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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.
| 701. |
Assertion : Through the same current flows through the line wires and the filament of the bulb but heat produced in the filament is much higher then that in line wires. Reason : The filament of bulbs is made of a material of high resistance and high melting point.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) As filament of bulb and line wire are in series, hence current through both is same. Now because `H = (i^(2) Rt)/(4.2)` and resistance of the filament of the bulb is much higher than that in line wires. |
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| 702. |
Assertion : In practial application, power rating of resistance is not important. Reason : Property of resistance remains same even at high temperature.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 - D (d) Because of heat production every resistance has a maximum power rating, the maximum power than can be disspated without overheating the device. When this rating is exceded, heat is produced, due to which resistance may change unpredictably. |
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| 703. |
A tungsten filament bulb is connected to a variable voltage supply. The potential difference, `V` is varied and the current, `I` and steady temperature `T` of the filament is recorded. A graph is plotted for ln `(VI) V_(s) ln (T)`. Find the slope of the graph. Assume that temperature of filament `T gt gt T_(0) =` atmospheric temperature |
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Answer» Correct Answer - 4 |
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| 704. |
A capacitor is charged using an external battery with a resistance x in series. The dashed line showns the variation of In I with respect to time. If the resistance is changed to 2x, the new graph will be A. PB. QC. RD. S |
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Answer» Correct Answer - B (b) `I = (epsilon)/(R ) e^(-t//RC)` `log_(e) I = log_(e) (epsilon)/(R ) - (t)/(RC) log_(e) e implies 1n I = 1n ((epsilon)/(R )) = (t)/(RC)` Slope `= - (t)/(RC) y` intercept `= 1n((epsilo)/(r ))` If resistance is made doubled, both of the above decrease in magnetude. |
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| 705. |
In the given circuit, no current is passing through the galvanometer. If the cross-sectional diamter of `AB` is doubled then for null point of galvanomter the value of `AC` would A. `x`B. `x//2`C. `2x`D. None |
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Answer» Correct Answer - A Null point does not depends upon area of cross sections of wire `AB`. |
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| 706. |
If `10^(6)` electrons/s are flowing through an area of cross section of `10^(-4) m^(2)` then the current will be:-A. `1.6xx10^(-7)A`B. `1.6xx10^(-13)A`C. ` 1xx10^(-6)A`D. `1xx1^(2)A` |
| Answer» Correct Answer - 2 | |
| 707. |
A `1Omega` voltmeter has range 1V find the additional resistance which has to join in series in voltmeter to increase the range of voltmeter to 100V :-A. `10Omega`B. `(1)/(99)Omega`C. `99Omega`D. `100Omega` |
| Answer» Correct Answer - 3 | |
| 708. |
The terminal voltage is `(E)/(2)` when a c urrent of 2A is flowing through `2Omega` resistance, then the internal resistance of cell is:-A. `1Omega`B. `2Omega`C. `3Omega`D. `4Omega` |
| Answer» Correct Answer - 2 | |
| 709. |
How a battery is to be connected so that the shown rheostat will behave like a potential divider? Also indicate the points about which output can be taken. . |
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Answer» Correct Answer - Battery should be connected across `A` and `B`. Ouput can be taken across the terminals `A` and `C` or `B` and `C` Battery should be connected across `AB` output can be taken across the terminal `A` and `C` or `B` and `C`. |
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| 710. |
For the post office arrangement to determine the value of unknown resistance, the unknown resistance should be connected between. A. A and BB. B and CC. C and DD. A and D |
| Answer» Correct Answer - D | |
| 711. |
For the post office arrangement to determine the value of unknown resistance, the unknown resistance should be connected between. A. `B` and `C`B. `C` and `D`C. `A` and `D`D. `B_(1)` and `C_(1)` |
| Answer» Correct Answer - C | |
| 712. |
In the figure shown the current through `2 Omega` resistor is A. `2A`B. `0A`C. `4A`D. `6A` |
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Answer» Correct Answer - B Resistance of `2 Omega` does not have any return path so its current through it will be zero. |
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| 713. |
Potential difference between the points P and Q in the electric circuit shown is A. `4.5V`B. `1.2V`C. `2.4V`D. `2.88V` |
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Answer» Correct Answer - D |
