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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.
| 1001. |
The number density of electron in copper is `8.5 xx10^(28)m^(-3)`. Find the current flowing through a copper wire of length 20 cm, area of cross section 1 mm square,when connected to a battery of 3 V. Given the electron mobility `=4.5xx10^(-6)m^(2)V^(-1) s^(-1)` and electron charge `=1.6 xx 10^(-19)C`. |
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Answer» Here, `m=8.5 xx 10^(28) m^(-3), I=?` `l=20 cm =0.20 m`, `A=1 mm "square" =10^(-3) xx 10^(-3) m^(2) =10^(-6) m^(2)`, `V=3V, mu =4.5xx10^(-6) m^(2) V^(-1) s^(-1)`, `e=1.6xx10^(-19) C`. Electric field set up in a copper wire `E=V/l =3/0.20 =15 Vm^(-1)` Current, `I=n A e mu E=(8.5 xx10^(28))xx10^(-6)` `xx (1.6 xx 10^(-19)) xx(4.5xx10^(-6))xx5 `=0.918 A` |
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| 1002. |
The thermistors are usually made ofA. metals with low temperature coefficient of resistivityB. metals with high temperature coefficient of resistivity.C. metal oxides with high temperature coefficient of resistivityD. semiconducting meterials having low temperature coefficient of resistivity. |
| Answer» Correct Answer - C | |
| 1003. |
The thermistors are usually made ofA. metals with low temperature coefficient of resistivityB. metals with high temperature coefficient of resistivityC. metal oxides with high temperature coefficient of resistivityD. semiconducting materials having low temperature coefficient of resistivity |
| Answer» Correct Answer - C | |
| 1004. |
Calculate the electric field in a copper wire of cross-sectional area 2.0 mm square carrying a current of 2 A. The resistivity of copper is `1.7 xx 10^(-8)Omega m`. |
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Answer» Here, `A=2.0 mm sq = 2xx2 mm^(2)` `=4xx10^(6) m^(2)`, `I= 2A , rho =1.7 xx 10^(-8) Omega m`. Electric field `E=V/l = (IR)/(l) =(I rho l//A)/(l) = (l rho)/(A)` `=(2xx(1.7 xx 10^(-8)))/(4xx10^(-6)) = 8.5 xx10^(-3) V//m` |
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| 1005. |
Current flows through a constricted conductor as shown in figure The radius and the current density to the left of constriction are 2 mm and `4.2xx10^(5)Am^(-2)`. (a) How much current flows through the constriction? (b) If the current density is doubled as emerges from the right side of the constriction, what is the radius `r_(2)`? |
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Answer» Here, `r_(1) =2 mm = 2xx10^(-3) m` , `J_(1)=4.2xx10^(5)Am^(-2)` (a) Current flowing through the constriction `I_(1)=J_(1)A_(1)=J_(1) xx pi r_(1)^(2)` `=(4.5 xx 10^(5))xx(22 //7) (2xx10^(-3))^(2)=5.66A` (b) For a steady flow of current, `I_(1)=I_(2)` or `J_(1)A_(1) =J_(2) A_(2)=2J_(1) A_(2)` or `A_(2)=A_(1)/2 or pi r_(2)^(2)=1/2 pi r_(1)^(2) or r_(2)=r_(1)/(sqrt2)` or `r_(2)=2xx10^(-3)/(sqrt2)=sqrt(2)xx10^(-3) m=1.414 mm` |
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| 1006. |
An aluminium wire of diameter 2.5 mm is connectedf in series with a copper wire of diameter1.6 mm. A current of 2.0 A is passed through them. Find (a) current density in aluminium wire, (b) drift velocity of electrons in copper wire. Given the number density of conduction electrons in copper is `10^(29) m^(-3)`. |
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Answer» For A1, `r_(1)=2.5/2 mm =1.25 mm` `= 1.25 xx 10^(3) m , I=2.0A` Area of cross- section, `A_(1) = pi r_(1)^(2)` `=3.14 (1.25 xx10^(-3))^(2)=4.9 xx 10^(-6) m^(2)` For Cu, `r_(2)=1.6/2=0.8 mm=0.8xx10^(-3) m` , `n=10^(29) m^(-3)` Area of cross-section, `A_(2)=pi r_(2)^(2)` `=3.14(0.8 xx10^(-3))^(2)=2.01xx10^(-6) m^(2)` (a) Current density in aluminium wire, `J=I/A_(1)=2/(4.9xx10^(-6))=4.08 xx10^(5) A m^(-2)` (b) Drift velocity in copper wire, `v_(d)=(I)/("ne" A_(2))=2.0/(10^(29) xx 1.6 xx 10^(-19)xx2.01xx10^(-6))` `=6.2 xx10^(-5) ms^(-1)` |
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| 1007. |
An aluminium wire `7.5 m` long is connected in parallel with a copper wire `6 m` long. When a current of `5 A` is passed through the combination, it is found that the current in the aluminium wire is `3 A`. The diameter of the aluminium wire is `1 mm`. Determine the diameter of the copper wire. Resistivity of copper is `0.017 muOmega-m` and that of the aluminium is `0.028 muOmega-m`. |
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Answer» In parallel current distributes in inverse ratio of resistance `1rarr "Aluminimum" 2rarr Copper` `R_1/R_2=i_2/i_1` `(rho_1l_1//A_1)/(rho_2l_2//A_2)=2/3 ` `:. (rhi_1l_1d_2^2)/(rho_2l_2d_1^2)=2/3` `:. D_2=(sqrt((2rho_2l_2)/(3rho_1l_1)))d_1` `=(sqrt((2xx0.017xx6)/(3xx0.028xx7.5)))(1mm)` |
