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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.
| 2001. |
A wire has resistance of `10Omega`. If it is stretched by 1/10th of its length, then its resistance is nearlyA. `9Omega`B. `10Omega`C. `11Omega`D. `12Omega` |
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Answer» Correct Answer - D Resistance of wire `R = rho l//A =10 Omega,` New length, `l_(1)=l+l//10=11//10` `therefore` New are, `A_(1)=Al // l_(1)=(10A)/(11)` `therefore` New resistance, `R_(1)=rho l_(1)//A_(1)=rho(11//10)//(10//11)A` `= (121 rho l)/(100 A)=(121)/(100)xx10` `= 12.1 Omega` |
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| 2002. |
The resistance of a conductor increases withA. Increase in lengthB. Increase in temperatureC. Decrease in cross–sectional areaD. All of these |
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Answer» Correct Answer - D |
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| 2003. |
If `V_(A)-V_(B)=V_(0)` and the value of each resistance is R, then I. net resistance between AB is `(R)/(2)` II. Net resistance between AB is `(3R)/(5)` III. Current through CD is `(V_(0))/(R)` IV. current through EF is `(2V_(0))/(3R)` Which of the option/options are correct?A. I and IIB. I and IIIC. Only ID. All of these |
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Answer» Correct Answer - B |
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| 2004. |
A `3Omega` resistor and a `6Omega` resistor are connected in parallel and the combination is connected in series to a battery of 5 V and a `3Omega` resistor. The potential difference across the `6Omega` resistorA. `2V`B. `4V`C. `3V`D. `1V` |
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Answer» Correct Answer - A combination of resistors |
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| 2005. |
The resistance of a 5 cm long wire is `10Omega` . It is uniformly stretched so that its length becomes 20 cm . The resistance of the wire isA. `160 Omega`B. `80 Omega`C. `40 Omega`D. `20 Omega` |
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Answer» Correct Answer - A |
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| 2006. |
A wire of radius r has resistance R. If it is stretched to a radius of `(3r)/(4)`, its resistance becomesA. `(9R)/(16)`B. `(16R)/(9)`C. `(81R)/(256)`D. `(256R)/(81)` |
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Answer» Correct Answer - D |
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| 2007. |
When a wire is stretched and its radius becomes `r//2` then its resistance will beA. 16 RB. 4 RC. 2 RD. `R//2` |
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Answer» Correct Answer - A `(R_(2))/(R_(1))=((r_(1))/(r_(2)))^(4)` `R_(2)=(((r_(1))/(r_(1)))/(2))^(4)xxR_(1)=16.R_(1)=16R` |
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| 2008. |
n equal cell having emf E and internal resistance r, are connected in circuit of a resistance R. Same current flows in circuit either they connected in series or parallel, if:A. R=nrB. `R=r/n`C. `R=n^(2)r`D. R=r |
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Answer» Correct Answer - D Current when cells are connected in series `i_(s)=(nE)/(nr+R)` in parallel `i_(p)=(nE)/(r+nR)` According to question `i_(s)=i_(p)` `(nE)/(nr+R) =(nE)/(r+nR) rArr r+nR =nr+R rArr (n-1) R=r(n--1) rArr R=r` |
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| 2009. |
A heating element using nichrome connected to a 230 V supply draws a initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating elemtn if the room temperature is `27.0^(@)C`? Temperature coefficient of resistance of nichrome of nichrome averaged over the temperature range involved is `70xx10^(-4)C^(-1)` |
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Answer» Here, `R_(27) = 230//3.2 = 2300//32 Omega , R_(t) = 230//2.8 = 2300//28 Omega , alpha = 1.7 xx 10^(-4) ^@C^(-1)` `alpha = (R_(t)-R_(27))/(R_(27) xx ( t - 27)) or t - 27 = R_(t)-R_(27)/R_(27) xx alpha` or ` t= (R_(t)-R_(27))/(R_(27) xx alpha) + 27 = ((2300//28) - (2300//32))/((2300//32) xx 1.7 xx 10^(-4)) + 27 = 840.2 + 27 = 867.2 ^@C`. |
