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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.
| 1551. |
Find the charge on the capacitor in the circuit shown in fig. |
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Answer» Correct Answer - `24 muc` |
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| 1552. |
In the circuit shown in fig. the switch is kept closed in position 1 for a long time. At time `t = 0` the switch is moved to position 2. Write the dependence of voltage `(V_C)` across the capacitor as a function of time (t). Take `V_(C)` to be positive when plate a is positive. Given: `R_(1) = 20 Omega, R_(2) = 60 Omega, R_(3) = 400 kOmega, V_(1) = 40 V, V_(2)= 90 V and C = 0.5 mu F`. |
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Answer» Correct Answer - `V_(C)=(90-120 e^(-5t))` volt |
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| 1553. |
The capacitor A shown in fig. has a capacitance `C_(1) = 3 mu F`. The dielectric filled in it has a breakdown voltage of `40 V` and it has a resistance of `3 MOmega`. The capacitor `B` has a capacitance of `C_(2) = 2 mu F` and dielectric in it has a resistance of `2 MOmega`. Breakdown voltage for B is `50 V`. The switch is closed at `t = 0`. Will there be breakdown of any capacitor after the switch is closed ? If yes, which will breakdown first and at what time? |
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Answer» Correct Answer - Capacitor B, `6 l n 6` sec |
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| 1554. |
A cylindrical rod is reformed to half of its original length keeping volume constant. If its resistance before this change were R, then the resistance after reformation of rod will beA. RB. R/4C. 3R/4D. R/2 |
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Answer» Correct Answer - B The resistance of rod before reformation `R_(1)=R=(rhol_(1))/(pir_(1)^(2)) [therefore R=(rhol)/(A)=(rhol)/(pir^(2))]` Now the rod is reformed such that `l_(2)=(l_(1))/(2)` `therefore pir_(1)^(2)l_(1)=pir_(2)^(2)l_(2) (therefore "Volum remains constant")` or ` (pir_(1)^(2))/(pir_(2)^(2))=(l_(2))/(l_(1))..(i)` Now the resistance of the rod after reformation `R_(2)=(rhol_(2))/(pir_(2)^(2)) therefore (R_(1))/(R_(2))=(rhol_(1))/(pir_(1)^(2))/(rhol_(2))/(pir_(2)^(2))=(l_(1))/(l_(2))=(r_(2)^(2))/(r_(1)^(2))` `or (R_(1))/(R_(2))=(l_(1))/(l_(2))xx(l_(1))/(l_(2))=((l_(1))/(l_(2)))^(2)=(2)^(2)..("using"(i))` `therefore R_(2)=(R)/(4)` |
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| 1555. |
A piece of wire is cut into four equal parts and the pieces are bundled together side by side to from a thicker wire. Compared with that of the original wire, the resistance of the bundle isA. The sameB. 1/4 as muchC. 1/8 as muchD. 1/16 as muc |
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Answer» Correct Answer - D |
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| 1556. |
A cylindrical copper rod is reformed to twice its original length with no change in volume. The resistance between its ends before the change was R. Now its resistance will be :A. 8RB. 6RC. 4RD. 2R |
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Answer» Correct Answer - C |
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| 1557. |
In the given circuit (as shown in figure), each capacitor has a capacity of `3 mu F`. What will be the net charge on each capacitor ? A. ` 48 mu C`B. `24 mu C`C. `12 mu C`D. None of these |
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Answer» Correct Answer - C Net resistance between ABCD is `R=4+1=5Omega` `thereforeCurrent" "I=(V)/(R)=(10)/(5)=2A` Potential difference across A and B `1xx4=2xx4=8V` therefore Two capcitors of `3muF` each are in series therefore Potential difference across each capacitor `=(8)/(2)=4V` Charge on each capacitor, `q=CV=3xx4=12muC` |
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| 1558. |
`V-i` graphs of anichrome wire at three different temeperatures `t_(1)`, `t_(2)` and `t_(3)` are shown. From the graph. A. `t_(1)`B. `t_(2)`C. `t_(3)`D. `t_(1)=t_(2)=t_(3)` |
| Answer» Correct Answer - A | |
| 1559. |
A `5^(@)C` rise in the temperature is observed in a conductor by passing some current. When the current is doubled, then rise in temperature will be equal toA. `5^(@)C`B. `10^(@)C`C. `20^(@)C`D. `40^(@)C` |
