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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.
| 801. |
The voltage of clouds is `4 xx 10^(6) V` with respect to ground. In a lighting strike lasting `100 ms`, a charge of `4 C` is delivered to the ground. The power of ligthing strike isA. `20 MW`B. `80 MW`C. `160 MW`D. `500 kW` |
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Answer» Correct Answer - C (c ) Work done in delivering `q` coulomb of charge from clouds to ground. `W = V_(q) = 4 xx 10^(6) xx 4 = 16 xx 10^(6) J` The power of lighting strike is `P = (W)/(t) = (16 xx 10^(6))/(0.1) = 160 xx 10^(6) W = 160 MW` |
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| 802. |
Assume that clouds are distributed around the entire earth at a height of `3000 m` above the ground. The atmosphere can be modeled as a spherical capacitor with the earth as one plate and the cloud as other. When the electric field between the earth and the cloud becomes large, the air begins to conduct and the phenomena is called lightning. On a typical day `4 xx 10^(5) C` of positive charge is spread over the surface of the earth and equal amount of negative charge is there on the cloud. Resistivity of the air is `rho = 3 xx 10^(13) Omega m` and radius of the earth `= 6000 km`. (a) Find the resistance of the air gap between the earth’s surface and cloud. (b) Estimate the potential difference between the surface of the earth and the cloud (c) In how much time the capacitor formed between the earth and the cloud will lose 63 % of the charge ? |
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Answer» Correct Answer - (a) `199 Omega` (b) `3 xx 10^(5)V` (c) `165.3 s` |
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| 803. |
Lamps used for household lighting are connected inA. SeriesB. ParallelC. Mixed circuitD. None of the above |
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Answer» Correct Answer - B |
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| 804. |
In a post office box if the position os the cell and the galvenometer are interchangd, then the:A. null point will not changeB. null point will changeC. post office box will not workD. Nothing can be said. |
| Answer» Correct Answer - A | |
| 805. |
The drift velocity of electrons in a conducting wire is of the order of 1 mm/s, yet the bulb glows very quickly after the switch is put on beauseA. The random speed of electons is very high, of the order of `10^(6)m//s`.B. The electrons transfer their energy very quickly through collisionsC. Electric field is set up in the wire very quickly, producing a current through each cross section, almost instantaneously.D. All of above |
| Answer» Correct Answer - C | |
| 806. |
The example of an ohmic conductor isA. diodeB. germaniumC. tungsten wireD. torch bulb |
| Answer» Correct Answer - C | |
| 807. |
The velocity of charge carries of current (about 1 A) in a metal under normal conditions is of the order ofA. a fraction of mm/sB. velocity of lightC. several thousand metres/secondD. A few hundred metres per second. |
| Answer» Correct Answer - A | |
| 808. |
An electron is revolving n times per second. The charge passing in t second isA. netB. `(n e)/(t)`C. `(n t)/(e)`D. `(et)/(n)` |
| Answer» Correct Answer - A | |
| 809. |
If the voltage V applied across a conductor is increased to 2V, how will the drift velocity of the electrons change? |
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Answer» The drift velocity, `V_(d) = (eE)/(m) tau = (eV)/(mL) tau` `(V_(d_(1)))/(V_(d_(2))) = (V_(1))/(V_(2))` Here `V_(1) = V, V_(2)= 2V` `(V_(d_(1)))/(V_(d_(2))) = (V)/(2V)` `:. V_(d_(2)) = 2 V_(d_(1))` `:.` Drift velocity is increased by twice. |
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| 810. |
If A is the area of cross section of conductor, `v_(d)` is the drift velocity of electrons and n is the number of electrons per unit volume, then current density through conductor isA. ne `v_(d)`B. `(n e A)/(v_(d))`C. `(n e v_(d))/(A)`D. `(n/a)A v_(d)` |
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Answer» Correct Answer - A `J=(I)/(A)=(q)/(tA)=(nA l e )/(tA)= (n l e )/(t)=n e v_(d)` |
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| 811. |
At room temperature copper has free electron density of `8.4xx10^(28) per m^(3)` . The copper conductor has a cross-section of `10^(-6)m^(2)` and carries a current of 5.4 A. What is the electron drift velocity in copper? |
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Answer» Here, `n=8.4 xx 10^(28) m^(-3)` , `A=10^(-6) m^(2), I=5.4 A` `v_(d)=(I)/(n Ae) =5.4/((8.4xx10^(28))xx(10^(-6))xx(1.6xx10^(-19)))` `=0.4 xx 10^(-3) m//s =0.4mm//s` |
