InterviewSolution
Saved Bookmarks
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.
| 1501. |
In the given circuit, the switch S is closed at time t = 0 . The charge Q on the capacitor at any instant t is given by `Q(t) = Q(1-e^(-alphat))`. Find the value of `Q_0` and `alpha` in terms of given parameters as shown in the circuit. . |
|
Answer» Correct Answer - A::B::C Given `Q = Q_0 [1-e^(alphat)]` Here `Q_0 = Maximum charge and `alpha = 1/(tau_c) = 1/(C_(Req))` Now the maximum charge `Q_0 = C[V_(0)]` where `V_0` = max potential difference across C `= C[V/(R_1+R_2) xx R_2]` and `tau_c = C R_(eq)` `=C[(R_1R_2)/(R_1+R_2)] :. alpha = (1/tau_c) = (R_1 + R_2)/(CR_1R_2)`. |
|
| 1502. |
The current through a wire depends on time as, `i=(10+4t)` Here , `i` is ampere and `t` in seconds. Find the charge crossed through section in time interval between `t=0` to `t=10s`. |
|
Answer» Correct Answer - C `/_q=int_0^10 idt= int_0^10(10+4t)dt` `=300 C` |
|
| 1503. |
Two electric bulbs rated `P_(1)` and `P_(2)` watt at `V` volt are connected in series across `V` volt mains then their total power consumption `P` isA. `(P_(1)+P_(2))`B. `sqrt(P_(1)P_(2))`C. `P_(1)P_(2)//(P_(1)+P_(2))`D. `(P_(1)+P_(2))//P_(1)P_(2)` |
|
Answer» Resistance of `I^(st)=` bulb `R_(1)=(V^(2))/(P_(1))` Resistance of `II^(nd)` bulb `R_(2)=(V^(2))/(P_(2))` When both bulb are connected in series `R_(eq)=V^(2)[(1)/(P_(1))+(1)/(P_(2))]=(V^(2)(P_(1)+P_(2)))/(P_(1)P_(2))` Hence power consumed `P=(V^(2))/(R )=(V^(2))/(V^(2)((P_(1)+P_(2))/(P_(1)P_(2))))=(P_(1)P_(2))/(P_(1)+P_(2))` |
|
| 1504. |
The potentiometer wire AB shown in figure is 50cm long.When AD=30cm, no deflection occurs in the galvanmeter.Find R. A. `1Omega`B. `2Omega`C. `3Omega`D. `4 Omega` |
| Answer» Correct Answer - D | |
| 1505. |
The potentiometer wire AB shown in figure is 50cm long.When AD=30cm, no deflection occurs in the galvanmeter.Find R. |
|
Answer» `(6)/( R) = (AD)/(DB) = (30)/(50 - 30) = (30)/(20) = (3)/(2)` `R = 4 Omega`. |
|
| 1506. |
The current through a wire depends on time as `I = i_(0) + alphat`, where `i_(0)=10 A and alpha=4 As^(-1)`.Find the charge crossed through a section of the wire in 10 second |
|
Answer» `i=i_(0)+alphat,buti=(dq)/(dt)` `impliesdq=(i_(0)+alphat)dt` `q=int_(t=0)^(t=10)dqimpliesq=[i_(0)t+(alphat^(2))/(2)]_(0)^(10)` `=(10i_(0)+50alpha)=300` coloumb |
|
| 1507. |
In the circuit shown in the adjacent figure A. the potential at `P` is `-7.5V`B. the potential at `Q` is `-1V`C. the potential at `R` is zeroD. the potential at `S` is zero |
|
Answer» The current in the circuit `I=((20-10))/((7.5+.5+1+1))=(10)/(10)=1Amp` The potential difference across `PS V_(PS)=7.5xx1=7.5V` The potential difference across `Q R`, `V_(QR)=1xx1=1V` As point `G` is connected to earth hence potentials of `R` and `S` is zero. The direction of the current in the circuit is from `P` to `S`, hence point `P` is at higher potential. `:. V_(p)=7.5V` similarly `V_(Q)= -1V` |
|
| 1508. |
Refer .for what value of `R` (in ohm) will the current in galvanmeter `G` be zero |
|
Answer» Correct Answer - 7 When the currect through galvanometer is zero, then pot diff across `R Omega = 7 V` pot diff across `5 Omega= 12 - 7= 5 V` current in `5 Omega` = (5)/(5) = 1A` Resistance, `R = (7)/(1)= 7 Omega ` |
|
| 1509. |
A and B are two points on a uniform ring of resistance `15Omega`. The `ltAOB=45^(@)` The equivalent resistance between A and B is A. `1.64Omega`B. `2.84Omega`C. `4.57Omega`D. `2.64Omega` |
|
