2492 + Interview Questions in CURRENT-ELECTRICITY IN GENERAL AWARENESS Page 47 InterviewSolution

2301.	In a neon discharge tube `2.9 xx 10^(18) Ne^(+)` ions move to the right each second while `1.2 xx 10^(18)` eletrons move to the left per second. Electron charge is `1.6 xx 10^(-9) C`. The current in the discharge tubeA. 1A towards rightB. `0.66A` towards rightC. `0.66A` towards leftD. Zero
Answer» Correct Answer - B

Discussion

2302.	The resistance of a discharge tube isA. OhmicB. Non-ohmicC. Both (a) and (b)D. Zero
Answer» Correct Answer - B

Discussion

2303.	In a neon discharge tube `2.9 xx 10^(18) Ne^(+)` ions move to the right each second while `1.2 xx 10^(18)` eletrons move to the left per second. Electron charge is `1.6 xx 10^(-9) C`. The current in the discharge tubeA. 0.27 A towards rightB. 0.66 A towards rightC. 0.66 A towards leftD. zero
Answer» Correct Answer - B Given, number of ions, i.e., `n_(1)=2.9xx10^(18)` `n_(2)=1.2xx10^(18)` Charge of an electron, i.e., `q=1.6xx10^(-19)C` So,net electric current, i.e, `l=((n_(1))/(t)+(n_(2))/(t))xxq` `= ((2.9xx10^(18))/(1)+(1.2xx10^(18))/(1))xx1.6xx10^(-19)=0.66 A` The net electric current 0.66 towards right.

Discussion

2304.	In the circuit `P != R, ` the reading of the galvanometer is same with switch S open or closed. Then A. `I_(R)=I_(G)`B. `I_(P)=I_(G)`C. `I_(Q)=I_(G)`D. `I_(Q)=I_(R)`
Answer» Correct Answer - A As there is no change in the reading of galvanometer with switch S open or closed. It implies that bridge is balanced current through S is zero and `I_(R)=I_(G),I_(P)=I_(Q)`

Discussion

2305.	In the circuit `P != R, ` the reading of the galvanometer is same with switch S open or closed. Then A. `I_(R)=I_(G)`B. `I_(P)=I_(G)`C. `I_(Q)=I_(G)`D. `I_(Q)=I_(R)`
Answer» Correct Answer - A

Discussion

2306.	In the following Wheatstone bridge `P//Q = R//S`. If key `K` is closed, then the galvanometer will show deflection A. In left sideB. In right sideC. No deflectionD. In either side
Answer» Correct Answer - D

Discussion

2307.	In the circuit `P != R, ` the reading of the galvanometer is same with switch S open or closed. Then A. `I_(R) = I_(G)`B. `I_(P) = I_(G)`C. `I_(Q) = I_(G)`D. `I_(Q) = I_(R)`
Answer» Correct Answer - A (a) Reading of galvanometer remains same whether switch `S` is open or closed, hence no current will flow through the switch i.e., `R` and `G` will be in series and same current will flow through them. `I_(R) = I_(G)`.

Discussion

2308.	A wire connected in the left gap of a meter bridge balance a `10Omega` resistance in the right gap to a point, which divides the bridge wire in the ratio 3:2. If the length of the wire is 1m. The length of one ohm wire isA. 0.057mB. 0.067mC. 0.37mD. 0.134m
Answer» Correct Answer - B `"Here", (R )/(S) =(3)/(2),S=10Omega` `therefore R=(3)/(2) xxS=(3)/(2)xx10=15Omega` As the length of wire is 1m. length of one ohm wire=1//15=0.067m

Discussion

2309.	A cell develops the same power across two resistances `R_(1)` and `R_(2)` separately. The internal resistance of the cell isA. `sqrt(R_(1)R_(2))`B. `sqrt(2R_(1)R_(2))`C. `R_(1)+R_(2)`D. `R_(1)-R_(2)`
Answer» Correct Answer - A `P_(1)=i^(2)R_(1)=((E)/(R_(1)+r))^(2)R_(1)` `P_(2)=i^(2)R_(2)=((E)/(R_(2)+r))^(2)R_(2)`

