Saved Bookmarks
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.
| 2251. |
Two wires of resistance `R_(1)` and `R_(2)` have temperature coeficient of resistance `alpha_(1)` and `alpha_(2)` respectively. These are joined in series. The effeictive temperature coefficient of resistance isA. `(alpha_(1) + alpha_(2))/(2)`B. `sqrt(alpha_(1) alpha_(2))`C. `(alpha_(1) R_(1) + alpha_(2) R_(2))/(R_(1) + R_(2))`D. `(sqrt(R_(1) R_(2) alpha_(1) alpha_(2)))/(sqrt(R_(1)^(2) + R_(2)^(2)))` |
|
Answer» Correct Answer - C (c ) `R_(t_(1)) = R_(1) (1 + alpha_(1) r)` and `R_(t_(2)) = R_(2) (1 + alpha_(2) t)` Also `R_(eq.) = R_(t_(1)) + R_(t_(2)) implies R_(eq) = R_(1) + R_(2) + (R_(1) alpha_(1) + R_(2) alpha_(2))t` `implies R_(eq) = (R_(1) + R_(2)) {1 + ((R_(1) alpha_(1) + R_(2) alpha_(2))/(R_(1) + R_(2))) t}` So `alpha_(eff) = (R_(1) alpha_(1) + R_(2) alpha_(2))/(R_(1) + R_(2))` |
|
| 2252. |
For what value of R in the circuit as shown, current passing through `4 Omega` resistance will be zero A. `1 Omega`B. `2 Omega`C. `3 Omega`D. `4 Omega` |
|
Answer» Correct Answer - A |
|
| 2253. |
In the circuit shown, when switch `S_(1)` is closed and `S_(2)` is open, the ideal voltmeter shows a reaiding of `18 V`. When switch `S_(2)` is closed and `S_(1)` is open, the reading of voltmeter is `24 V`. When `S_(1)` and `S_(2)` both are closed, the voltmeter reading will be A. `14.4 V`B. `20.6 V`C. `24.4 V`D. `10.8 V` |
|
Answer» Correct Answer - A (a) Terminal potential difference across the terminals of cell `E` `V = E - ir = E - ((E)/(r + R)) r = (ER)/(R + r)` According to question `18 = (6 E)/(6 + r)` and `24 = (12 E)/(12 + r)` Solving (i) and (ii), we get `r = 6 Omega, E = 36 V` When both `S_(1)` and `S_(2)` are closed, the effective external resistance in the circuit will be. `R = (6 xx 12)/(18) = 4 Omega` So, `V = (ER)/(r + R) = (36 xx 4)/(6 + 4) = (144)/(10) = 144` volt `V = 14.4` volt Thus, choice (a) is correct. |
|
| 2254. |
An electron of mass `9xx10^(-31)kg` moves around a nucleus in a circular orbit of radius `2A^(@)` under the action of centripetal force 3.2N. Then the equivalent electric current isA. `(32)/(3pi)`B. `(3pi)/(32)`C. `(16)/(3pi)`D. `(3pi)/(16)` |
|
Answer» Correct Answer - A `i=(q)/(t)=(qv)/(2pir)` `[becauseF=(mv^(2))/(r)]` |
|
| 2255. |
In the circuit shown, when switch `S_(1)` is closed and `S_(2)` is open, the ideal voltmeter shows a reaiding of `18 V`. When switch `S_(2)` is closed and `S_(1)` is open, the reading of voltmeter is `24 V`. When `S_(1)` and `S_(2)` both are closed, the voltmeter reading will be A. 14.4 VB. 20.6 VC. 24.2 VD. 10.8 V |
|
Answer» Correct Answer - A |
|
| 2256. |
The current in a conductor varies with time `t` as `I=2-0.02t` ampers. The electric charge that passes from `t=0` to `t=100` sec isA. 50 CB. 100 CC. 25 CD. 75 C |
|
Answer» Correct Answer - B `q=int_(t_(1))^(t_(2))Idt` |
|
| 2257. |
The equivalent resistance and potential difference between A and B for the circuit is respectively A. `4Omega,8V`B. `8Omega,4V`C. `2Omega,2V`D. `16Omega,8V` |
| Answer» Correct Answer - 1 | |
| 2258. |
