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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.
| 51. |
Four spheres A, B, C and D of different metals but all same radius are kept at same temperature. The ratio all their densities and specific heats are `2:3:5:1` and `3:6:2:4`. Which sphere will show the fastest rate all cooling (initially) (assume black body radiation for all of them)A. AB. BC. CD. D |
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Answer» Correct Answer - D |
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| 52. |
The figure shows a meter bridge circuit with AB=100 cm, X=`12 Omega andR=18 Omega` and the jockey J in the position of balance. If R is now made `8Omega` through what distance will J have to be moved to obtain balance?A. 10 cmB. 20 cmC. 30 cmD. 40 cm |
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Answer» Correct Answer - D |
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| 53. |
One mole of an ideal gas is kept enclosed under a light piston (area `=10^(-1)m^(2)`) connected by a compressed spring (spring constant 100 N/m). The volume of gas is `0.83m^(3)` and its temperature is 100K. The gas is heated so that it compresses the spring further by `0.1m`. The work done by the gas in the process is : (Take R =`8.3`J/K-mole and suppose there is no atmosphere). A. 3 JB. 6 JC. 9 JD. `10.5J` |
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Answer» Correct Answer - D |
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| 54. |
If the temperature of a good conductor increases, how does the relaxation time of electrons in the conductor change? |
| Answer» With the increase in temperature, the free electrons collide more frequently with the ions/atoms of the conductor, resulting decrease on relaxation time. | |
| 55. |
If temperature is increases, then relaxation time of electrons in metals willA. increaseB. decreaseC. fluctuateD. remain constant |
| Answer» Correct Answer - B | |
| 56. |
Calculate the relaxation time and mean free path at room temperature (i.e. `27^(@)C`). If the number of free electrons per unit volume is `8.5 xx 10^(28)//m^(3)` and resistivity `rho = 1.7 xx 10^(8) Omega-m`. Given that mass of electron `= 9.1 xx 10^(-31) kg` `e = 1.6 xx 10^(-19)C and k = 1 .36 xx 10^(-23) JK^(-1)` |
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Answer» Correct Answer - `2.46 xx 10^(-14)s ;28.7Å` Relaxation time, ` tau =(m)/(n e^(2)rho)` `= (9.1 xx 10^(-31))/( (8.5 xx 10^(28)) (1.6 xx 10^(-19))^(2) xx (1.7 xx 10^(-8)))` `= 2.46 xx 10^(-14)s` From kinetic theory `upsilon_(rms) = sqrt((3kT)/(m)) = sqrt((3 xx (1.38 xx 10^(-23)) xx 300)/(9.1 xx 10^(-31)))` ` = 1.17 xx 10^(5)ms^(-1)` Therefore mean free path `lambda = upsilon_(rms) xx tau` `= 1.17 xx 10^(5) xx 2.46 xx 10^(-14) = 2.87 xx 10^(-9)m = 28.7Å` |
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| 57. |
A copper wire of cross sectional area 2.0 `mm^(2)`, resistivity `=1.7xx10^(-8)Omegam`, carries a current of 1 A. The electric field in the copper wire isA. `8.5xx10^(-5)(V)/(m)`B. `8.5xx10^(-4)(V)/(m)`C. `8.5xx10^(-3)(V)/(m)`D. `8.5xx10^(-2)(V)/(m)` |
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Answer» Correct Answer - C `E=(irho)/(A)` |
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| 58. |
Find the time of relaxation between collsion and free path of electrons in copper at room temperature. Given resistivity of copper`=1.7xx10^(-8) Omega`m, number density of electrons in copper `=8.5xx10^(28) m^(-3)`, charge on electron`=1.6xx10^(-19)` C, mass of electron `=9.1xx10^(-3)` kg and drift velocity of free electrons`=1.6xx10^(-4) ms^(-1)` |
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Answer» Correct Answer - `2.5xx10^(-14)s, 4.0xx10^(-8) m` |
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| 59. |
