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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.
| 2051. |
In meter brigde of Wheatstone bridge for measurment of resistance, the known and the unknown resistance are interchanged. The error so removed isA. End correctionB. Index errorC. Due to temperature effectD. Randon error |
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Answer» Correct Answer - A |
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| 2052. |
For measurement of potential difference, potentiometer is perferred in comparison to voltmeter becauseA. potentiometer is more sensitive than voltmeterB. the resistance of potentiometer is less than voltmeterC. potentiometer is cheaper than volt meterD. potentiometer does not take current from the circuit |
| Answer» Correct Answer - D | |
| 2053. |
For measurement of potential difference, potentiometer is perferred in comparison to voltmeter becauseA. potentiometer is more sensitive than voltmeterB. the resistance of potentiometer is less than volumeterC. potentiometer is cheaper than voltmeterD. potentiometer does not take current from the circuit |
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Answer» Correct Answer - D |
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| 2054. |
For measurement of potential difference, potentiometer is perferred in comparison to voltmeter becauseA. potentiometer is more sensitive than voltmeterB. the resistance of potentiometer is less than voltmeterC. potentiometer is cheaper than voltmeterD. potentiometer does not take current from the circuit |
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Answer» Correct Answer - D Potentiometer works on null deflection method. In balance condition no current flow in secondary circuit. |
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| 2055. |
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 resistanceB. When the jocky 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 point shifts to left |
| Answer» At neutral point, potential at B and neutral point are same. When jockey is place at to the right of D, the potential drop across AB, which brings the potential of point D less than that of B, hence current shows from B to D. | |
| 2056. |
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. `995 Omega`B. `1995 Omega`C. `2995 Omega`D. None of these |
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Answer» Correct Answer - A |
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| 2057. |
Each resistor shown in figure is an infinite network of resistance`1Omega`. The effective resistance Between points A and B is A. less than `1Omega`B. `1 Omega`C. more thn `1Omega` but less than `3Omega`D. `3Omega` |
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Answer» Correct Answer - C `R_(AB)=2` [Net resistance of infinite series ]+1 In parallel net resistance is always less than the smallest one. Hence., net resistance of infinite series is less than `1Omega` `:. 1 Omega lt R_(AB) lt 3 Omega` |
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| 2058. |
The effective resistance between A and B is the given circuit isA. `3Omega`B. `2Omega`C. `4Omega`D. `6Omega` |
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Answer» Correct Answer - B combination of resistors |
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| 2059. |
Equivalent resistance across A and B in the given circuit isA. `(2r)/(3)`B. `(8r)/(7)`C. `(7r)/(3)`D. `6r` |
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Answer» Correct Answer - B combination of resistors |
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| 2060. |
Determine the current drawn from a `12V` supply with internal resistance `0.5Omega`. By the infinite network shown in fig. Each resistor has `1Omega` resistance. |