Two equal resistance at `0^(@)C` with temperature coefficient of resistance `alpha_(1)` and `alpha_(2)` joined in series act as a single resistance in a circuit the temperature coefficient of their single resistance will beA. `alpha_(1)+alpha_(2)`B. `(alpha_(1)alpha_(2))/(alpha_(1)+alpha_(2))`C. `(alpha_(1)-alpha_(2))/(2)`D. `(alpha_(1)+alpha_(2))/(2)` |
|
Answer» Correct Answer - 4 At `o^(@)C`, let two resistance be `R_(0)` and `R_(0)`. At `t^(@)C` let the resistances be `R_(1)` and `R_(2)`. `therefore R_(1)=R_(0)(1+alpha_(1)t)` `R_( 2)=R_(0)(1+alpha_(2)t)` when combined is series. Temperature coefficient `(beta)=("change in resistance")/("Original resistance"xxt)` or `beta=("Final resistance"-"Original resistance")/("Original resistance"xxt)` `beta=((R_(1)+R_(2))-2R_(0))/((2R_(0))t)` `=R_(0)(1+alpha_(1)t)+R_(0)(1+alpha_(2)t)(-2)/(2R_(0)xxt)` `=(2+alpha_(1)t+alpha_(2)t-2)/(2t)=(alpha_(1)+alpha_(2))/(2)` |
|
| 2259. |
Two wires of resistance `R_(1)` and `R_(2)` have temperature coefficient of resistance `alpha_(1)` and `alpha_(2)` respectively. These are joined in series. The effective temperature coefficient of resistance isA. `(alpha_(1)+alpha_(2))/(2)`B. `sqrt(alpha_(1)alpha_(2))`C. `(alpha_(1)R_(1)+alpha_(2)R_(2))/( R_(1)+R_(2))`D. `(sqrt(R_(1)R_(2)alpha_(1)alpha_(2)))/(sqrt(R_(1)^(2)+R_(2)^( 2)))` |
|
Answer» Correct Answer - 3 `R_(1)=R_(1)(1+prop_(1) theta),R_(2)=R_(2)(1+prop_(2)theta)` `R_("net")=R_(1)+R_(2)=R_(1)(1+prop_(1)theta)+R_(2)(1+prop_(2)theta)` `=R_(1)+R_(2)+theta(R_(1)prop_(1)+R_(2)prop_(2))` `=( R_(1)+R_(2))[1+(R_(1)prop_(1)+R_(2)prop_(2))/(R_(1)+R_(2))theta]` compair this equation `R=R_(0)(1+prop_("net")theta)` `therefore prop_("net")=(R_(1)prop_(1)+R_(2)prop_(2))/(R_(1)+R_(2))` |
|
| 2260. |
At the temperature `0^(@)C` the electron of conductor `B` is `n` times that of condoctor A temperaturte coefficient of resistance are equal to `alpha_(2) and alpha_(1)` respectively find the resistance and A temperaturte coefficient of resistance of a segment of these two condustore when they are conected in series |
|
Answer» Let `R_(0)` be the resistance of coefficient `A` at `0^(@)C` resistance of conductor `B` at `0^(@)C = nR_(0)` If `R_(t_1),R_(t_2)` are the resistance of coefficient A and B at at `t^(@)C` then `R_(t_1)=R_(0)(1 + alpha_(2) t) and R_(t_2)= nR_(0)(1 + alpha_(2) t)` If `R_(g)` is the resistance at `t^(@)C` then they are commected in series then `R_(s) =R_(t_1)+ R_(t_2)( 1+ alpha_(1) t) +nR_(0) (1 + alpha_(2) t)= R_(0) (1+ n) + R_(0) (alpha_(1) +n alpha_(2))t` `= R_(0) (1 + n) [ 1+ ((alpha_(1) +n alpha_(2)))/(( 1+ n))t]`....(i) If `R_(0)` is the resistance of combination of two condactor at `0^(@)C` and `alpha ` is the temperature coefficient of these two conducators in series then `R_(s) = R_(0) (1 + alpha t)` ....(ii) Comparing (i) and (ii) we have `R_(0) (1 + n) and alpha = ((alpha_(1) + n alpha_(2))/(1 + n))` |
|
| 2261. |
Resistance of a resistor at temperature `t^@C` is `R_t =R_0 (1+alphat + betat^2)`, where `R_0` is the resistance at `0^@C`. The temperature coefficient of resistance at temperature `t^@C` isA. `(a+2bt)/(1+at+bt^(2))`B. `(a+2bt)`C. `(1+at+bt^(2))/(a+2bt)`D. constant |
|
Answer» Correct Answer - A |
|
| 2262. |
