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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.
| 101. |
A wire of resistance `12Omega` is stretched uniformly till its length becomes three times original length. The change in resistance of wire isA. `96 Omega`B. `108 Omega`C. `150Omega`D. `208 Omega` |
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Answer» Correct Answer - B `(R_(2))/(R_(1))=((l_(2))/(l_(1)))^(2)` `:.R_(2)=9xx12=108 Omega` |
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| 102. |
A wire of resistance `4 Omega` is stretched to four times of its original length resistance of wire now becomesA. `4Omega`B. `8Omega`C. `64Omega`D. `16Omega` |
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Answer» Correct Answer - C `R_(1)=4 Omega, l_(2)=4 l_(1), R_(2)=?` `(R_(2))/(R_(1))=((l_(2))/(l_(1)))^(2)` `:.R_(2)=((4l_(1))/(l_(1)))^(2)xxR_(1)=16xx4=64Omega`. |
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| 103. |
Current supplied by the cell in the adjoining figure is A. 1.5 AB. `1 A`C. `0.1A`D. `0.5A` |
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Answer» Correct Answer - C `I=(E)/(R+r)` Where R is total external resistance `= 20 Omega`. `:.R=(2)/(20+0)=(1)/(10)=0.1A` |
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| 104. |
An incandescent bulb has a thin filament of tungsten that is heated to high temperature by passing an electric current. The hot filament emits black - body radiation. The filament is observed to break up at random to break up at random locations after a sufficiently long time of operation due to non-uniform evaporation of tungsten from the fialment. If the bulb is powered at constant voltage, which of the following statement (s) is (are) true?A. The temperature distribution over the filament is uniformB. The resistance over small sections of the filament decrease with timeC. The filament emits more light at higher band of frequencies before at breaks upD. The filament consumes less electrical power towards the end of the lift of the bulb `2` |
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Answer» Correct Answer - c,d Since evaporation is given be non-uniform, hence temperature most be non-uniform Then option (a) is wrong Due to evaporasion, the cross sectional area of the filament at section decrease Hence resistance of the given section increases as `R prop 1//A`. Thus option (b) is wrong. The portion of the filament of the bulb having highest evaporation rate would have rapidly reduced area and would becomes break up area of cross - section At break up junction temperature would be highest Due, to it the light of highest head frequency would be emitted at those cross section ,Thus potion ( c) is true Due to continuous heating of filament the cross-sectional area of a section of filament decrease Hence the resistance of the As a result of it , the power decrease because `P = V^(2) or P prop 1//R` as `V` is constant This potion (d) is true. |
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| 105. |
Three different arrangemnets of matrials `1` and `2,3` to from a wall Thremal conductivities are `k_(1) gt k_(2) gt k_(3)` The left side of the wall is `20^(@)C` higher than the right side Temperature difference `DeltaT` across the material 1 has following relation in three cases A. `DeltaT_(a)gtDeltaT_(b)gtDeltaT_(c)`B. `DeltaT_(a)=DeltaT_(b)=DeltaT_(c)`C. `DeltaT_(a)=DeltaT_(b)gtDeltaT_(c)`D. `DeltaT_(a)=DeltaT_(b)ltDeltaT_(c)` |
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Answer» Correct Answer - B |
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| 106. |
In an aluminium (Al) bar of square cross section, a square hole is drilled and is filled with iron (Fe) as shown in the figure. The electrical resistivities of Al and Fe are `2.7 xx (10^-8) Omega m and 1.0 xx (10^-7) Omegam,` respectively. The electrical resistance between the two faces P and Q of the composite bar is A. `2475/64 mu Omega`B. `1875/64 muOmega`C. `1875/49 muOmega`D. `2475/132 muOmega` |
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Answer» Correct Answer - B (b) `R_(Fe) = (rho_(Fe) xx l_(Fe))/A_(Fe) = (10^(-7) xx 50 xx 10^(-3))/( 4 xx 10^(-6)) = 25/2 xx 10^(-4)` `R_(Al) = (rho_(Al) xx l_(Al))/(A_(Al)) = (2.7 xx 10^(-8) xx 50 xx 10^(-3)) /((49-4) xx 10^(-6)) = (2.7 xx 50)/(45) xx (10^(-5))` `= 0.3 xx 10^(-4)` `R_(t otal) = (R_(Fe) xx R_(Al))/(R_(Fe) + R_(Al)) = (12.5 xx 10^(-4) xx 0.3 xx 10^(-4))/(12.8 xx 10^(-4)) ~~ 29 muOmega`. |
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| 107. |