Discussion

2310.	`n` identical resistors are taken `(n)/(2)` resistors are connected in series and the remaining are connected in paralle. The series connected group is kept in the left gap of a meter bridge and the parallel connected group in the right gap. The distance of te balance point from the left end of the wire isA. `(100n^(2))/(n^(2)+4)`B. `(100n^(2))/(n^(2)+1)`C. `(400)/(n^(2)+4)`D. `(400)/(n^(2)+1)`
Answer» Correct Answer - A `(X)/(100-X)=((nr)/(2))/((2r)/(n))`

Discussion

2311.	In a meter bridge, the gaps are closed by resistances 2 and 3 ohm. The value of shunt to be added to 3 ohm resistor to shift the balancing point by 22.5 cm isA. `1Omega`B. `2Omega`C. `2.5Omega`D. `5Omega`
Answer» Correct Answer - B `(2)/(3)=(l)/(100-l)impliesl=40cm,(2)/((3r)/(3+r))=(62.5)/(100-62.5)`

Discussion

2312.	The resistance of a discharge tube isA. zeroB. ohmicC. non-ohmicD. infinity
Answer» Correct Answer - C

Discussion

2313.	In the measurement of resistance by a meter-bridge, the current is necessarily reversed through the bridge wire to eliminateA. end errorB. index errorC. error due to electric effectD. random error
Answer» Correct Answer - c

Discussion

2314.	In the circuit `P != R, ` the reading of the galvanometer is same with switch S open or closed. Then A. `I_(R) = I_(G)`B. `I_(P) = I_(G)`C. `I_(Q) = I_(G)`D. `I_(Q) = I_(R)`
Answer» Correct Answer - A

Discussion

2315.	In a meter bridge experiment the ratio of left gap resistance to right gap resistance is `2:3` the balance point from isA. `20 cm`B. `40 cm`C. `50 cm`D. `60 cm`
Answer» Correct Answer - b

Discussion

2316.	A metallic conductor at `10^(@)C` connected in the left gap of meter bridge gives balancing length 40 cm. When the conductor is at `60^(@)C`, the balancing point shifts by --- cm, (temperature coefficient of resistance of the material of the wire is `((1)/(220))/(.^(@)C))`A. 4.8B. 10C. 15D. 7
Answer» Correct Answer - A `(X)/(R)=(40)/(60)=(2)/(3),(X_(0)(1+alphat_(1)))/(R)=(2)/(3)` `(X_(0)(1+alphat_(2)))/(R)=(l)/(100-l)` `(1+alphat_(1))/(1+alphat_(2))=(2)/(3)xx[(100-l)/(l)]` `l=44.8` Balancing point shift by `=44.8-40=4.8`

Discussion

2317.	When unknown resistance and a resistance of `5 Omega` are used in left and rigt gaps of meter bridge the balance point is 50 cm. The balancing point of `5Omega` resistance is now connected in seriece to the resistor in right gapA. 20 cmB. 33.3 cmC. 60 cmD. 60 cm
Answer» Correct Answer - B `(X)/(R)=(l)/(100-l)`

Discussion

2318.	Metal wire is connected in the left gap, semi conductor is connected in the right gap of meter bridge and balancing point is found. Both are heated so that change of resistances in them are same. Then the balancing pointA. will shift toward rightB. will shift toward leftC. will be shiftD. depends on rise of temperature
Answer» Correct Answer - a

Discussion

2319.	In meter bridge , the balancing length from left is found to be 20 cm when standard connected of `1 Omega` is in right gap . The value of unknown resistance isA. `0.8 Omega`B. `0.5Omega`C. `0.4 Omega`D. `0.25 Omega`
Answer» Correct Answer - D

Discussion

2320.	In meter bridge , the balancing length from left is found to be 20 cm when standard connected of `1 Omega` is in right gap . The value of unknown resistance isA. `0.25 Omega`B. `0.4 Omega`C. `0.5 Omega`D. `4 Omega`
Answer» Correct Answer - A