Answer» Correct Answer - A Resistance per unit length of ring, `rho=(R )/(2pir)` Length of sections ADB and ACB are `rtheta and r theta and (2pir-theta)` Resistance of section ADD, `R_(1)=rhortheta=(R)/(2pir)rtheta=(Rtheta)/(2pi)` and resistance of section, ACB, `R_(2)=rhor(2pi-theta)=(R)/(2pir)r(2pi=theta)=(R(2pi-theta))/(2pi)` Now, `R_(1) and R_(2)` are connected in parallel between A and B then, `R_("eq")=(R_(1)R_(2))/(R_(1)+R_(2))=((Rtheta)/(2pi)xx(R(2pi-theta))/(2pi))/((Rtheta)/(2pi)+R(2pi-theta)/(2pi))=(Rtheta(2pi-theta))/(4pi^(2))` `"Putting" theta=45^(@)=(pi)/(4) "rad" and R=15Omega ` `R_("eq")=((15xx(pi))/(4)xx(2pi-(pi)/(4)))/(4pi^(2))=((15)/(4)((7pi)/(4)))/(4pi^(2))=1.64Omega` |
|
| 1510. |
The electron in a hydrogen atom moves in circular orbit of radius `5xx10^(-11)m` with a speed of `0.6pixx10^(6)m//s` thenA. the frequency of the electron is `6xx10^(15)rev//s`B. the electron carries `-1.6xx10^(-19)C` around the loopC. the current in the orbit is `0.96mA`D. the current flow is in the opposite direction to the direction of the motion of electron |
|
Answer» Electric current at a point on the circle `i=fe`, where `f=`frequency `=(omega)/(2pi)=(v)/(2pir)=(0.6xx10^(6)xxpi)/(2pixx5xx10^(-11))=6xx10^(15)rev//s` `i=6xx10^(15)xx1.6xx10^(-19)=0.96xx10^(-3)A=0.96mA` |
|
| 1511. |
Find the total linear momentum of the electrons in a conductor of length `l = 1000m` carrying a current `I = 70A`. |
|
Answer» Correct Answer - 4 Let a be the area of the cross-section of the wire and `n` be the number density of electrons in the wire. If `N` is the total no of electron crossing per second through a cross section of wire due to current `I` flowing then `N = (1)/(e ) = n a l or I = n a l e`.....(i) But `I = n a e upsilon_(d) or upsilon_(d) = (1)/(n a e)= ( na l e )/(n a e) = l` Total momentum of electrons in the stringht wire is `= Nm upsilon_(d) = (1)/(e ) xx mxxl = (70 xx (9xx 10^(-31)) xx 1000)/(16 xx 10^(-19))` `~~ 4 xx 10^(-7) Ns`. |
|
| 1512. |
In a hydrogen atom, electron moves in an orbit of radius `5xx10^(-11)m` with a speed of `2.2xx10^(6)(m)/(s)`. Calculate the equivalent current. |
| Answer» Correct Answer - 1.12mA | |
| 1513. |
Find the total momentum of electrons in a straight wire of length `l = 1000 m` carrying a current `I=70 A`. |
| Answer» Correct Answer - `0.4xx10^(-6)N` sec | |
| 1514. |
The potential difference applying to an X- ray tube is `5kV` and the current through it is `3.2mA`. Then the number of electrons striking the target per second is |
| Answer» Correct Answer - `2xx10^(16)` | |
| 1515. |
A point source of heat of power P is placed at the centre of a spherical shell of mean radius R. The material of the shell has thermal conductivity K. If the temperature difference between the outer and inner surface of the shell in not to exceed T, the thickness of the shell should not be less than .......A. `(2piR^(2)KT)/(P)`B. `(4piR^(2)KT)/(P)`C. `(piR^(2)KT)/(P)`D. `(piR^(2)KT)/(4P)` |
|
Answer» Correct Answer - B |
|
| 1516. |
In process `Tprop(1)/(V)`, pressure of the gas increases from `p_(0)` to `4p_(0)`. Match the following. |
|
Answer» Correct Answer - (A) R, (B) T, (C) Q, (D) P |
|
| 1517. |
Two batteries of emf `epsi_(1) and epsi_(2)(epsi_(2) gt epsi_(1))` and internal resistance `r_(1) and r_(2)` respectively are connected in parallel as shown in figure.A. The equivalent emf `epsi_(eq)` of the two cells is between `epsi_(1) and epsi_(2)`, i.e., `epsi_(1) lt epsi_(eq) lt epsi_(2)`B. the equivalent emf `epsi_(eq)` is smaller than `epsi_(1)`C. The `epsi_(eq)` is given by `epsi_(eq)=epsi_(1)+epsi_(2)` always.D. `epsi_(eq)` is independent of internal resistance `r_(1) and r_(2)`. |
|
Answer» Correct Answer - A The equivalent emf of this combination is given by `epsi_(eq)=(epsi_(2)r_(1)+epsi_(1)r_(2))/(r_(1)+r_(2))` This suggest that the equivalent emf `epsi_(eq)` of the two cells is given by `epsi_(1) lt epsi_(eq) lt epsi_(2)`. |