A copper wire iis stretched to make it `0.1%` longer. What is the percentage change in its resistance? |
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Answer» `R=rho 1/A=(rhol)/(V/l)` (`V=` volume of wire) `=(rhol^2)/V` `:. R propl^2` (`rho "and" V=` constant) For small percentage change % change `R=2(%"change in" l)=2(0.1%)=0.2%` Since `Rpropl^2`, with increase in the value of `l`, resistance will also increase. |
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| 60. |
Resistivity of a material of conductor in terms of relaxation time is given byA. `rho = (m)/(n e^(2)t)`B. `rho=m n e ^(2)t`C. `rho=(n e ^(2)t)/(m)`D. `rho=n e^(2)t` |
| Answer» Correct Answer - A | |
| 61. |
Find the resistance of a hollow cylindrical conductor of length 1.0m and inner and outer radii 1.0mm and 2.0mm respectively. The resistivity of the material is `2.0xx10^(-8)(Omega)m`. |
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Answer» Correct Answer - `2.1xx10^(-3) Omega` |
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| 62. |
As the temperature of a conductor increases,its resistivity and conductivity change.the ratio of resistivity to conductivityA. decreasesB. Remains sameC. IncreasesD. May increases or decreases |
| Answer» Correct Answer - A | |
| 63. |
A copper wire is stretched to make it 0.2% longer. What is the percentage change in its resistance? |
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Answer» The mass m of the wire of length l, area of cross section A and density d is givne by `m=Ald or A=(m)/(ld)` The resistance R of the wire of resistivity `rho` is given by `R= (rho l)/(A) = (rho l^(2)d)/(m)= k l^(2)` Where `k=(rho d)//(m)` is a constant of the wire. `:. (dR)/(R) xx 100 = (2 xx 0.2)/(100) xx 100=0.4%`. |
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| 64. |
Find the resistivity of a conductor in which a current density of 2.5 `A m^(-2)` is found to exist, when an electric field of `15 V m^(-1)` is applied on it. |
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Answer» Here, `J=2.5 Am^(-2) ,E=15 Vm^(-1)` Resistivity, `rho = (RA)/(l) = (V)/(I) (A)/(l) =(V//l)/(I//A) = (E)/(J) = (15)/(2.5)` `= 6 Omega m` |
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| 65. |
When 5 V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is `2.5 xx 10^(-4) ms ^(-1)`. If the electron density in the wire is ` 8 xx 10^(28) m^(-3)`, the resistivity of the material is close to :A. `1.6xx10^(-8)Omegam`B. `1.6xx10^(-7)Omegam`C. `1.6xx10^(-6)Omegam`D. `1.6xx10^(-5)Omegam` |
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Answer» Correct Answer - D `R=(rhol)/(A)` `(V)/(i)=(rhol)/(A)implies(V)/(nAeV_(d))=(rhol)/(A)impliesrho=(V)/(n eV_(d)l)` |
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| 66. |
When 5 V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is `2.5 xx 10^(-4) ms ^(-1)`. If the electron density in the wire is ` 8 xx 10^(28) m^(-3)`, the resistivity of the material is close to :A. `1.6 xx 10^(-6) Omega m`B. `1.6 xx 10^(-5) Omegam`C. `1.6 xx 10^(-8)Omega m `D. `1.6 xx 10^(-7) Omega m` |
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Answer» Correct Answer - B (b) `V= IR = (neAv_d)rho l/A` `:. rho = V/(V_dline)` Here V = potential difference l = length of wire n = no. of electrons per uint volume of conductor. e = no. of electrons Placing the value of above parameters we get resistivity . `rho = (5/(8 xx 10^(28) xx 1.6 xx 10^(-19) xx 2.5 xx 10^(-4) xx 0.1))` `= 1.6 xx 10^(-5) Omegam` |
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| 67. |
When 5 V potential difference is applied across a wire of length 0.1 m. The drift speed of electrons is `2.5 xx 10^(-4) ms^(-1)`. If the electron density in the wire is `8 xx10^(28) m^(-3)`, calculate the resistivity of the material of wire |