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| 812. |
For the resistance network shown in the figure, choose the correct options (s) A. The currnet through PQ is zeroB. `I_(1)=3A`C. The potential at S is less than that at QD. `I_(2)=3A` |
| Answer» Correct Answer - A::B::C::D | |
| 813. |
Heater of an electric kettle is made of a wire of length L and diameter d. It takes 4 minutes to raise the temperature of 0.5 kg water by 40K. This heater is replaced by a new heater having two wires of the same material, each of length L and diameter 2d. The way these wires are connected is given in the options. How much time in minutes will it take to raise the temperature of the same amount of water by 40 K?A. 4 it wires are in parallelB. 2 if wires are in seriesC. 1 if wires are in seriesD. 0.5 if wires are in parallel. |
| Answer» Correct Answer - B::D | |
| 814. |
A motor operating on `120 V` draws a current of `2 A` . If the heat is developed in the motor at the rate of `9 cal s^(-1)`, what is the efficiency? |
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Answer» Correct Answer - `80.9%` Here`V= 110 V, l = 2A` power supplied to motor `= 1//1 = 110 xx 2 = 220 W` power loss in the form of heat `= 10 cals^(-1)` `= 10 xx 4.2 Js^(-1) = 42 Js^(-1)` power used by motor `= 220 - 42 = 178 W` Effeciency `eta = ("power used by motor")/("power supplied to motor") xx 100` `= (178)/(220) xx 100 = 80.9%` |
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| 815. |
A room is lighted by `200W,124 V` incandescent lamps fed by a generator whose output voltage is `130 V`. The conducting wires from the generator to the user are made of aluminimum wire of total length 150 m and cross-sectional area `15 mm^(2)`. How many such length can be installed ? What is the total power consumed by the user ? sp. resistance of aluminium is `2.9 xx 10^(-4) Omega m ` |
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Answer» Correct Answer - `12,2.4kW` Resistance of aluminium wire `R_(Al) = (rho l)/(A) = (2.9 xx 10^(-8) xx 150)/( 15 xx 10^(-6)) = 0.29 Omega` current through main line `I = (130 - 124)/(0.29)= (20.69 )A` Current through one lamp `l_(1) = (200)/(124) = 1.613 A` ltb rgt No of lamps `= (20.69)/(1.613) = 12.83 = 12.85` Hence no of lamp to be used `= 12` power connected `= 12 xx 200 = 2400 W= 2.4kW` |
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| 816. |
The power of a heating coil is P. it is cut into two equal parts. The power of one of them across same mains isA. 2PB. 3PC. `(P)/(2)`D. `4P` |
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Answer» Correct Answer - A `P=(V^(2))/(R),R=(rhol)/(A)` |
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| 817. |
Two wires A and B are formed from the same material with same mass. Diameter of wire A is half of diameter of wire B. If the resistance of wire A is `32 Omega`, find the resistance of wire B. |
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Answer» Volume of A = Volume of B, `pi(D_(A)^(2)//4)l_(A) = pi (D_(B)^(2)//4)l_(B) or l_(A)/l_(B) = D_(B)^(2)/D_(A)^(2)=4` Resistance, `R=rhol/A= rho l/((pi D^(2)//4)) or R prop l/D^(2)` `:. R_(B)/R_(A) =l_(B)/l_(A) xx D_(A)^(2)/D_(B)^(2) = 1/4 xx 1/4=1/16` or `R_(B)=R_(A)/16 = 32/16=2 Omega`. |
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| 818. |
Explain how electron mobility changes for a good conductor, when (i) the temperature of the conductor is decrased at constant potential difference and (ii) applied potential difference is doubled at constant temperature. |
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Answer» Electron mobility in a conductor is given by `mu = v_(d)/E = ((eE)/(m)tau)/(E) = (e tau)/m` (i) When temperature of the conductor decreases, the relaxation time `tau` of the electrons in the conductor increases, so mobility `mu` increase. (ii) Mobility `mu` is independent of applied potential difference. |
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| 819. |
In a house there are four bulbs each of `50W` and 5 fans each of 60W. If they are used at the rate of 6 hours a day, the electrical energy consumed in a month of 30 days isA. 64 KWHB. 90.8KWHC. 72KWHD. 42KWH |
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Answer» Correct Answer - B 1 unit `=("no.watts"xx"no. of hours")/(1000)` |
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| 820. |
A cylindrical wire is stretched to increase its length by 10%. Calculate the percentage increase in resistance. |