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| 714. |
The current between B and D in the given figure is A. 1 ampB. 2 ampC. ZeroD. `0.5` amp |
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Answer» Correct Answer - C |
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| 715. |
A bridge circuit is shown in figure. The equivalent resistance between A and B will be A. `(14)/(3) Omega`B. `(3)/(14) Omega`C. `(9)/(14) Omega`D. `(14)/(9) Omega` |
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Answer» Correct Answer - A |
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| 716. |
A bridge circuit is shown in figure. The equivalent resistance between A and B will be A. `21Omega`B. `7Omega`C. `(252)/(85)Omega`D. `(14)/(3) Omega` |
| Answer» Correct Answer - D | |
| 717. |
Potentiometer measures the potential difference more accurately than a voltmeter, becauseA. it has a wire of high resistanceB. it has a wire of low resistanceC. it does not draw current from external circuitD. it draws a heavy current from external circuit |
| Answer» Correct Answer - C | |
| 718. |
Potentiometer measures the potential difference more accurately than a voltmeter, becauseA. it has a wire resistanceB. it has a wire of low resistanceC. it does not draw current from external circuitD. it draws a heavy current from external circuit |
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Answer» Correct Answer - A When we measure the emf of a cell by the potentiometer then no current flows in the circuit in zero-deflection condition i.e., cell is in open circuit. Thus, in this condition the actual value of a cell is found. In this way, potentiometer is equivalent to an ideal voltmeter of infinite resistance. |
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| 719. |
Potentiometer measures the potential difference more accurately than a voltmeter, becauseA. it does not draw current from external circuitB. it draws a heavy current from external circuitC. it has a wire of high resistanceD. it has a low resistance |
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Answer» Correct Answer - A |
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| 720. |
In the shown arrangement of the experiment of the meter bridge if AC corroesponding to null deflection of galvanometer is x, what would be its value if the radius of the wire AB is doubled? .A. `x`B. `(x)/(4)`C. `4x`D. `2x` |
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Answer» Correct Answer - A The ratio `(AC)/(CB)` will remain unchanged |
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| 721. |
What is the change in experiment if battery is connected between `B` and `D` and galvanometer is connected across `A` and `C` ? |
| Answer» Experiment can be done is similar manner but now `K_2` should be pressed first then `K_1`. | |
| 722. |
`R_1, R_2, R_3` are different values of `R, A, B` and `C` are the null points obtained corresponding to `R_1, R_2` and `R_3` respectively. For which resistor, the value of X will be the most accurate and why? |
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Answer» Correct Answer - A::B::C Slide wire bridge is most sensitive when the resistance of all the four arms of bridge is same. Hence , `B` is the most accurate answer. |
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| 723. |
In a balanced metre bridge, the segment of wire opposite to a resistance of `70Omega` is 70 cm. The unknown resistance isA. `30Omega`B. `60Omega`C. `90Omega`D. `15Omega` |
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Answer» Correct Answer - A `X=(l_(x))/(l_(r))R=(30xx70)/(70)=30 Omega` |
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| 724. |
A metal wire of resistivity `64 xx 10^-6 Omega cm` and length `198 cm` has a resistance of `7 Omega`. Calculate its radius.A. `2.4 cm`B. `0.24 cm`C. `0.024 cm`D. `24 cm` |
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Answer» Correct Answer - C |
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| 725. |
Find out the value of current through `2Omega` resistance for the given circuit. |
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Answer» Correct Answer - A `2Omega` resistor is in open circuit so current is 0 |
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| 726. |
Same mass of copper is drawn into 2 wires of 1 mm thick ad 3 mm thick. Two wires are connected in series and current is passed. Heat produced in the wires is the ratio ofA. `3:1`B. `9:1`C. `81:1`D. `1:81` |
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Answer» Correct Answer - C `Q=i^(2)Rt` `QalphaRalpha(1)/(A^(2))` when wire is stretched `(Q_(1))/(Q_(2))=(r_(2)^(4))/(r_(1)^(4))=(3^(4))/(1^(4))=81.1` |
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| 727. |
A copper wire of length `1 m` and radius is joined in series with an iron wire of length `2 m` and radius `3 mm` and a current is passed through the wire. The ratio of the current density in the copper and iron wires isA. `18 :1`B. `9 :1`C. `6:1`D. `2 :3` |
