives a balance point at `66 cm` length of wire. The standard cell is then replaced by a cell of unknows emf `e` (internal resistance `r`), and the balance. Point found similarly turns out to be `88 cm` length of the wire. The length of potentiometer wire `AB` is `1 m`. The value of `e` isA. `1.36 V`B. `2.63 V`C. `1.83 V`D. `none |
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Answer» Correct Answer - A `(e)/(1.02) = (88)/(66)` or `e = 1.36 V` `4V` is greater than applied emf `2 V`, hence no balance point is obtained. On connecting the resistance across `e`, current will flow in `e` due to which terminal potential difference will be less than emf and the balancing length will decreases. |
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| 32. |
In a meter bridge the point D is a neutral point (Fig. 2(EP).4). A. The meter bridge can have no other neutral A point for this set of resistancesB. When the jockey contacts a point on meter wire left of D, current flows to B from the wireC. When the jockey contacts a point on the meter wire to the right of D, current flows from B to the wire through galvanometerD. When R is increased the neutral points shifts to left |
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Answer» Correct Answer - A::C |
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| 33. |
In Fig. `6.16`, `AB` is a `1 m` long uniform wire of `10 Omega` resistance. Other data are shows in the figure. Calculate (i) potential gradient along `AB` and (ii) length of AO when galvanometer shows no deflection. |
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Answer» (i) Current in wire `Ab` is `l = 2//(15 + 10) = 2//25 A` potential difference across `AB` is `V = lR = 2//25 xx 10 = 0.8 V` Potential gradient along `AB` is `k = V//l = 0.8//1 = 0.8 V` (ii) Current through `0.3 Omega` is `(1.5)/(1.2 + 0.3) = 1 A` Potential difference acros `0.3 Omega`is `1 xx 0.3 = 0.3 V` Let `l_(1)` be the length`AO`, then `0.3 = 0.8 xx l_(1)` or `l_(1) = 0.3//0.8 = 0.375 m` |
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| 34. |
Figure `6.51` shows a simple a potentiometer circuit for measuring a small emf produced by a thermocouple. The meter wire `PQ` has a resistance of `5 Omega`, and the driver cell has an emf of `2.00 V`. If a balance point is obtained `0.600 m` along `PQ` when measuring an emf of `6.00 mV`, what is the value of resistance `R`?A. `95 Omega`B. `995 Omega`C. `195 Omega`D. `1995 Omega` |
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Answer» Correct Answer - B The voltage per unit length on the meter wire `PQ` is `(6.00 mV)/(0.60 m)`or `10 mVm^(-1)` Hence, potential across the meter wire `PQ` is `10 mVm^(-1) (1 m)` ` = 10 mV`. Current draw from the driver cell is `I = (10mV)/(5 Omega) = 2mA` Resistance of the resistor `R` is `R = (2V - 10 mV)/(2mA) = (1990 mV)/(2mV) = 995 Omega` |
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| 35. |
In the experiment of calibration of voltmeter, a standard cell of emf `11 V` is balanced against `440 cm` of potentiometer wire. The potentail difference across the ends of a resistance is found to balance aginst `220 cm` of the wire. The corresponding reading of the voltmeter is `0.5 V`. Find theerror in the reading of voltmeter. |
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Answer» `k= (E_(B))/(L) = (1.1)/(440) = 0.0025 Vcm^(-1)` potential difference across `R` is `0.0025 xx 220 = 0.550 V` Error in the reading of voltmeter is reading of voltemetre - reading of potentional `= 0.5 - 0.55 = - 0.05 V` |
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| 36. |
Figure `6.46`, shows a wheatstone net, with `P = 1000 Omega`,`Q = 10.0 Omega`,`R`(unknows), `S` variable and near `150 Omeag` for balance. If the connections across `A,C` and `B,D` are interchanged, the error range in `R determination would A. remain unaffectedB. increase substantiallyC. increase marginallyD. decrease substantially |
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Answer» Correct Answer - D `R` is the order of `15,000 Omega` The junctions of the heighest two an dthe lowest two resistances are `A` and `C`, and for better sensitivity, the galvanometer will be between these. So, error will decrease with the suggested interchange. Alternatively : Let `R` be not exactly `15,000 Omega`. Then some potential difference will be creates between points `B` and `D`. Let it be `Delta V_(1)`. Now suppose the connections are interchanged, then the same type of potential difference will be created between points `A` and `C`. Let it be `Delta V_(2)`. So larger current will flow through galvanometer in the second case telling us that value of `R` needs to be changed. And we will try to put more accurate value of `R`. In this way, error will decrease. |
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| 37. |