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| 1008. |
The potential difference between two points in a wire `75.0 cm` apart is `0.938 V`, when the current density is `4.40 xx 10^7 A// m^2`. What is(a) the magnitude of `E` in the wire?(b) the resistivity of the material of which the wire is made? |
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Answer» Correct Answer - A::B::D a. `E=V/l=0.938/0.75=1.25V/m` b. `E=Jrho` `:. rho=E/J=1.25/(4.4xx10^7)` `=2.84xx10^-8Omega-m` |
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| 1009. |
A typical thermistor can easily measure a change in temperature of the order ofA. `10^(3) .^(@)C`B. `10^(2) .^(@)C`C. `10^(-2) .^(@)C`D. `10^(-3) .^(@)C` |
| Answer» Correct Answer - D | |
| 1010. |
The range of voltmeter is 10 V and its internal resistance is `50 Omega`. To convert it to voltmeter of range 15 V, how much resistance is to be added ?A. Add `25 Omega` resistor in parallelB. Add `25 Omega` resistor in seriesC. Add `125 Omega` resistor in parallelD. Add `125 Omega` resistor in series |
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Answer» Correct Answer - B |
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| 1011. |
Identify the wrong statement.A. Charge is a vector quantityB. Current is a scalar quantityC. Charge can be quantityD. Charge is additive in nature |
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Answer» Correct Answer - A |
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| 1012. |
(A) Consider circuit How much energy is absorted by electrons from the initial state of no current to the state of draft velocity? (b) Electrons give up energy at the rate of `Rt^(2)` per second in the thermal energy What time state would one assoclate energy with energy in problem (a) ? n = no of electron /volume `= 10^(29)//m^(3)` length of ciruit `= 10 cm` cross section `A = (1 mm)^(2)` |
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Answer» (a) current `I = (V)/(R) = (6V)/(6 Omega) = 1A`, driff velocity `v_(d) = (I)/(nA e) = (1)/(10^(29) xx (10^(-4)) xx (1.6 xx 10^(-19)) = (1)/(1.6) xx 10^(-4) m//s` No of electron in the wire `= nAI` KE of all the electron `= (1)/(2) m_(e) u_(d)^(2) xx nAI = (1)/(2) xx (9.1 xx 10^(-31)) xx ((1)/(16) xx 10^(-4))^(2) xx 10^(29) xx(10^(-4)) xx 10^(-1)` `= 2 xx 10^(-17) J` (b) power loss `= I^(2) xx 6 = 6 J//s` All the `KE` of electron would be lost in time `= (2 xx 10^(-17)J)/(6 J//s) = 0.33 xx10^(-17) s ~~ 10^(-17) s` |
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| 1013. |
The electric potential variation around a single closed loop contining an ideal battery and one of more resistors is shown in figure. If current of `1 A` flows in the circuit, A. two resistor and two batteriesB. one resistor and three batteriesC. maximum net emf of voltD. three resistor |
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Answer» Correct Answer - D (d) We notice that potential increases twice and drops twice. If the circuit had only one battery and three resistance, the currents in all three batteries would definitely be in same direction. Either potential would increase along all three resistances or it would decreases along each. This is different from the situation given in graph. |
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| 1014. |
A cell of emf E having an internal resistance R varies with R as shown in figure by the curve A. AB. BC. CD. D |
| Answer» Correct Answer - B | |
| 1015. |
An electric current is passed through a circuit containing two wires of the same material connected in parallel. If the length and radii of the wires are in the ratio `2//3 and 4//3`, then find the ratio of the current passing through the wires |
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Answer» `R_(1)/R_(2) = (rho l_(1)//pi r_(1)^(2))/(rho l_(2) // pi r_(2)^(2)) = (l_(1))/(l_(2)) xx r_(2)^(2)/r_(1)^(2) = 2/3 xx (3/4)^(2) = 3/8` For wires connected in parllel `V_(1) = V_(2) or I_(1)R_(1) = I_(2)R_(2)or I_(1)/I_(2)=R_(2)/R_(1) = 8/3` |
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| 1016. |
Figure represents a part of closed circuit. What is the potential differnce between points A and B? |
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Answer» Potential difference across `2 Omega = 2 xx 2 = 4V`, Potential differnce across `1 Omega =1 xx 2 = 2 V`. Given the e.m.f. of the cell is 3 V. Let the positive electrode of cell be + 3 V and - ve electrode of cell at 0V. Therefore potential of `A = 4+ 3= 7V` : Potential of `B =- 2V`, as current is flowing from negative terminal of cell to point B, hence B is at lower potential . Therefore potential differnce between A and B = ` 7 -( -2 )= 9V` . |