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Answer» Here, `r = 6.0 mm = 6 xx 10^(-3)m`, `l = 3.0 xx 10^(-2)m`, `rho = 3.5 xx 10^(-5)Omega m , P = 2.0W`. (a) As `P = I^(2)R = (JA)^(2)R = J^(2)A^(2)R` Current density , `J = 1/A sqrt(P/R) = 1/A sqrt(P/(rho l//A)) = sqrt(P/(rho l//A)` ` = sqrt(2.0/((3.5xx10^(-5)) xx (3.0 xx 10^(-2)) xx pi xx (6 xx 10^(-3))^(2))` ` = 1.3 xx 10^(5)Am^(-2)` (b) As `P = IV =JAV` or `V = P/(AJ) = P/(pi r^(2)J)` `20/(3.14 xx (6 xx 10^(-3))^(2)xx(1.3 xx 10^(5))` `=0.136 V` |
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| 68. |
A battery is of emf `E` is being charged from a charger such that positive terminal of the battery is connected to terminal `A` of charger and negative terminal of the battery is connected to terminal `B` of charger. The internal resistance of the battery is `r`.A. Potential difference across points `A` and `B` must be more than `E`.B. A must be at higher potential than `B`C. In battery, current flows from positivity terminal to the negative terminalD. No current flows through battery |
| Answer» Correct Answer - A::B::C | |
| 69. |
The freezing point on a thermometer is marked as `-20^(@)` and the boiling point as `130^(@)`. A temperature of human body `(34^(@)C)` on this thermometer will be read asA. `31^(@)`B. `51^(@)`C. `20^(@)`D. None of these |
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Answer» Correct Answer - A |
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| 70. |
Two moles of a monoatomic gas are taken from a to c, via three paths abc, ac and adc. Work done by the gas in process ac isA. 1000 RB. 900 RC. 600 RD. 1500 R |
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Answer» Correct Answer - C |
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| 71. |
Assertion A potentiometer is preferred over that of a voltmeter for measurement of emf of a cell Reason potentiometer does not draw any current from the cell.A. 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 | |
| 72. |
In figure shows a rectangular block with dimensions x , 2 x and 4 x . Electrical contacts can be made to the block between opposite pairs of faces (for example, between the faces labelled A - A , B - B and C - C ). Between which two faces would the maximum electrical resistance be obtained ( A - A : Top and bottom faces, B - B : Left and right faces, C - C : Front and rear faces) A. A-AB. B-BC. C-CD. Same for all three pairs |
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Answer» Correct Answer - C |
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| 73. |
Conductivity increases in the order ofA. `Al, Ag, Cu`B. `Al, Cu, Ag`C. `Cu, Al, Ag`D. `Ag, Cu, Al` |
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Answer» Correct Answer - B |
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| 74. |
The size of a carbon block is `1.0xx10 cm xx50 cm`. Find its resistance (i) between the opposite square faces (ii) between the opposite rectangular faces of the block. The resistivity of carbon is `3.5xx10^(-5) Omega cm`. |
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Answer» Correct Answer - `0.175 Omega, 7.0xx10^(-5) Omega` |
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| 75. |
The massses of the three wires of copper are in the ratio 1 : 3 : 5. And their lengths are in th ratio 5 : 3 : 1. the ratio of their electrical resistance isA. `1: 3:5`B. `5:3:1`C. `1:15:125`D. `125:15:1` |
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Answer» Correct Answer - D |
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| 76. |
The massses of the three wires of copper are in the ratio 1 : 3 : 5. And their lengths are in th ratio 5 : 3 : 1. the ratio of their electrical resistance isA. `1:3:5`B. `5:3:1`C. `1:15:125`D. `125:15:1` |
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Answer» Correct Answer - D |
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| 77. |
