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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.
| 851. |
A wire has a resistance R. Find new resistance, (i) if radius of cross-section of a cylindrical wire is doubled, then find ratio of initial to final resistance. (ii) if length of wire is increased by `10%`, then find the percentage increase in its resistance. (iii) if length of wire is increased by `20%`, then find the percentage increase in its resistance. A. 0.1B. 0.2C. 0.4D. 0.21 |
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Answer» Correct Answer - D `l_(2)=1.1 l_(1),R_(2)-R_(1)=?` `(R_(2))/(R_(1))=((l_(2))/(l_(1)))^(2)` `R_(2)=((1.1 l_(1))/(l_(1)))^(2)xxR_(1)=1.21R_(1)` Now increase in resistance `=R_(2)-R_(1)` and % increase in resistance `=((R_(2)-R_(1))/(R_(1)))xx100` `=((1.21R_(1)-R_(1))/(R_(1)))xx100` `=21%` |
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| 852. |
A metal wire of circular cross-section has a resistance `R_(1)`. The wire is now stretched without breaking, so that its length is doubled and the density is assumed to remain the same. If the resistance of the wire now becomes `R_(2)`, then `R_(2) : R_(1)` isA. `1 : 1`B. `1 : 2`C. `4 : 1`D. `1 : 4` |
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Answer» Correct Answer - C |
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| 853. |
When the length and area of cross-section both are doubled, then its resistanceA. halvedB. unchangedC. doubledD. none of these |
| Answer» Correct Answer - B | |
| 854. |
A metallic block has no potential difference applid across it, then the mean velocity of free electrons is `(T=` absolute temperature of the block)A. Proportional to TB. Proportional to .`sqrt(T)`C. ZeroD. Finite but independent of temperature |
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Answer» Correct Answer - A |
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| 855. |
The resistance of a wire is `20 ohm`. It is so stretched that the length becomes three times, then the new resistance of the wire will beA. `6.67 ohms`B. `60.0 ohms`C. `120 ohms`D. `180.0 ohms` |
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Answer» Correct Answer - D |
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| 856. |
When the length and area of cross-section both are doubled, then its resistanceA. will become halfB. will be doubledC. will remain the sameD. will become four time |
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Answer» Correct Answer - C (c ) `R_(1) prop (1)/(A) implies R_(2) prop (21)/(2A)` i.e., `R_(2) prop (1)/(A)` `:. R_(1) = R_(2)` |
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| 857. |
When the length and area of cross-section both are doubled, then its resistanceA. Will become halfB. Will be doubledC. Will remain the sameD. Will become four times |
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Answer» Correct Answer - C |
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| 858. |
Metal haveA. Zero resistivityB. High resistivityC. Low resistivityD. Infinite resistivity |
| Answer» Correct Answer - A | |
| 859. |
A metallic block has no potential difference applid across it, then the mean velocity of free electrons is `(T=` absolute temperature of the block)A. Propotional to `T`B. Proportional to `sqrt(T)`C. zeroD. Finite but independent of temperature. |
| Answer» Correct Answer - B | |
| 860. |
A wire of resistance 0.1 `cm^(-1)` bent to form a square ABCD of side 10 cm. A simailar wire is connected between the corners B and D to form the diagonal BD. Find the effective resistance of this combination between cornaers A and C. If a 2V battery of negligible internal resistance is connected across A and C calculate the total power dissipated.A. `1Omega,3W`B. `1Omega, 4W`C. `2Omega, 3 W`D. `2 Omega, 4W` |
| Answer» Correct Answer - B | |
| 861. |
Current through XY of circuit shown is A. 1AB. 4AC. 2AD. 3A |
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Answer» Correct Answer - C |
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| 862. |
When a resistance of 2 ohm is connected across terminals of a cell, the current is `0.5 A`. When the resistance is increased to 5 ohm, the current is `0.25 A` The e.m.f. of the cell isA. 1.0VB. 1.5VC. 2.0VD. 2.5V |
| Answer» Correct Answer - 2 | |
| 863. |