Discussion

2321.	Current is flowing from a conductor of nonuniform cross sectional area. If `A_1 gt A_2,` then find relation between a. `i_1 and i_2` b. `j_1 and j_2` c. `(v_d)_1 and (v_d)_2` (drift velocity)where I is current, j is current density, and V is drift velocity.
Answer» Correct Answer - `i_(1)=i_(2),V_(1)ltV_(2), j_(1)ltj_(2)` (a) I = charge flowing through a cross-section per unit time. `therefore i_(1)=i_(2)` (b) `j=i/A"as "A_(1)gtA_(2)" then "j_(1)ktj_(2)` ( c) `j="nev"_(d)` `V_(d)=j/("ne")"as "j_(1)gtj_(2)" then ,"V_(1)ltV_(2)`

Discussion

2322.	The balancing point in meter bridge experiment is obtained at 30 cm from the left. If the right gap contains `3.5 Omega`, what is the resistance in the left gap?
Answer» `(x)/(R ) = (l_(1))/(100 - l_(1)) implies (x)/(3.5) = (30)/(70) implies x = (30 xx 3.5)/(70) = 1.5` `:. X = 1.5 Omega` The resistance in the left gap in meter bridge is `x = 1.5 Omega`

Discussion

2323.	In the circuit shown in figure the emf E of the battery is increased linearly from zero to 28 V in the interval `0 le t le 14 s`. (a) find the energy gained by the `10 V` cell in the interval `0 le t le 14 s`. (b) At what time the `10 V` cell begin to charge?
Answer» Correct Answer - (a) zero (b) 7 s

Discussion

2324.	Three electric bulbs of 200 W, 200 W and 400 W are connected as shown in figure. The resultant power of the combination is A. 800 WB. 400 WC. 200 WD. 600 W
Answer» Correct Answer - A

Discussion

2325.	An ammeter and a voltmeter are joined in sereis to a cell. Their readings are `A` and `V` respectively. If a resistance is now joinding parallel with the voltmeter. ThenA. both A and V will decreaseB. both A and V will increase.C. A will increase V will dec rease.D. A will decrease V will increase.
Answer» Correct Answer - 3

Discussion

2326.	By mistake, a voltmeter is connected in series and an ammeter is connected in parallel, with a resistance in an electrical circuit. What will happen to the instruments?A. The main currnt in the circuit will be very low and almost all current will flow through the ammeter, if resistance of ammeter is much smaller than the resistance in parallel.B. If the devices are ideal, a large current will flow through the ammeter and it will be damagedC. If the devices (inculuding battery ) are idal, ammeter will read zero current and voltmeter will read the emf of cellD. The devices may get damaged if emf of the cell is very high and the meters are nonideal.
Answer» Correct Answer - A::C::D

Discussion

2327.	An ammeter and a voltmeter are connected in series to a battery of emf `E=6.0V`. When a certain resistance is connected in parallel with the voltmeter, the readding of the voltmeter decreases two times, whereas the reading of the ammeter increases the same number of times. Find the voltmeter reading after the connection of the resistance.A. 2 VB. 4 VC. 8 VD. 18 V
Answer» Correct Answer - A

Discussion

2328.	An ammeter and a voltmeter are joined in sereis to a cell. Their readings are `A` and `V` respectively. If a resistance is now joinding parallel with the voltmeter. ThenA. both A and V will increaseB. both A and V will decreaseC. A will decrease, V will increaseD. A will increase, V will decrease
Answer» Correct Answer - D