In the above question, the resistance between the square faces is.A. `3 xx 10^-9 ohm`B. `3 xx 10^-7 ohm`C. `3 xx 10^-5 ohm`D. `3 xx 10^-3 ohm` |
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Answer» Correct Answer - D `l = 100 cm, A = 1 cm^2` `R = (rho l)/(A) = (3 xx 10^-7 xx 100)/(1 xx 10^-2) = 3 xx 10^-3 ohm`. |
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| 78. |
The massses of the three wires of copper are in the ratio 1 : 3 : 5. And their lengths are in th ratio 5 : 3 : 1. the ratio of their electrical resistance isA. `1 : 3 : 5`B. `5 : 3 : 1`C. `1 : 15 : 125`D. `125 : 15 : 1` |
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Answer» Correct Answer - D `m = d A_1 5 l rArr A_1 = k//5` `3 m = d A_2 3l rArr A_2 = k` `5 m = d A_3 l rArr A_3 = 5 k` `R_1 : R_2 : R_3 = (5l)/(A_1) : ( 3l)/(A_2) : (l)/(A_3)` =`(5l)/(k//5) : (3 l)/(k) : (l)/(5 k)` =`125 : 15 : 1`. |
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| 79. |
Dimensions of a block are `1 cm xx 1 cm xx 100 cm`. If specific resistance of its material is `3 xx 10^(-7)`ohm-m, then the resistance between the opposite rectangular faces isA. `3 xx 10^-9 ohm`B. `3 xx 10^-7 ohm`C. `3 xx 10^-5 ohm`D. `3 xx 10^-3 ohm` |
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Answer» Correct Answer - B Between rectangular faces, `l = 1 cm, A = 100 cm^2` `R = (rho l)/(A) = (3 xx 10^-7 xx 1)/(100 xx 10^-2) = 3 xx 10^-7 ohm`. |
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| 80. |
An electric power station ( 10 MW) transmits power to a distant load through long and thin cables. Which of the two modes of transmission would result in lessere power wastage : Power transmission of : (i) 20,000 V or (ii) 200 V? |
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Answer» Let R be the resistance cable. Here, `P = 10 MW = 10 xx 10^(6)W = 10^(7)W`. (i) `V_(1) = 20,000V`, Current `I_(1) = P/V_(1) = 10^(7)/(20,000) = 500 A` Rate of heat dissipated, `P_(1) = I_(1)^(2) R = (500)^(2)R = 25 xx 10^(4)R W` (ii) `V_(2) = 200 V`, Current `I_(2) = P/V_(2) = 10^(7)/200 = 5 xx 10^(4)A` Rate of heat dissipated, `P_(2) = I_(2)^(2) R = (5 xx 10^(4))^(2) R= 25 xx 10^(8) R W` Clearly `P_(1) lt P_(2)`, so there will be lesser power wastage during transmission at `20,000 V` than at `200 V`. |
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| 81. |
What is the law defines heat produced by an electric current ? |
| Answer» Joule law of heating. It state that the amount of heat produced in a conductor when a constant current flows through it is directly proportional to (i) the square of the current, (ii) the resistance of the conductor and (iii) the time for which current is passed. | |
| 82. |
A line having a total resistance of `0.2` `omega` delivers `10 KW` at `220 V` to a small factory . Calculate the efficiency of transmission . |
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Answer» Power loss in the transmission line in the form of heat `P = I^(2) R= (P/V)^(2) R = ((20,000)/(220))^(2) xx 0.4` `= 3306 W = 3.31 kW` Efficiency of transmission line `eta = ("power delivered by line")/("power supplied to line")` `= ("power delivered")/("power delivered + power loss")` `= 20/(20+3.31) = 0.858 = 85.8%` |
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| 83. |
A boiler is made of a copper plate `2.4mm` thick with an inside coating of a `0.2mm` thick layer of tin The surface area exposed to gases at `700^(@)C` is `400cm^(2)` The maximum amount of steam that could be generated per hour at atmospheric pressure is `({:(K_(cu)=0.9cal//cm-s-^(0)&k_("tin"=0.15cal//cm-s-^(0)C),,,,),(andL_(steam)=540cal//g,,,,):})` .A. 5000 kgB. 1000 kgC. 4000 kgD. 200 kg |
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Answer» Correct Answer - C |
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| 84. |