When connected across the terminals of a cell, a voltmeter measures 5 V and a connected ammeter measures 10 A of current. A resistance of 2 ohm s is connected across the terminals of the cell. The current flowing through this resistance will beA. 7.5AB. 5.0AC. 2.5AD. 2.0A |
| Answer» Correct Answer - 4 | |
| 864. |
A moving coil galvanometer has 150 equal divisions. Its current sensitivity is 2 divisions per millivolt. In ordre that each division reads 1 volt, the resistance in ohms needed to be connected in series with the coil will beA. 99995B. 9995C. `10^(3)`D. `10^(5)` |
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Answer» Correct Answer - B (b) Voltage sensitivity `= ("Current sensitvity")/("Resistance of galvanometer G")` `implies G = (10)/(2) = 5 Omega` Here `i_(g) =` Full scale deflection current `= (150)/(10) = 15 mA`. `V = ` Voltage to be measured ` = 150 xx 1 = 150 V` Hence `R = (V)/(i_(g)) - G = (150)/(15 xx 10^(-3)) - 5 = 9995 Omega` |
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| 865. |
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 `porp 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 `3 R` is `2 R` `:. V_(3) = ((2)/(3)) E = 0.67 E implies V_(2) gt V_(1) gt V_(3)` |
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| 866. |
A brass disc and a carbon disc of same radius are assembled alternatively to make a cylindrical conductor. The resistance of the cylinder is independent of the temperature. The ratio of thickness of the brass disc to that of the carbon disc is [`alpha` is temperature coefficient of resistance & Neglect linear expansion]A. `|(alpha_(C)rho_(C))/(alpha_(B)rho_(B))|`B. `|(alpha_(C)rho_(B))/(alpha_(B)rho_(C))|`C. `|(alpha_(B)rho_(C))/(alpha_(C)rho_(B))|`D. `|(alpha_(B)rho_(B))/(alpha_(C)rho_(C))|` |
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Answer» Correct Answer - A `DeltaR_(B)+DeltaR_(C)=0` `R_(B)alpha_(B)Deltatheta+R_(C)alpha_(C)Delta theta=0` `(rho_(B)t_(B)alpha_(B))/(A)+(rho_(C)t_(C)alpha_(C))/A=0 implies (t_(B))/(t_(C))=(rho_(C)alpha_(C))/(rho_(B)alpha_(B))` |
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| 867. |
The equivalent resistance between the points A and B is: A. `(36)/(7)Omega`B. `10 Omega`C. `(85)/(7)Omega`D. `18 Omega` |
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Answer» Correct Answer - C |
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| 868. |
In the figure shown, each resistance is R. Match the following |
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Answer» Correct Answer - `(A)Q, (B)P,(C)R` |
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| 869. |
Six batteris of increasing emf and increasing internal resistance are as shown in figure. Match the following |
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Answer» Correct Answer - `(A)P, (B)P,(C)P,(D)P` |
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| 870. |
In the potentiometer arrangement shown in figure null point is obtained l. match the following |
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Answer» Correct Answer - `(A)Q, (B)P,(C)P,` |
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| 871. |
In the figure shown the current flowing `2 R` is: A. `(V)/(2R)`B. from right to leftC. both are correctD. both are wrong |
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Answer» Correct Answer - C |
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| 872. |
A current in a circuit is due to a potential difference of 30 V applied to a resistor of resistance `300 Omega`. What resistance would permit the same current to flow, if the supply voltage was 300 V ?A. `3 k Omega`B. `6 k Omega`C. `9 k Omega`D. `300 Omega` |
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Answer» Correct Answer - A When V = 30 V and `R = 300 Omega` The current is `I=(V)/(R )=(30)/(300)=0.1A` If now the voltage is increased to 300 V and same current has to be maintained, then the resistance required given by `R=(V)/(I) rArr R = (300)/(0.1)=3 k Omega` |
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| 873. |
In the figure, a carbon resistor has band of different colours on its body. The resistanceof the following body is A. `2.2 k Omega`B. `3.3 k Omega`C. `5.6 k Omega`D. `9.1 k Omega` |
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Answer» Correct Answer - D According to colours indicated in the figure we have First band = White = 9 Second band = Brown = 1 Third band = Red `= 10^(2)` `R = 91xx10^(2)pm 10%` `R = 9.1 k Omega` |