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Answer» Correct Answer - B |
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| 728. |
A copper wire of length `1 m` and radius is joined in series with an iron wire of length `2 m` and radius `3 mm` and a current is passed through the wire. The ratio of the current density in the copper and iron wires isA. `3//1`B. `1//3`C. `9//1`D. `1//9` |
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Answer» Correct Answer - C In series circuit, current is same `J=(I)/(A) prop (1)/(r^(2))` `:.J_(1)r_(1)^(2)= J_(2)r_(2)^(2) :.(J_(1))/(J_(2))=(9)/(1)` |
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| 729. |
Four identical resistances are joined as shown in the figure. The equivalent resistance between points A and B is `R_(1)` and that between A anc C is `R_(2)` then ratio `R_(1)//R_(2)` is A. `1:1`B. `4:3`C. `3:4`D. `1:2` |
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Answer» Correct Answer - C `(R_(AB))/(R_(AC))=(R_(1))/(R_(2))=((3xx1)/(3+1))/((2xx)/(2+2))=3/4` |
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| 730. |
A strip of copper and another of germanium are cooled from room temperature to 80 K. The resistance ofA. Each of these increasesB. Each of these decreasesC. Copper strip increases and that of germanium decreasesD. Copper strip decreases and that of germanium increases |
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Answer» Correct Answer - D |
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| 731. |
If n, e, `tau`, m, are representing electron density charge, relaxation time and mass of an electron respectively then the resistance of wire of length 1 and cross sectional area A is given byA. `(ml)/(n e^(2)tauA)`B. `(2mA)/(n e^(2)tau)`C. `n e^(2)tauA`D. `(n e^(2)tauA)/(2m)` |
| Answer» Correct Answer - A | |
| 732. |
Three resistors each of resistance 10 ohm are connected, in turn, to obtain (i) minimum resistance (ii) maximum resistance. Complete (a) The effective resistance in each case (b) ratio of minimum to maximum resistance so obtained. |
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Answer» Given, Resistance of each resistor `R = 10 Omega` no. of resistor, n = 3 (i) If three resistors are connected in parallel, we get minimum resistance. `:.` Minimum resistance `R_("min") = R_(P) = (R )/(n) = (10)/(3) Omega = 3.33 Omega` (ii) If three resistors are connected in series, we get maximum resistance. `:.` Maximum resistance `R_("max") = R_(s) = n R = 3 xx 10 = 30 Omega` The effective resistance to get maximum resistance `R_(eff) = nR = 3 xx 10 = 30 Omega` (in series) (b) `(R_("min"))/(R_("max")) = (((10)/(3)))/((3 xx 10)) = (10)/(90) :. (R_("min"))/(R_("max")) = (1)/(9)`. |
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| 733. |
Three resistances each of `3Omega` are connected as shown in fig. the resultant resistance between A and F isA. `9Omega`B. `2Omega`C. `4Omega`D. `1Omega` |
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Answer» Correct Answer - D The 3 resistances are parallel |
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| 734. |
Two wires made of same material have lengths in the ratio `1:2` and their volumes in the same ratio. The ratio of their resistances isA. `4:1`B. `2:1`C. `1:2`D. `1:4` |
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Answer» Correct Answer - D `Rpropl^(2)` `therefore` V constant |
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| 735. |
There are five equal resistors. The minimum resistance possible by their combination is 2 ohm. The maximum possible resistance we can make with them isA. 25 ohmB. 50 ohmC. 100 ohmD. 150 ohm |
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Answer» Correct Answer - B `(R)/(5)=2` `R_(max)=5R` `R_(min)=(R)/(5)` |
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| 736. |
Two wires made of same material have their electrical resistances in the ratio `1:4` if their lengths are in the ratio `1:2`, the ratio of their masses isA. `1:1`B. `1:8`C. `8:1`D. `2:1` |
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Answer» Correct Answer - A `Rprop(l^(2))/(m)` |
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| 737. |
In the part of circuit shown in figure-3.364 A. current will flow form A to BB. current may flow from A to BC. current will flow B to AD. the direction of curent will depend on r. |
| Answer» Correct Answer - B | |
| 738. |
A resistor of resistance R is connected to a cell internal resistance `5 Omega`. The value of R is varied from ` 1 Omega` to `5 Omega`. The power consumed by RA. increases then increasesB. decreases continuouslyC. first decreases then increasesD. first increases then decreases. |
| Answer» Correct Answer - A | |
| 739. |
There are n similar conductors each of resistance R . The resultant resistance comes out to be x when connected in parallel. If they are connected in series, the resistance comes out to beA. `x//n^(2)`B. `n^(2)x`C. `x//n`D. nx |