|
| 1518. |
Find out the value of current through `2Omega` resistance for the given circuit. A. `5A`B. `2A`C. zeroD. `4A` |
|
Answer» Correct Answer - c We know in a closed circuit, the amount of current coming out of positive terminal of a cell is equal to the amount of current entering the negative terminal of the cell. Therefore, the current in the resistance closed circuits will remain confined in the respective circuits. Due to it, the current in `2 Omega` resistance is zero |
|
| 1519. |
Two sources of equal emf are connected to an external resistance R. The internal resistance of the two sources are` R_(1) and R_(2) (R_(1) gt R_(1))`. If the potential difference across the source having internal resistance `R_(2)` is zero, thenA. `R = R_(1)R_(2)//(R_(1)+R_(2))`B. `R = R_(1)R_(2)//(R_(2)-R_(1))`C. `R = R_(2) xx (R_(1) +R_(2))// (R_(2)-R_(1))`D. `R = R_(2) -R_(1)` |
|
Answer» Correct Answer - D |
|
| 1520. |
The length of a potentiometer wire is 10m and a potential difference of 2 volt is applied to its ends. If the length of its wire is increased by 1m, the value of potential gradient in volt/m will be (potential difference across potentiometer wire remain same)A. 0.18B. 0.22C. 1.3D. 0.9 |
|
Answer» Correct Answer - A `x=(2)/(11) =0.18 V//m` |
|
| 1521. |
In the shown circuit, what is the potential difference across A and B A. 50 VB. 45 VC. 30 VD. 20 V |
|
Answer» Correct Answer - D `V_(AB)=20 V` |
|
| 1522. |
When a battery connected across a resistor of `16 Omega`, the voltage across the resistor is 12 V. When the same battery is connected across a resistor of `10 Omega`, voltage across it is 11 V. The internal resistance of the battery (in ohm) isA. `(10)/(7)`B. `(20)/(7)`C. `(25)/(7)`D. `(30)/(7)` |
|
Answer» Correct Answer - B Here, VltE `lt`E = V + lr For first case, `E=12+(12)/(16)r ` For second case, `E=11+(11)/(10)r From Eqs. (i) and(ii),we get `12+(12)/(16)r=11+(11)/10rrArr r==20/7Omega` |
|
| 1523. |
n mole of an ideal gas undergo an isothernal process at temperature T. Further, P-V graph of the process is as shown in the figure. Tangent at point A, cuts the V-axis at point D. AO is the line joining the point A to the origin O of P-V diagram. Choose from following the correct option(s). A. coordinates of point D is `((3V_(1))/(2),0)`B. coordinates of point D is `(2V_(1),0)`C. area of the triangle AOD is nRTD. area of the triangle AOD is `(3)/(4)` nRT |
|
Answer» Correct Answer - B::C |
|
| 1524. |
A 1250 W heater operates at `115V`. What is the resistance of the heating coil?A. `1.6Omega`B. `13.5Omega`C. `1250Omega`D. `10.6Omega` |
|
Answer» Correct Answer - D |
|
| 1525. |
To send 10% of the main current through a moving coil galvanometer of resistance `99omega`, the shunt required is –A. `9Omega`B. `11Omega`C. `10Omega`D. `9.9Omega` |
|
Answer» Correct Answer - B |
|
| 1526. |
Masses of 3 wires of same metal are in the ratio 1 : 2 : 3 and their lengths are in the ratio 3 : 2 : 1. The electrical resistances are in ratioA. `1:1:1`B. `1:2:3`C. `9:4:1`D. `27: 6:1` |
|
Answer» Correct Answer - 4 `rho` is same so `Rprop(l)/(A)` Now m d `rArr m=Ald` `Aprop(m)/(l) rArrR prop(l^(2))/(m)` `R_(1):R_(2):R_(3)=(9)/(1):(4)/(2):(1)/(3)=27:6:1` |
|
| 1527. |
Masses of 3 wires of same metal are in the ratio `1 : 2 : 3` and their lengths are in the ratio `3 : 2 : 1`. The electrical resistances are in ratioA. `1 : 4 : 9`B. `9 : 4 : 1`C. `1 : 2 : 3`D. `27 : 6 : 1` |
|
Answer» Correct Answer - D (d) `R prop (l^(2))/(m) implies R_(1) : R_(2) : R_(3) = (l_(1)^(2))/(m_(1)) : (l_(2)^(2))/(m_(2)) : (l_(3)^(2))/(m_(3))` `implies R_(1) : R_(2) : R_(3) = (9)/(1) : (4)/(2) : (1)/(3) = 27 : 6 : 1` |
|
| 1528. |