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| 874. |
Find equivalent resistance between A and B A. `(R)/(3)`B. `(R)/(4)`C. `R`D. `(R)/(2)` |
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Answer» Correct Answer - B |
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| 875. |
RMS speed of a monoatomic gas is increased by 2 times. If the process is done adiabatically then the ratio of initial volume to final volume will beA. 4B. `(4)^(2//3)`C. `2^(3//2)`D. 8 |
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Answer» Correct Answer - D |
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| 876. |
A piece of copper wire has a resistance of `25 Omega` at `10^(@)C`. What is the maximum operating temperature, if the resistance of the wire is to be increased by 20 % ? (Assume `alpha` at `10^(@)C=0.0041// .^(@)C^(-1)`).A. `60.38^(@)C`B. `58.78^(@)C`C. `40.73^(@)C`D. `20.23^(@)C` |
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Answer» Correct Answer - B `R_(1)=50 Omega, t_(1)=10^(@)C, alpha_(1)=0.0041^(@)C` `R_(2)=50+0.2xx50=60 Omega` `t_(2)` = the unknown temperature at which R will be `60 Omega`. Since, `" " R_(t_(2))=R_(t_(1))[1+alpha_(t_(1))(t_(2)-t_(1))]` Here, `" " R_(2)=R_(1)[1+alpha_(t_(1))(t_(2)-t_(1))]` `" " 60=50[1+0.0041xx(t_(2)-10)]` `rArr " " t_(2)-10=((60)/(50)-1)xx(1)/(0.0041)` or `" " t_(2)=58.78 .^(@)C` |
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| 877. |
The value of equivalent resistance for the circuit shown below is, A. `2.123 Omega`B. `5.123 Omega`C. `4.23 Omega`D. `6.283 Omega` |
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Answer» Correct Answer - B Step I Find out the resistance between the nodes P and Q. `R_(eq_(1))=R_(1)+(1)/(((1)/(R_(2))+(1)/(R_(3))))` or `R_(eq_(1))=1+(1)/(((1)/(2)+(1)/(3)))` or `R_(eq_(1))=1+(2xx3)/(5)` or `R_(eq_(1))=2.2 Omega` Step II Now, find the resistance between nodes Q and T. `R_(eq_(2))=R_(4)+(1)/(((1)/(R_(5))+(1)/(R_(6))+(1)/(R_(7))))+R_(8)` `= 1+(1)/(((1)/(2)+(1)/(3)+(1)/(4)))+1` `= 1+0.923+1` `rArr R_(eq_(2))=2.923 Omega` Step III As from circuit diagram, the equivalent resistance must be the series combination of `R_(eq_(1))` and `R_(eq_(2))`. `therefore " " R_(eq)=R_(eq_(1))+R_(eq_(2))` `= 2.2+2.923` `R_(eq)=5.123 Omega` |
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| 878. |
The resistance of eureka wire is `2.5 Omega`. What is the value of specific resistance of write of 14 m length and diameter of 0.14 cm ?A. `27.5xx10^(-6)Omega - cm`B. `20.6xx10^(-6)Omega - cm`C. `25.3xx10^(-6)Omega - cm`D. None of the above |
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Answer» Correct Answer - A Length of wire, `l=14 mxx100=1400 cm` Diameter of wire = 0.14 cm `therefore " "` Radius of wire `r=(0.14)/(2)=0.07 cm` Cross-sectionl area of wire `= pi r^(2)=pi(0.07)^(2)` Resistance of wire `= 2.5 Omega` By formula, `R = rho (l)/(A)` `rArr " " rho=(AR)/(l)=(pixx(0.07)^(2)xx2.5)/(1400)` `" " = (pixx0.07xx0.07xx2.5)/(1400)` or `" " rho = 27.5xx10^(-6)Omega - cm` |
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| 879. |
An ideal gas is taken from the state A (pressure p, volume V) to the state B (pressure `p/2`, volume 2V) along a straight line path in the p-V diagram. Select the correct statement(s) from the following.A. The work done by the gas in the process A to B exceeds the work that would be done by it if the system were taken from A to B along an isothermB. In the T-V diagram, the path AB becomes a part of a parabolaC. In the p-T diageam, the AB becomes a part of a hyperbolaD. In going from A to B, the temperature T of the gas decreases |
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Answer» Correct Answer - A::B::D |
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| 880. |
A wire is broken in four equal parts. A packet is formed by keeping the four wires together. The resistance of the packet in comparison to the resistance of the wire will beA. EqualB. One fourthC. One eightD. `(1)/(16)` th |
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Answer» Correct Answer - D |
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| 881. |
Four resistances are connected in a circuit in the given figure. The electric cirrent flowing through `4 ohm` and `6 ohm` resistance is respectively A. 2 amp and 4 ampB. 1 amp and 2 ampC. 1 amp and 1 ampD. 2 amp and 2 amp |