At `0^(@)C` the resistance of conductor of a conductor `B` is `n` times that of condoctor A temperature coefficient of resistance for A and B are `alpha_(1) and alpha_(2)` respectively the temperature coefficient of resistance of a circiut segment constant A and B in series is |
|
Answer» Let `R_(0)` be the resistance of the conducor at `0^@C`. Then resistance of conductor B at `0^@C= n R_(0)`. Resistance of conductor A at `theta^@C`, `R_(1)=R_(0)(1+alpha_(1)theta)` Resistance of conductor B at `theta^@C`, `R_(2)=n R_(0)(1+alpha_(2)theta)` Thus the total resistance of the series combinatio at `theta^@C` is `R_(s)=R_(1)+R_(2)= R_(0)(1+alpha_(1)theta) + n R_(0)(1+alpha_(2)theta)` `= R_(0)[(1+n)+(alpha + n alpha_(2))theta]` `=(1+n)R_(0) [1+ (alpha_(1) + n alpha_(2))/(1+n) theta]` Comparing this relation with `R_(s)=R_(s0)[1+alpha_(s) theta]` We have resistance of series combination at `0^@C` `R_(s0)=(1+n)R_(0)` Tempeature coefficinet of resistance of the series combination is `alpha_(s) = (alpha_(1) _ n alpha_(2))/(1+n)` |
|
| 2263. |
The resistance t R of a conductor varies with temperature t as shown in the figure. If the variation is represented by `R_(t) = R_(0)[1+ alphat +betat^(2)]`, then A. `alpha` and `beta` are both negativeB. `alpha` and `beta` are both positiveC. `alpha` is positive and `beta` is negativeD. `alpha` is negative and `beta` are positive |
|
Answer» Correct Answer - B |
|
| 2264. |
When the rate of flow of charge through a metallic condoctor of non-uniform cross-section is uniform, then the quantity that remains constant along the conductor isA. current densityB. electric fieldC. electric potentialD. current |
|
Answer» Correct Answer - D |
|
| 2265. |
Resistance as shown in figure is negative at A. AB. BC. CD. None of these |
|
Answer» Correct Answer - A |
|
| 2266. |
The resistance of a carbon resistor of colour code Red-Red Green Silver is (in `k Omega`)A. `22 00 pm 5%`B. `2200 pm 10%`C. `220 pm 10%`D. `220 pm 5%` |
|
Answer» Correct Answer - B |
|
| 2267. |
Variation of current and voltage in a conductor has been shown in the diagram below. The resistance of the conductor is. A. 4 ohmB. 2 ohmC. 3 ohmD. 1 ohm |
|
Answer» Correct Answer - D |
|
| 2268. |
The slope of the graph showing the variation of potential difference V on X-axis and current Y-axis gives conductorA. resistanceB. resistivityC. reciprocal of resistanceD. conductivity |
|
Answer» Correct Answer - C |
|
| 2269. |
Variation of current and voltage in a conductor has been shown in the diagram below. The resistance of the conductor is. A. `4Omega`B. `2Omega`C. `3Omega`D. `1Omega` |
|
Answer» Correct Answer - B |
|
| 2270. |
A battery is used to charge a parallel plate capacitor till the potential differece between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and the work done by the battery will beA. `1//2`B. 1C. 2D. `1//4` |
|
Answer» Correct Answer - A (a) Required ratio `= (Energy stored in capacitor )/(Workdone by the battery) = (1/2CV^2)/(Ce^2)` where C = Capacitance of capacitor V = Potential difference e = emf of battery. `= (1/2Ce^2)/(Ce^2) = 1/2` `(.: V=e)`. |
|
| 2271. |
The resistance 4R,16R,64R....`oo` are connected in series. Find their equivalent resistance. |
|
Answer» Resultant of the given combination. `R_(eq)=4R+16R+ 64R+..... oo= oo` |