A resistance R is stretched to four times its length. Its new resistance will beA. 4RB. 64RC. `R//4`D. `16R` |
|
Answer» Correct Answer - D |
|
| 1529. |
What is the resistance of a carbon resistance which has bands of colours brown, black and brownA. `100 Omega`B. `1000Omega`C. `10 Omega`D. `1 Omega` |
|
Answer» Correct Answer - A |
|
| 1530. |
When a potential difference is applied across the ends of a linear metallic conductorA. the free electrons are accelrated continously from the lower potential end to the higher potential and of the conductorB. the free electrons are accelerated continously from the higher potential end to the lower potential end of the conductorC. the free electrons acqurie a costant drift velocity from the lower potential end to the higher potential end of the conductor.D. the free electrons are set in motion from their position of rest. |
| Answer» Correct Answer - 3 | |
| 1531. |
When a potential difference is applied across the ends of a linear metallic conductorA. The free electrons are accelerated continuously from the lower potential end to the higher potential end of the conductorB. The free electrons are accelerated continuously from the higher potential end to the lower potential end of the conductorC. The free electrons acquire a constant drift velocity from the lower potential end to the higher potential end of the conductorD. The free electrons are set in motion from their position of rest |
|
Answer» Correct Answer - C |
|
| 1532. |
The alloys constantan and manganin are used to make standard resistance due to they haveA. Low resistivityB. High resistivityC. Low temperature coefficient of resistanceD. Both (b) and (c) |
|
Answer» Correct Answer - D |
|
| 1533. |
A wire of length 100 cm is connected to a cell of emf 2 V and negligible internal resistance. The resistance of the wire is `3 Omega`. The additional resistance required to produce a potential drop of 1 milli volt per cm isA. `60 Omega`B. `47 Omega`C. `57 Omega`D. `35 Omega` |
|
Answer» Correct Answer - C |
|
| 1534. |
A galvanometer can ber converted into an ammeter by connectingA. Low resistance in seriesB. High resistance in parallelC. Low resistance in parallelD. High resistance in series |
|
Answer» Correct Answer - C (c ) To converts a galvanometer into an ammeter a low value resistance is to be connected in parallel to it called shunt. |
|
| 1535. |
A cell of internal resistance `5.1 Omega` and of e.m.f. 1.5 volt balances 500 cm on a potentiometer wire. If a wire of `15Omega` is connected between the balance point and the cell, then the balance point will shiftA. To zeriB. By 500 cmC. By 750 cmD. None of the above |
|
Answer» Correct Answer - D |
|
| 1536. |
The net resistance of a volmeter should be large to ensure thatA. It does not get overheatedB. It does not draw excessive currentC. It can measure large potential differenceD. It does not appreciably change the potential difference to be measured |
|
Answer» Correct Answer - D (d) The resistance of voltmeter is too high, so that it draws negligible current from the circuit,hence potential drop in the external circuit is also negligible. |
|
| 1537. |
A cell of itnernal resitance `1.5 Omega` and of e.m.f. `1.5` volt balances `500 cm` on a potentiomter wire. If a wirr of 15 `Omega` is connected between the balance point and the cell, then tha balance point will shiftA. To zeroB. By `500 cm`C. By `750 cm`D. None of these |
|
Answer» Correct Answer - D (d) Balance point has some fixed position on potentiometer wire. It is not affect by the additions of resistance between balance point and cell. |
|
| 1538. |