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| 2010. |
Two electric bulbs marked 25W-220V and 100W-220V are connected in series to a 440 V supply. Which of the bulbs will fuse ?A. bothB. 100WC. 25WD. neither |
| Answer» Correct Answer - C | |
| 2011. |
Two bulbs 25W, 220V and 100W, 220V are given. Which has higher resistance?A. 25 w bulbB. 100 W bulbC. both bulbs will have equal resistanceD. resistance of bulbs cannot be compared |
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Answer» Correct Answer - A Power of electric bulb `P=(V^(2))/R` So, resistance of electric bulb `R=(V^(2))/P` Given `P_(1)=25 W , P_(2)=100 W, V_(1)=V_(2)=200V` Therefore, for same potential difference `V, R prop 1/P` Thus, we observe that for minimum power, resistance will be maximum and vice-versa. hence, resistance of 25W bulb is maximum and 100 W bulb is minimum. |
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| 2012. |
A carbon and an aluminimum wire are connected in series.If the combination has resistance of `27` ohm at `0^(@)C`,what is the resistance of each wire at `0^(@)C` so that the resistance of the combination does not change with temperature ? `[alpha_(C ) = -0.5 xx 10^(-3)(.^@C)^(-1)` and `alpha_(A) = 4 xx 10^(-3)(.^@C)^-1]`. |
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Answer» The equivalent resistance in series `R = R_(Cu) + R_(A 1)` =`(R_(Cu))_0[1 + alpha_(Cu) Delta theta] + (R_(A1))_0[1 + alpha_(A1) Delta theta]` =`(R_(Cu))_0 +(R_(A1))_0 + [(R_(Cu))_0 alpha_(Cu) + (R_(A1))_0 alpha_(A1)]Delta theta` `(R_(Cu))_0 + (R_(A1))_0 = 27` ...(i) `(R_(Cu))_0 alpha_(Cu) + (R_(A1))alpha_(A1) =0` `4 xx 10^-3(R_(Cu))_0 -0.5 xx 10^-3(R_(A1))_0 = 0` `(R_(A1))_0 = 8 (R_(Cu))_0` ....(ii) Solving `(R_(Cu))_0 = 3 Omega, (R_(A1))_0 = 24 Omega`. |
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| 2013. |
Six resistors of each 2 ohm are connected as shown in the figure. The resultant resistance between A and B is.A. `4Omega`B. `2Omega`C. `1Omega`D. `10Omega` |
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Answer» Correct Answer - C K.V.L |
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| 2014. |
Six resistors each of 10 ohm are connected shown. The equivalent resistance between points X and Y is A. `20 Omega`B. `5 Omega`C. `25//3 Omega`D. `10 Omega` |
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Answer» Correct Answer - B |
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| 2015. |
If all the resistors shown have the value 2 ohm each, the equivalent resistance over AB is A. 2 ohmB. 4 ohmC. `1(2)/(3)ohm`D. `2(2)/(3)ohm` |
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Answer» Correct Answer - D |
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| 2016. |
An ideal gas mixture filled inside a balloon expands according to the relation `PV^(2//3)=` constant. What will be the temperature inside the balloonA. increasingB. decreasingC. constantD. Cannot be defined |
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Answer» Correct Answer - A |
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| 2017. |
A galvanometer has resistance `1 Omega`. Maximum deflection current through it is 0.1 AA. To measure a current of 1A a resistance of `(1)/(10)Omega` is put in parallel with galvanometerB. To measure a current of 1A a reistance of `(1)/(9) Omega` is put in parallel with galvanometerC. To measure a potential difference of 10 V a resistance of `99 Omega` is put in series with galvanometerD. To measure a potential difference of 10 V a resistance of `100 Omega` is put in series with galvanometer |
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Answer» Correct Answer - B::C |
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| 2018. |
`n` moles of a monoatomic gas undergoes a cyclic process ABCDA as shown. Process AB is isobaric, process BC is adiabatic process CD is isochoric and process DA is isothermal the maximum temperature and minimum temperature in cycle are `4T_(0)` and `T_(0)` respectively. Then:A. `T_(B)gt T_(C )gt T_(D)`B. heat is released by the gas in the process CDC. heat is supplied to the gas in the process ABD. total heat supplied to the gas is `2nRT_(0)ln(2)` |