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Answer» Correct Answer - B |
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| 740. |
Equivalent resistance between A and B will be A. 2 ohmB. 18 ohmC. 6 ohmD. 3.6 ohm |
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Answer» Correct Answer - D |
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| 741. |
A wire has resistance 12 ohms. If it is bent in the form of a equilateral triangle. The resistance between any two terminals in ohms is:A. 9 ohmsB. 12 ohmsC. 6 ohmsD. `8//3` ohms |
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Answer» Correct Answer - D |
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| 742. |
A charge is moving across a junction, thenA. momentum will be conservedB. momentum will not be conservedC. at some places momentum be conserved and at other places momentum will not be conservedD. none of these. |
| Answer» Correct Answer - D | |
| 743. |
The direction of the flow of current through electric circuit isA. from low potential to high potential.B. from high potential to low potential.C. does not depend upon potential valueD. current cannot flow through circuit. |
| Answer» Correct Answer - B | |
| 744. |
The resistivity of alloy manganin isA. Nearly independent of temperatureB. Increases rapidly with increase in temperatureC. Decrease with increase in temperatureD. Increase rapidly with decrease in temperature |
| Answer» Correct Answer - A | |
| 745. |
The p.d between the terminals A & B isA. 2 VB. 3.6 VC. 1.8 VD. 2.8 |
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Answer» Correct Answer - C `i=((E_(1))/(r_(1))+(E_(2))/(r_(2)))/(1+R((1)/(r_(1))+(1)/(r_(2))))=((5)/(20)+(2)/(10))/(1+10((1)/(20)+(1)/(10)))` `V=iR=ixx10=1.8v` |
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| 746. |
The potential drop across the `3 Omega` resistor is A. `1 V`B. `1.5 V`C. `2 V`D. `3 V` |
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Answer» Correct Answer - A (a) Equivalent resistance `R = 4 + (3 xx 6 )/(3 + 6) = 6 Omega` and main current `i= (E)/(R ) = (3)/(6) = 0.5 A` Now potential difference across the combination of `3 Omega` and `6 Omega, V = 0.5 xx ((3 xx 6)/(3 + 6)) = 1`volt The same potential difference, also develops across `3 Omega` resistance. |
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| 747. |
An electric bulb is marked to `100 W, 220 V`. If the supply voltage drops to `115 V`, what is the heat end light energy produced by the bulb in `20` minutes. Calculate the current flowing through it. |
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Answer» Correct Answer - `30 kJ, (5)/(23)A` Resistance of the bulb `R = ((230)^(2))/(100) = 529 Omega` when the voltage drops to `V = 115V` , the total heat and light energy produced by the bulb in `20 min will be `H = (V^(2))/(R ) l = (115)^(2))/(529) xx (20 xx 60) = 30000 J = 30kJ` Current `l = (V)/(R ) = (115)/(529) = (5)/(23)A` |
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| 748. |
Two electric bulbs (60 W and 100 W respectively) are connected in series. The current passing through them isA. more in 100 W bulbB. More in 60W bulbC. Same in bothD. none of these |
| Answer» In series combination , same current passes through all components. | |
| 749. |
The coil of a 1 000W electric heater is cut into two equal parts. If the two parts are used separately as heaters, their combined wattage willA. 500WB. 1000WC. 2000WD. 4000W |
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Answer» Power `=(V^(2))/(R)` where R denotes the resistance of the coil `therefore R=(V^(2))/(P)=(V^(2))/(1000)Omega` Resistance of each part `=(R)/(2)=(V^(2))/(2000)Omega` `therefore` power of each part `=(V^(2)xx2000)/(V^(2))` =2000W `therefore` combined wattage `=2xx200=4000` W |
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| 750. |
A heating coil of `2000 W` is immersed in water . How much time will it take in raising the temperature of `1 L` of water from `4^(@) C "to" 100^(@) C`? Only `80%` of the thermal energy produced is used in raising the temperature of water. |
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Answer» Here, `P = 2000 W, t = ?` (in seconds) Volume of water = 1 litre = 1000 c.c. Mass of water, m = volume `xx` density `1000 xx 1 = 1000` gram Rise in temperature, ` = theta_(2) - theta_(1) = 100 - 4` `96^@C` We know sp. Heat of water, `s= 1 cal g^(-1) @C^(-1)` `:.` Heat taken by water `= ms (theta_(2) - theta_(1))` `= 1000 xx 1 xx 96 = 96000` cal. Energy spent in heating coil = pt `= 2000 xx t` Useful thermal energy produced `= 80 %` ` = 2000 xx t xx 80/100 = ( 2000 xx t xx 80)/( 100 xx 4.2)` cal As this thermal energy is taken by water, therefore, `( 2000 xx t xx 80)/(100 xx 4.2) = 96000` or `t (96000 xx 100 xx 4.2)/(2000 xx 80)` `= 252 seconds`. |
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