Discussion

2329.	A storage battery of emf 8.0 V and internal resistance `0.5 Omega` is being charged by a 120V dc supply using a series resistor of `15.5Omega`. what in the terminal voltage of the battery during charging ? What is the purpose of having a series resistor in the charging circuit?
Answer» Emf of the storage battery, `E = 8.0 V` Internal resistance of the battery, `r - 0.5 Omega` DC supply voltage, `V = 120 V` Resistance of the resistor, `R = 15.5 Omega` Effective voltage in the circuit `= V^(1)` R is connected to the storage battery in series. Hence, it can be written as `V^(1)=V-E` `V^(1) = 120 - 8 = 112 V` Current flowing in the circuit `= I`, which is given by the relation, `I = (V^(1))/(R+r)` `=(112)/(15.5 + 5)=(112)/(16)=7A` Voltage across resistor R given by the product, `IR - 7 xx 15.5 = 108.5 V` DC supply voltage = Terminal voltage of battery + Voltage drop acorss R Terminal voltage of battery `= 120 - 108.5 = 11.5 V` A series resistor in a charging circuit limits the current drawn from the external source. The current will be extremely high in its absence. This is very dangerous.

Discussion

2330.	(i) A storage battery of emf `8V`, internal resistance `1 Omega` is being charged by a `120 V` d.c. source using a `15 Omega` resistor in series in the circuit. Calculate the current in the circuit (ii) terminal voltage across the battery during charging and (ii) chemical energy stored in the battery in `5` minutes.
Answer» Correct Answer - `(i)7A,(ii)15V,(ii)16800 J` Effective emf of the circuit `epsilon = 120 - 8 = 112V` Total resistance of circuit `R = 1 + 15 = 10 Omega` ltbr gt (i) current `I = (epsilon)/(R ) = (112)/(10) = 7A` (ii) Terminal voltage during charging `V = epsilon + Ir = 8 + 7 xx xx 1 = 15 V` (iii) Chemical energy stored in the battery in `5` minutes ` = epsilon It = 8 xx 7 xx (5 xx 60)= 16800 J`

Discussion

2331.	In the circuit shown, total power supplied by an ideal battery is 80 W. if `R = 10 Omega`. Then If the resistance R is removed from the circuit. ThenA. power consumed by 2R will increaseB. power consumed by 6R will increaseC. both (a) and (b) are correctD. both (a) and (b) wrong
Answer» Correct Answer - A

Discussion

2332.	Each resistor shown in the figure has a resistance of `10 Omega` and the battery has the emf 6 V. What will be the current supplied by the battery ? A. `0.6 A`B. `1.2 A`C. `1.8 A`D. `0.3 A`
Answer» Correct Answer - A

Discussion

2333.	In the circuit shown, total power supplied by an ideal battery is 80 W. if `R = 10 Omega`. Then EMF of the battery isA. 20 VB. 10 VC. 40 VD. 60 V
Answer» Correct Answer - C

Discussion

2334.	In India electricity is supplied for domestic use at `220 V`. It is supplied at `110 V` in USA. If the resistance of a `60 W` bulb for use in India is `R`, the resistance of a `60 W` bulb for use in USA will beA. R/4B. R/2C. `R`D. 2R
Answer» Correct Answer - A

Discussion

2335.	In the circuit shown, total power supplied by an ideal battery is 80 W. if `R = 10 Omega`. Then Ratio of power developed across R, 2R and 6R will beA. `2:4:3`B. `1:2:6`C. `2:4:6`D. `1:2:3`
Answer» Correct Answer - A

Discussion

2336.	two resistors R and 2R are connected in series in an electric circuit. The thermal energy developed in R and 2R are in the ratioA. `1:2`B. `2:1`C. `1:4`D. `4:1`
Answer» Correct Answer - A

Discussion

2337.	Assertion : Inside a conductor, electrons have no motion in the absence of some potential difference across it. Reason : In the absence of potential difference no electrostatic force will act on the electrons inside the conductor.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 both Assertion and Reason are false.
Answer» Correct Answer - D

