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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.
| 1051. |
A conductor has a cross-section of `15 mm^(2)` and resistivity of `7.6 xx 10^(-8)Omega` m at `0^@C`. If the temperature coefficient of resistance of the material of the conductor is `5 xx 10^(-3).^(@)C^(-1)`, calculate its resistance for 2 km length of conductor when its temperature is `60^@C`. |
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Answer» Here, `A= 15 mm^(2) = 15xx10^(-6) m^(2)`, `rho=7.6xx10^(-8) Omega m,alpha =5xx10^(-3).^(@)C^(-1)`, `l_(0)=2 km = 2000m, t_(1)=0^@C,t_(2)=60^@C`, `R_(0)=rhol/A=((7.6xx10^(-8))xx2000)/(15xx10^(-6)) = 10.12Omega` `R_(60)=R_(0)[1+alpha (t_(2)-t_(1))]` `=10.12 [1+5xx10^(-3) xx60]` `=10.12 xx1.3 =13.16 Omega` |
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| 1052. |
The heating element of an electirc toaster is of nichrome. When a vary small current passes through it, at room temperrature `27^@C`, its resistance is `75.3Omega`. When toaster is connected to a 230V supply, the current settles after a few seconds to a steady value of 2.68A. What is the steady temperature of nichrome element? The temperature coefficient of resistance of nichrome over averged temperature range is `1.7xx10^(-4)^(@)C^(-1)`. |
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Answer» Here, `R_(27) =75.3 Omega`, `alpha=1.7xx10^(-4).^(@)C^(-1),V=230 V, I=2.68A` `R_(t) =230/2.68 =85.82 Omega`, As, `t_(2)-t_(1) =(R_(1)-R_(1))/(R_(1)xxalpha)` `:. t -27 = (R_(1)-R_(27))/(R_(27) xx alpha)=(85.82-75.3)/(75.3xx1.7xx10^(-4))=822^@C` or `t=822 +27 =849^@C` |
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| 1053. |
(a) In Example 3.1, the electron drift speed is estimated to be only a few mm `s^(-1)` for currents in the range of a few amperes? How then is current established almost the instant a circuit is closed? (b) The electron drift arises due to the force experienced by electrons in the electric field inside the conductor. But force should cause acceleration. Why then do the electrons acquire a steady average drift speed? (c) If the electron drift speed is so small, and the electron’s charge is small, how can we still obtain large amounts of current in a conductor? (d) When electrons drift in a metal from lower to higher potential, does it mean that all the ‘free’ electrons of the metal are moving in the same direction? (e) Are the paths of electrons straight lines between successive collisions (with the positive ions of the metal) in the (i) absence of electric field, (ii) presence of electric field? |
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Answer» (a) Electric field is established throughout the circuit, almost instantly (with the speed of light) causing at every point a local electron drift. Establishment of a current does not have to wait for electrons from one end of the conductor travelling to the other end. However, it does take a little while for the current to reach its steady value. (b) Each ‘free’ electron does accelerate, increasing its drift speed until it collides with a positive ion of the metal. It loses its drift speed after collision but starts to accelerate and increases its drift speed again only to suffer a collision again and so on. On the average, therefore, electrons acquire only a drift speed. (c) Simple, because the electron number density is enormous, `~10^(29) m^(-3)`. (d) By no means. The drift velocity is superposed over the large random velocities of electrons. (e) In the absence of electric field, the paths are straight lines, in the presence of electric field, the paths are, in general, curved. |
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| 1054. |
The energy of the rotational motion of the molecules in n moles of nitrogen at temperature of T K is x nRT. Find the value of x. |
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Answer» Correct Answer - A |
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| 1055. |
P-V diagram of a diatomic gas is a straight line passing through origin. The molar heat capacity of the gas in the process will beA. 4RB. `2.5R`C. 3RD. `(4R)/(3)` |