|
| 2272. |
In the given circuit the power dissipated through resistance `R_(1) = 9 Omega` is `P_(1)`, when switch `S_(1)` is closed and `S_(2)` is opened and the power dissipated through resistance `R_(2) = 4 Omega` is `P_(2)`, when switch `S_(2)` is closed and `S_(1)` is opened. If `P_(1) = P_(2)`, then r is equl to : A. `2 Omega`B. `6 Omega`C. `4 Omega`D. `3 Omega` |
|
Answer» Correct Answer - B |
|
| 2273. |
A perosn decides to use his bath tub water to generate electric power to run a 40W bulb. The bath tub is located at a height of 10m from the ground and it holds 200 litres of water. He instals a water driven wheel generator on the ground. At what rate should the water drain from the bath tub to light the bulb? How long can he keep the bulb on , if bath tub was full initially ? Efficiency of generator is `90%`. Take `g=9.8m//s^(2)` |
|
Answer» Correct Answer - `4//9kg//sec., 450 sec` `40=0.9xxmxxgxxhimplies40=0.9xxmxx` `10xx10impliesm=4/9 kg//s 200=4/5xxt` `t=450 sec` |
|
| 2274. |
If a cell of constant `E.M.F.` produces the same amount of the heat during the same time in two independent resistors `R_1` and `R_2`, when they are separately connected across the terminals of the cell, one after the another, find the internal resistance of the cell. |
|
Answer» Correct Answer - `sqrt(R_(1)R_(2))` `I_(1)^(2)R_(1)=I_(2)^(2)R_(2)` `(E/(R_(1)+r))^(2)xxR_(1)=(E/(R_(2)+r))^(2)xxR_(2)` `R_(1)(R_(2)+r)^(2)=R_(2)(R_(1)+r)^(2)` `R_(1)(R_(2)^(2)+r^(2)+2R_(2)r)=R_(2)(R_(1)^(2)+r^(2)+2R_(1)r)` `R_(1)R^(2)_(2)+R_(1)r^(2)+2R_(1)R_(2)r=R_(2)R_(1)^(2)+R_(2)r^(2)+2R_(1)R_(2)r` `R_(1)R_(2)(R_(2)-R_(1))=r^(2)(R_(2)-R_(1))` `r=sqrt(R_(1)R_(2))` |
|
| 2275. |
A galvanometer has resistance `G` and Current `I_(g)` produces full scale deflection. `S_(1)` is the value of the shunt which converts it into an ammeter of range `0-I` and `S_(2)` is the value of shunt for the range `0-2I`. The ratio of `S_(1)` and `S_(2)` isA. 1B. 2C. `(1)/(2)((i-i_(a))/(2iu-i_(a)))`D. ` ((2i-i_(a))/(i-i_(a)))` |
| Answer» Correct Answer - 4 | |
| 2276. |
It is advised that the jockey is not to be rubbed on potentiometer wire while using it? |
| Answer» If jockey is rubbed on the potentiometer wire, it will affect the uniformity of the cross-sectional area of the wire and hence will affect the potential gradient of the wire. | |
| 2277. |
In a potentiometer experiment, the galvanometer shows no deflection when a cell is connected across `60 cm` of the potentiometer wire. If the cell is shunted by a resistance of `6 Omega`, the balance is obtained across `50 cm` of the wire. The internal resistance of the cell isA. `0.5Omega`B. `0.6Omega`C. `1.2Omega`D. `1.5Omega` |
|
Answer» Correct Answer - C |
|
| 2278. |
The galvanometer, in each of the two given circuits does not show any deflection. Find the ratio of the resistors `R_(1)` and `R_(2)` used in these two circuits. |
|
Answer» In circuit 1, the Wheatstone bridge is in balanced state so, `4/R_(1) = 6/9` or `R_(1) = (4xx9)/6 = 6 Omega` In circuit 2, the positions of galvanometer and battery are interchanged but the galvanometer shows no defelction, means the bridge is in balanced state, so `6/12 = R_(2)/8 or R_(2) = (6xx8)/12 = 4 Omega` `R_(1)/R_(2) = 6/4 = 3/2` or `R_(1) : R_(2) = 3:2` |