Discussion

2338.	Find the heat developed per minute in each of the resistors.
Answer» Correct Answer - `540 J, 480J, 960J` Equivalent resistance of the circuit is `(8 xx 4)/(8 + 4) xx 1 = (8)/(3) xx 1 = (11)/(3) Omega` current drawn from the battery is `1 = (11V)/((11//3) Omega) = 3A` Since the current through `1 Omega` tresistor is `3A` so heat produced in this resistor in `1` minut `(= 60` second ) is `H = l^(2)Rt = (3)^(2) xx 1 xx 60 = 540 J` current through `8 Omega` resistor `l_(1) = (3)/(8 + 4) xx 4 = 1A` Head produced in `8 Omega` resistor in `1` minut `H_(1) = l_(1)^(2) R_(1)t = (1)^(2) xx 8 xx 60 = 480 J` Current through `4 Omega` resistor `l_(2) = 1 - l_(1) = 3 - 1 = 2A` Current through `4 Omega` resistor in `1` minute `H_(2) = l_(2)^(2) R_(2) t = (2)^(2) xx 4xx 60 = 960 J`

Discussion

2339.	Assertion : Potential difference across the terminals of a battery is always less than the emf of the battery. Reason : During discharging of a battery potential differnce across to terminals of a battery is `V = E -ir` `therefore V lt E`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.
Answer» Correct Answer - D

Discussion

2340.	The resistance of the filament of an electric bulb changes with temperature. If an electric bulb rated 220 volt and 100 watt is connected `(220 xx 8)` volt sources, then the actual power would beA. `100 xx 0.8 watt`B. `100 xx (0.8)^(2) watt`C. Between `100 xx 0.8 `watt and `100 watt`D. Between `100 xx (0.8)^(2) `watt `100 xx 0.8` watt`
Answer» Correct Answer - D (d) `P_(1) = ((220)^(2))/(R_(1))` and `P_(2) =((220 xx 0.8)^(2))/(R_(2))` `(P_(2))/(P_(1)) = ((220 xx 0.8)^(2))/((220)^(2)) xx (R_(1))/(R_(2)) implies (P_(2))/(P_(1)) = (0.8)^(2) xx (R_(1))/(R_(2))` `R_(2) lt R_(1)` because voltage decreases from `220 V rarr 220 xx 0.8 V`. It means heat produced rarr decrease) So `(R_(1))/(R_(2)) gt 1 implies P_(2) gt (0.8)^(2) P_(1) implies P_(2) gt (0.8)^(2) xx 100 W` Also `(P_(2))/(P_(1)) = ((220 xx 0.8)i_(2))/(220 i_(1))`, Since `i_(2) lt i_(1)` (we expect) So `(P_(2))/(P_(1)) lt 0.8 implies P_(2) lt (100 xx 0.8)` Hence, the actual power would be between `100 xx (0.8)^(2) W` and `(100 xx 0.8) W`

Discussion

2341.	A battery of emf `epsilon` and internal resistance r sends currents `I_(1)` and `I_(2)`, when connected to external resistance `R_(1)` and `R_(2)` respectively. Find the emf and internal resistance of the battery.
Answer» In first case, current is `I_(1)=epsilon/R_(1)+r or epsilon = I_(1)(R_(1)+r)`....(i) In second case, current, `I_(2)=epsilon/(R_(2)+r)` or `epsilon = I_(2)(R_(2)+r)`....(ii) From (i) and (ii), we have `I_(1)(R_(1) +r)=I_(2)(R_(2) +r)` or `r (I_(1)- I_(2))=I_(2)R_(2) - I_(1)R_(1) or r =(I_(1)R_(2)-I_(1)R_(1))/(I_(1)-I_(2))` Putting this value of r in equation (i), we get, `epsilon= I_(1)[R_(1) +(I_(2)R_(2) - I_(1)R_(1))/(I_(1)-I_(2))]` `=I_(1)[(I_(1)R_(1)-I_(2)R_(1)+I_(2)R_(2)- I_(1)R_(1))/(I_(1)-I_(2))]` `=(I_(1)I_(2)(R_(2)-R_(1)))/(I_(1)-I_(2))`

Discussion

2342.	if the electrc current in a lamp decreases by `5%` then the power output decreases byA. `20%`B. `10%`C. `5%`D. `2.5%`
Answer» Correct Answer - B `P=i^(2)RimpliesPalphai^(2)implies(DeltaP)/(P)xx100=2xx(DeltaI)/(I)xx100`