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| 1017. |
A cross a metallic conductor of non-uniform cross-section, a constant potential difference is applied. The quantity which remain (s) constant along the conductor isA. current densityB. currentC. drift velocityD. electric field |
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Answer» Correct Answer - B |
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| 1018. |
The two ends of a uniform conductor are joined to a cell of e.m.f. `E` and some internal resistance. Starting from the midpoint `P` of the conductor, we move in the direction of current and return to `P`. The potential `V` at every point on the path is plotted against the distance covered `(x)`. which of the follwoing graphs best represent the resulting curve ?A. B. C. D. |
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Answer» Correct Answer - B (b) When we move in the direction of the current in a uniorm conductor, the potential difference decreases linearly. When we pass through the cell, from its negative to its positive terminal, the potential increase by an amount equal to its potential difference. This is less than its emf, as there is some potential drop across its internal resistance when the cell is driving current. |
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| 1019. |
The two ends of a uniform conductor are joined to a cell of e.m.f. `E` and some internal resistance. Starting from the midpoint `P` of the conductor, we move in the direction of current and return to `P`. The potential `V` at every point on the path is plotted against the distance covered `(x)`. which of the following graphs best represent the resulting curve ?A. B. C. D. |
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Answer» Correct Answer - B |
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| 1020. |
The two ends of a uniform conductor are joined to a cell of e.m.f. `E` and some internal resistance. Starting from the midpoint `P` of the conductor, we move in the direction of current and return to `P`. The potential `V` at every point on the path is plotted against the distance covered `(x)`. which of the following graphs best represent the resulting curve ?A. B. C. D. |
| Answer» Correct Answer - B | |
| 1021. |
In the circuit shown in figure potential difference between point `A` and `B` is `16 V`. Find the current passing through `2 Omega` resistance. A. `3.5A`B. `3A`C. `4.5A`D. `5.5A` |
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Answer» Correct Answer - A `V_(A)-V_(B)=16` `4i_(1)+2(i_(1)+i_(2))-3+4i_(1)=16` ..(i) `9-i_(2)-2(i_(1)+i_(2))=0` Solving eqs (i) and (ii) `i_(1)=1.5A` and `i_(2)=2A` |
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| 1022. |
What is the momentum acquired by the electron in a wire of length `1` meter when a current of `1` ampere start floeing in wire ? The mass and charge of electron are `m` and arespectively |
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Answer» Let A be the area of the cross section of the wire Then volume of the wire `= AL` If `n` is the no density (i.e. no of electron per unit volume ) then number of electron in the wire is `N = A 1n` If `v_(d)` is the dreft velocity of electron in meter wire , then momentum acquired by the electron `= mv_(d) ` momentum acquired by wire = momentum acquiredd by all the electron `= (mv_(d) ) xx A //m` `= m xx((1)/(nAe)) xx Aln = (iml)/(e)` |
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| 1023. |
Cell A has emf 2 E ad internal resistance 4 r. Cell B has emf E and internal resistance r. The negative of A is connected to the positive of B and a load resistance of R is connected across the battery formed. If the terminal potential difference across A is zero, then R is equal toA. `3r`B. `2r`C. `r`D. `5r` |
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Answer» Correct Answer - C `i=(E_(1)+E_(2))/(R+r_(1)+r_(2)),V_(1)=E_(1)-ir_(1)` |
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| 1024. |
A group of N cells where e.m.f. varies directly with the internal resistance as per the equation `E_(N)=1.5r_(N)` are connected as shown in the figure. The current I in the circuit is:A. 0`.51` ampB. `5.1` ampC. `0.15`ampD. `1.5 `amp |
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Answer» Correct Answer - D |
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| 1025. |