The thermo emf E of a thermocouple is found to vary wit temperature T of the junction (cold junction is `0^(@)C`) as `E=40T-(T^(2))/(20)` The temperature of inversion for the thermocouple isA. `20^(@)C`B. `400^(@)C`C. `-200^(@)C`D. `-100^(@)C` |
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Answer» Correct Answer - C `E=40t+(1)/(10)t^(2)` At inversion temperature E will be minimum i.e. `(dE)/(dt)=0` `(d)/(dt)[40t+ (1)/(10) t^(2)]=0` `:.t=-200^(@)C`. |
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| 85. |
The temperature of the cold junction of the thermocouple is `^(@)C` and the temperature of the hot junction is `T^(@)C`. The relation for the thermo emf is given by, `E = AT = (1)/(2) BT^(2)` ltbRgt (when `A = 16` and `B = 0.08)`. The temperature of inversion will beA. `500^(@)C`B. `400^(@)C`C. `600^(@)C`D. `460^(@)C` |
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Answer» Correct Answer - B (b) Let temperature of cold junction be `0^(@)C` and the of hot junction be `(T^(@)C)`. The relation for thermo emf is given as `E = AT = (1)/(2) BT^(2)` Given, `A = 16, B = 0.08` `:. E = 16 T = (1)/(2) xx 0.08 xx T^(2)` Since emf is zero at temperature of inversion we have `0 = 16 T - 0.04 T^(2) implies T = (16)/(0.04) = 400^(@)C` |
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| 86. |
The thermo emf E of a thermocouple is found to vary wit temperature T of the junction (cold junction is `0^(@)C`) as `E=40T-(T^(2))/(20)` The temperature of inversion for the thermocouple isA. `200^(@)C `B. `400^(@)C`C. `800^(@)C`D. `1000^(@)C` |
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Answer» The temperature of inversion `T_(i)=2T_(n)-_(c)` whre `T_(c)=0^(@)C ` and `T_(n)` is netural temperature. At `T_(n) ,(dE)/(dt)=0` `therefore E=40T-(T^(2))/(20)rArr(dE)/(dt)=40-(2T)/(20)` or `0 =40-(2T_(n))/(20)rArr T_(n)=400^(@)C` `therefore T_(i)=2T_(n)-T_(c)` `=(2xx400)-0=800^(@)C` |
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| 87. |
Two wires of equal diameters of resistivies `rho_(1)` and `rho_(2)` and lengths l and l respectively are joined in series. The equivalent resistivity of the combination isA. `(rho_(1)l_(1) +rho_(2)l_(2))/(l_(1)+l_(2)`B. `(rho_(1)l_(2)+rho_(2)l_(1))/(l_(1)-l_(2))`C. `(rho_(1)l_(2)+rho_(2)l_(1))/(l_(1)+l_(2))`D. `(rho_(1)l_(1)-rho_(2)l_(2))/(l_(1)-l_(2))` |
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Answer» Correct Answer - A |
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| 88. |
Two conducting wires of the same material and of equal length and equal diameters are first connected in series and then in parallel in an electric circuit. The ratio of the heat produced in series and parallel combinations would be : |
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Answer» Total resistance in series, `R_(1) = R + R = 2R`, Total resistance in parallel, `R_(2) = (R xx R)/(R + R) = R/2` For a given potential difference `H prop 1//R` So, `H_(1)/H_(2) = R_(2)/R_(1) = (R//2)/(2R)=1/4` |
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| 89. |
Two wires A and B of the same material have their lengths in the ratio `1 : 5` and diameters in the ratio ` 3 : 2`. If the resistance of the wire B is `180 Omega`, find the resistance of the wire A. |
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Answer» Correct Answer - `16 Omega` |
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| 90. |
Two wires of the same material but of different diameters carry the same current `i`. If the ratio of their diameters is `2:1` , then the corresponding ratio of their mean drift velocities will beA. `4:1`B. `1:1`C. `1:2`D. `1:4` |
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Answer» Correct Answer - D Drift velocity `v_(d)=(l)/(nAe)=(lxx4)/(n pi D^(2)e) " " [because A = pi((D)/(2))^(2)]` i.e., `v_(d) prop (1)/(D^(2)) therefore (v_(d_(1)))/(v_(d_(2)))=(D_(2)^(2))/(D_(1)^(2))=((1)/(2))^(2)=(1)/(4)` |