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Answer» Correct Answer - C |
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| 1056. |
In which of the following process, convection does not take place primarlily?A. Sea and land breezeB. Boiling of waterC. Warming of glass of bulb due to filamentD. Heating of air around a furnace |
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Answer» Correct Answer - C |
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| 1057. |
Two resistance `R_(1)` and `R_(2)` when connected in series and parallel with `120 V` line, power consumed will be `25 W` and `100 W` respectively. . Then the ratio of power consumed by `R_(1)` to that consmed by `R_(2)` will beA. `(1)/(4)`B. `(1)/(3)`C. `(1)/(2)`D. `1` |
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Answer» Correct Answer - D `P_(S)=(P_(1)P_(2))/(P_(1)+P_(2)),P_(P)=P_(1)+P_(2)` |
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| 1058. |
Two resistance `R_(1)` and `R_(2)` when connected in series and parallel with `120 V` line, power consumed will be `25 W` and `100 W` respectively. . Then the ratio of power consumed by `R_(1)` to that consmed by `R_(2)` will beA. `1:1`B. `1:2`C. `2:1`D. `1:4` |
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Answer» Correct Answer - A (a) `P = (V^(2))/(R ) = (P_(P))/(P_(S)) = (R_(S))/(R_(P)) = ((R_(1) + R_(2)))/(R_(1) R_(2) // (R_(1) + R_(2))) = ((R_(1) + R_(2))^(2))/(R_(1) R_(2))` `implies (100)/(25) = ((R_(1) + R_(2))^(2))/(R_(1) R_(2)) implies (R_(1))/(R_(2)) = (1)/(1)` |
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| 1059. |
A cell of constant e.m.f. first connected to a resistance `R_(1)` and then connected to a resistance `R_(2)`. If power delivered in both cases is then the internal resistance o f the cell isA. `sqrt(R_(1) + R_(2))`B. `sqrt((R_(1))/(R_(2)))`C. `(R_(1) - R_(2))/(2)`D. `(R_(1) + R_(2))/(2)` |
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Answer» Correct Answer - A (a) Power dissipated `= i^(2) R = ((E)/(R + r))^(2) R` `:. ((E)/(R_(1) + r))^(2) R_(1) = ((E)/(R_(2) + r))^(2) R_(2)` `implies R_(1) (R_(1)^(2) + r_(2) + 2 R_(2) r) = R_(2) (R_(1)^(2) + r^(2) + 2 R_(1) r)` `implies R_(2)^(2) R_(1) + R_(1) r^(2) + 2 R_(2) r = R_(1)^(2) R_(2) + R_(2) r^(2) + 2 R_(1) R_(2) r` `implies (R_(1) - R_(2)) r^(2) = (R_(1) - R_(2)) r^(2) = (R_(1) - R_(2)) R_(1) R_(2)` `implies r = sqrt(R_(1) R_(2))` |
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| 1060. |
Four resistances carrying a current shown in Fig. `7.41` are immersed in a box containing ice at `0^(@)C`. How much ice must be put in the box every `10 min` to keep the average quantity of ice in the box constant? Latent heat of ice is `80 calg^(-1)`. |
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Answer» The equivalent resistance of the circuit is `R = ( 10 xx 5)/(10+5) + (5 xx 10)/(5+ 10) = 10/3 + 10/3 = 20/3 Omega`. Heat produced in 10 min. `H = (i^(2)Rt)/J = ((10 xx 10) xx 20 xx (10 xx 60))/(3xx4.2) cal` Let m be the mass of the ice melted in 10 minutes Then, `m xx 80 = (10 xx 10 xx 20 xx10xx 60)/(3xx4.2)` or ` m = (10 xx 10 xx 20 xx10xx 60)/(80 xx 3xx4.2)` ` = 1190` gram |
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| 1061. |
Two heater `A` and `B` are in parallel across the supply voltage. Heater `A` produces `500 kJ` in 20 minutes and `B` produces `1000 kJ` in 10 minutes. The resistance of `A` is `100 Omega`. If the same heaters are connected in series across the same voltage, then total heat produced in 5 minutes will beA. `200 kJ`B. `100 kJ`C. `50 kJ`D. `10 kJ` |
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Answer» Correct Answer - B (b) For heater `A: 500 xx 10^(3) = (V^(2))/(R_(1)) (20 xx 60)` Where `R_(1) = 100 Omega` For heater `B`: `100 xx 10^(3) = (V^(2))/(R_(2)) (10 xx 60)` From (i) and (ii) `R_(2) = 25 Omega` When heaters are connected in series : `R_(eq) = R_(1) + R_(2)` Hat produced : `H = (V^(2))/(R_(1) + R_(2)) (5 xx 60)` Form (i) and (iii) solve to get `H = 100 kJ`. |
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| 1062. |