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Answer» Let `x` be the equivalent resistance of infinite net work. Since the net work is infinite, therefore, the addition of one more unit of three resistance each of value of `1Omega` across the terminals will not alter the total resistance of net work, i.e., it should remain `x`. Therefore, the network would appear as shown in Fig.`2(NCT)`.4 and its total resistance should remain `x`. Here the parallel combination of `x` and `1Omega`is in series with the two resistors of `1Omega`each. The resistance of paralle combination is `1/R_(p) =1/x+1/1 =(1+x)/x or R_(p) = x/(x+1)` `:.` Total resistance of net work will be given by `x=1+1+(x)/(x+1) =2+(x)/(x+1) or x(x+1)= 2(x+1)+x or x^(2)+x=2x+2+x or x^(2)-2x-2=0 or x=(2+-sqrt(4+8))/2 =(2+-sqrt(12))/2 =(2+-2sqrt3)/2 =1+-sqrt3` The value of resistance can not be negative, therefore, the resistance of net work `=1+sqrt3=1+1.73Omega =2.73Omega` Total resistance of the circuit `=2.73 + 0.5 = 3.23Omega .:` Current drawn, `l=12/3.23 =3.72` amp |
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| 2061. |
In the circuit shown `E, F, G` and `H` are cells of emf `2V, 1V, 3V` and `1V` respectively and their internal resistance are `2Omega, 1Omega, 3Omega` and `1Omega` respectively. A. `V_(D) - V_(B) = -(2)/(13)V`B. `V_(D) - V_(B) = (2)/(13) V`C. `V_(G) = (21)/(13) V =` potential difference across GD. `V_(H) = (19)/(13) V =` potential difference across H |
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Answer» Correct Answer - B::C::D |
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| 2062. |
Each resistor shown in figure is an infinite network of resistance`1Omega`. The effective resistance Between points A and B is A. Less than` 1Omega`B. `1Omega`C. More than` 1Omega` but less than` 3Omega`D. `3Omega` |
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Answer» Correct Answer - A |
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| 2063. |
The equivalent resistance between points A and B of an infinite network of resistance each of `1Omega` connected as shown isA. `(1+sqrt(5))/(2)`B. `(2+sqrt(5))/(4)`C. `(3+sqrt(5))/(2)`D. `(1+sqrt(7))/(3)` |
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Answer» Correct Answer - A combination of resistors |
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| 2064. |
A current of 1 A is passed through two resistances `1Omega` and `2Omega` connected in parallel. The current flowing through `2Omega` resistor will beA. `(1)/(3)A`B. `1A`C. `(2)/(3)A`D. `3A` |
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Answer» Correct Answer - A `i_(2)=(iR_(1))/(R_(1)+R_(2))` |
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| 2065. |
Two cells of e.m.f. 2.5 V and 2.0 V having internal resistance of `1Omega` and `2Omega` respectively are connected in parallel with similar poles connected together so as to send the current in the same direction through an external resistance of `2Omega`. The current in the external resistance isA. 0.87 AB. 1.29 AC. 1.00 AD. 2.29 A |
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Answer» Correct Answer - A `I=(E_(1)r_(2)+E_(2)r_(1))/(r_(1).r_(2)+R(r_(1)+r_(2)))=(1.5xx2+2xx1)/(1xx2+5( 1+2))` `=0.29A` |
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| 2066. |
Two cells each of emf 10 V and each `1Omega` internal resistance are used to send a current through a wire of `2Omega` resistance. The cells are arranged in parallel. Then the current through the circuitA. 2AB. 4AC. 3AD. 5A |
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Answer» Correct Answer - B `i=(E)/((r)/(n)+R)` |
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| 2067. |
The equivalent resistance between points A and B of an infinite network of resistances each of `1Omega` connected as shown, is A. InfiniteB. `2 Omega`C. `(1+sqrt(5))/(2)Omega`D. Zero |
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Answer» Correct Answer - C |
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| 2068. |
Six cells each of e.m.f. 2 V and internal resistance `0.1Omega` are connected to three resistances as shown in figure. The reading of a low resistance ammeter A in the circuit is, A. `2.1A`B. `3.0A`C. `1.2A`D. `1.5A` |
| Answer» Correct Answer - C | |