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Answer» Correct Answer - A::B::C |
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| 2019. |
Two identical beakers with negligible thermal expansion are filled with water to the same level at `4^(@)C`. If one says `A` is heated while the other says `B` is cooled, then:A. water level in A must riseB. water level in B must riseC. water level in A must fallD. water level in B must fall |
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Answer» Correct Answer - A::B |
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| 2020. |
A process is shown in the diagram. Which of the following curves may represent the same process ? BC is rectangular hyperbola. A. B. C. D. |
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Answer» Correct Answer - C |
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| 2021. |
The temperature of the filament of an electric bulb is `2700^@C` when it glows. It is not burnt up at such a high temperature. Why? |
| Answer» It is so because the filament of the electric bulb has high melting point and it remains safe in an atmosphere of inert gases which protect its oxidation. | |
| 2022. |
If the electron in a Hydrogen atom makes `6.25xx10^(15)` revolutions in one second, the current isA. `1.12mA`B. `1mA`C. `1.25mA`D. `1.5mA` |
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Answer» Correct Answer - B `i=qf` |
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| 2023. |
Five resistor are connected as shown in the diagram. The equivalent resistance between `A` and `B` is A. 6 ohmB. 9 ohmC. 12 ohmD. 15 ohm |
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Answer» Correct Answer - A |
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| 2024. |
All bulbs consume same power. The resistance of bulb `1` is ` 36 Omega` . Answer the following questions: What is the resistance of bulb `3`?A. `4 Omega`B. `9 Omega`C. `12 Omega`D. `18 Omega` |
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Answer» Correct Answer - B |
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| 2025. |
The temperature drop through each layer of a two layer furnace wall is shown in Assume that the external temperature `T_(10` and `T_(3)` are mainitained constant and `T_(1)gtT_(3)` If the thichness of the layers `x_(1)` and `x_(2)` are the same which of the following statements are correct .A. `k_(1)gtk_(2)`B. `k_(1)ltk_(2)`C. `k_(1)=k_(2)` but heat flow through material (1) is larger then through(2)D. `k_(1)=k_(2)` but heat flow through material (1) is less than that through (2) |
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Answer» Correct Answer - A |
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| 2026. |
The temperature drop through a two layer furnace wall is `900^@C`. Each layer is of equal area of cross section. Which of the following actions will result in lowering the temperature `theta` of the interface?A. By increasing the thermal conductivity of outer layerB. By increasing thermal conductivity of inner layerC. By increasing thickness of outer layerD. By increasing thickness of inner layer |
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Answer» Correct Answer - A::D |
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| 2027. |
There are three voltmeters of the same range but of resistance `10000 Omega, 8000 Omega` and `4000 Omega` respectively. The best voltmeter among these is the one whose resistance isA. `10000Omega`B. `8000Omega`C. `4000Omega`D. all are equally good |
| Answer» Correct Answer - 1 | |
| 2028. |
There are three voltmeters of the same range but of resistance `10000 Omega, 8000 Omega` and `4000 Omega` respectively. The best voltmeter among these is the one whose resistance isA. `10000 Omega`B. `8000 Omega`C. `4000 Omega`D. All are equally good |
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Answer» Correct Answer - A |
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| 2029. |
8 cells, each of internal resistance `0.5 Omega` and emf 1.5V are used to send a current through an external ressistor of (a) `200 Omega` (b) `0.002 Omega` (c ) `1.0 Omega`. How would you arrange them to get the maximum current in each case? Find the value of current in each case. |