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| 91. |
A carbon film resistor has colour green, black, violet and gold . The value of the resistor isA. `50 M Omega`B. `500 M Omega`C. `(500 pm 5%) M Omega`D. `(500pm 10%)M Omega` |
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Answer» Correct Answer - C Corresponding to the colours of the first and second bands green and black, the figure are 5 and 0. Corresponding to the colour of third band, violet, the multiplier is `10^(7)`. Therefore, the value of the resistance is `50xx10^(7)Omega`. The gold colour of the fourth band indicates the tolerance of 5%. So, that the value of the resistor is written as, `50xx10^(7) Omega pm 5% = 500 xx 10^(6)Omega+5%=500 pm 5% M Omega` |
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| 92. |
A resistor is marked with the rings coloured brown, balck, green, and gold. The resitance in ohm isA. `(3.5xx10^(5)pm5%)`B. `(1.10xx10^(5)pm10%)`C. `(8xx10^(6)pm5%)`D. `(1xx10^(6)pm5%)` |
| Answer» Correct Answer - D | |
| 93. |
The resistance of a 10 m long wire is 10 `Omega`. Its length is increased by 25% by stretching the wire uniformly . The resistance of wire will change toA. `12.5Omega`B. `14.5Omega`C. `15.6Omega`D. `16.6Omega` |
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Answer» Correct Answer - C Given, `l_(1)=l+(25)/(100)l=(5l)/(4)`. Since, volume of wire remains unchanged on increasing length, hence `A_(1)l_(1)=Al` `A_(1)xx(5l)/(4)=Al` or `A_(1)=4A//5` Given, `R= rho l//A= 10 Omega` and `R_(1)=(rho l_(1))/(A_(1))=(rho5l //4)/(4A//5)=(25 rho l)/(16 A)` `therefore " " R_(1)=(25)/(16)xx10=(250)/(16)=15.6 Omega` |
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| 94. |
The resistance of a wire is R . If the length of the wire is doubled by stretching, then the new resistance will beA. `2R`B. `4R`C. RD. `(R)/(4)` |
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Answer» Correct Answer - B |
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| 95. |
A carbon film resistor has cololur code green, black, violet, gold. The value of the resistor isA. `50M Omega`B. `500M Omega`C. `500 pm 5%M Omega`D. `500 pm 10%M Omega` |
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Answer» Correct Answer - C |
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| 96. |
The resistance of a wire is `10 Omega`. Its length is increased by `10%` by stretching. The new resistance will now beA. `12Omega`B. `1.2Omega`C. `13Omega`D. `11Omega` |
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Answer» Correct Answer - A |
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| 97. |
The resistance of a wire is `10 Omega`. Its length is increased by `10%` by stretching. The new resistance will now beA. `12 Omega`B. `1.2 Omega`C. `13 Omega`D. `11 Omega` |
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Answer» Correct Answer - A |
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| 98. |
A wire of resistance `4 Omega` is stretched to twice its original length. The resistance of stretched wire would beA. `2 Omega`B. `4 Omega`C. `8 Omega`D. `16 Omega` |
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Answer» Correct Answer - D |
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| 99. |
A wire of resistance `4 Omega` is stretched to twice its original length. In the process of stretching, its area of cross-section gets halved. Now, the resistance of the wire isA. `8 Omega`B. `16 Omega`C. `1 Omega`D. `4 Omega` |
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Answer» Correct Answer - B |
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| 100. |
A wire of resistance `4 Omega` is stretched to twice its original length. The resistance of stretched wire would beA. `2Omega`B. `4Omega`C. `8Omega`D. `16Omega` |
| Answer» Correct Answer - D | |