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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.
| 1251. |
Figure shows an experimental set up to find the value of an unknown resistance `(R_x)` using a meter bridge. `AB` is the uniform meter bridge wire of length `L = 100 cm`. When the sliding jockey is placed at `J (AJ = x)`, the galvanometer shows zero deflection. `AJ = x` is known as balancing length and is measured using a scale having 1 mm as least count. (a) In one experiment known resistance R was taken to be `20 Omega` and balancing length was measured as `x = (20.0 pm 0.1) cm`. find the value of `R_(x)`. (b) Show that fractional error in calculated value of `R_(x)` is least when `x =(L)/(2)`. What shall we do to ensure that `x` is close to `L//2` ? |
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Answer» Correct Answer - (a) `(5.00 pm 0.03)Omega` |
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| 1252. |
Two heaters designed for the same voltage `V` have different power ratings. When connected individually across as source of voltage` V`, they produce `H` amount of heat each in time `t_1` and `t_2` respectively. When used together acros the same source, they produce H amount of heat in time tA. if they are in series `t=t_1+t_2`B. if they are in series `t=2(t_1+t_2)`C. if they are in parallel `t=(t_1t_2)/((t_1+t_2))`D. if they are in parallel `t=(t_1t_2)/(2(t_1+t_2))` |
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Answer» Correct Answer - A::C `H=V^2/R_1 t_1implies R_1=(V^2t_1)/H` Similarly, `R_2=(V^2t_2)/H` In series, `H=(V^2/(R_1+R_2))t` `t=(H(R_1+R_2))/V^2` substituting the value of `R_1` and `R_2` we get `t=t_1+t_2` In parallel, `H=V^2/R_("net")t=V^2t(1/R_1+1/R_2)` `=V^2t(H/(V^2t_1)+H/(V^2t_2))` Solving we get, `t=(t_1t_2)/(t_1+t_2)` |
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| 1253. |
The external diameter of a 314 m long copper tube is 1.2 cm and the internal diameter is 1 cm. calculate its resistance if the specific resistance of copper is `2.2xx10^(-8)` ohm-metre.A. `2.0xx10^(-1)ohm`B. `4.4xx10^(-2)ohm`C. `3.14xx10^(-2)ohm`D. `2.2xx10^(-2)ohm` |
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Answer» Correct Answer - 1 Given `rho=2.2xx10^(-8)` ohm-metre `l=314m,D=1.2xx10^(-2)m,D_(1)=1.0xx10^(-2)m`. `R=(rhol)/((pi)/(4)(D^(2)-D_(1)^(2)))rArrR=(4xx2.2xx10^(-8)xx314)/(pi(1.2^(2)-1.0^(2))xx10^(-4))` `=4 xx5xx1 0^(-2)Omega=2xx10-1Omega`. |
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| 1254. |
In the shown arrangement of the experiment of the meter bridge if AC corresponding to null deflection of galvanometer is x, what would be its value if the radius of the wire AB is doubled?A. 4xB. 2xC. xD. x/2 |
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Answer» Correct Answer - 3 Potential across `R_(1)`=potential across x. From wheatston e Bridge principle . `(R_(1))/(R_(2))=(R_(AC))/(R_(C B))=(AC)/(CB)` If radius is doubled, then using `(R=(rhol)/(A))` `R_(AC)= 4R_(AC)` `R_(CB)=4R_(CB)` From (i) and (ii), `(R_(1))/(R_(2))=(4R_(AC))/(4R_(CB))=(R_(AC))/(R_(CB))=(AC)/(CB)` Thus new balancing length AC=x is same as b efore. |
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| 1255. |
A ring consisting of two parts `ADB` and `ACB` of same conductivity k carries an amount of heat `H` The `ADB` part is now replaced with another metal keeping the temperature `T_91)` and `T_(2)` constant The heat carried increases to `2H` What should be the conductivity of the new `ADB` Given `(ACB)/(ADB)=3` .A. `(7)/(3)K`B. 2KC. `(5)/(2)K`D. 3K |
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Answer» Correct Answer - A |
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| 1256. |
An electric heater operating at 220 volts boils 5 litre of water in 5 minutes. If it is used on 110 volts, it will boil the same amount of water inA. 10 minutsB. 20 minutesC. 15 minutesD. 25 minutes |
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Answer» Correct Answer - B `H=(V^(2))/(R)t` `therefore(t_(2))/(t_(1))=(V_(1)^(2))/(V_(2)^(2))` |
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| 1257. |
There are two separate coils in a heater . If one col is used then it takes 20 minutes to boil a given amount of water completely. But when its other coil is used then it takes 5 minutes to boil the same amount of water. Calculate the time taken to boil the same amount of water using the same source when the two coils are connected (1) in series and (2) in parallel. |