| 2069. |
An electric bulb is rated 220 volt - 100 watt. The power consumed by it when operated on 110 volt will beA. `50` wattB. `75` wattC. `40` wattD. `25` watt |
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Answer» Correct Answer - D `H_(1)=(V^(2))/Rt & H_(2)=(V^(2))/(R//2)t` `:. (H_(2))/(H_(1))=2implies H_(2)=2H_(1)` |
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| 2070. |
Five identical resistors each of resistance `1Omega` are initially arranged as shown in the figure by clear lines. If two resistances are added as shown by the dashed lines then change in resistance in final and initial arrangement is A. `2Omega`B. `1Omega`C. `3Omega`D. `4Omega` |
| Answer» Correct Answer - A | |
| 2071. |
The current `i` in the circuit (see figure) isA. `1/45amp`B. `1/15amp`C. `1/10amp`D. `1/5amp` |
| Answer» Correct Answer - C | |
| 2072. |
An electric bulb is rated 220 volt - 100 watt. The power consumed by it when operated on 110 volt will beA. 25 wattB. 50 wattC. 75 wattD. 40 watt |
| Answer» Correct Answer - A | |
| 2073. |
An electron in potentiometer experiences a force `2.4xx10^(-19)N`. The length of potentiometer wire is 6m. The emf of the battery connected across the wire is (electronic charge `=1.6xx10^(-19)C`)A. 6 VB. 9 VC. 12 VD. 15 V |
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Answer» Correct Answer - B |
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| 2074. |
Four identical bulbs each rated 100 watt, 220 volts are connected across a battery as shown. The total electric power consumed by the bulbs is: A. 75 wattB. 400 wattC. 300 wattD. `400/3watt` |
| Answer» Correct Answer - A | |
| 2075. |
Four identical bulbs each rated 100 watt, 220 volts are connected across a battery as shown. The total electric power consumed by the bulbs is: A. 75 wattB. 400 wattC. 300 wattD. 150 watt |
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Answer» Correct Answer - A |
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| 2076. |
If two bulbs of `25 W and 100 W` rated at `220 V` are connected in series across a `440 V` supply, will both the bulbs fuse ? If not which one ?A. 25 W bulbB. 100 W bulbC. both of theseD. none of these |
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Answer» Correct Answer - a Resistance of one bulb, `R_(1) = (220)^(2)//25` ohm Resistance of other bulb, `R_(2) = (220)^(2)//100` ohm Total resistance when two bulb are in series will be `R = (220)^(2)//25 + (220)^(2)//100 = (220)^(2)//20 Omega` Given `I = (V)/(R ) = (440)/((220)^(2)//20) = (2)/(11)` Amp Potential difference across `25W` bulb `= IR_(1) = ((220)^(2))/(25) xx (2)/(11) = 352V` Potential difference across `100W` bulb `= ((220)^(2))/(100) xx (2)/(11) = 88 V` The bulb of `25` watt will be fixed become if can tolerate only `220V` while the voltage across it is `352V` |
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| 2077. |
The potential difference between points `A` and `B` is A. `20/7V`B. `40/7V`C. `10/7V`D. zero |
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Answer» Correct Answer - D `(8Omega)/(6Omega)=(4Omega)/(3Omega)` `:. V_A=V_B` or `V_(AB)=0` |
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| 2078. |
A potentiometer wire of length 1000 cm has a resistance of 10 ohms. It is connected in series with a resistance and a cell of e.m.f 2 volts and of negligible internal resistance. A source of e.m.f 10 millivolts is balanced against a length of 40 cm of the potentiometer wire. What is the value of the external resistance?A. 1.5VB. 3.0VC. 0.67VD. 1.33V |
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Answer» Correct Answer - 2 `because E=((V)/(L))xxlrArrE=((IR_(W))/(LL))xxl` `rArr E=(E_(0))/(R+r+R_(W))xx(R_(W))/(L)xxl` `E=(10)/(4+1+5)xx(5)/(3)xx3=3` volt |
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| 2079. |
Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wires is lighter ? Hence explain why aluminium wires are prefered for overhead power cables ? Given `rho_(Al)=2.63xx10^(-8) Omega m`, `rho_(Cu)=0.72xx10^(-8) Omega m` , relative density of `Al=2.7` and that of `Cu =8.9`. |