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Answer» Correct Answer - 0.21 |
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| 821. |
Two copper wires A and B of equal masses are taken. The length of A is double the length of B. If the resistance of wire A is `160 Omega `, then calculate the resistance of the wire B. |
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Answer» Correct Answer - `40 Omega` |
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| 822. |
Three materials A, B and C have electrical conductivities `sigma , 2 sigma` and `2 sigma` respectively. Their numbers densities of free electrons are 2 n, n and 2n respectively. For which material is a average collision time of free electrons maximum? |
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Answer» Conductivity `sigma=1/rho = (n e^(2) tau)/m` Relaxation time, `tau = (m sigma)/(n e^(2)`. i.e., `tau prop sigma/n` `:. tau_(A) : tau_(B):tau_(C) = sigma/2n : 2sigma/n : 2sigma/2n = sigma/2n : 2sigma/n : sigma/n` Thus, `tau_(B) gt tau_(C ) gt tau_(A)`. So average collision time for material B is maximum. |
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| 823. |
Assertion : The connecting wires are made of copper. Reason : The electrical conductivity of copper is highA. If both assertion and reason are ture and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason both are false. |
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Answer» Correct Answer - A (a) Due to high electrical conductivity of copper, it conducts the current without offering much resistance. The copper being diamagnetic material does not get magnetised due to current through it and hence does not disturb the current in the circuit. |
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| 824. |
The resistance of each of the three wires joined as shown in figure is `4 Omega` and each one can have a maximum power of 20 watt (otherwise it will melt). What maximum power will the whole circuit dissipate ? |
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Answer» Let `I` be the current through a resistance wire of maximum power 20 watt. `:. I^(2)R = 20 or I^(2) xx 4 = 20 or I^(2) = 5` Effective resistance `R_(p)` between A and C will be `R_(p) = ( 4 xx 4)/(4 +4 ) = 2 Omega` Electric power dissipated as heat by the resistance between A and C `= I^(2)R_(p) = 5 xx 2 = 10` watt. Electric power dissipated as heat by the resistance between C and B `= I^(2)R = 5 xx 4 = 20` watt. `:.` Total electric power dissipated as heat by the whole circuit `= 10 +20 = 30` watt. |
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| 825. |
While making a standard resistance. The coil is made of maganin. The coil is doubled folded and is wound over non- conducting frame.Why? |
| Answer» For manganin, the temperature coefficient of resistance is very low. Due to it, the resistance of manganin wire remains almost unchanged with change in temperature. The resistivity of manganin is high. Therefore, for making a standard resistance of the given value, the smaller lenght of wires is needed. It is due to these facts, the wire of manganin is used for making stnadard resistance coil. The coil is double folded on itself to avoid the inductive effect and it is wound over the non-conducting frame in order to avoid the conductive effect and the leakage of current. | |
| 826. |
The car-battery is of 12 volt. 8 simple cells connected in series can give 12 volt. But such cells are not used in starting a car , why ? |
| Answer» To start a car, high current is required which cannot be obtained from the series combination of 8 simple cells, because their internal resistance is of the order of `10 Omega`, while the resistance of the car battery is only of the order of `0.1 Omega`. | |
| 827. |
When cells are connected in parallel, what will be the effect on (i) current capacity (ii)e.m.f. of the cells |
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Answer» (i) Current capacity from the combination of cells is the total current available from the cell. Current capacity increase for the parallel combination of cells. (ii) The effective e.m.f. of the cells of equal e.m.f. in parallel will be equal to e.m.f. of one cell. |
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| 828. |
Is it possible that there is no potential difference between the paltes of a cell ? If yes, under what condition? |
| Answer» When cell is short circuited, it gets totally dischargred. The potential difference between the two plates of the cell become zero. | |
| 829. |