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Answer» Let H=quantity of heat required to boil water than for `1^(st)` coil `H=(V^(2))/(R_(1))(20xx 60)`, for `2^(nd)` coil `H=(V^(2))/(R_(2))(5xx60)` When coils are connected in series `H=(V^(2))/((R_(1)+R_(2)))xx(t_( s)xx60) rArr H=(HV^(2)xx(t_(s)xx60))/((V^(2))/(4)xx(40xx60)+(V^(2)(5xx 60))/(4))rArrt_(s)=25` second When coils are connected in parallel `H=V^(2)(Deltatxx60)xx((1)/(R_(1))+(1)/(R_(2)))=V^(2)(Deltatxx60)xx[(H)/(V^(2)(20xx60))+(H)/(V^(2)(15xx 60))]rArr Deltat=4` second. |
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| 1258. |
A heater takes 50 minute to boil a given amount of water, its coil is broken in five equal parts. If one part is used, then how much time it will take to boil the same amount of water using the same source? |
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Answer» For heater `H=(V^(2))/(R)xx50` for one part of coil `H=(V^(2))/(R//5)xxt` By equation (1) and (2) `(V^(2))/(R)xx50=(V^(2))/(R//5)xxt` t=10 minute |
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| 1259. |
The current in `C` is `2 A`. The rate of heat of loss in `A,B` and `C are 1.20, 0.60 and `1.20 J//sec respectively. The emf of the cell [for cell internal resistance = 0] is A. `3 V`B. `1.5 V`C. `3.6 V`D. None |
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Answer» Correct Answer - B (b) Power `= EI` `implies (1.20 + 0.60 + 1.20) = E xx 2 implies E = 1.5 V` |
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| 1260. |
`AB` is a potentiometer wire of length `100 cm` and its resistance is `10 ohms`. It is connected in series with a resistance `R = 10 ohms` and a battery of e.m.f. `2 V` and negligible internal resistance. If a source of unknown e.m.f. `E` is balanced by `40 cm` length of the potentiometer wire, the value of `E` is A. `0.8V`B. `1.6V`C. `0.08V`D. `0.16V` |
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Answer» Correct Answer - D |
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| 1261. |
`AB` is a potentiometer wire of length `100 cm` and its resistance is `10 ohms`. It is connected in series with a resistance `R = 10 ohms` and a battery of e.m.f. `2 V` and negligible internal resistance. If a source of unknown e.m.f. `E` is balanced by `40 cm` length of the potentiometer wire, the value of `E` is A. `0.8 V`B. `1.6 V`C. `0.08 V`D. `0.16 V` |
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Answer» Correct Answer - D (d) `E = (e)/((R + R_(h) = r)) (R )/(L) xx l` `= (2)/((10 + 40 + 0)) xx (10)/(1) xx 0.4 = 0.16 V` |
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| 1262. |
Temperature of 1 mole of an ideal gas is increased from 300K to 310K under isochoric process. Heat supplied to the gas in this process is Q=25R, where R=universal gas constant. What amount of work has to be done by the gas if temperature of the gas decreases from 310K to 300K adiabatically?A. 10 RB. 50 RC. 25 RD. `(25)/(2)R` |
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Answer» Correct Answer - C |
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| 1263. |
Two nonideal batteries are connected in parallel. Consider the following statements: (A)The equivalent emf is smaller than either of the two emfs. (B) The equivalent internal resistance is smaller than either of the two internal resistances.A. Both A and B are correctB. A is correct but B is wrongC. B is correct but A is wrongD. Both A and B are correct |
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Answer» Correct Answer - C |
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| 1264. |
In the circuit shown here, what is the value of the unknown resistor R so that the total resistance of the circuit between points P and Q is also equal to R A. `3Omega`B. `sqrt(39)Omega`C. `sqrt(69)Omega`D. `10Omega` |
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Answer» Correct Answer - 3 `((R+3).10)/((R+3)+10)+3=R` (given) solve it `(10R+30)=(R-3)(R+13)` `10R+30=R^(2)-39+ 10R` or `R=sqrt(69)Omega` |
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| 1265. |
In the circuit shown in the adjoining figure, the current between `B` and `D` is zero, the unknown resistance of A. `4 Omega`B. `2 Omega`C. `3 Omega`D. e.m.f of a cell is required to find the value of X |
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Answer» Correct Answer - B |