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Answer» Resistivity of aluminium, `rho_(Al) = 2.63 xx 10^(-8) Omega` m Relative density of aluminium, `d_(1)-2.7` Let `l_(1)` be the length of aluminium wire and `m_(1)` be its mass. Resistance of copper, `rho_(Cu)=1.72 xx 10^(-8) Omega` m Relative density of copper, `rho_(Cu) =1.72 xx 10^(-8) Omega` m Relative density of copper, `d_(2)=8.9` Let `l_(2)` be the length of copper wire and `m_(2)` be its mass Resistance of the copper wire `= R_(2)` Area of cross-section of the copper wire `=A_(2)` The two relation can be written as `R_(1) = rho_(1)(l_(1))/(A_(1))`....(1) `R_(2) = rho_(2) (l_(2))/(A_(2))` It is given that `R_(1)=R_(2)` ltbr. `rho_(1)(l_(2))/(A_(1))=rho_(2)(l_(2))/(A_(2))` And, `l_(1)=l_(2)` `:. (rho_(1))/(A_(1))=(rho_(2))/(A_(2))` `=(2.63 xx 10^(-8))/(1.72 xx 10^(-8))=(2.63)/(1.72)` Mass of the aluminium wire, `m_(1)` = Volume `xx` Density `= A_(1)l_(1)xxd_(1)=A_(1)l_(1)d_(1)`...(3) Mass of the copper wire, `m_(2)`= Volume `xx` Density `= A_(2)l_(2)xxd_(2)=A_(2)l_(2)d_(2)`....(4) Dividing equation (3) by equation (4), we obtain `(m_(1))/(m_(2))=(A_(1)l_(1)d_(1 ))/(A_(2)l_(2)d_(2))` For `l_(1)=l_(2)` `(m_(1))/(m_(2))=(A_(1)d_(1))/(A_(2)d_(2))` For `(A_(1))/(A_(2))=(2.63)/(1.72)` `(m_(1))/(m_(2))=(2.63)/(1.72) xx(2.7)/(8.9) = 0.46` It can be inferred this ratio that `m_(1)` is less than `m_(2)`. Hence, aluminium is lighter than copper Since aluminium is lighter, it is preferred for overhead power cables over copper. |
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| 2080. |
Eight cells marked 1 to 8, each of emf `5 V` and internal resistance `0.2 Omega` are connected as shown. What is the reading of ideal voltmeter ? A. `40 V`B. `20 V`C. `5 V`D. Zero |
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Answer» Correct Answer - D (d) Current flowing in circuit `i= (nE)/(nr) = (5)/(0.2) = 25 A` Terminal potential difference across the terminals of battery numbered 8, `V = E - ir = 5 - 25 xx 0.2 = 0` `V = 0` Hence, voltmeter reading is zero. So, choice (d) is correct |
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| 2081. |
Assetion : In the following circuit emf is `2 V` and internal resistance of the cell is `1 Omega` and `R = 1 Omega`, then reading of the voltmeter is `1 V`. Reason : `V = E - ir` where `E = 2 v, i= (2)/(2) = 1 A` and `R = 1 Omega` A. If both the assertion and reason are true and reason is a true explanation of the assertion.B. If both the assertion and reason are true but the reason is not the correct explanation of assertion.C. If the assertion and reason are falseD. If both the assertion and reason are false. |
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Answer» Correct Answer - A (a) Here, `E = 2 V, 1 = 2//2 = 1 A` and `r = 1 Omega` Therefore, `V = E - ir = 2 - 1 xx 1 = 1 V` |
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| 2082. |
The figure shows a potentiometer using a cell `E` of emf `2.0 V` and internal resistance `0.40 Omega` connected to a resistor wire `AB`. A standard cell of constant emf of `1.02 V` gives a balance point a t `67.3 cm` length of the wire. A very high resistance `R = 100 k Omega` is put in with the standard cell. This resistance is shorted by inserting switch `S` when close to the balance point. The standard cell is then replaced by a cell of unknown emf `E` and the null point turns out to be `82.3 cm` length of the wire. (a) What is the value of `E` ? (b) What is the purpose of using the high resistance `R` ? ( c) Is the null point affected by this high resistance ? (d) Is the null point affected by the internal resistance of the cell `E` ? ( e) Would this method work if (i) the internal resistance of cell `E` were higher than the resistance of wire `AB` and (ii) the emf of cell `E` were `1.0 V` instead of `2.0 V` ? |