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Answer» Correct Answer - C `(Deltat_(2))/(Deltat_(2))=(i_(1)^(2))/(i_(2)^(2))impliesDeltat_(2)=20^(@)C` |
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| 1560. |
A source which gives constant potential difference and hence flow of charge isA. source of e.m.f.B. source of lightC. source of heatD. none of these |
| Answer» Correct Answer - A | |
| 1561. |
A piece of silver and another of silicon are heated from room temperature The resistance ofA. each of them incresesB. each of them decreasesC. silver increases and silicon decreasesD. silver decreases and silivon increases |
| Answer» Correct Answer - C | |
| 1562. |
A `5^(@)C` rise in the temperature is observed in a conductor by passing some current. When the current is doubled, then rise in temperature will be equal toA. `20^(@)C`B. 16^(@)C`C. `12^(@)C`D. `10^(@)C` |
| Answer» Correct Answer - A | |
| 1563. |
A `1^(@) C` rise in temperature is observed in a conductor by passing a certain current . If the current is doubled , then the rise in temperature is approximatelyA. `10^(@)C`B. `12^(@)C`C. `16^(@)C`D. `20^(@)C` |
| Answer» Correct Answer - d | |
| 1564. |
Resistance of a conductor increases with the rise of temperature, becauseA. relaxation time decreasesB. relaxation time increasesC. electron density increasesD. electron density decreases |
| Answer» Correct Answer - A | |
| 1565. |
In which property of free electrons causes increase in the resistance of a conductor with rise in temperature ?A. Number densityB. relaxation timeC. MassD. none of these |
| Answer» Correct Answer - D | |
| 1566. |
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. `(m l)/(n e ^(2) tA)`B. `(2 ntA)/(n e^(2)l)`C. `( n e ^(2)t)/(2m) * (A)/(l)`D. `( n e^(2) m)/(2 t)*(l)/(A)` |
| Answer» Correct Answer - A | |
| 1567. |
Why is nichrome used as a heating element ?A. nichromeB. copperC. silverD. maganin |
| Answer» Correct Answer - A | |
| 1568. |
Three conductors draw respectively currents of 1 A, 2 A and 4 A when connected in turn across a battery. If they are connected in series across the same battery, the current drawn will beA. `(2)/(7) A`B. `(3)/(7) A`C. `(4)/(7) A`D. `(5)/(7) A` |
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Answer» Correct Answer - C Let the potential of the battery be V `R_(1)=(V)/(1),R_(2)=(V)/(2),R_(3)=(V)/(4)` `rArrR_(s)=V(1+(1)/(2)+(1)/(4))rArrR_(s)=(7)/(4)V` `because` By definition `I=(V)/(R_(s))rArrI=(4)/(7)A` |
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| 1569. |
Why is resistance less in parallel combination of resistors ? |
| Answer» In parallel combination of resistore, the effective area of cross-section of the conductor incereases. As `R prop 1//A`, therefore, resistance decreases. | |
| 1570. |
How will you join three resistances, each of `2 Omega` so that the effective resistance is `3 Omega` ? |
| Answer» Two resistances in parallel and the third resistance is connected in series with them. | |
| 1571. |
What is the difference between electromotive force and terminal voltage of a cell? How are they related with each other. |
| Answer» Electromotive force is the maximum potential difference between two electrodes of a cell in the open circuit and terminal voltage of a cell is the potential differnce between two electrodes of a cell in closed circuit. In a closed circuit, when cell supplies a current in the circuit, then electromotive force = terminal voltage + voltage drop across the internal resistance of the cell. | |
| 1572. |
What is the internal resistance of a cell due to? |
| Answer» Internal resistance of a cell depends upon, (i) the nature, concentrartion and temperature of electrolyte, (ii) the nature of electrodes, (iii) the distnace between the electrodes and (iv) area of the electrodes immersed in the electrolyte. | |
| 1573. |
On increasing the current drawn from a cell, the potential difference of its terminlas is lowered. Why? |
| Answer» This is due to internal resistance r of the cell. We know that terminal potential difference `V = epsilon - I r` . If I is increased, V will be lowered. | |
| 1574. |