Four resistances carrying a current shown in Fig. `7.41` are immersed in a box containing ice at `0^(@)C`. How much ice must be put in the box every `10 min` to keep the average quantity of ice in the box constant? Latent heat of ice is `80 calg^(-1)`. |
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Answer» Total resistance `R=2xx(5xx10)/(5+10)=(20)/(3)Omega` heat produced in 5 minute `=I^(2)Rt=5xx5xx(20)/(3)xx5xx 60=50000J` Latent heat `L=80xx4.2xx10^(3)J//kg`, mass of ice melted , `m=(H)/(L)=(50000)/(80xx4.2xx10^(3))=0.1488` kg=148.8g. |
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| 1063. |
How much time heater will take to increase the temperature of 100g water by `50^(@)C` if resistance of heating coat is `484Omega` and supply voltage is 220V a.c. |
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Answer» Heat given by heater=heat taken by water `rArr (V^(2))/(R)t=ms J Delta theta rArr t=(ms J Delta thetaR)/(V^(2))rArr t=((100xx10^(-3))(4.2xx10^(3))(50)xx484)/(220xx22 0)=210s` |
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| 1064. |
Two 120 V light bulbs, one of 25 W other of 200 W were connected in series across a 240 V line. One bulb burnt out almost instantaneously. Which one was burnt and why ? |
| Answer» since the b ulbs are connected in series, th erefore same current will flow in them. The resistance of 200W bulb is less than 25W. Therefore more potential difference than 120V will be developed in 25W bulb hence it will fuse. | |
| 1065. |
In the circuit shown in figure potential difference across |
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Answer» Correct Answer - A::B |
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| 1066. |
Potentiometer is superior to voltmeter becauseA. voltmeter has high resistanceB. resistance of potentiometer wire is quite lowC. potentiometer does not draw any current from the unknown source of emf. To be measured.D. sensititvity of potentiometer is higher than that of a voltmeter. |
| Answer» Correct Answer - C | |
| 1067. |
Assertion: meterbridge wire is made up of manganin Reason: The temperature coefficient of resistance is very small for manganinA. If both Assertion `&` Reason are True `&` the Reason is a correct explanantion of the AssertionB. If both Assertion `&` Reason True but Reason is not a correct explanation of the AssertionC. If Assertion is True but the Reason is FalseD. If both Assertion `&` Reason are False. |
| Answer» Correct Answer - 1 | |
| 1068. |
Assertion: meterbridge wire is made up of manganin Reason: The temperature coefficient of resistance is very small for manganinA. Both (A) and (R) are true and (R) is the correct explanantion of A.B. Both (A) and (R) are true but (R) is not the correct explanation of A.C. (A) is true but (R) is falseD. (A) is false but (R) is true. |
| Answer» Correct Answer - A | |
| 1069. |
Potentiometer wire is made of manganin because it hasA. If both Assertion `&` Reason are True `&` the Reason is a correct explanantion of the AssertionB. If both Assertion `&` Reason True but Reason is not a correct explanation of the AssertionC. If Assertion is True but the Reason is FalseD. If both Assertion `&` Reason are False. |
| Answer» Correct Answer - 2 | |
| 1070. |
In the following figure, the reading of an ideal voltmeter V is zero. Then, the relation between R, `r_(1) " and " r_(2)` is A. `R=r_(2)-r_(1)`B. `R=r_(1)-r_(2)`C. `R=r_(1)+r_(2)`D. `R=(r_(1)r_(2))/(r_(1)+r_(2))` |
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Answer» Correct Answer - B `l=(2E)/(r_(1)+r_(2)+R)` ` V_(1)=E-lr_(1)=0` `implies " " l=E//r_(1)` `E=(2Er_(1))/(r_(1)+r_(2)+R)` `r_(1)+r_(2)+R=2r_(1)` `implies " " R=r_(1)+r_(2)` |
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| 1071. |
The effective resistance between points A and B is A. `10 Omega`B. `20 Omega`C. `40 Omega`D. None of the above three values |
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Answer» Correct Answer - A |
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| 1072. |
In the circuit shown in the figure, the current flowing in `2 Omega` resistance A. `1.4A`B. `1.2A`C. `0.4A`D. `1.0A` |
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Answer» Correct Answer - D |