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| 1266. |
In the circuit shown in the adjoining figure, the current between `B` and `D` is zero, the unknown resistance of A. `4 Omega`B. `2 Omega`C. `3 Omega`D. e.m.f. of a cell is required to find will be, when |
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Answer» Correct Answer - B (b) By balanced Wheatstone bridge condition `(16)/(X) = (4)/(0.5)` `implies X = (8)/(4) = 2 Omega` |
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| 1267. |
`AB` is a potentiometer wire Fig. If the value of `R` is increased, in which direction will the balance point `J` shift ? |
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Answer» With the increase of R, the current in main circuit decreases which in turn, decreases the potential difference across AB and hence potential gradient (k) across AB decreases. Since, at neutral point, for given emf of cell, I increases as potential gradient (k) across AB has decreased because. `E=kI` ltBrgt thus, with the increase of I, the balance point neutral point will shift towards B. |
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| 1268. |
AB is wire of potentiometer with the increase in the value of resistance R, the shift in the balance point J will be A. toward BB. toward AC. remains constantD. first towards B then back towards A |
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Answer» Correct Answer - A Due to increase in resistance R the current through the wire will decrease and hence the potentail gradient also decreases, which results in increase in balancing length. So J will shift towards B. |
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| 1269. |
Which of the following can cause the null point of a potentiometer to shift beyond the wire ?A. e.m.f. of driving cell is lowB. e.m.f. of accumulator is highC. length of wire is smallD. length of wire is large |
| Answer» Correct Answer - A | |
| 1270. |
For a copper-iron and a chromel-alumel thermocouple, the plots between thermoelectric emf and the temperature `theta` of the hot junction (when the cold junction is at `0^(@)C`)are found to satisfy approximately the parabola equation `V=alpha theta+(1)/(2)betatheta^(2)` with `alpha=14muV^(@)C^(-1) beta=-0.04V^(@)C^(-2)` (copper-iron) `alpha=41muCV^(@)C^(-1),beta=-0.002V^(@)C^(-2)` (chromel-alumel) Which of the two thermocouples would you use to measure temperature in the range of a about `500^(@)C` to `600^(@)C`? |
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Answer» The temperature `theta_(n)` (neutral temperature) corresponding to maximum emf is given by `(dV)/(dtheta)=0,i.e., alpha+beta theta_(n)=0` or `theta_(n)=-(alpha)/(beta)` For a copper-iron therm ocouple, `theta_(n)=-(14)/(-0.04).^(@)C=350^(@)C` |
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| 1271. |
The mass of product liberated on anode in an electrochemical cell depends on (where t is the time period for which the current is passed. )A. `(It)^(1//2)`B. `It`C. `I//t`D. `I^(2)t` |
| Answer» Correct Answer - B | |
| 1272. |
An electric toaster uses nichrome for its heating element. When a negligibly small current passes through it, its resistance at room temperature `(27.0^(@)C)` is found to be `75.3 Omega`. When the toaster is connected to a 230 V supply, the current settles, after a few seconds, to a steady value of 2.68 A. what is the steady temperature of the nichrome element? The Temperature coefficient of resistance of nichrome averaged over the temperature range involved is `1.70 xx 10^(-4) .^(@)C^(-1)`. |
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Answer» When the current through the element is very small, heating effects can be ignored and the temperature `T_(1)` of the element is the same as room temperature. When the toaster is connected to the supply, its initial current will be slightly higher than its steady value of 2.68 A. But due to heating effect of the current, the temperature will rise. This will cause an increase in resistance and a slight decrease in current. In a few seconds, a steady state will be reached when temperature will rise no further, and both the resistance of the element and the current drawn will achieve steady values. The resistance `R_(2)` at the steady temperature `T_(2)` is `R_(2) = (230 V)/(2.68 A) = 85.8 Omega` Using the relation `R_(2) = R_(1) [1 + alpha (T_(2) - T_(1))]` With `alpha = 1.70 xx 10^(-4) .^(@)C^(-1)`, we get `T_(2) - T_(1) = ((85.8 - 75.3))/(75.3( xx 1.70 xx 10^(-4)) )= 820^(@)C` that is, `T_(2) = (820 + 27.0) .^(@)C = 847^(@)C` Thus, the steady temperature of the heating element (when heating effect due to the current equals heat loss to the surroundings ) is `847^(@)C` |