A 10 V battery of negilgible internal resistance is connected across a 200V battery and a resistance of `38 Oemga` as shown in figure. . Find the value of current in the circuit. |
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Answer» Here, `epsilon_(1)=200V, epsilon_(2)=10 V , R=38 Omega , I=?` `I=(epsilon_(1)-epsilon_(2))/R = (200-10)/(38)=5A` |
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| 830. |
In the arrangement shown in figure when the switch `S_2` is open, the galvanometer shown no deflection for `l=L//2`. When the switch `S_2` is closed, the galvanometer shown no deflection for `l=5L//12`. The internal resistance `(r)` of `6V` cell, and the emf `E` of the other battery are respectively A. `3 Omega,8V`B. `2 Omega,12 V`C. `2 Omega,24 V`D. `3 Omega,12 V` |
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Answer» Correct Answer - B Internal resistance `r=(l_(1)-l_(2))/(l_(2))xxR` `((L//2)-(5L//12))/(5L//12)xx10=2 Omega` Emf of cell By keeping `S_(2)` open `6=E/LxxL/2` `E=12 V` |
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| 831. |
In the given potentiometer circuit length of the wire `AB` is `3 m` and resistance is `R = 4.5 Omega`. The length `AC` for no deflection in galvanometer is A. `2m`B. `1.8 m`C. dependent on `r_(1)`D. None of these |
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Answer» Correct Answer - D Since cell connected in secondary circuit is in wrong way so we will not get balancing point. |
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| 832. |
The length of a potentiometer wire is `l`. A cell of emf `E` is balanced at a length l/3 from the positive end of the wire. If the length of the wire is increased by l/2. At what distance will the same cell give a balance point.A. `2l//3`B. `l//2`C. `l//6`D. `4l//3` |
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Answer» Correct Answer - B `E=x_(1)l_(1)=x_(2)l_(2)` `(E_(0))/lxxl/(3)=E_(0)/(3l//2)xxl_(2)impliesl_(2)=l/2` |
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| 833. |
In the circuit shown in figure, reading of voltmeter is `V_1` when only `S_1` is closed, reading of voltmeter is `V_2` when only `S_2` is closed, and reading of voltmeter is `V_3` when both `S_1 and S_2` are closed. Then . A. `V_(3) gt V_(2) gt V_(1)`B. `V_(2) gt V_(1) gt V_(3)`C. `V_(3) gt V_(1) gt V_(2)`D. `V_(1) gt V_(2) gt V_(3)` |
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Answer» Correct Answer - B (b) In series , potential difference `prop R` When only `S_(1)` is closed, `V_(1) = (3)/(4) E = 0.75 E` When only `S_(2)` is closed, `V_(2) = (6)/(7) E = 0.86 E` and when both `S_(1)` and `S_(2)` are closed combined resistance of `6R` and `3R` is `2R`. `V_(3) = ((2)/(3)) E = 0.67 E` Hence `V_(2) gt V_(1) gt V_(3)`. |
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| 834. |
In the circuit shown in figure , `V` must be A. 50 VB. 80 VC. 100 VD. 1290 V |
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Answer» Correct Answer - B |
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| 835. |
Current through XY of circuit shown is A. 1AB. 4AC. 2AD. 3A |
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Answer» Correct Answer - C |
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| 836. |
The resistance of an electrolyte varies with temperature `T` as `R_(T) = (R_(O))/(1 + alpha T)`, where `R_(0)` and are constants. A constant potential difference `V` is applied to the two electrodes (each of surface area `A`) which are dipped in this electrolyte. If `T_(0)` is the temperature of the surroundings, the loss of heat to the surroundings per unit surface area put unit time is governed by the same relation `H = k (T - T_(0)) A`, the steady state temperature isA. `((V^(2) + k A R_(O) T))/((k R_(O) A - alpha V^(2)))`B. `(V^(2))/(kR_(O) A - alpha V^(2))`C. `(V^(2))/(RkA)`D. None of these |
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Answer» Correct Answer - A (a) Power generated in the resistance is being lost in the environment. At steady state `(V^(2))/(R ) = k (T - T_(0)) A` Given `R = (R_(0))/(1 + alpha T) , (V^(2))/(R_(0)) (1 + alpha T) = k (T - T_(0)) A`, `T (- (V^(2) alpha)/(R_(0)) + k A) = (- (V^(2))/(R_(0)) + k AT_(0)), T = ((V^(2) + k AR_(0) T)/(kR_(0) A - alpha V^(2)))` |
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| 837. |
In the circuit diagram shown in figure, potential difference across `3Omega` resistance is 20 V. Then, match the following two columns. |
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Answer» Correct Answer - `(Ararrr;Brarrp;,Crarrp;Drarrs)` |
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| 838. |