The equivalent resistance of the arrangement of resistances shown in adjoining figure between the points `A` and `B` is A. 54 ohmB. 18 ohmC. 36 ohmD. 9 ohm |
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Answer» Correct Answer - B |
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| 1575. |
Can the terminal potential difference of a cell exceed its e.m.f. ? |
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Answer» Yes, when cell itself is being charged, because terminal potential difference, `V = epsilon - (- Ir) = epsilon + Ir`. |
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| 1576. |
The current-voltage graph for a resistor is as shown in the figure. Is it right to say that resistance decreases as the current through it increases ? |
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Answer» Correct Answer - No |
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| 1577. |
The potential difference between points `A` and `B` adjoining figure is A. `(2)/(3)V`B. `(8)/(9)V`C. `(4)/(3)V`D. `2V` |
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Answer» Correct Answer - C |
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| 1578. |
A (i) series (ii) parallel combination of two given resistors is connected one by one across a cell. In which case will the terminal potential difference across the cell have higher value? |
| Answer» Terminal potential differnce is equal to the voltage drop across the external resistor connected to cell. As resistance of two resistors in series is more than in parallel, so the terminal potential difference will have higher value in series combination of two resistors. | |
| 1579. |
Two wires that are made up of two different materials whose specific resistance are in the ratio 2 : 3, length 3 : 4 and area 4 : 5. The ratio of their resistances isA. `6 :5`B. `6:8`C. `5: 8`D. `1 :2` |
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Answer» Correct Answer - C |
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| 1580. |
A cylindrical conductor has length `l` and area of cross section A. Its conductivity changes with distance `(x)` from one of its ends as `sigma = sigma_(0) (l)/(x). [sigma_(0) "is a constant"]`. Calculate electric field inside the conductor as a function of `x`, when a cell of emf V is connected across the ends. |
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Answer» Correct Answer - `(2 V_(x))/(l^(2))` |
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| 1581. |
In the circuit shown in the figure, switch S is closed at time `t = 0`. Charge on positive plate of capacitor is `q` at time `t`. (a) Derive a differential equation for q at time t. (b) Solve the equation to write q as a function of time. (c) Put t = 0 and `t = infty` in your equation to get charge on the capacitor at these times. |
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Answer» Correct Answer - (a) `14 (dq)/(dt)+(5)/(4)q=12` (b) `q=(48)/(5)[1-e^(-(5t)/(56))]` (c) `0 ; (48)/(5) muC` |
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| 1582. |
A thick wire is stretched so that its length become two times. Assuming that there is no change in its density, then what is the ratio of change in resistance of wire to the initial resistance of wireA. `2:1`B. `4 :1`C. `3:1`D. `1:4` |
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Answer» Correct Answer - C |
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| 1583. |
The electric field E, current density j and conductivity `sigma` of a conductor are related asA. `sigma = E//j`B. `sigma = j//E`C. `sigma = jE`D. `sigma = 1//jE` |
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Answer» Correct Answer - B |
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| 1584. |
A thick wire is stretched so that its length become two times. Assuming that there is no change in its density, then what is the ratio of change in resistance of wire to in initial resistance of wireA. `2 : 1`B. `4 : 1`C. `3 : 1`D. `1 : 4` |
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Answer» Correct Answer - C (c ) In stretching `R prop l^(2) implies (R_(2))/(R_(1)) = (l_(2)^(2))/(l_(1)^(2)) implies (R_(2))/(R_(1)) = ((2)/(1))^(2)` `implies R_(2) = 4 R_(1)`. Change in resistance `= R_(2) - R_(1) = 3R_(1)` Now, `("Changein resistance")/("Original resistance") = (3R_(1))/(R_(1)) = (3)/(1)` |