The measurement of an unknown resistance R is to be carried out using Wheatstone bridge (see Fig. 2(EP).3). Two students perform an experiment in two way. The first student takes `R_(2)=10 Omega` and `R_(1)=5Omega`. The other student takes `R_(2)=1000Omega` and `R_(1)=500Omega`. In the standard arm, both take `R_(3)=5Omega`. Both find `R=(R_(2))/(R_(1))R_(3)=10Omega` within errors.A. The errors of measurement of the two students are the sameB. Errors of measurement do depends on the accuracy with which `R_2 and R_1` can be measuredC. If the student uses large values of `R_2 and R_1` The currents through the arms will be feeble. This will make determination of null point accurately more difficultD. Wheatstone bridge is a very accurate instrument and has no errors of measurement |
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Answer» Correct Answer - B::C |
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| 38. |
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 - 0.5i))/(100)` |
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Answer» Correct Answer - A Using the principle of potentiometer, `V prop l.` So `(V)/(E) = (l)/(L)` or `V = (l)/(L) E = (30)/(100) E = (30E)/(100)` |
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| 39. |
Assertion : The e.m.f. of the drivercell in the potentiometer experiment should be greater than the e.m.f. of the cell to determined.A. Statement `1` is true, Statement `2`is True , Statement `2` is correct explanation for Statement `1`.B. Statement `1` is True, Statement `2` is True, Statement `2` is NOT a correct explantion for Statement `1`.C. Statement `1` is True, Statement `2` is False.D. Statement `1` is False, Satement `2` is True. |
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Answer» Correct Answer - A If either the emf of the driver cell or the potential difference across the whole potentiometer wire is lesser than the emf of the experimental cell, the balance point will not be obtained. |
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| 40. |
Potentiometer wire `PQ` of `1 m` length is connected to a standard cell `E_(1)`. Another cell `E_(2)` of emf `1.02 V` is connected with a resistance `r` and a switch `S` as shown in the circuit diagram. With switch `S` open, the null position is obtained at a distance of `51 cm` from `P`. (i) Calculate the potential gradient of the potentiometer wire. (ii) Find the emf of cell `E_(1)`. (iii) When switch `S` is closed, will the null point move toward `P` or toward `Q`? Give reason for your answer. |
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Answer» (i) Potential gradient is `k = (V)/(l) = (1.02)/(0.51) = 2 Vm^(-1)` (ii) The emf of the cell is `E_(1) = k_(l) = 2 xx 1= 2 V` (iii) When the swich `S` is closed, there is no shift in the position of the null point because it depends on the potential gradient along the poteniometer wire (which depends on the emf of battery `E_(1)` and the resistance of the potentiometer wire circul and length of potentiometer) and the emf of cell `E_(2)`, which does not change when the swich `S` is closed. |
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| 41. |
In an experiment with a post office box, the radio of arms are `1000 : 10`. If the value of the third resistance is `999 Omega`, find the unknows resistance. |
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Answer» Correct Answer - `9.99 Omega` In the given case, the ratio of arms is `1000: 10`. `(P)/(Q) = (1000)/(10) = 100` Third resistance is `R = 999 V`. Lat `X` be the unknows resistance. Then, `(P)/(Q) = (R )/(X)` or `X = (Q)/(P) xx R = (1)/(100) xx999= 9.99 Omega` |
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| 42. |
In the connection shown in the figure, initially the switch K is open and the capacitor is uncharged. Then the switch is closed, and the capacitor is charged up to the steady state and the switch is opened again. Determine the values indicated by the values indicated by the ammeter. `[Given V_0 = 30 V, R_1 = 10k Omega, R_2 = 5kOmega`] . just after closing the switchA. `2 mA`B. `3 mA`C. `0 mA`D. none of these |
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Answer» Correct Answer - C Just after closing, capacitor behaves as a short circuit and all current flows through it, hence ammeter reads zero. After a long time, capacitor behaves like an open circuit and no current flows through it. Therefore, `I = (V_(0))/(R_(1) + R_(2)) = (30)/(10 + 5) = 2 mA` Just after reopening, potential difference across `R_(2)` remains the same as charge on the capacitor does not change intially, hence current remains the same. |
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| 43. |
`AB` is a potentiometer wire of length `100 cm`. When a cell `E_(2)` is connected across `AC`, where `AC = 75 cm`, no current flows from `E_(2)`. The internal resistance of the cell `E_(1)` is negligible. Find emf of the cell `E_(2)`.A. `2 V`B. `1.5 V`C. `1 V`D. `1.75 V` |
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Answer» Correct Answer - B `k = (E_(1))/(100) = (2)/(100) 0.02 V cm^(-1)` `E_(2) = k(75) = 0.02 xx 75 = 1.5 V` |