In the circuit shown the cells `A` and `B` have negligible resistance. For `V_(A) = 12 V, R_(1) = 500 Omega` and `R = 100 Omega`, the galvanometer `(G)` shows no deflection. The value of `V_(B)` is A. 12VB. 6VC. 4VD. 2V |
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Answer» Correct Answer - 4 `V_(B)=`potential drop across R `rArr V_(B)=((R)/(R+R_(1)))V_(A)=((100)/(100+500))(12)=2V` |
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| 839. |
Potential diffference across the terminals of the batery shown in Figyre is `(r =` internal resistance of battey) A. `8V`B. `10V`C. `6V`D. Zero |
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Answer» Correct Answer - D |
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| 840. |
If voltage across a bulb rated 22 volt 100 watt drops by 2.5% of its rated value, the percentage of the rated value by which the power would decrease isA. 0.05B. 0.1C. 0.2D. 0.025 |
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Answer» Correct Answer - 0.01 `PpropV^(2)` So `(DeltaP)/(P)=2(DeltaV)/(V)=5%` |
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| 841. |
Assertion : Two 60 W bulbs are first connected in series and then in parallel with same battery. Total power produced in second case will become four times. Rated voltage is same for two bulbs. Reason : In series total power produced will be 30 W and in parallel 120 W.A. If both Assertion and Reason are true and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are true and Reason is not correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true. |
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Answer» Correct Answer - A |
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| 842. |
Consider the circuit shown in the figure. Both the circuits are taking same current from battery but current through `R` in the second circuit is `(1)/(10) th` of current through `R` in the first circuit. If `R` is `11 Omega`, the value of `R_(1)` A. `9.9 Omega`B. `11 Omega`C. `8.8 Omega`D. `7.7 Omega` |
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Answer» Correct Answer - A |
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| 843. |
Assertion : The `200 W` bulbs glows with more brightness then `100 W` bulbs. Reason : A `100 W` bulb has more resistance than a `200 W` bulb.A. If both assertion and reason are ture and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason both are false. |
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Answer» Correct Answer - A (a) The resistance, `R = (V^(2))/(P) implies R prop 1//P` i.e., higher is the wattage of a bulb, lesser is the resistance and so it will glow bright. |
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| 844. |
Assertion : Two resistance wires shown in figure are of same material. They have equal length. More heat is generated in wire A. Reason : In series `H prop R` and resistance of wire A is more.A. If both Assertion and Reason are true and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are true and Reason is not correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true. |
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Answer» Correct Answer - A |
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| 845. |
Two bulbs are fitted in a room in the domestic electric installation. If one of them glows brighter than the other, thenA. the brighter bulb has smaler resistanceB. the brighter bulb has larger resistanceC. both the bulbs have the same resistanceD. nothing can be said about the resistance unless other factors are known |
| Answer» Correct Answer - A | |
| 846. |
The potential difference across a conductor is doubled, the rate of generation of heat willA. become one fourthB. be halvedC. be doubledtimesD. become four times |
| Answer» Correct Answer - D | |
| 847. |
A conductor of length l and area of cross-section A has a resistance R then its specific resistance isA. `rho=(RA)/(l)`B. `rho=RA l`C. `rho=(l)/(RA)`D. `rho=(R^(2)A)/(l)` |
| Answer» Correct Answer - A | |
| 848. |
The resistance of the material of unit length and unit area of cross section isA. specific resistanceB. conductanceC. inductanceD. capacitance |
| Answer» Correct Answer - A | |
| 849. |
Consider the combination of resistor, The equivalent resistance between a and b isA. `(R )/(6)`B. `(2R)/(3)`C. `(R )/(3)`D. `3R` |
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Answer» Correct Answer - C |
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| 850. |
S.I. unit of specific resistance isA. `Omega cm`B. `Omega m`C. `Omega//m`D. `Omega//cm` |
| Answer» Correct Answer - B | |