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| 1585. |
The length of the resistance wire is increased by 10%. What is the corresponding change in the resistance of wireA. `10%`B. `25%`C. `21%`D. `9%` |
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Answer» Correct Answer - C |
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| 1586. |
The length of the resistance wire is increased by `10%`. What is the corresponding change in the resistance of wire?A. `10%`B. `25%`C. `21%`D. `9%` |
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Answer» Correct Answer - C (c ) `(R_(1))/(R_(2)) = ((l_(1))/(l_(2))^(2))`, If `I_(1) = 100` then `l_(2) = 110` `implies (R_(1))/(R_(2)) = ((100)/(110))^(2) implies R_(2) = 1.21 R_(1)` `%` change `(R_(2) - R_(1))/(R_(1)) xx 100 = 21 %` |
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| 1587. |
Current flows through a metallic conductor whose area of cross-section increases in the direction of the current. If we move in this direction.A. the current will changeB. the carrier densiy will changeC. the drift velocity will increaseD. the drift velocity will decrease |
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Answer» Correct Answer - D (d) `I = ne Av_(d) implies v_(d) = (I)/(ne A)` `v_(d) prop (I)/(A)` so as `A` increases `v_(d)` decreases |
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| 1588. |
A current `I` flows through a uniform wire of diameter `d` when the mean electron drift velocity is `v`. The same current will flow though a wire of diameter `d//2` made of the same material if the mean drift velocity of the electron isA. `v//4`B. `v//2`C. `2v`D. `4v` |
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Answer» Correct Answer - D `I="ne"A_(1)v_(1)="ne"A_(2)v_(2)` `"ne"pi(d/2)^(2)v="ne"pi(d/4)^(2)v_(2)implies v_(2)=4v` |
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| 1589. |
A wire has non-uniform cross-section as shown in fig. steady current flows through it. The drift speed of electrons at point `P` and `Q` is `v_(P)` and `v_(Q)` A. `v_(P)=v_(Q)`B. `v_(P)ltv_(Q)`C. `v_(P)gtv_(Q)`D. Data insufficient |
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Answer» Correct Answer - C `I="ne"Av_(d)implies v_(d) prop1/A implies v_(P)gtv_(Q)` |
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| 1590. |
Two wires each of radius of cross section `r` but of different materials are connected together end to end (in series). If the densities of charge charge carries in the two wires are in the ratio `1:4`, the drift velocity of electrons in the two wires will be in the ratio:A. `1:2`B. `2:1`C. `4:1`D. `1:4` |
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Answer» Correct Answer - C `I=n_(1) eAv_(d_(1))=n_(2)eAv_(d_(2))` `implies (v_(d_(1)))/(v_(d_(2)))=(n_(2))/(n_(1))=4/1` |
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| 1591. |
In a wire of cross-section radius `r`, free electrons travel with drift velocity `v` when a current `I` flows through the wire. What is the current in another wire of half the radius and of the same material when the drift velocity is `2v`?A. `2I`B. `I`C. `I//2`D. `I//4` |
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Answer» Correct Answer - C `I_(1)="ne" A_(1)v_(d_(1)) & I_(2)="ne" A_(2)v_(d_(2))` `I_(2)/I_(1)=A_(2)/(A_(1))xxv_(d_(2))/v_(d_(1))=1/4xx(2v)/vimplies I_(2)=I/2` |
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| 1592. |
A wire has non-uniform cross-section as shown in fig. steady current flows through it. The drift speed of electrons at point `P` and `Q` is `v_(P)` and `v_(Q)` A. `v_(P) = v_(Q)`B. `v_(P) lt v_(Q)`C. `v_(P) gt v_(Q)`D. data insufficient |
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Answer» Correct Answer - C (c ) `v_(d) = (i)/(A n e)` As `A uarr` so `v_(d) darr implies v_(P) gt v_(Q)` |
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| 1593. |
The area of cross-section length and density of a piece of a metal of atomic weight `60` are `10^-6 m^2`, `1.0 m` and `5xx10^3 kg//m^3` respectively,every atom contributes one free electron. (Given Avogadro number `=6xx10^23//mol)`. Find the drift velocity of electrons in the metal when the current of `16A` passes through:A. `2mm//s`B. `2 cm//s`C. `20 cm//s`D. none of these |