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| 44. |
A battery of e.m.f. 10 V and internal resistance `2 Omega` is connected in primary circuit with a uniform potentiometer wire and a rheostat whose resistance is fixed at `998 Omega`.A battery of unknown e.m.f. is being balanced on this potentiometer wire and balancing length is found to be 50 cm.When area of cross section of potentiometer wire is doubled, then balancing length is found to be 75 cm. (i)Calculate e.m.f. of the battery. (ii)Calculate resistivity of potentiometer wire if length of wire is 100 cm and area of cross-section (initially) is `100 cm^2`. |
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Answer» Correct Answer - (a)2.5 volt , (b)`10 Omega-m` |
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| 45. |
A battery is connected to a potentiometer and a balance point is obtained at `84 cm` along the wire. When its terminals are connected by a `5 Omega` resistor, the balance point changes to `70 cm`. Find the new position of the balance point when `5 Omega` resistor is changed by `4 Omega` resistor.A. `26.5 cm`B. `52 cm`C. `67.2 cm`D. `83.3 cm` |
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Answer» Correct Answer - C `r = (l_(1) - l_(2))/(l_(2))R = (84 - 70)/(70) xx 5 = 1 Omega` `1 = (l_(1) - l_(3))/(l_(3)) xx 4`, put `l_(1) = 84 cm` get `l_(3) = 67.2 cm` |
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| 46. |
A battery is connected to a potentiometer and a balance point is obtained at `84 cm` along the wire. When its terminals are connected by a `5 Omega` resistor, the balance point changes to `70 cm`. Calculate the internal resistance of the cell.A. `4 Omega`B. `2 Omega`C. `5 Omega`D. `1 Omega` |
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Answer» Correct Answer - D `r = (l_(1) - l_(2))/(l_(2))R = (84 - 70)/(70) xx 5 = 1 Omega` `1 = (l_(1) - l_(3))/(l_(3)) xx 4`, put `l_(1) = 84 cm` get `l_(3) = 67.2 cm` |
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| 47. |
The circuit shows in Fig . `6.33` shows the use of potentiometer to measure the internal resistance of a cell. (a) When the key is open, how does the balance point change, if the driver cell decreases ? (b) When the key is closed, how does the balance point change, if `R` is increased, keeping the current from the driver cell constant ? |
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Answer» Correct Answer - a. Balance point will shift towards right. (b) Balance point will shift towards right. If current from driver cell decreases, then potential gradient decreases. So the balancing length should increase. Hence, balanced point shifs toward right. b. If `R` is increased, current in the lower circuit decreases, due to which terminal potential difference of battery increases. To balance increased terminal potential difference, balancing length should increase, hence balance point shifts toward right. |
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| 48. |
A potentiometer wire of length `10 m` and resistance `30 Omega` is connected in series with a battery of emf `2.5 V`, internal resistance `5 Omega` and an external resistance `R`. If the fall of potential along the potentiometer wire is `50 mu V mm^(-1)`, then the value of `R` is found to be `23 n Omega`. What is `n`? |
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Answer» Correct Answer - `(5)` Potential gradient, `k = (I R_(P))/(L) = (E R_(P))/((R + R_(P) + r)L)` (where `R_(P)` is resistance of the wire) or `50 xx 10^(-3) = (2.5 xx 30)/((R + 30 + 5) xx 10)` or R = 115 Omega` |
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| 49. |
A potentiometer arrangement is shows in Fig. `6.62`. The driver cell has emf `e` and internal resistance `r`. The resistance of potentiometer wire `AB` is `R.F` is the cell of emf `e//3` and internal resistance `r//3`. Balance point `(J)` can be obtained for all finite value of A. `R gt r//2`B. `R lt r//2`C. `R gt r//3`D. `R lt r//3` |
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Answer» Correct Answer - A Current in `AB = I = (e)/(R + r)` potential difference across `AB = IR = (eR)/(R + r)` To obtain balanced point, `(eR)/(R + r) gt (e)/(3)` or `R gt r//2` |
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| 50. |
The emf of the driver cell of a potentiometer is `2 V`, and its internal resistance is negligible. The length of the potentiometer wire is `100 cm`, and resistance is `5 omega`. How much resistance is to be connected in series with the potentiometer wire to have a potential gradient of `0.05 m V cm^(-1)`?A. `1990 Omega`B. `2000 Omega`C. `1995 Omega`D. none of these |
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Answer» Correct Answer - C `k = 0.05 m vcm^(-1) = 5mVcm^(-1)` `I = (kl)/(r ) = (5 xx 10^(-3) xx 1)/(5) = 10^(-3) A` Now, `2 = I(R + r) = 10^(-3) (R + 5)` or `R = 1995 Omega` |
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