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| 1073. |
In the figure given the value of X resistance will be, when the p.d. between B and D is zero A. 4 ohmB. 6 ohmC. 8 ohmD. 9 ohm |
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Answer» Correct Answer - C |
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| 1074. |
In a potentiometer experiment, the balancing length is found to be 1.8 m for a cell of e.m.f. 1.5 V. What is the balancing length for a cell of e.m.f. 1 V ?A. 1.2 mB. 0.5mC. 2.2mD. 0.2 |
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Answer» Correct Answer - A `(E_(2))/(E_(1))=(l_(2))/(l_(1)) " " :.l_(2)=(1xx1.8)/(1.5)=1.2m` |
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| 1075. |
Two cells are connected in series and e.m.f. of combination is found to balance against length of 450 cm of potentiometer wire. Whent the two cells are connected in parallel emf of combination balances against 50 cm length of wire. The ratio of e.m.f. of two cells isA. 2.25B. 1.55C. 1.25D. 2.55 |
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Answer» Correct Answer - C `(E_(1))/(E_(2))=(l_(1)+l_(2))/(l_(1)-l_(2))=(450+50)/(450-50)=5/4=1.25` |
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| 1076. |
Two students A and B were asked to pick a resistor of `15 k Omega` from a collection of carbon resistors. A picked a resistor with bands of colour: brown, green, orange while B choose a resistor with bands black, green, red. Who picked the correct resistor? Explain. |
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Answer» For student A, the numbers attached to brown, green and orange are 1,5 and 3 . So the value of resistor is, `R_(A)=15 xx10^(3) Omega = 15 k Omega`. For student B, the numbers attached to black, green and red are 0,5 and 2 respectively. So the value of resistor is, `R_(B) = 05 xx 10^(2) Omega = 500 Omega`. Thus, student A picked up the correct resistor of `15 k Omega`. |
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| 1077. |
A resistance of `4 Omega` and a wire of length 5 meters and resistance `5 Omega` are joined in series and connected to a cell of e.m.f. `10 V` and internal resistance `1 Omega`. A parallel combination of two identical cells is balanced across `300 cm` of wire. The e.m.f. `E` of each cell is A. `1.5V`B. `3.0V`C. `0.67V`D. `1.33V` |
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Answer» Correct Answer - B |
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| 1078. |
The drift velocity of electrons is given byA. `v_(d)=(e)/(m)(V)/(l)t`B. `v_(d)=(eV t)/(l)`C. `v_(d)=(e)/(m)t`D. `v_(d)=(V)/(l)(m)/(e)t` |
| Answer» Correct Answer - A | |
| 1079. |
Which of the following is correct regarding drift velocity.A. When there is an electric field inside a conductor, the electrons drift slowly opposite to field with constant velocity. `v = (e tau)/(m) E = kE` `tau` : relaxtion time (average time taken by electrons in successive collision with atoms of material) depends on temperature `k` : depends on material of conductor and its temperature.B. The drift velocity is very small `(~= 10^-4 m//sec)` as compared to thermal speed of electrons at room temperature `(~= 10^6 m//sec)`C. Drift velocity depends on the nature of metal, i.e., relaxation time `tau`. With rise in temperature relaxation time and hence drift velocity will decreaseD. All options are correct |
| Answer» Correct Answer - D | |
| 1080. |
The internal energy of a gas is given by `U=2pV`. It expands from `V_0` to `2V_0` against a constant pressure `p_0`. The heat absorbed by the gas in the process is |
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Answer» Correct Answer - C |
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| 1081. |
Sensitivity of potentiometer can be increased byA. Increasing the e.m.f. of the cellB. Increasing the length of the potentiometer wireC. Decreasing the length of the potentiometer wireD. None of the above |
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Answer» Correct Answer - B |
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| 1082. |