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Answer» Correct Answer - A `I="ne"Av_(d)…..(i)` `m=5xx10^(3)xx10^(-6)xx1=5xx10^(-3) kg=5 gm` putting all values is equation `(1)` `16=5xx10^(28)xx1.6xx10^(-19)xx10^(-6)xxV_(d)` `V_(d)=2 mm//s` |
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| 1594. |
An insulating pipe of cross-section area `A` contains an electrolyte which has two types of ions: their charges being-e and `+ 2e`. A potential difference applied between the ends of the pipe result in the drifting of the two types of ions, having drift speed `= v (- ve ion)` and `v//4(+ ve ion)`. Both ions have the same number per unit volume `= n`. The current flowing through the pipe isA. nev`A//2`B. nev `A//4`C. `5` nev `A//2`D. `3` nev `A//2` |
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Answer» Correct Answer - D `I=I_((-e))+I_((+2))` `I="ne"Av+n(2e)A(v/4)` `="ne"Av+("ne"Av)/2=3/2 "ne"Av` |
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| 1595. |
Resistances of 6 ohm each are connected in the manner shown in adjoining figure. With the current 0.5 ampere as shown in figure, the potential difference `V_(P) - V_(Q)` is A. `3.6 V`B. `6.0 V`C. `3.0 V`D. `7.2 V` |
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Answer» Correct Answer - C |
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| 1596. |
An isulating pipe of cross-section area `A` contains an electrolyte which has two types of ions: their charges being-e and `+ 2e`. A potential difference applied between eht ends of the pipe result in the drifting of the two types of ions, having drift speed `= v (- ve ion)` and `v//4(+ ve ion)`. Both ions have the same number per unit volume `= n`. The current flowing through the pipe isA. `(1)/(2) n e v A`B. `(1)/(5) n e v A`C. `(5)/(2) n e v A`D. `(3)/(2) n e v A` |
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Answer» Correct Answer - D (d) `i= n e v_(d) A, I = (2 n e v A)/(4) - (- n e vA) = (3)/(2) n e v A` |
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| 1597. |
A certain amount of ice is supplied heat at a constant rate for 7 min. For the first one minute the temperature rises uniformly with time. Then it remains constant for the next 4 min and again the temperature rises at uniform rate for the last 2 min. Given `S_("ice")=0.5cal//g^(@)C,L_(f)=80cal//g`. Final temperature at the end of 7 min isA. `10^(@)C`B. `20^(@)C`C. `30^(@)C`D. `40^(@)C` |
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Answer» Correct Answer - D |
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| 1598. |
A certain amount of ice is supplied heat at a constant rate for 7 min. For the first one minute the temperature rises uniformly with time. Then it remains constant for the next 4 min and again the temperature rises at uniform rate for the last 2 min. Given `S_("ice")=0.5cal//g^(@)C,L_(f)=80cal//g`. The initial temperature of ice isA. `-10^(@)C`B. `-20^(@)C`C. `-30^(@)C`D. `-40^(@)C` |
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Answer» Correct Answer - D |
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| 1599. |
Find the equivalent resistance of the networks shown in Fig. 4.51 between the points A and B. |
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Answer» Correct Answer - `(4)/(3) r (b) (r )/(4) ( c) r` (a) \(\frac 1R = \frac1r + \frac1 r + \frac1r\) \(\frac 1R = \frac3r\) \(R = \frac r3\) \(R = \frac r3 + r\) \(R = \frac{4r}3\) (b) \(\frac 1R = \frac1r + \frac1 r \) \(\frac 1R = \frac2r\) \(R = \frac r2\) \(\frac 1R = \frac1r + \frac1 r \) \(\frac 1R = \frac2r\) \(R = \frac r2\) then \(\frac 1R = \frac1R + \frac1 R\) \(\frac 1R = \frac2r + \frac2r \) \(\frac 1R = \frac4r\) \(R = \frac r4\) (c) \(R = r + r + r + r\) \(R_1 = 4r\) \(\frac 1R = \frac1r + \frac1 r \) \(R_2 = \frac r2\) then \(R = R_1 + R_2\) \(R = 4r + \frac r2\) \(R = \frac{9r}2\) |
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| 1600. |
Calculate the resistance between points A and B for the following networks : |
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Answer» Correct Answer - `(a) (2)/(3) Omega (b) (4)/(3) Omega ( c) (R )/(3) Omega (d) 6 Omega` |
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