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Answer» Here, total number of cells= 8, `r=0.5 Omega , epsilon=1.5V`. (a) When `R=200 Omega` , then R gtgt r, so for maximum current, the cells are to be connected in series in circuit. Total internal resistance of 8 cells = 8r Current in circuit, `I= (8 epsilon)/(R +8r)= (8xx1.5)/(200+8 xx 0.5)=12.0/204 = 0.059A` (b) When `R=0.002 Omega` , then R ltlt r, so for maximum current, the cells are to be connected in parallel in circuit. Total internal resistance of 8 cells = `r//8` Total resistance of circuit `= R + r//8` `0.002 +0.5 //8 = 0.0645 Omega` Effective emf of all cells = emf of each cell 1.5V Current in circuit, `I=1.5/0.0645=23.26A` (c ) When `R= 1.0 Omega`, then R is comparable to r. For a mximum current, the cells are to be connected in mixed grouping. Let there be m rows of cells in parallel with n cells in sereis, in each row. Then, `m n=8` For maximum current, `R= (nr)/(m)` or `1.0 = nxx0.5/m or n=2 m` From (i), `m xx 2 m = 8 or m^(2) = 4 or m=2`. So `n = 2xx2=4` Thus, 4 cells in series in a row and 2 such rows of cells in parallel. Max. current, `I=( n epsilon)/(R + nr//m) = ( m n epsilon)/(mR + nr)` `= ( 8 xx 1.5)/(2 xx 1 + 4 xx 0.5)=3 A` |
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| 2030. |
A potentiometer is an ideal device of measuring potential difference becauseA. It uses a sensitive galvanometerB. It does not disturb the potential difference it measuresC. It is an elaborate arrangementD. It has a long wire hence heat developed is quickly radiated |
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Answer» Correct Answer - B |
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| 2031. |
A steady current flows in a metallic conductor of non-uniform cross-section i.e., `i` : constant. `A ne` constant.A. `j = (i)/(A), j` vary with area of `X-section`B. `j = sigma E`, `(i)/(A) = sigma E rArr (i)/(sigma A) = E` `E` : changes with `A`C. `j = n e v _d` `(i)/(A) = n e v_d` `v_d = (i)/(n e A), v_d` varies with `A`D. All options are correct |
| Answer» Correct Answer - D | |
| 2032. |
A wire of resistance `10 Omega` is drawn out so that its length is thrice its original length. Calculate its new resistance (resistivity and density of the material remain uncharged). |
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Answer» Correct Answer - `90 Omega` |
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| 2033. |
Potentiometer wire of length `1 m` is connected in series with `490 Omega` resistance and `2 V` battery. If `0.2 mV/cm` is the potential gradient, then resistance of the potentiameter wire is approximatelyA. `4.9 Omega`B. `7.9 Omega`C. `5.9 Omega`D. `6.9 Omega` |
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Answer» Correct Answer - A |
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| 2034. |
In the circuit shown in figure find the potentials of `A,B,C` and `D` and the current through `1Omega` and `2Omega` resistance. |
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Answer» Correct Answer - A::B::C::D `V_A=0V` (as it is earthed) `V_C-V_A=5V` `:. V_C=5V` `V_B-V_A=2V` `:. V_B=2V` `V_D-V_C=10V` `:. V_D=10+V_C=15V` `i_(2) Omega = (V_(C ) - V_(B))/(1) = 3A` from `C` to `B` as `V_(C ) gt V_(B)` `i_(1Omega)=(V_C-V_B)/2=7.5A` from `C` to `B` is `V_CgtV_B` `i_(2Omega)=(V_D-V_A)` |
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| 2035. |
If in the circuit shown below, the internal resistance of the battery is `1Omega` and `V_(P)` and `V_(Q)` are the potentials at P and Q respectively, the potential difference between te point P and Q isA. 9 VB. 11 VC. 7 VD. 6 V |
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Answer» Correct Answer - A `i=(V)/(R)` |
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| 2036. |
In the circuit shown below, the power developed in `6Omega` resistor is 6 watt. The power in watts developed in the `4Omega` resistor is A. 16B. 9C. 6D. 4 |
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Answer» Correct Answer - 2 Power `P=(V^(2))/(R)` `therefore (P_(1))/(P_(1))=(R_(2))/(R_(1))` or `(6)/(P_(2))=(4)/(6)` or `P_(2)=9` watt. |