Two identical cells , whether joined together in series or in parallel give the same current, when connected to external resistance of `1 Omega`. Find the internal resistance of each cell. |
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Answer» Let `epsilon`, r be the emf and internal resistance of each cell. External `R= 1 Omega`. When the two cells are connected in series Total emf of cells `= epsilon + epsilon = 2 epsilon` Total resistance of circuit `= R +r+r=1+2r` Currnet in the circuit, `I_(1)= (2epsilon)/(1+2r)` When the two cells are connseted in parallel Effective emf of two cells = emf of single cell=`epsilon` Total internal resistance of two cells `(r xx r)/(r+r)=r/2` Total resistance of the circuit `= R + r//2 = 1+r//2`. Current in the circuit, `I_(2)=(epsilon)/(1+r//2)` As per question, `I_(1) = I_(2)` So, `(2 epsilon)/(1+2r) = (epsilon)/(1+r//2) = (2 epsilon)/(2+r)` or `1 + 2r = 2+r or r = 1 Omega` |
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| 1083. |
In the given circuit as shown in figure, in the steady state, obtain the expression for (i) potential drop (ii) the charge and (iii) the energy stored in the capacitor C. |
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Answer» In the steady state, the capacitor C is the fully charged. So no current flows through the arm BE of the circuit. `:.` Effective emf of the circuit `= 2V- V=V` Effective resistance of the circuit `= 2 R +R = 3R` Current in the circuit, `I= (V)/(3R)` Potential difference between across B and E `= 2 V - I xx 2R= 2V - (V)/(3R) xx 2R = 4/3 V` (i) potential difference across C `=4/3 V - V= V/3` (ii) Charege on capacitor `Q= CV = C xx V/3 = (CV)/(3)` (iii) Energy stored in the capacitor `=1/2 C(V/3)^(2)= (CV^(2))/(18)` |
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| 1084. |
How would you arrange 64 similar cells each having an emf of 2.0 V and internal resistance `2 Omega` so as to send maximum current through an external resistance of `8 Omega`. |
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Answer» Correct Answer - `n=16, m=4,2.0 A` |
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| 1085. |
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. |
| Answer» Correct Answer - 1 | |
| 1086. |
Find the equivalent resistance across `AB` : A. `1Omega`B. `2Omega`C. `3Omega`D. `4Omega` |
| Answer» Correct Answer - 3 | |
| 1087. |
Two copper wire of length l and 2l have radii, r and 2r respectively. What si the ratio of their specific resistance.?A. `1:2`B. `2:1`C. `1:1`D. `1:3` |
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Answer» Correct Answer - C The specific resistance `(rho)` is the characteristic of the material of conductor. Its value depends only on the material of conductor and its temperature. Its value does not depend on the length and area of cross-section of the conductor. If a wire is stretched or doubled on itself, it resistance will change, but its specific resistance will remain unaffected. |
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| 1088. |
Which of the following materials is the best conductor of electricity ?A. PlatinumB. GoldC. SiliconD. copper |
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Answer» Correct Answer - B Gold |
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| 1089. |
For a cell terminal potential difference is 2.2 V when circuit is open and reduces to 1.8V when cell is connected to a resistance of R=`5Omega` then determine internal resistance of cell is:-A. `(10)/( 9)Omega`B. `(9)/(10)Omega`C. `(11)/(9)Omega`D. `(5)/(9)Omega` |
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Answer» Correct Answer - 1 T.P.D(V)=E-Ir(Remember it) `V=E-((E)/(R+r))r=(ER)/((R+r))` from given condition `E=2.2 &` when R=5 then TPD V=1.8V therefore `1.8=(2.2xx5)/( 5+r)rArrr=(10)/(9)Omega` |
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| 1090. |
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. 0.825B. 0.8C. 0.9D. 0.875 |
| Answer» Correct Answer - D | |
| 1091. |
Two resistances are connected in two gaps of a meter bridge. The balance point is `20 cm` from the zero end. A resistance of `15 Omega` is connected in series with the smaller of the two. The null point shifts to `40 cm`. The value of the smaller resistance in `Omega` isA. 3B. 6C. 9D. 12 |