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Answer» (a) `E = (1.02)/(67.3) xx 82.3 = 1.25 V` (b) The high resistance `R` keeps the current from standard cell within permissible limit and prevent a large current to flow through the galvanometer when far away from the balance point. ( c) Al null point, no current flow through `R`, hence no effect no null point. (d) The null point depends on terminal voltage of `E_0` and the emf of `E` only, hence no effect. ( e) If `p.d` across `AB` due to the driver cell `E_0` become less than the emf of cell `E_1`, method will fail, there would be no null point on the wire `AB` (i) `p.d` across `AB` due to cell `E_0` of emf `2 V` will be less than `1 V`. Since emf of the standard cell is greater than this value, no null point on `AB` (ii) No null point on `AB`. |
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| 2083. |
Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wire is lighter? Hence explain why aluminium wires are preferred for overhead power cables. `(rho_(Al) = 2.63 xx 10^(-8) Omega m, rho_(Cu) = 1.72 xx 10^(-8) Omega m` Relative density of Al = 2.7 of Cu = 8.9) |
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Answer» Given for aluminium wire, `R_(1) = R l_(1) = l` Relative density `d_(1) = 2.7` Let `A_(1), A_(2)` be the area of cross section for aluminium wire and copper wire. We know, `R_(1) = rho_(1) (l_(1))/(A_(1)) = (2.63 xx 10^(-8) xx l)/(A_(1)) ` and mass of the aluminium wire `m_(1) = A_(1) l_(1) xx d_(1) = A_(1) l xx 2.7` `R_(2) = rho_(2) = (l_(2))/(A_(2)) = (1.72 xx 10^(-8) xx l)/(A_(2))` Mass of copper wire `m_(2) = A_(2) l_(2) xx d_(2) = A_(2) l xx 8.9` Since two wires are of equal resistance `R_(1) = R_(2)` `(2.63 xx 10^(-8) xx 1)/(A_(1)) = (1.72 xx 10^(-8) xx l)/(A_(2)) "or" (A_(2))/(A_(1)) = (1.72)/(2.63)` From (ii) and (iv) we have `(m_(2))/(m_(1)) (A_(2) l xx 8.9)/(A_(1) l xx 2.7) = (8.9)/(2.7) xx (A_(2))/(A_(1))` `= (8.9)/(2.7) xx (1.72)/(2.63) = 2.16` It shows that copper wire is 2.16 times heavier than aluminium wire since for the same value of length and resistance aluminium wire has lesser mass then copper wire, therefore aluminium wire is prederred for overband power cables. A heavy cable may sag down owing to its own weight. |
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| 2084. |
In the circuit cell are of equal `emf E` but of difference internal resistance `r_(1) = 6 Omega` reading of the ideal voltmeter connected across cell `1` is zero The value of the external resistance `R` is ohm is equal to A. `10 Omega`B. `4 Omega`C. `2 Omega`D. `2.4 Omega` |
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Answer» Correct Answer - c Current in the circuit `1 = (E+E)/(r_(1) + (r_(2) + R)) = (2E)/(6+4+R) = (2E)/(10+R)` Potential difference across `(R+ r_(2))` `= (2E)/((10+R)) (R+r_(2))` As voltmeter connecting cell `1` records zero reading therefore the e.m.f of cell `2` is equal to voltage drop across `(R+r_(2))` hence `E = (2E)/((10+R))(R+r_(2))= (2E(R+4))/(10+R)` On solving `R = 2 Omega` |
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| 2085. |
Assetion : In the following circuit emf is `2 V` and internal resistance of the cell is `1 Omega` and `R = 1 Omega`, then reading of the voltmeter is `1 V`. Reason : `V = E - ir` where `E = 2 v, i= (2)/(2) = 1 A` and `R = 1 Omega` A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is falseD. If the assertion and reason both are false |
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Answer» Correct Answer - A |
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| 2086. |
A voltmeter of resistance `1000 Omega` is connected across a resistance of `500 Omega` in the given circuit. What will be the reading of voltmeter A. `1V`B. `2V`C. `6V`D. `4V` |
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Answer» Correct Answer - D |
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| 2087. |
The current, in a potentiometer wire of 100 cm length, is adjusted to give a null point at 5 cm with standard cell of e.m.f. 1.018 V. The e.m.f. of cell which gives null point of 60 cm isA. 1.221VB. 2.22VC. 3.22VD. 4.22V |