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Answer» Correct Answer - D |
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| 882. |
Four resistances are connected in a circuit in the given figure. The electric cirrent flowing through `4 ohm` and `6 ohm` resistance is respectively A. 2A and 4AB. 1A and 2AC. 1A and 1AD. 2A and 2A |
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Answer» Correct Answer - D Resistance of `4 Omega` and `4 Omega` are in parallel, their equivalent resistance `= 2 Omega` Similarly, equivalent resistance of `6 Omega` and `6Omega = 3 Omega` Now `2 Omega` and `3 Omega` are in series, hence their total resistance `R = 2+3=5 Omega` From `V = iR rArr i = (V)/(R )=(20)/(5)=4 A` This current will be equally divided in `4 Omega` and `6 Omega` resistance So, current through `4 Omega` and `6 Omega` resistance is 2A respectively. |
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| 883. |
A group of identical cells (all in parallel) are connected to an external resistance `R_(1)`. Current through `R_(1)` is `i_(1)`. Another group of identical cells (all in series) are connected to some other resistance `R_(2)`. Current through `R_(2)` is `i_(2)`. Now one cell is removed from both the groups. ThenA. `i_(1)` will decreaseB. `i_(2)` may decrease or increase depending on the value E and rC. both (a) and (b) are correctD. both (a) and (b) wrong |
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Answer» Correct Answer - A |
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| 884. |
In the circuit shown, the potential difference between x and y will be A. zeroB. 120 VC. 60 VD. 20 V |
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Answer» Correct Answer - B |
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| 885. |
The current in the adjoining circuit will be A. `(1)/(45) ampere`B. `(1)/(15) ampere`C. `(1)/(10) ampere`D. `(1)/(5) ampere` |
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Answer» Correct Answer - C (c ) `R_(e q u i v a l e n t) = ((30 + 30) 30)/((30 + 30) 30) = (60 xx 30)/(90) = 20 Omega` `:. i= (V)/(R ) = (2)/(20) = (1)/(10)` ampere |
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| 886. |
For the circuit shown in figure, A. resistance `R=46Omega`B. current through `20 Omega` resistance is `0.1A`C. potential difference across the middle resistance is 2 VD. All of the above are true |
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Answer» Correct Answer - D |
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| 887. |
The reading of the ammeter as per figure shown is A. `(1)/(2)A`B. `(3)/(4)A`C. `(1)/(8)A`D. `2A` |
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Answer» Correct Answer - B |
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| 888. |
A current of `2 A` flows in a system of conductor as shown. The potential difference `(V_(A) - V_(B))` A. `+ 2 V`B. `+ 1 V`C. `- 1 V`D. `-2 V` |
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Answer» Correct Answer - B (b) Current through each arm `DAC` and `DBC = 1 A` `V_(D) - V_(A) = 2` and `V_(D) - V_(B) = 3 implies V_(A) - V_(B) = + 1V` |
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| 889. |
A current of `2 A` flows in a system of conductor as shown. The potential difference `(V_(A) - V_(B))` A. `-1V`B. `+1V`C. `-2V`D. `+2V` |
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Answer» Correct Answer - D |
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| 890. |
A current of `2 A` flows in a system of conductor as shown. The potential difference `(V_(A) - V_(B))` A. `+1V`B. `-1V`C. `+2V`D. `-2V` |
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Answer» Correct Answer - A |
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| 891. |
A box with two terminals is connected in series with a 2 V battery, an ammeter and a switch. When the switch is closed the needle of the ammeter moves quickly across the scale and drops back to zero. The box containsA. `20 Omega` resistorB. a strip of copperC. a diodeD. a short length of fuse wire |
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Answer» Correct Answer - D When a box contains fuse wire, a strong current flows through fuse when circuit is closed. Due to which a very strong heating effect takes place, resisting the breakage of fuse wire. |