A group of N cells where e.m.f. varies directly with the internal resistance as per the equation `E_(N)=1.5r_(N)` are connected as shown in the figure. The current I in the circuit is:A. 0.51 AB. 5.1 AC. 0.15 AD. 1.5 A |
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Answer» Correct Answer - D `i=(E_(N))/(r_(N))=1.5` |
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| 1026. |
Potential different across the terminals of the battery shown in figure is A. 8 VB. 10 VC. 6 VD. zero |
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Answer» Correct Answer - D |
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| 1027. |
A group of N cells whose emf varies directly with the internal resistance as per the equation `E_(N)=1.5r_(N )` are connected as shown in the figure. The current in the circuit is A. 5.1AB. 0.51AC. 1.5AD. 0.15A |
| Answer» Correct Answer - 3 | |
| 1028. |
If length of a conductor is doubled by keeping volume constant, then what is its new resistance if initial were `4Omega` ?A. `16Omega`B. `8Omega`C. `4Omega`D. `2Omega` |
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Answer» Correct Answer - A `R_(1)=4 Omega, R_(2)=? ` if `l_(2)=2l_(1)` by keeping volume constant. `(R_(2))/(R_(1))=((l_(2))/(l_(1)))^(2)` `R_(2)=((2l_(1))/(l_(1)))^(2) R_(1)=4xx4=16 Omega`. |
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| 1029. |
A galvanometer has resistance `36Omega` if a shunt of `4Omega` is added with this, then fraction of current that passes through galvanometer is:A. `(1)/(4)`B. ` (1)/(9)`C. `(1)/(10)`D. `(1)/(40)` |
| Answer» Correct Answer - 3 | |
| 1030. |
Two non-ideal identical batteries are connected in parallel. Consider the following statements (i) The equivalent e.m.f. is smaller than either of the two e.m.f.s (ii) The equivalent internal resistance is smaller than either of the two internal resistancesA. Both (i) and (ii) are correctB. (i) is correct but (ii) is wrongC. (ii) is correct but (i) is wrongD. Both (i) and (ii) are wrong |
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Answer» Correct Answer - C |
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| 1031. |
In the circuit shown, if a wire is connected between points A and B. How much current will flow through that wire? A. 5AB. `(10)/(3)A`C. `(20)/(3)A`D. `(5)/(3)A` |
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Answer» Correct Answer - B |
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| 1032. |
In the two circuits shown in figure A. `R_(AB) = R_(CD) = (sqrt(3) + 2)Omega`B. `R_(AB) = (sqrt(3) + 1) Omega`C. `R_(CD) = (sqrt(5) + 1) Omega`D. `R_(AB) gt R_(CD)` |
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Answer» Correct Answer - D |
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| 1033. |
The net resistance between points P and Q in the circuit shown in figure is A. R/2B. 2R/5C. 3R/5D. R/3 |
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Answer» Correct Answer - B |
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| 1034. |
Current in `3 Omega` resistance is A. `1 A`B. `(1)/(7) A`C. `(5)/(7) A`D. `(15)/(7) A` |
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Answer» Correct Answer - C |
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| 1035. |
In the given figure, the current through the 20 V battery is A. `11A`B. `12A`C. `7 A`D. `14A` |
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Answer» Correct Answer - A |
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| 1036. |
the current in resistance `R_(3)` in the given circuit is `2/x A`. Find the value of x. A. `1A`B. `2//3A`C. `0.25A`D. `0.50A` |
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Answer» Correct Answer - B |
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| 1037. |
In the circuit shown in figure, the resistance R has a value that depends on the current. Specifically R is 20 `Omega` when `i` is zero and the amount of increase in resistance is numerically equal to one-half of the current. What is the value of current I in circuit? A. 15 AB. `10A`C. 20 AD. `5A` |
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Answer» Correct Answer - B |
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| 1038. |
The charge flowing in a conductor varies with times as `Q = at - bt^2.` Then, the currentA. reaches a maximum and then decreasesB. falls to zero after `t=(a)/(2b)`C. changes at a rate of `(-2b)`D. Both (b) and (c) |
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Answer» Correct Answer - B |
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| 1039. |