Discussion

2343.	An electric bulb has the folloiwng specifications 100 watt, 220 volt. The resistance of bulbA. `384Omega`B. `484Omega`C. `344Omega`D. `584Omega`
Answer» Correct Answer - B `R=(V^(2))/(P)`

Discussion

2344.	A cell sends a current through a resistance `R_(1)` for time `t`, next the same cell sends current through another resistance `R_(2)` for the time `t` If the same amount of heat is developed in both the resistance then find the internal resistance of the cell
Answer» `[(E)/(R_(1)+r)]^(2)R_(1)=[(E)/(R_(2)+r)]^(2)R_(2)rArr r=sqrt(R_(1)R_(2))`

Discussion

2345.	Two electric bulbs whose resistances are in the ratio of `1:2` are connected inparallel to a constant voltage source. The powers dissipated in them have the ratioA. `1:2`B. `1:1`C. `2:1`D. `1:4`
Answer» Correct Answer - C `P=(V^(2))/(R)`

Discussion

2346.	A heater coil is rated 100W,200V. It is cut into two identical parts. Both parts are connected together in parallel, to the same sources of 200V. Calculate the energy liberated per second in the new combination.
Answer» `because P= =(V^(2))/(R) therefore R=(V^(2))/(P)=((200)^(2))/(100)=400Omega` Resistance of half piece `=(400)/(2)=200Omega` Resistance of pieces connected in parallel `=(200)/(2)=100Omega` Energy liberated/sec ond `=P=(V^(2))/(R)=(200xx200)/(100)=400W`

Discussion

2347.	Two electric bulbs whose resistances are in the ratio of `1:2` are connected inparallel to a constant voltage source. The powers dissipated in them have the ratio
Answer» Let resistance of bulbs are R and 2R Power dissipated `P=(V^(2))/(R) therefore P_(1)=(V^(2))/(R)` and `P_(2)=(V^(2))/(2R) therefore (P_(1))/(P_(2))=(2)/(1)`

Discussion

2348.	The power of a heater is 500W at `800^(@)C`. What will be its power at `200^(@)C` it `alpha=4xx10^(-4)` per `.^(@)C`?
Answer» `P=(V^(2))/(R) therefore (P_(200))/(P_(800))=(R_(800))/(P_(200))=(R_(0)(1+4xx10^(-4)xx800))/(R_(0)(1+4xx10^(-4)xx200))rArrP_(200)=(500xx 1.32)/(1.08)=611W`

Discussion

2349.	A battery has e.m.f. 4 V and internal resistance r . When this battery is connected to an external resistance of 2 ohms , a current of 1 amp . flows in the circuit. How much current will flow if the terminals of the battery are connected directlyA. 1AB. 2AC. 4AD. infinite.
Answer» Correct Answer - 2

Discussion

2350.	In a typical wheatstone network the resistance in cyclic order are A=`10Omega,B=5O,ega,C=4Omega` and `D=4Omega` for the bridge to be balanced. A. `10 Omega` should be connected in parallel with AB. `10Omega` should be connected in series with AC. `5Omega` should be connected in series with BD. `5Omega` should be connected in parallel with B.
Answer» Correct Answer - 3

Discussion

Explore topic-wise InterviewSolutions in .

In a neon discharge tube `2.9 xx 10^(18) Ne^(+)` ions move to the right each second while `1.2 xx 10^(18)` eletrons move to the left per second. Electron charge is `1.6 xx 10^(-9) C`. The current in the discharge tubeA. 1A towards rightB. `0.66A` towards rightC. `0.66A` towards leftD. Zero

The resistance of a discharge tube isA. OhmicB. Non-ohmicC. Both (a) and (b)D. Zero

In a neon discharge tube `2.9 xx 10^(18) Ne^(+)` ions move to the right each second while `1.2 xx 10^(18)` eletrons move to the left per second. Electron charge is `1.6 xx 10^(-9) C`. The current in the discharge tubeA. 0.27 A towards rightB. 0.66 A towards rightC. 0.66 A towards leftD. zero

In the circuit `P != R, ` the reading of the galvanometer is same with switch S open or closed. Then A. `I_(R)=I_(G)`B. `I_(P)=I_(G)`C. `I_(Q)=I_(G)`D. `I_(Q)=I_(R)`