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Answer» Correct Answer - C `P/Q=20/(100-20) or Q=4P`….i `:. `PltQ` Now,` (P+15)/Q=40/(100-40)` `or (P+15)/Q=2/3` Solving these two equations we get `P=9Omega` |
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| 1092. |
In the given circuit, the voltmeter records 5 V. The resistance of the voltmeter in `Omega` is A. 200B. 100C. 10D. 50 |
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Answer» Correct Answer - B |
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| 1093. |
For a cell, the terminal potential difference is `2.2 V`, when circuit is open and reduces to `1.8 V `. When cell is connected to a resistance `R=5Omega`, the internal resistance of cell `(R)` isA. `10/(9Omega)`B. `9/(10Omega)`C. `11/(9Omega)`D. `5/(9Omega)` |
| Answer» Correct Answer - A | |
| 1094. |
In the given circuit, the voltmeter records 5 volt. The resistance of the voltmeter in `Omega` is A. 200B. 100C. 10D. 50 |
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Answer» Correct Answer - B Total potential of `10 V` equally distributes between`50 Omega` and other parallel combination of `100 Omega` and voltmeter. Hence, their net resistance should be same. Or `(100xxR)/(100+R)=50` `:. R=100 Omega=`resistance of voltmeter |
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| 1095. |
By error, a student places moving-coil voltmeter `V` (nearly ideal) in series with the resistance in a circuit in order to read the current, as shown. The voltmeter reading will be A. `0`B. `4V`C. `6V`D. `12V` |
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Answer» Correct Answer - D Current is nearly zero so voltmeter reading is `12V` |
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| 1096. |
A lab voltmeter has scale as shown in figure. A student uses this nearly ideal voltmeter to record potential difference between points A and B in the circuit shown. He could not see any deflection in the pointer and thinks that the voltmeter must be faulty. He replaces the voltmeter with another similar one. What reading will he record this time? Can you give some suggestion to record the potential difference across A and B ? |
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Answer» Correct Answer - Zero. He shall use a milli voltmeter |
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| 1097. |
A 10 wire potentiometer has first five wires of cross sectional radius `r` and next five wires of radius `2r`. The wires are uniform and made of same material. An ideal cell of emf `2 V` is connected across the potentiometer wire. What length of the potentiometer wire will balance the emf of– (a) a Daniell cell (emf = 1.0 V) (b) a Lechlanche cell (emf = 1.5 V) (c) a cell of emf 1.8 V ? Length of each wire is 100 cm. |
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Answer» Correct Answer - (a) 3.125 (b) 4.688 m (c) 7.5 m |
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| 1098. |
If specific resistance of a potentiometer wire is `10^(–7 ) Omega m`, the current flow through it is `0.1 A` and the cross-sectional area of wire is `10^(–6) m^(2)` then potential gradient will beA. `10^(-2)` volt/mB. `10^(-4)` volt/mC. `10^(-6)`volt/mD. `10^(-8)`volt/m |
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Answer» Correct Answer - 1 Potential gradient `x=(V)/(l)=(IR)/(l)=(Irho)/(A)` |
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| 1099. |
The expression for potential gradent of potentiometer wire in terms of specific resistance p of the wire area of its cross -section A and current `I` flowing wire is…….. |
| Answer» Correct Answer - `lp//A | |
| 1100. |
In the circuit element given here, if the potential at point `B = V_(B) = 0`, then the potentials of `A` and `D` are given as A. `V_(A)=-1.5V, V_(D)=+2V`B. `V_(A)=-1.5V, V_(D)=+0.5V`C. `V_(A)=+1.5V, V_(D)=+0.5V`D. `V_(A)=+1.5V, V_(D)=-0.5V` |
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Answer» Correct Answer - D |
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