|
| 2279. |
A galvanometer of resistance `100Omega` gives full scale deflection for `10mA` current. What should be the value of shunt, so that it can measure a current of 100mA?A. `11.11Omega`B. `9.9Omega`C. `1.1Omega`D. `4.4Omega` |
| Answer» Correct Answer - 1 | |
| 2280. |
A 5V battery with internal resistance `2Omega` and a 2V battery with internal resistance`1Omega` are connected to a `10Omega` resistor as shown in the figure. The current in the `10Omega` resistor isA. `0.27A` `P_(2)` to `P_(1)`B. 0.03 A `P_(1)` to `P_(2)`C. 0.03 A `P_(2)` to `P_(1)`D. 0.27 A `P_(1)` to `P_(2)` |
|
Answer» Correct Answer - C `(X)/(R)=(l)/(100-l)` |
|
| 2281. |
A 5V battery with internal resistance `2Omega` and a 2V battery with internal resistance`1Omega` are connected to a `10Omega` resistor as shown in the figure. The current in the `10Omega` resistor isA. `0.27A`B. `0.05A`C. `0.25A`D. `0.3A` |
|
Answer» Correct Answer - C `I_(S)=(2E)/(2r+R)` and `I_(P)=(2E)/(r+2R)` `r=R` |
|
| 2282. |
The potentiometer wire `AB` is `600` cm long. a. At what distance from A should be jockey `J` touch the wire to get zero deflection i the galvanometer. b. If the jockey touches the wire at a distance `560 cm` from A, what will be the current through the galvanometer. |
|
Answer» Correct Answer - (a)` 320 cm`(b)` (3E)/(8r)` Current through potentiometer wire `I = (E )/(15 r + r) = (E )/(16 r)` potential drop across potentiometer wire `= (E )/(16 r) xx 15 r = (15E )/(16)` potential gradient of potentiometer wire `K = (15 E//16)/(600) "volt"//cm` (a) UIf galvanometer shown zero deffiection at length `l` of potentiometer wire , then `(E )/(2) = K1 = (15 E)/(16 xx 600)I` `or l = (16 xx 600)/(15 xx 2) = 320 cm` (b) Fall of potential across the length `560 cm = K xx 560` volt .If `l` is the current through galvanometer arm , then `K xx 560 - (E )/(2) = I r or (15 E)/(16 xx 600) xx 560 - (E )/(2) = I r ` `or I = (3E)/(8r)` |
|
| 2283. |
In the following diagram the deflection in the galvanometer in a potentiometer circuit is zero. Then:- A. `E_(1)gtE_(2)`B. `E_(2)gtE_(1)`C. `E_(1)=E_(2)`D. `E_(1)+E_(2)=E` |
| Answer» Correct Answer - 2 | |
| 2284. |
One of the circuits for the measurement of resistance by potentiometer is shown. The galvanometer is connected at point `A` and zero deflection is observed is observed at length `PJ = 30 cm`. In second case the secondary cell is changed. Take `E_(s) = 10 V` and `r = 1 Omega` in `1^(st)` reading and `E_(s) = 5 V` and `r = 2 Omega` in `2^(nd)` reading. In second case, the zero deflection is observed at length `PJ = 10 cm`. What is the resistance `R` (in ohm)?A. `1 Omega`B. `2 Omega`C. `3 Omega`D. `4 Omega` |
|
Answer» Correct Answer - A (a) When there is no current through galvanometer, current throuhg `R` is `I = (E_(S))/(R + r)` Potential difference `= IR = (E_(S) R)/(R + r)` Now `(E_(S) R)/(R + r) = K l` For first case, `(10 R)/(R + 1) = K (30)` Divide (i) and (ii) to get `R = 10 Omega` |
|
| 2285. |