In the circuit `P != R, ` the reading of the galvanometer is same with switch S open or closed. Then A. `I_(R)=I_(G)`B. `I_(P)=I_(G)`C. `I_(Q)=I_(G)`D. `I_(Q)=I_(R)`

In the following Wheatstone bridge `P//Q = R//S`. If key `K` is closed, then the galvanometer will show deflection A. In left sideB. In right sideC. No deflectionD. In either side

In the circuit `P != R, ` the reading of the galvanometer is same with switch S open or closed. Then A. `I_(R) = I_(G)`B. `I_(P) = I_(G)`C. `I_(Q) = I_(G)`D. `I_(Q) = I_(R)`

A wire connected in the left gap of a meter bridge balance a `10Omega` resistance in the right gap to a point, which divides the bridge wire in the ratio 3:2. If the length of the wire is 1m. The length of one ohm wire isA. 0.057mB. 0.067mC. 0.37mD. 0.134m

A cell develops the same power across two resistances `R_(1)` and `R_(2)` separately. The internal resistance of the cell isA. `sqrt(R_(1)R_(2))`B. `sqrt(2R_(1)R_(2))`C. `R_(1)+R_(2)`D. `R_(1)-R_(2)`

In a meter bridge, the gaps are closed by resistances 2 and 3 ohm. The value of shunt to be added to 3 ohm resistor to shift the balancing point by 22.5 cm isA. `1Omega`B. `2Omega`C. `2.5Omega`D. `5Omega`

The resistance of a discharge tube isA. zeroB. ohmicC. non-ohmicD. infinity

In the measurement of resistance by a meter-bridge, the current is necessarily reversed through the bridge wire to eliminateA. end errorB. index errorC. error due to electric effectD. random error

In the circuit `P != R, ` the reading of the galvanometer is same with switch S open or closed. Then A. `I_(R) = I_(G)`B. `I_(P) = I_(G)`C. `I_(Q) = I_(G)`D. `I_(Q) = I_(R)`

In a meter bridge experiment the ratio of left gap resistance to right gap resistance is `2:3` the balance point from isA. `20 cm`B. `40 cm`C. `50 cm`D. `60 cm`

A metallic conductor at `10^(@)C` connected in the left gap of meter bridge gives balancing length 40 cm. When the conductor is at `60^(@)C`, the balancing point shifts by --- cm, (temperature coefficient of resistance of the material of the wire is `((1)/(220))/(.^(@)C))`A. 4.8B. 10C. 15D. 7

When unknown resistance and a resistance of `5 Omega` are used in left and rigt gaps of meter bridge the balance point is 50 cm. The balancing point of `5Omega` resistance is now connected in seriece to the resistor in right gapA. 20 cmB. 33.3 cmC. 60 cmD. 60 cm

In meter bridge , the balancing length from left is found to be 20 cm when standard connected of `1 Omega` is in right gap . The value of unknown resistance isA. `0.8 Omega`B. `0.5Omega`C. `0.4 Omega`D. `0.25 Omega`

In meter bridge , the balancing length from left is found to be 20 cm when standard connected of `1 Omega` is in right gap . The value of unknown resistance isA. `0.25 Omega`B. `0.4 Omega`C. `0.5 Omega`D. `4 Omega`

Current is flowing from a conductor of nonuniform cross sectional area. If `A_1 gt A_2,` then find relation between a. `i_1 and i_2` b. `j_1 and j_2` c. `(v_d)_1 and (v_d)_2` (drift velocity)where I is current, j is current density, and V is drift velocity.

The balancing point in meter bridge experiment is obtained at 30 cm from the left. If the right gap contains `3.5 Omega`, what is the resistance in the left gap?

In the circuit shown in figure the emf E of the battery is increased linearly from zero to 28 V in the interval `0 le t le 14 s`. (a) find the energy gained by the `10 V` cell in the interval `0 le t le 14 s`. (b) At what time the `10 V` cell begin to charge?