Two wires `A` and `B` made of same material and having their lengths in the ratio `6:1`a re connected in series The potential difference ascross the wire `3V` and `2V` respectively. If `r_A` and `r_B` are the radii of `A` and `B` respectively, then `r_B/r_A` isA. `1/4`B. `1/2`C. 1D. 2 |
|
Answer» Correct Answer - B `V=iR` `:. VpropR` (as i= constant) `:. V_A/V_B=((rhoI_A)/(pir_A^2))((pir_B^2)/(rhol_B))` or `r_B/r_A=sqrt(V_A/V_BxxI_B/l_A)` `=sqrt(3/2xx1/6)=1/2` |
|
| 1539. |
Two wires of different materials P and Q have resistance per unit lengths `50 Omega km^(-1)` and `25 Omega km^(-1)` and temperature coefficients of resistance, `0.0025 ^@C^(-1)` and `0.00075^@C^(-1)` respectively. If it is desired to make a coil having `700 Omega` resistance and a temperature coefficient of resistance `0.001^@C^(-1)` by using suitable length of two wires in series. Calculate their respective length. |
|
Answer» Let `R_(P) and R_(Q)` be the resistance of the two wires at the given temperature and `l_(P) and l_(Q)` be their respective lengths. If `Delta T` be the change in temperature, then the total resistance of two wires in series is `R = R_(P) ( 1 + 0.00025 Delta T) + R_(Q) (1 + 0.00075 Delta T)` As `0.0001^@C^(-1)` is the temperature coefficient of resistance of the combination of two wires in series, so `R = (R_(P) + R_(Q)) [1 + 0.001 Delta T]` From (i) and (ii), we have `= R_(P) ( 1 +0.0025 Delta T) +R_(Q) (1 +0.00075 Delta T) = (R_(P) + R_(Q)) [1 + 0.001 Delta T]` or `R_(P) (0.0015 Delta T) = R_(Q) (0.00025 Delta T)` or `R_(Q) = 150/25 R_(P) = 6 R_(P)` Since `R_(P) + R_(Q) = 700` `:. R_(P) + 6 R_(P) = 700 or R_(P) = 100 Omega` and `R_(Q) = 600 Omega` `:.` length of wire `P, l_(P) = 1/50 xx 100 = 2 km` length of wire Q, `l_(Q) = 1/25 xx 600 = 24 km` |
|
| 1540. |
A steady beam of `alpha`-particles travelling with kinetic energy `E=83.5 keV` carries a current of `I=0.2 mu A`. Mass of `alpha`-particle `=6.68 xx10^(-27)kg`. (i) If this beam strikes a plane surface at an angle `theta=60^@` with normal to the surface, how many `alpha` -particles strike the surface in t=4 second? |
|
Answer» Here, current `I = 0.25 mu A = 0.25 xx 10^(-6)A` Kinetic energy of `alpha`-particle, `K = 84 KeV = 84 xx 10^(3) xx 1.6 xx 10^(-19) J` `= 84 xx 1.6 xx 10^(-16)J` (a) Charge on `alpha`-particles striking the plane surface per second `n = I/q = ( 0.25 xx 10^(-6))/( 2 xx 1.6 xx 10^(-19)) = 7.81 xx 10^(11) s^(-1)` Number of `alpha`-particles striking the surcface in `4 s = nt = (7. 81 xx 10^(11)) xx 4` `= 31.24 xx 10^(11) = 3.1 xx 10^(12)` (b) Let v be the velocity of `alpha`-particle when they are travelling towards the surface. Then `K = 1/2 mv^(2)` or `v = sqrt((2 K)/(m)) = sqrt((2 xx 84 xx 1.6 xx10^(-16))/(6.68 xx 10^(-27))) = 2 xx 10^(6) ms^(-1)` It shows that a beam of length `2 xx 10^(6)` m crosses a section in one second. But numbere of `alpha`-particle passing through the section in one second is `n = 7.81 xx 10^(11) s^(-1)`. Therefore, no. of `alpha`-particles in unit length of beam ` =n/v` Hence no. `alpha`-particles in length l of the beam `= (nl)/(v) = ((7. 81 xx 10^(11)) xx 0.30)/(2 xx 10^(6)) = 1.17 xx 10^(5)` |
|
| 1541. |
A charged belt 60 cm wide, travels at `20 ms^(-1)` between a source of charge and a sphere. The belt carries charges into the sphere at a rate corresponding to `120 mu A`. Calculate the surface charge density of the belt. |
|
Answer» Here, width of belt, `w = 60 cm = 0.60 m` , Speed of belt, `v = 20 ms^(-1)` Current `I =` change added per second `= 120 mu A = 120 xx 10^(-6)A` Area of the belt moved per second, `A = wv` Charge density, `sigma = I/A = I/w v = (120 xx 10^(-6))/(0.60 xx 20)` `= 10^(-5) Cm^(-2)` |
|
| 1542. |