A potentiometer circuit is setup as shown. The potential gradient across the potentiometer wire is `k "volt"//cm` and the ammeter present in the circuit reads. `1.0 A` When two-way key is switched off. The balance point, when the key between the terminals (i) 1 and 2 (ii) 1 and 3, is plugged in, are found to be at lengths `l_(2) cm` and `l_(2) cm` respectively. The magnitudes, of the resistors `R` and `X`, in ohm, are then, equal, respectively, to A. `K(l_(2)-l_(1)) and K l_(2)`B. `K l_(1) and K(l_(2)-l_(1))`C. `K(l_(2)-l_(1)) and K l_(1)`D. `K l_(1) and K l_(2)` |
|
Answer» Correct Answer - b When the key between terminals `1 and 2` is plugged is ,`R` is in the circuit `V_(1)=IR=1 R=Kl_(1)" "(because I=1A)` when the key between terminal 1 and 2 is plugged in `R` and `X` both are in the circuit `V_(2)=U(R+X)=1(R+X)=Kl_(2)` `V_(2)-V_(1)=R+X-R=X=K(l_(2)-l_(1))` |
|
| 2286. |
`5xx10^(-3)A` current gives a full scale deflection in a galvanometer of `1Omega` resistance. To measure `5V` with this galvanometer, a resistance of……….is connected in ……….. of the galvanometer. |
|
Answer» Correct Answer - `999 Omega`, series `R=V/(I_(s))-4=5/(5xx10^(-3))-1=99 Omega` in series |
|
| 2287. |
A potentiometer wire `AB` is `100 cm` long and has total resistance of `10Omega`. If the galvanometer shows zero deflection at the position `C`, then find value of unknown resistance |
|
Answer» Correct Answer - `4 Omega` Potentiometer will give terminal potential. `V=E-Ir impliesV=5-5/((R+1))xx1=x xx40` `5-5/((R+1))=10/100xx40impliesR=4 Omega` |
|
| 2288. |
A battery of emf `epsilon_0=10V` is connected across a `1m` long uniform wire having resistance `10Omega//m.` Two cells of emf `epsilon_1=2V` and `epsilon_2=4V` having internal resistance `1Omega` and `5Omega` respectively are connected as shown in the figure. If a galvanometer shows no deflection at the point `P`, find the distance of point `P` from the point `a`. |
|
Answer» Correct Answer - `46.67 cm` `E_(eq)=(E_(1)r_(2)+E_(2)r_(1))/(r_(1)+r_(2))=(10+4)/(5+1)=7/3` volt `E_(eq)x.l implies 7/3=10/(10+10)xx10xxl` `l=700/(3xx5)m=46.67 cm` |
|
| 2289. |
`V_A - V_B = ?` .A. `-1 V`B. `-2 V`C. `-3 V`D. `-4 V` |
|
Answer» Correct Answer - D `i= (5 -2)/(1 + 2) = 1 A` `V_B - V_A = 5 - ixx 1 = 5 - 1 xx = 4 V` `V_A - V_B = -4 V`. |
|
| 2290. |
You are given two resistances `R_(1)` and `R_(2)`. By using them singly, in series and in parallel, you can obtain four resistances of `1.5Omega, 2Omega, 6Omega` and `8Omega`. The value of `R_(1)` and `R_(2)` areA. `1Omega, 7Omega`B. `1.5Omega, 6.5Omega`C. `3Omega,5Omega`D. `2Omega, 6Omega` |
|
Answer» Correct Answer - D |
|
| 2291. |
A battery of emf `E_0=12V` is connected across a `4 m` long uniform wire having resistance `4Omega//m`. The cell of small emfs `epsilon_1=2V` and `epsilon_2=4V` having internal resistance `2Omega` and `6Omega` respectivley are connected as shown in the figure. If galvanometer shows no diflection at the point `N` the distance of points `N` from the point `A` is equal to A. `(5)/(3)m`B. `(4)/(3)m`C. `(3)/(2)m`D. None of these |
|
Answer» Correct Answer - A |
|
| 2292. |
A battery of emf `E_0=12V` is connected across a `4 m` long uniform wire having resistance `4Omega//m`. The cell of small emfs `epsilon_1=2V` and `epsilon_2=4V` having internal resistance `2Omega` and `6Omega` respectivley are connected as shown in the figure. If galvanometer shows no diflection at the point `N` the distance of points `N` from the point `A` is equal to A. `5/3m`B. `4/3m`C. `3/2m`D. none of these |