In an aluminium (Al) bar of square cross section, a square hole is drilled and is filled with iron (Fe) as shown in the figure. The electrical resistivities of Al and Fe are `2.7 xx (10^-8) Omega m and 1.0 xx (10^-7) Omegam,` respectively. The electrical resistance between the two faces P and Q of the composite bar is A. `(2475)/(64)muOmega`B. `(1875)/(64)muOmega`C. `(1875)/(49)muOmega`D. `(2475)/(132)muOmega` |
| Answer» Correct Answer - B | |
| 108. |
In fig. battery E is balanced on 55cm length of potentiometer wire but when a resistance of `10Omega` is connected in parallel with the battery then it balances on 50cm length of the potentiometer wire then internal resistance r of the battery is:- A. `1Omega`B. ` 3Omega`C. `10Omega`D. `5Omega` |
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Answer» Correct Answer - 1 `r=(( l_(0)-l_(C))/(l_(C)))R=((55-50)/(50))10=1Omega` Here `l_(0)=55cm:l_(c)=50cm:R=10Omega` |
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| 109. |
How can the resistances of `2Omega, 3 Omega " and " 6 Omega` be connected to give an effective resistance of `4 Omega` ? |
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Answer» Correct Answer - `2 Omega` resistance should be connected in series with parallel combination of `3 Omega` and `6 Omega` resistances |
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| 110. |
Resistance of each `10Omega` are connected as shown in the fig. The effective resistance between A and G isA. `16Omega`B. `20Omega`C. `12Omega`D. `8Omega` |
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Answer» Correct Answer - A Solving for effective resistance by series and parallel combination |
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| 111. |
If four resistances are connected as shown in the fig. between A and B the effective resistance isA. `4Omega`B. `8Omega`C. `2.4Omega`D. `2Omega` |
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Answer» Correct Answer - D combination of resistors |
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| 112. |
Which arrangement of four identical resistances should be sued to draw maximum energy from cell of voltage VA. B. C. D. |
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Answer» Correct Answer - B combination of resistors |
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| 113. |
Which arrangement of four identical resistances should be sued to draw maximum energy from cell of voltage VA. B. C. D. |
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Answer» Correct Answer - B |
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| 114. |
The graph shown in the figure represents change in the temperature of 5 kg of a substance as it absorbs heat at a costant rate of 42 kJ `"min"^(-1)`. The latent heat of vaporization of the substance is : A. `96"kJ kg"^(-1)`B. `126"kJ kg"^(-1)`C. `84"kJ kg"^(-1)`D. `12.6kJ kg"^(-1)` |
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Answer» Correct Answer - C |
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| 115. |
Two resistance `R_(1)` and `R_(2)` are made of different material. The temperature coefficient of the material of `R_(1)` is `alpha` and of the material of `R_(2)` is `-beta`. Then resistance of the series combination of `R_(1)` and `R_(2)` will not change with temperature, if `R_(1)//R_(2)` will not change with temperature if `R_(1)//R_(2)` equalsA. `(alpha)/(beta)`B. `(alpha+beta)/(alpha-beta)`C. `(alpha^(2)+beta^(2))/(alphabeta)`D. `(beta)/(alpha)` |
| Answer» Correct Answer - 4 | |
| 116. |
Two resistance `R_(1)` and `R_(2)` are made of different material. The temperature coefficient of the material of `R_(1)` is `alpha` and of the material of `R_(2)` is `-beta`. Then resistance of the series combination of `R_(1)` and `R_(2)` will not change with temperature, if `R_(1)//R_(2)` will not changee with temperature if `R_(1)//R_(2)` equalsA. `(alpha)/(beta)`B. `(alpha + beta)/(alpha - beta)`C. `(alpha^(2) + beta^(2))/(alpha beta)`D. `(beta)/(alpha)` |
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Answer» Correct Answer - D (d) `R_(1) + R_(2) = R_(1) (1 + alpha t) + R_(2) (1 - beta t)` `implies R_(1) + R_(2) = R_(1) + R_(2) + R_(1) alpha r = R_(2) beta t implies (R_(1))/(R_(2)) = (beta)/(alpha)` |
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| 117. |
A wire of resistance `10 Omega` is bent to form a circle. P and Q are points on the circumference of the circle dividing it into a quadrant and are connected to a Battery of 3 V and internal resistance `1 Omega` as shown in the figure. The currents in the two parts of the circle are A. `(6)/(23)A` and `(18)/(23)A`B. `(5)/(26)A` and `(15)/(26)A`C. `(4)/(25)A` and `(12)/(25)A`D. `(3)/(25)A` and `(9)/(25)A` |