A battery of e mf 10 V and internal resistance `3Omega` is connected to a resistor as shown in the figure. If the current in the circuit is 0.5 A . then the resistance of the resistor will be A. `13Omega`B. `15Omega`C. `17Omega`D. `19Omega` |
|
Answer» Correct Answer - C `I=(E)/(R+r)` `0.5=(10)/(R+3)` `10=0.5R+1.5` `0.5R=8.5` `R=(8.5)/(0.5)=17Omega`. |
|
| 1543. |
A battery of e mf 10 V and internal resistance `3Omega` is connected to a resistor as shown in the figure. If the current in the circuit is 0.5 A . then the resistance of the resistor will be A. `19 Omega`B. `17 Omega`C. `10 Omega`D. `12 Omega` |
|
Answer» Correct Answer - B |
|
| 1544. |
A galvanometer of resistance `50 Omega ` is connected to a battery of `3 V` along with resistance of `2950 Omega` in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 division the above series resistance should beA. `4450 Omega`B. `5050 Omega`C. `5550 Omega`D. `6050Omega` |
|
Answer» Correct Answer - A Current decreases `20/30` times or `2/3` times. Therefore, net resistance should become `3/2` times. `:. R+50=3/2(2950+50)` Solving we get, `R=4450Omega` |
|
| 1545. |
In an experiment on the measurement of internal resistance of as cell by using a potentionmeter, when the key `K` is kept open then balancing length is obtained at y metre. When the key `K` is closed and some resistance `R` is inserted in the resistance box, then the balancing length is found to be `x` metre. Then the internal resistance is A. `((x-y))/y R`B. `((y-x))/xR`C. `((y-x))/yR`D. `((x-y))/xR` |
|
Answer» Correct Answer - B `r=R(l_1/l_2-1)` `=R=(y/x=1)=((y-x)/x)R` |
|
| 1546. |
There are three copper wires of length and cross-sectional area (L,A),(2L,A/2)(L/2,2A). In which case in the resistance minimum?A. it is the same in all three casesB. Wire of cross-sectional are 2AC. Wire of cross-sectional area AD. Wire of cross-sectional are `(1)/(2)`A |
| Answer» Correct Answer - B | |
| 1547. |
n identical cells each of e.m.f. E and internal resistance r are connected in series. An external resistance R is connected in series to this combination. The current through R isA. `(nE)/(R + nr)`B. `(nE)/(nR +r)`C. `(E)/(R +nr)`D. `(nE)/(R+r)` |
|
Answer» Correct Answer - A |
|
| 1548. |
`E` denotes electric field in a uniform conductor, `I` corresponding current through it, `v_(d)` velocity of electrons and `P` denotes thermal power produced in the conductor, then which of the following graph is correct?A. B. C. D. |
|
Answer» Correct Answer - C |
|
| 1549. |
`I-V` characterstic of a copper wire of length `L` and area fo cross-section `A` is shown in Fig. The slope of the curve becomes A. More if the experiment is performed at higher temperatureB. More if a wire of steel of same dimension is usedC. More if the length of the wire increasedD. Less if the length of the twice is increased |
|
Answer» Correct Answer - D (d) Slope of `V-i` curve `= R(=(rho l)/(A))`. But in given curve axis of `i` and `V` are interchanged. So slope of given curve `= (1)/(R ) (=(A)/(rho l))` i.e., with the increase in length of the wire. Slope of the curve will decrease. |
|
| 1550. |
A parallel plate capacitor has its two plates connected to an ideal spring of force constant K. Relaxed length of spring is L and it is made of non conducting material. The area of each plate is A. The capacitor has a charge `q_(0)` on it. To discharge the capacitor through the resistance R, switch S is closed. If the time constant of the circuit is very large and discharge process is very slow, how much heat will be dissipated in the resistance? Assume that there is no friction and the plates always remains parallel to each other. |
|
Answer» Correct Answer - `(q_(0)^(2)L)/(2 in_(0)A)-(q_(0)^(4))/(8 in_(0)^(2)A^(2)k)` |
|