The charge flowing through a resistance `R` varies with time `t as Q = at - bt^(2)`. The total heat produced in `R` isA. `(a^(3)R)/(3b)`B. `(a^(3)R)/(2b)`C. `(a^(3)R)/(b)`D. `(a^(3)R)/(6b)` |
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Answer» Correct Answer - D |
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| 1040. |
Under the section of electric field, a material is said to be a conductor of electricity if there is flow ofA. same type of charge in or opposite direction of fieldB. opposite type of charge in the same direction of fieldC. no flow of chargeD. none of these |
| Answer» Correct Answer - A | |
| 1041. |
In conductors, current conduction take place due toA. electrons moving in the direction of fieldB. electrons moving opposite to the direction of fieldC. positive ions moving in the direction of fieldD. none of these |
| Answer» Correct Answer - B | |
| 1042. |
How many lamps each of `50W`and `100W`can be connected in parallel across a `120V` battery of internal resistance `10 Omega` so the each glows to full power ?A. 2B. 4C. 6D. 8 |
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Answer» Correct Answer - b Resistance of each lamp `R = V^(2)//P = (100)^(2)//50` `= 200 Omega` , current through each lamp `= P//V= 50//100 = 1//2A` If there n lamps in parallel to a battery of `120V` current each lamp glown to fall power then total current in n cell `= n xx 1//2 A ,` Total resistance of circuit `= 10 + 200//n` As, current `= ("emf")/("total resistance")` So `(n)/(2) = (120)/(10+200//n)` On solving we get , `n = 4` |
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| 1043. |
A givne mass of a gas expands from a state A to the state B by three paths 1, 2 and 3 as shown in T-V indicator diagram. If `W_(1),W_(2)` and `W_(3)` respectively be the work done by the gas along the three paths, then A. `W_(1)gtW_(2)gtW_(3)`B. `W_(1)ltW_(2)ltW_(3)`C. `W_(1)=W_(2)=W_(3)`D. `W_(1)ltW_(2),W_(1)gtW_(3)` |
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Answer» Correct Answer - A |
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| 1044. |
In the absence of applied potential, the electric current flowing through a metallic wire is zero becauseA. The average velocity of electron is zeroB. The electrons are drifted in randon direction with a speed of the order of `10^(-2)(cm)/(s)`.C. The electrons move in random direction with a speed of the order close to that of velocity of light.D. Electrons and ions move in opposite direction. |
| Answer» Correct Answer - A | |
| 1045. |
The example for non- ohm ice resistance isA. diodeB. copper wireC. filament lampD. carbon resistance |
| Answer» Correct Answer - A | |
| 1046. |
The resistivity of a wireA. Increases with the length of the wireB. Decreases with the area of cross-sectionC. Decreases with the length and increases with the cross-section of wireD. None of the above statement is correct |
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Answer» Correct Answer - D |
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| 1047. |
On increasing the temperature of a conductor, its resistance increases becauseA. the collisions of the conducting electrons with the electrons increasesB. the collisions of the conducting electrons with the lattice consisting of the ions of the metal increasesC. the number of the conduction electrons decreases.D. The number of conduction electrons increase. |
| Answer» Correct Answer - B | |
| 1048. |
The example for non- ohm ice resistance isA. Copper wireB. Carbon resistanceC. DiodeD. Tungston wire |
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Answer» Correct Answer - C |
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| 1049. |
Drift velocity `v_(d)` varies with the intensity of electric field as per their relationA. `v_(d) prop E`B. `v_(d) prop (1)/(E)`C. `v_(d) =` constantD. `v_(d) prop E^(2)` |
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Answer» Correct Answer - A |
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| 1050. |
On increasing the temperature of a conductor, its resistance increases becauseA. Relaxation time decreasesB. Mass of the electrons increasesC. Electron density decreasesD. None of the above |
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Answer» Correct Answer - A (a) Resistance of conductor depends upon relation as `R prop (1)/(tau)`. With rise in temperature ems speed of free electron inside the conductor increase. So relaxation time decrease and hence resistance increase |
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