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| 2037. |
Two cells of emf 4 V and 8 V are connected to two resistor `4Omega` and `6Omega` as shown. If 8 V cell is short circuited. Then current through resistance `4Omega` and `6Omega`A. 2AB. 1AC. 2.5AD. 3A |
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Answer» Correct Answer - B `V=iR` |
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| 2038. |
A meter bridge is set up as shown, to determine an unknown resistance `X` using a standard 10 ohm resistor. The galvanometer shows null point when tapping -key is at 52 cm mark. The end-corrections are 1 cm and 2 cm respectively for the ends A and B. The determine value of `X` is A. `10.2 ohm`B. `10.6 ohm`C. `10.8 ohm`D. `11.1 ohm` |
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Answer» Correct Answer - b Using the condition for balanced Wheatstone bridge, we get `(X)/(10) = ((52+1)"cm")/((100- 52+2)"cm") = (53)/(50)` or `X = (53 xx 10)/(50) = 10.6 Omega` |
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| 2039. |
A meter bridge is set up as shown, to determine an unknown resistance `X` using a standard 10 ohm resistor. The galvanometer shows null point when tapping -key is at 52 cm mark. The end-corrections are 1 cm and 2 cm respectively for the ends A and B. The determine value of `X` is A. `10.2` ohmB. `10.6` ohmC. `10.8` ohmD. `11.1` ohm |
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Answer» Correct Answer - B Apply condition of wheatstone bridge, `x/(52+1)=10/(48+2)impliesx=10/50xx53xxx10.6 Omega` |
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| 2040. |
A meter bridge is set up as shown, to determine an unknown resistance `X` using a standard 10 ohm resistor. The galvanometer shows null point when tapping -key is at 52 cm mark. The end-corrections are 1 cm and 2 cm respectively for the ends A and B. The determine value of `X` is A. `10.6 ohm`B. `10.8 ohm`C. `10.2 ohm`D. `11.1 ohm` |
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Answer» Correct Answer - A (a) Applying the condition of balanced Wheatstone bridge, after applying end correction we have, `(X)/(10 Omega) = ((52 + 1) cm)/((78 + 2) cm) = (53)/(50), X = 10 Omega xx (53)/(50) = 10.6 Omega` |
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| 2041. |
A `120 V` house circuit has the following light bulbs switched on : `40 W, 60 W` and `75 W`. Find the equivalent resistance of these bulbs. |
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Answer» Correct Answer - A::B `P=V^2/R` `:. R=V^2/P` `R_1=((120)^2)/40=360Omega` `R_2=((120)^2)/60=240Omega` `R_3=((120)^2)/75=192Omega` Now, all these resistors are in parallel. |
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| 2042. |
A meter bridge is set up is as shown determine as unknown resistance `X` using a standerd `5 ohm` resistor The galvanometer shown nall point when topping key is at `44cm` mark The end corections see `1cm` and `2cm` respectively for the ends `A` and `B` The determine value of `X` is A. `4.12 ohm`B. `3.88ohm`C. `4.22 ohm`D. `4.77 ohm` |
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Answer» Correct Answer - b Using the condition for balanced Wheatstone bridge we get `(X)/(5) = ((44+1))/((100- 44+2)) = (45)/(58)` or `X = (45 xx 5)/(58)= 3.88 Omega` |
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| 2043. |
In the given circuit a bridge is shown in the balanced state.The meter bridge wire has a resistance of `1 Omega cm^(-1)` .Calculate the unknown resistance `X` and the current drawn from the battery of negligible internal resistence. If the magnitude the position of galvanometer and the cell, how it will affect the position of the galvanometer ? |