Three electric bulbs of 200 W, 200 W and 400 W are connected as shown in figure. The resultant power of the combination is A. 800 WB. 400 WC. 200 WD. 600 W

An ammeter and a voltmeter are joined in sereis to a cell. Their readings are `A` and `V` respectively. If a resistance is now joinding parallel with the voltmeter. ThenA. both A and V will decreaseB. both A and V will increase.C. A will increase V will dec rease.D. A will decrease V will increase.

An ammeter and a voltmeter are joined in sereis to a cell. Their readings are `A` and `V` respectively. If a resistance is now joinding parallel with the voltmeter. ThenA. both A and V will increaseB. both A and V will decreaseC. A will decrease, V will increaseD. A will increase, V will decrease

A storage battery of emf 8.0 V and internal resistance `0.5 Omega` is being charged by a 120V dc supply using a series resistor of `15.5Omega`. what in the terminal voltage of the battery during charging ? What is the purpose of having a series resistor in the charging circuit?

In the circuit shown, total power supplied by an ideal battery is 80 W. if `R = 10 Omega`. Then If the resistance R is removed from the circuit. ThenA. power consumed by 2R will increaseB. power consumed by 6R will increaseC. both (a) and (b) are correctD. both (a) and (b) wrong

Each resistor shown in the figure has a resistance of `10 Omega` and the battery has the emf 6 V. What will be the current supplied by the battery ? A. `0.6 A`B. `1.2 A`C. `1.8 A`D. `0.3 A`

In the circuit shown, total power supplied by an ideal battery is 80 W. if `R = 10 Omega`. Then EMF of the battery isA. 20 VB. 10 VC. 40 VD. 60 V

In India electricity is supplied for domestic use at `220 V`. It is supplied at `110 V` in USA. If the resistance of a `60 W` bulb for use in India is `R`, the resistance of a `60 W` bulb for use in USA will beA. R/4B. R/2C. `R`D. 2R

In the circuit shown, total power supplied by an ideal battery is 80 W. if `R = 10 Omega`. Then Ratio of power developed across R, 2R and 6R will beA. `2:4:3`B. `1:2:6`C. `2:4:6`D. `1:2:3`

two resistors R and 2R are connected in series in an electric circuit. The thermal energy developed in R and 2R are in the ratioA. `1:2`B. `2:1`C. `1:4`D. `4:1`

Find the heat developed per minute in each of the resistors.

A battery of emf `epsilon` and internal resistance r sends currents `I_(1)` and `I_(2)`, when connected to external resistance `R_(1)` and `R_(2)` respectively. Find the emf and internal resistance of the battery.

if the electrc current in a lamp decreases by `5%` then the power output decreases byA. `20%`B. `10%`C. `5%`D. `2.5%`

An electric bulb has the folloiwng specifications 100 watt, 220 volt. The resistance of bulbA. `384Omega`B. `484Omega`C. `344Omega`D. `584Omega`

A cell sends a current through a resistance `R_(1)` for time `t`, next the same cell sends current through another resistance `R_(2)` for the time `t` If the same amount of heat is developed in both the resistance then find the internal resistance of the cell

Two electric bulbs whose resistances are in the ratio of `1:2` are connected inparallel to a constant voltage source. The powers dissipated in them have the ratioA. `1:2`B. `1:1`C. `2:1`D. `1:4`

A heater coil is rated 100W,200V. It is cut into two identical parts. Both parts are connected together in parallel, to the same sources of 200V. Calculate the energy liberated per second in the new combination.

Two electric bulbs whose resistances are in the ratio of `1:2` are connected inparallel to a constant voltage source. The powers dissipated in them have the ratio

The power of a heater is 500W at `800^(@)C`. What will be its power at `200^(@)C` it `alpha=4xx10^(-4)` per `.^(@)C`?

A battery has e.m.f. 4 V and internal resistance r . When this battery is connected to an external resistance of 2 ohms , a current of 1 amp . flows in the circuit. How much current will flow if the terminals of the battery are connected directlyA. 1AB. 2AC. 4AD. infinite.