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Answer» Correct Answer - A |
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| 118. |
Two resistance `R_(1)` and `R_(2)` are made of different material. The temperature coefficient of the material of `R_(1)` is `alpha` and of the material of `R_(2)` is `-beta`. Then resistance of the series combination of `R_(1)` and `R_(2)` will not change with temperature, if `R_(1)//R_(2)` will not change with temperature if `R_(1)//R_(2)` equalsA. `(alpha)/(beta)`B. `(alpha+beta)/(alpha-beta)`C. `(alpha^(2)+beta^(2))/(alpha beta)`D. `(beta)/(alpha)` |
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Answer» Correct Answer - D |
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| 119. |
An ionization chamber with parallel conducting plates as anode and cathode has `5 xx 10^(7)` electrons and the same number of singly-charged positive ions per `cm^(2)`. The electrons are moving at `0.4 m//s`. The current density from anode to cathodes `4 mu A//m^(2)`. The velocity of positive ions moving towards cathode isA. `0.4 m//s`B. `16m//s`C. ZeroD. `0.1 m//s` |
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Answer» Correct Answer - D |
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| 120. |
Volume versus temperature graph of two moles of helium gas is as shown in figure. The ratio of heat absorbed and the work done by the gas in process `1-2` is A. 3B. `5//2`C. `5//3`D. `7//2` |
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Answer» Correct Answer - B |
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| 121. |
One mole of a diatomic gas undergoes a process `P = P_(0)//[1 + (V//V_(0)^(3))]` where `P_(0)` and `V_(0)` are constant. The translational kinetic energy of the gas when `V = V_(0)` is given byA. `(5p_(0)V_(0))/(4)`B. `(3p_(0)V_(0))/(4)`C. `(3p_(0)V_(0))/(2)`D. `(5p_(0)V_(0))/(2)` |
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Answer» Correct Answer - B |
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| 122. |
Copper has one conduction electron per atom. Its density is `8.89 g//cm^3` and its atomic mass. `63.54 g//mol`. If a copper wire of diameter `1.0 mm` carries a current of `2.0 A`, what is the drift speed of the electrons in the wire? |
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Answer» Correct Answer - A::D `rho=8.89xx10^3kg/m^3` Mass of `1m^3=8.89xx10^3=8.89xx10^6g` `:.` Number of gram moles `=(8.89xx10^6)/63.54=1.4x10^5` Number of atoms `=1.4xx10^5xx6.02xx10^23` `=8.42xx10^28` One atom emits one condition electron. therefore number of free electrons in unit volume (or `1m^3` volume) `n=8.42xx10^28per m^3` `No2, i="ne"Av_d` `:. v_d=1/("ne"A)=i/("ne"pi r^2)` `=2.0/((8.42xx10^28)(1.6xx10^-19)(pi)(0.5xx10^-3)^2)` `=1.9xx10^-4m//s` |
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| 123. |
In Figure if the potential at Point P is `100 V`, what is the potential at point Q? |
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Answer» Correct Answer - A `i=(150-50)/(2+3)=20A` (anti clockwise) `V_Q+150=20xx2=V_P` `:. V_Q=V_P-110=-10V` |
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| 124. |
`A_(1), A_(2) and A_(3)` Are the ammeters and `A_(2)` reads 0.5A. (i) What are the readings of ammeters `A_(1)` and `A_(3)` ? (ii) What is the total resistance of the circuit ? |
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Answer» (i) P.D. across `3 Omega` `6 xx0.5 = 3 xx I_(3) =1.0A` Current through `A_(1)=0.5+1.0 =1.5A` Effective resistance of circuit `=(6xx3)/(6+3) + 2=4 Omega` |
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| 125. |
X, Y and Z are ammeters and Y reads 0.5 A. (i) What are the readings in ammeters X and Z ? (ii) What is the total resistance of the circuit ? |
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Answer» Correct Answer - `(i) 1.5 A , 1.0 A (ii) 4 Omega` |
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| 126. |
Two capacitors A and B of capacitances `2muF` and `5 mu F` are connected to two battery as shown in figure The potential difference in volt between the plate of A is A. 2B. 5C. 7D. 18 |
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Answer» Correct Answer - b Let `Q` b ethe charge stored in each capactor. Then `18 - 11 = (Q)/(2 mu F) + (Q)/(5 mu F) = (7Q)/(10mu F)` or `Q = 10 mu C` Pot. Diff. across capactior A `V_(A) = (Q)/(C_(A))= (10muC)/(2 muF) = 5V` |
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| 127. |
A heater coil is cut into two equal parts and only one part is now used in the heater. The heat generated will now beA. becomes one fourthB. halvedC. doubledD. become four times |
| Answer» Correct Answer - C | |
| 128. |