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Answer» Correct Answer - A `(E_(1))/(E_(2))=(l_(1))/(l_(2))` `:.E_(2)=(l_(2))/(l_(1))E_(1)=(60xx10^(-2)xx1.018)/(50xx10^(-2))=1.221V` |
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| 2088. |
A voltmeter has a resistance of `50 Omega` is connected across a cell of e.m.f. 2V and internal resistance `10 Omega`. The reading of voltmeter isA. 1.667VB. 16.7VC. 167VD. 0.167V |
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Answer» Correct Answer - A `V=I R=((E)/(R+r))R=((2)/(50+10))50=1.667 V` |
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| 2089. |
A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380 `Omega`. What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car? |
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Answer» Here `E = 1.9 V, r = 380 Omega` `I_("max") = (epsilon)/(r ) = (1.9)/(380) = 0.005 A` This amount of current cannot start a car because to start the motor, the current required is 100 A for few seconds. |
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| 2090. |
A voltmeter having a resistance of 998 ohms is connected to a cell of e.m.f. 2 volt and internal resistance 2 ohm. The error in the measurment of e.m.f. will beA. `4 xx 10^(-1)` voltB. `2 xx 10^(-3)` voltC. `4 xx 10^(-3)` voltD. `2 xx 10^(-1)` volt |
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Answer» Correct Answer - C |
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| 2091. |
In the adjoining circuit, the e.m.f. of the cell is 2 volt and the internal resistance is negligible. The resistance of the voltmeter is `80 ohm`. The reading of the voltmeter will be A. 0.80 voltB. 1.60 voltC. 1.33 voltD. 2.00 volt |
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Answer» Correct Answer - C (c ) Total resistance of the circuit `= (80)/(2) + 20 = 60 Omega` `implies` Main current `i= (2)/(60) = (1)/(30)` Combination of voltmeter and `80 Omega` resistance is connected in series with `20 Omega`, so current through `20 Omega` and this combination will be same `= (1)/(30) A`. Since the resistance of voltmeter is also `80 Omega`, so this current is equally distributed in `80 Omega` resistance and voltmeter (i.e., `1//60 A` through each). `P.D.` across `80 Omega` ressitance `= (1)/(60) xx 80 = 1.33 V` |
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| 2092. |
In the adjoining circuit, the e.m.f. of the cell is 2 volt and the internal resistance is negligible. The resistance of the voltmeter is `80 ohm`. The reading of the voltmeter will be A. 0.80 voltB. 1.60 voltC. 1.33 voltD. 2.00 volt |
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Answer» Correct Answer - C (c ) Total resistanceof the circuit `= (80)/(2) + 20 = 60 Omega` `implies` Main current `i= (2)/(60) = (1)/(30) A` Combination of voltmeter and `80 Omega` resistance is connected in series with `20 Omega`, so current through `20 Omega` and this combination will be same `= (1)/(30) A` Since the resistance of voltmeter is also `80 Omega`, so this current is equally distributed in `80 Omega` , so this current is equally distributed in `80 Omega` resistance and voltmeter (i.e., `(1)/(60) A` through each) `P.D.` across `80 Omega` resistance `= (1)/(60) xx 80 = 1.33 V` |
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| 2093. |
(a) Six lead-acid type of secondary cells each of emf `2-0` V and internal resistance `0-015Omega` are jouned in series to provide a supply to a resistance of `8-5Omega`. What are the current drawn from the supply and its terminal voltage ? (b) A secondary cells after long use has an emf of `1-9` V and a large internal resistance of `380 Omega`. What maximum current can be drawn from the cell ? Could the cell drive the starting motor of a car ? |
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Answer» (a) Here, `epsilon = 2.0V , n = 6, r = 0.015 Omega , R = 8.5 Omega` Current, `I = (n epsilon)/(R+ nr) = (6xx2.0)/(8.5 + 6 xx 0.015) = 1.4A` Terminal voltage, `V = I R = 1.4 xx 8.5 = 11.9 V` (b) Here, `epsilon = 1.9 V , r =380 Omega` `I_(max) = epsilon/r = 1.9/380 = 0.005A` This amount of current cannot start a car because to start a car beacuse to start the motor, the current required is 100 A for few seconds. |