|
Answer» Correct Answer - D Equivalent emf of two batteires `epsilon_1` and `epsilon_2` is `epsilon=(epsilon_1//r_1+epsilon_2//r_2)/(1//r_1+1//r_2)=((2//2)+(4//6))/((1//2)+(1//6))` `2.5V` Now, `V_(AN)=epsilon` `:. I_(AN)=(R_(AN))=epsilon` or `((12)/(4+4xx4))(4)(l)=2.5` Solving this equation we get `l=2.5/24m` |
|
| 2293. |
The potential differece `V_A-V_B` between points `A` and `B` for the circuit segment shown in figure at the given instant is `A. 12VB. `-12V`C. 6VD. `-6V` |
|
Answer» Correct Answer - B,C,D |
|
| 2294. |
A silver wire has a resistance of `2.1Omega` at `27.5^(@)C&2.7Omega` at `100^(@)C` Determine the temperature coefficient of resistivity of silver. |
|
Answer» `R_(1)=R_(0)(1+alphatheta)` `2.1=R_(0)(1+alphaxx27.5)` ….(1) `2.7=R_(0)(1+alphaxx100)` ….(2) Solve equation (1) and (2) `alpha=0.0039^(@)C^(-1)` |
|
| 2295. |
By using only two resistance coils-singly, in series, or in parallel one should be able to obtain resistances of 3, 4, 12 and 16 ohms . The separate resistances of the coil areA. 3 and 4B. 4 and 12C. 12 and 16D. 16 and 3 |
|
Answer» Correct Answer - B |
|
| 2296. |
A technician has onlytwo resistance coils. By using them singly, in series or in parallel, he is able to obtain the the resistance `3 Omega,4 Omega,12 Omega and 16 Omega`. The resistance of two coils are:A. `6Omega and 10 Omega`B. `4 Omega and 12 Omega`C. `7 Omega and 9 Omega`D. `4 Omega and 16 Omega` |
|
Answer» Correct Answer - A,C |
|
| 2297. |
If the resistance of a conductor is `5 Omega` at `50^(@)C` and `7Omega` at `100^(@)C` then the mean temperature coefficient of resistance of the material isA. `0.008//C`B. `0.006//C`C. `0.004//C`D. `0.001//C` |
|
Answer» Correct Answer - A |
|
| 2298. |
We are able to obtain fairly large currents in a conductor becauseA. The electron drift speed is usually very largeB. The number density of free electrons is very high and this can compensate for the low values of the electron drift speed and the very small magnitude of the electron chargeC. The number density of free electrons as well as the electron drift speeds are very large and these compensate for the very small magnitude of the electron chargeD. The very small magnitude of the electron charge has to be divided by the still smaller product of the number density and drift speed to get the electric current |
|
Answer» Correct Answer - B |
|
| 2299. |
If the deflection of galvanometer in Wheatstone circuit is zero, the value of resistance will be A. `2 Omega`B. `4 Omega`C. `6 Omega`D. `8 Omega` |
| Answer» Correct Answer - C | |
| 2300. |
Two cells of unequal emfs `epsi_(1)" and "epsi_(2)`, and internal resistances `r_(1)" and "r_(2)` are joined as shown. `V_(A)" and "V_(B)` are the potentials as A and B respectively. A. One cell will supply energy to the otherB. The potential difference across both the cells will be equalC. The potential difference across one cell will be greater then its emfD. `D_(A)-V_(B)=((epsi_(1)r_(2)+epsi_(2)r_(1)))/(r_(1)+r_(2))` |
| Answer» Correct Answer - A::B::C::D | |