A heater coil is cut into two equal parts and only one part is now used in the heater. The heat generated will now beA. One fourthB. HalvedC. DoubledD. Four times |
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Answer» Correct Answer - C (c ) `H = (V^(2) t)/(R ) implies (H_(Half))/(H_(F u l l)) = ((R_(F u l l))/(R_(Half))) = (R )/(R // 2) = 2` `implies H_(Half) = 2 xx H_(Full)`. |
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| 129. |
A heater coil is cut into two equal parts and only one part is now used in the heater. The heat generated will now beA. one fourthB. halvedC. doubledD. four time |
| Answer» Correct Answer - C | |
| 130. |
Consider a thin square sheet of side L and thickness t, made of a material of resistivity `rho`. The resistance between two opposite faces, shown by the shaded areas in the figure is A. directly proportional to LB. directly proportional to tC. independent of LD. independent of t |
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Answer» Correct Answer - C `R = (rho(L))/(A)=(rho L)/(tL)=(rho)/(t)` i.e., R is independent of L Hence, the correct option is (c ). |
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| 131. |
Two batteries A and B whose emf is 2V are connected in series with external resistance `R=1Omega`. Internal resistance of battery A is `1.9Omega` and that of B is `0.9 Omega`. A. 2VB. 3.8VC. zeroD. None of these |
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Answer» Correct Answer - C Total resistance `r = 1.9+0.9+1=3.8 Omega` Total emf `E = 2+=4` `therefore " " E = iR` `" " i=(E )/(R )=(4)/(3.8)` Potential difference across battery A `V = E -ir` `= 2-(4)/(3.8)xx1.9=0` |
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| 132. |
When two identical batteries of internal resistance `1Omega` each are connected in series across a resistor R, the rate of heat produced in R is `J_1`. When the same batteries are connected in parallel across R, the rate is ``J_2 = 2.25 J_2` then the value of R in `Omega` is |
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Answer» Correct Answer - 4 Given `J_(1)//J_(2) = 2.25` `((2E)/(2r+R))^(2) R = J_(1) and ((2E)/(r+2R))^(2) R = J_(2)` `:. ((r +2R)^(2))/((2r+R)^(2)) = (J_(1))/(J_(2)) = 2.25` or `(r + 2R)/(2r+R) = 1.5 = (3)/(2)` On solving we get `R = 4r = 4 xx 1 = 4 Omega` |
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| 133. |
Two identical batteries, each of EMF 2V and internal resistance `r =1Omega` are connected as shown. the maximum power that can be developed across R using these batteries is : A. 3.2WB. 8.2WC. 2WD. 4W |
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Answer» Correct Answer - A |
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| 134. |
If an ammeter is to be used in place of a voltmeter, then we must connect with the ammeter aA. low resistance in parallelB. high resistance in parallelC. high resistance in seriesD. low resistance in series. |
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Answer» Correct Answer - C (c) KEY CONCEPT : To convert a galvanometer into a voltmeter we connect a high resistance in series with the galvanometer. The same procedure needs to be done if ammeter is to be used as a voltmeter. |
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| 135. |
A galvanometer gives full scale deflection with 0.006 A current. By connecting in to a `4990 Omega` resistance, it can be converted into a voltmeter of range 0-30V. If connected to a `(2n)/(249)Omega` resistance, it becomes an ammeter of range `0-1.5A`. The value of n is |
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Answer» `((I-I_g)/(I_g))S = V/(I_g) - R` `(1.5 - 0.006)/(0.006) xx (2n)/(249) = 30/(0.006) - 4990` `:. n~~5`. |
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| 136. |
When the current i is flowing through a conductor, the drift velocity is v . If 2i current is flowed through the same metal but having double the area of cross-section, then the drift velocity will beA. `4v`B. `(v)/(2)`C. `(v)/(4)`D. `v` |
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Answer» Correct Answer - D |
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| 137. |
When the current i is flowing through a conductor, the drift velocity is v . If 2i current is flowed through the same metal but having double the area of cross-section, then the drift velocity will beA. v/4B. v/2C. vD. 4v |
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Answer» Correct Answer - C |
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| 138. |
Two wires of the same material are given. The first wire is twice as long as the second and has twice the diameter of the second. The resistance of the first will beA. Twice of the secondB. Half of the secondC. Equal to the secondD. Four times of the second |