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| 2094. |
24 cells of emf `1.5 V` each having internal resistance of 1 ohm are connected to an external resistance of `1.5` ohms. To get maximum current,A. all cells are connected in series combinationB. all cells are connected in parallel combinationC. 4 cells in each row are connected in series and 6 such rows are connected in parallelD. 6 cells in each row are connected in series and 4 such rows are connected in parallel |
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Answer» Correct Answer - D |
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| 2095. |
In the adjoining circuit, the e.m.f. of the cell is 2 volt and the internal resistance is negligible. The resistance of the voltmeter is `80 ohm`. The reading of the voltmeter will be A. `0.80V`B. `1.60V`C. `1.33V`D. `2.00V` |
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Answer» Correct Answer - C |
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| 2096. |
Twenty four cells each of emf 1.5 V and internal resistance 0.5 ohms are to be connected to a 3 ohm resistance. For maximum current through this resistance the nuber of rows and number of columns that you connect these cells is.A. 12 cells in series 2 ros in parallelB. 8 cells in series 3 rows in parallelC. 4 cells in series 6 rows in parallelD. 6 cells in series 4 rows in parallel |
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Answer» Correct Answer - A `i=(mnE)/(mR+nr)&mxxn=24` |
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| 2097. |
Resistors of 1, 2, 3 ohm are connected in the form of a triangle. If a 1.5 volt cell of negligible internal resistance is connected across 3 ohm resistor, the current flowing through this resistance will beA. 0.25AB. 0.5AC. 1.0AD. 1.5A |
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Answer» Correct Answer - B The effective resistance of `1Omega, 2 Omega` and `3 Omega` is , `R=1.5 Omega` Now, `I=(E)/(R+r)=(1.5)/(1.5+0)= 1 A` Thus, current through `3 Omega` resistor is 0.5 A |
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| 2098. |
A conducting wire of length `l` and cross sectional area A is used to short the terminals of a cell having emf of `epsilon` and internal resistance `r`. The resistively, density and Molar mass of the material of the wire are `r, d and M` respectively. Calculate the average time needed for a free electron to travel from positive terminal of the cell to its negative terminal. |
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Answer» Correct Answer - `t=(l edN_(A)(rhol+rA))/(epsilonM)` |
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| 2099. |
In an electrical cable there is a single wire of radius `9 mm` of copper. Its resistance is `5 Omega`. The cable is replaced by 6 different insulated copper, wires the radius of each wire is `3 mm`. Now the total resistance of the cable will beA. `7.5 Omega`B. `5.5 Omega`C. `6Omega`D. `8Omega` |
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Answer» Correct Answer - A Resistance of 9 mm cable is `5 Omega` . Now,`R prop (1)/(A) :.R prop (1)/(r^(2))` `:.` Resistance of 3 mm cable `= 9 xx 5 = 45 Omega`. In second case, six wires are connected in parallel. Thus, its resistance is given by, `R_(p)=(45)/(6)=7.5Omega`. |
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| 2100. |
In the figure given below find the resistance between points A and B. Both the circle and the diatmeter are made of uniform wire of resistance `r Omega` per meter. The length AB is 2 metre. A. 0.88 rB. 0.68 rC. rD. 2r |
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Answer» Correct Answer - A Let a be the radius of circle. Now, resistances `pi` a r, 2 ra and ` pi a r`, are connected in parallel. Given ` 2 a = 2m :.a=1m` `:.(1)/(R_(p))=(1)/(pi r)+(1)/(2 r)+(1)/(pi r) rArr R= 0.88 r` |
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