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| 892. |
In the meter bridge experiment, the length Ab of the wire is 1m. The resistors X and Y have values `5Omega` and `2Omega` respectively. When a shunt resistance S is connected to X, the balacing point is found to be 0.625 m from A. Then the resistance of the shunt isA. `5Omega`B. `10Omega`C. `7.5Omega`D. `12.5Omega` |
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Answer» Correct Answer - B `(X)/(R)=(l)/(100-l)` |
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| 893. |
A new flashlight cell of e.m.f. 1.5 volts given a current of 15 amps. When connected directly to an ammeter of resistance `0.04 Omega`. The internal resistance of cell isA. `0.04 Omega`B. `0.06 Omega`C. `0.10 Omega`D. `10 Omega` |
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Answer» Correct Answer - B |
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| 894. |
See the electrical circuit shown in this figure. Which of the following equations is the correct equation for it ? A. `epsilon_(1)-(i_(1)+i_(2))R+i_(1)r_(1)=0`B. `epsilon_(1)-(i_(1)+i_(2))R-i_(1)r_(1)=0`C. `epsilon_(2)-i_(2)r_(2)-epsilon_(1)-i_(1)r_(1)=0`D. `-epsilon_(2)-(i_(1)+i_(2))R+i_(2)r_(2)=0` |
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Answer» Correct Answer - 2 ` epsilon_(2)=i_(2)r_(2)+(i_(1)+i_(2))R` `epsilon_(1)=i_(1)r_(1)+(i_(1)+i_(2))R` ` epsilon_(2)-epsilon_(1)=i_(2)r_(2)-i_(1)r_(1)` |
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| 895. |
See the electrical circuit shown in this figure. Which of the following equations is the correct equation for it ? A. `epsi_(1)(i_(1)+i_(2))R-i_(1)r_(1)=0`B. `epsi_(2)-i_(2)r_(2)-epsi_(1)-i_(1)r_(1)=0`C. `epsi_(2)(i_(1)+i_(2))R+i_(2)r_(2)=0`D. `epsi_(1)(i_(1)+i_(2))R-i_(1)r_(1)=0` |
| Answer» Correct Answer - A | |
| 896. |
The V - I graphs for two resistors and their series combination are shown in Fig 5.23. Which one of these graphs represents the series combination of the two resistors? Given reason for your answer. |
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Answer» As slope of v-I graph `= V/I`= resistance R. From the given graphs, the slope of B is greater than that of A. So the resistance of B is greater than that of A. As resistance in series combination is more and in parallel combination is less, so B represents series combination and A represents parallel combination of resistors. |
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| 897. |
In the network shown in figure, the ring has zero resistance. Find the resistance between A and B. |
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Answer» Here, three resistance each of value 6 R are in parallel and resistance R is in series between A and B. The equivalent circuit is shown in figure. The effective resistance of three resistances in parallel, `1/R = (1)/(6R) + (1)/(6R) +(1)/(6R) = (1)/(2R)` `:. R_(p) = 2R` Total resistance between A and B `= 2 R + R= 3R` |
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| 898. |
Two resistance `R_(1)` and `R_(2)` are joined as shown in figure to two batteries of emf `E_(1)` and `E_(2)`. If `E_(2)` is short circuited, what is the current through `R_(1)` ? |
| Answer» When `E_(2)` is short circuited, it means positive pole of battery `E_(2)` is connected to negative pole of battery `E_(2)`. The battery gets discharged. The resistance of the arm become zero. The current from cell `E_(1)`, flowing through `R_(1)` will not pass through `R_(2)`, but will go through the short circuited path. Therefore, `R_(2)` becomes inffective. Now current through `R_(1)` is due to emf `E_(1)` with resistance `R_(1)` in circuit, i.e., `I= E_(1)//R_(1)` | |
| 899. |
If six identical cells each having an e.m.f. of 6 V are connected in parallel, the e.m.f. of the combination isA. `1V`B. `36V`C. `(1)/(6)V`D. `6V` |
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Answer» Correct Answer - D |
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| 900. |
The number of dry cells, each of e.m.f. `1.5` volt and internal resistance `0.5 omega` that must be joined in series with a resistance of `20` ohm so as to send a current of `0.6` ampere through the circuit is -A. 2B. 8C. 10D. 12 |
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Answer» Correct Answer - C |
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