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Answer» Correct Answer - B |
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| 139. |
A current of 5 ampere is passing through a metallic wire of cross-sectional area `4xx10^(-6) m^(2)`. If the density of the charge-carriers in the wire is `5xx10^(26) m^(-3)`, find the drift speed of the electrons.A. `1.5625xx10^(-2)m//s`B. `1.5625xx10^(-23)m//s`C. `1.5625xx10^(3)m//s`D. `1.5625xx10^(-4)m//s` |
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Answer» Correct Answer - A `v_(d)=(I)/(n A e)` `= (5)/(5 xx10^(26)xx4xx10^(-6)xx1.6xx10^(-19))` `=1.5625xx10^(-2)m//s` |
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| 140. |
When the current i is flowing through a conductor, the drift velocity is v . If 2i current is flowed through the same metal but having double the area of cross-section, then the drift velocity will beA. `v//4`B. `v//2`C. vD. `4v` |
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Answer» Correct Answer - C |
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| 141. |
5 amperes of current is passed through a metallic conductor. The charge flowing in one minute in coulombs will beA. 5B. 12C. `1//2`D. 300 |
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Answer» Correct Answer - D |
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| 142. |
There is a current of 20 amperes in a copper wire of `10^(-6)` square metre area of cross-section. If the number of free electrons per cubic metre is `10^(29)` then the drift velocity isA. `125 xx 10^(-3) m//sec`B. `12.5 xx 10^(-3)m//sec`C. `1.25 xx 10^(-3)m//sec`D. `1.25 xx 10^(-4)m//sec` |
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Answer» Correct Answer - C |
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| 143. |
A potential difference V is aplied across a conductor of length `l`. How is the drift velocity affected when V is doubled and `l` is halved ? |
| Answer» When a potential differnce is applied across the two ends of a conductor, the random motion of free electrons is partially directed towards higher potential side, i.e., positive end of the conductor. | |
| 144. |
A conducting wire of cross-sectional area `1 cm^(2)` has `3 xx 10^(23)` charge carriers per m3. If wire carries a current of `24 mA`, then drift velocity of carriers isA. `5xx10^(-2)ms^(-1)`B. `0.5ms^(-1)`C. `5xx10^(-3)ms^(-1)`D. `5xx10^(-6)ms^(-1)` |
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Answer» Correct Answer - C |
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| 145. |
Current of 4.8 amperes is flowing through a conductor. The number of electrons per second will beA. `3 xx 10^(19)`B. `7. 68 xx 10^(21)`C. `7.68 xx 10^(20)`D. `3 xx 10^(20)` |
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Answer» Correct Answer - A |
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| 146. |
A potential difference of 10 V is applied across a conductor of resistance `1 k Omega`. Find the number of electrons flowing through the conductor in 5 minutes. |
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Answer» Correct Answer - `1.875xx10^(19)` |
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| 147. |
The drift velocity of free electrons in a conductor is v, when a current i is flowing in it, Ifboth the radius and current are doubled, then the drift velocity will be :A. vB. v/2C. v/4D. v/8 |
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Answer» Correct Answer - B |
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| 148. |
A potential difference of 3V is applied across a conductor of resistance `1.5 Omega`. Calculate the number of electrons flowing through it in one second. Given charge on electron, `e=1.6 xx10^(-19)C`. |
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Answer» Here, `V=3 "volt" , R=1.5 Omega`, `e=1.6 xx10^(-19)C, t=1s`. Now, `I=V/R=3/1.5=2A` and `I=Q/t= ("ne")/(t)` or `n= It/e =2xx1/1.6xx10^(-19)=1.25xx10^(19)` |
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| 149. |
In which of the following the carriers of electric current are electrons only?A. a super conductorB. a voltalic cellC. a semiconductorD. a hydrogen discharge tube |
| Answer» Correct Answer - 2 | |
| 150. |
A current of 1.6 A is flowing in a conductor. The number of electrons flowing per second through the conductor isA. `10^(9)`B. `10^(19)`C. `10^(16)`D. `10^(31)` |
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Answer» Correct Answer - B `i=(n e)/(t)impliesn=(it)/(e)` |
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