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Answer» Correct Answer - ` 0.33A` Here resistance of arm `AB p = 30 xx 1 = 30 Omega` Resistance of arm `BC = Q = 70 xx 1 = 70 Omega` when bridge is balance then `(P)/(Q)= (X)/(Y) or X = (P)/(Q) Y = (30)/(70) xx 7 = 3 Omega ` When bridge is balance ,no current flow through galvanmeter arm .Now the effective resistance of arm `ADC = X = + Y = 3 + 7 = 10 Omega` Resistance of arm `ABC = 30 + 70 = 100 Omega` These two arms in parallel , the effective resistance of the circuit is ` R = (10 xx 100)/(10 + 100) = (100)/(11) Omega ` Currect draw from the battery `I = (3)/(100//11) = 0.33A` If the balancing position , we interchange the position of galvanometer and the cell , the bridge remains in balance wire .The galvanmeter show no deffective |
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| 2044. |
In the circuit a meter bridge a shown in the balanced state .The meter bridge wire has a resistance of `1 Omega //cm`. Calculate the unknown resistance `Y` and the current drawn from the battery of negiligible internal resistance |
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Answer» Correct Answer - ` 0.66 A` When is balance no current flow throgh galvameter so `(X)/(6) = (40)/(60) = (2)/(3) or X = (2)/(3) xx 6 = 4 Omega ` Total resistance of bridgh wire `AB = 1 xx 100 = 100 Omega` Total resistance of resistance `X (= 4 Omega and 6 Omega` connected in series `= 4 + 6 = 10 Omega ` The resistance is in parallel with the resistance of wire `AB` Effective resistance of circuit `R = (10 xx 100)/(10 + 100) = (100)/(11) Omega` current draw from the battery `I = (6)/(100//11) = 0.66 A` |
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| 2045. |
If resistance `R_1` in resistance box is `300 Omega` then the balanced length is found to be `75.0 cm` from end `A`. The diameter of unknown wire is 1 mm and length of the unknown wire is `31.4 cm`. Find the specific resistance of the unknown wire |
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Answer» `R/X=1/(100-l)` `rarr X=((100-l)/l)R` `=((100-75)/75) (300)=100Omega` Now, `X=(rhol)/A=(rhol)/((pid^2//4))` `rho=(pid^2X)/(4l)` `=((22//7)(10^-3)^2(100))/((4)(0.314))` `=2.5xx10^-4Omega-m` |
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| 2046. |
The charge on a capacitorofcapacitance `10muF` connected as shown in the figure-3.336 is: A. `20muC`B. `15muC`C. `10muC`D. Zero |
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Answer» Correct Answer - A |
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| 2047. |
The temperature of cold junction and neutral temperature of a thermocouple are `15^(@)C` and `280^(@)C` respectively. The temperature of inversion isA. `295^(@)C`B. `265^(@)C`C. `545^(@)C`D. `575^(@)C` |
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Answer» `theta_(n)-theta_(c)=theta_(1)-theta_(n)` `280-15=theta_(i)-280` `theta_(i)=560-15` `=545^(@)C`. |
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| 2048. |
In meter brigde of Wheatstone bridge for measurment of resistance, the known and the unknown resistance are interchanged. The error so removed isA. End correctionB. Index errorC. Due to temperature effectD. Random error |
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Answer» Correct Answer - A (a) In meter bridge experiment,it is assumed that the resistance of the `L` shaped plate is negligible, but actually it is not so, The error created due to this is called, end error. To remove this the resistance box and the unknown resistance must be interchanged and then the mean reading must be taken. |
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| 2049. |
If the temperature of cold junction of thermocouple is lowered, then the neutral temperatureA. increasesB. decreasesC. remains constantD. may increase or may decrease |
| Answer» Correct Answer - C | |
| 2050. |
When a resistance of `100Omega` is connected in series with a galvanometeer of resistance R, its range is V. To double its range, a resistance of `1000Omega` is connected in series. Find R.A. `700Omega`B. `800Omega`C. `900Omega`D. `100Omega` |
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Answer» Correct Answer - C When a resistance of `100 Omega` is connected in series current, `i=(V)/(100+R)` ….(i) When a resistance of `100 Omega` is connected in series, the its range double current, `i=(2V)/(1100+R)` …(ii) From Eqs. (i) and (ii), `(V)/(100+R)=(2V)/(1100+R)` `rArr " " R = 900 Omega` |
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