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| 1273. |
In an `RC` circuit while charging, the graph of `1n i` versis time is as shown by the dotted line in the diagram figure. Where `i` is the current. When the value of the resistance is doubled, which of the solid curve best represents the variation of `1n i` versus time A. PB. QC. RD. S |
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Answer» Correct Answer - C |
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| 1274. |
Find equivalent resistance between A and D A. `2 Omega`B. `3 Omega`C. `1 Omega`D. `5 Omega` |
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Answer» Correct Answer - C |
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| 1275. |
Three identical resistors are connected in parallel and total resistance of the circuit is R/3. Find the value of each resistance. |
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Answer» Given three resistances are identical. Hence, `R_(1) = R_(2) = R_(3) = x` (say) Total resistance in parallel, `R_(P) = (R )/(3)` If three identical resistances are connected in parallel, then `(1)/(R_(P)) = (1)/(R_(1)) + (1)/(R_(2)) + (1)/(R_(3))` `(1)/(((R )/(3))) = (1)/(x) + (1)/(x) + (1)/(x) implies (3)/(R ) = (1 + 1 + 1)/(x)` `:. x = R` |
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| 1276. |
In the simple Wheatstone bridge circuit, where the length `AB` of bridge circuit wire is `1m` the resistance `X` and `Y` have value `5 Omega` and `2 Omega` respectively, When `X` is shuted by a length of a wire the balance point to found to be `0.625 m` from A. What is the resistance of the shunt ? If the shunt is `0.75 m` long end 0.25 mm in diameter, what is the respectivity of the material of the wire ? |
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Answer» Correct Answer - `6.54 xx 10^(-7) Omega m ` Let `R` be the resistance of shunted wire , the effective resistance of `R and 5 Omega` is parallel ` =- (5 xx R) (% + R)` As balance pouint `(5R//(5 + R))/(2) = (0.625)/(1 - 0.625) = (0.625)/(0.375) = (5)/(3)` On solving we get `R = 10 Omega` Now `p = (RA)/(1) = (R pir^(2))/(l) ` `= (10 xx (22//7) xx (0.125 xx 10^(-3))^(2))/(0.75)` `= 6.54 xx 10^(-7) Omega m ` |
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| 1277. |
In comparing the resistance of two cells `P and Q` with a sides wire bridge, a balance point is obtained when the sliding contact is `30 cm` from the zero end of the wire. The resistances `P and Q` are interchanged and the balance is obtained at `120 cm` from the same emf. Find the ratio of the resistance `P and Q` and the length of the bridge wire |
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Answer» Correct Answer - `150 cm` Let `1` be the total length of bright wire AS per question `(P)/(Q) = (30)/(l - 30)`…..(i) and `(Q)/(P) = (120)/(l - 120) or (P)/(Q)= (l - 120)/(120)`……(ii) Solving (i) and (ii) we get `(P)/(Q) = (1)/(4) and l = 150 cm` |
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| 1278. |
A rotary potentiometer has a circular resistance (C) and a conducting wiper (W) can rotate over it. An ideal battery and an ideal voltmeter are connected to it as shown in the figure. As the wiper is rotated on the circular resistance the reading of the voltmeter changes from zero to `20 V`. What is reading of the voltmeter when wiper is at angular position `theta = 120^(@)` ? Assume that resistance C is almost a complete circle and resistance of all connecting wires and knobs in the circuit is zero. |
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Answer» Correct Answer - 13.33 volt |
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| 1279. |
In the circuit shown in figure. AB is a uniform wire of length L and resistance R. P is a sliding contact. Take the ratio of lengths PB to AB as `alpha`. (a) Find the ratio `(V_(0))/(V)` in terms of `R_(0),R` and `alpha` (b) Predict the value of `(V_(0))/(V)` when `R_(0) rar infty`. Use the result obtained in (a) to show that your prediction is correct. (c) Find the ratio `(V_(0))/(V)` when `R_(0) = 2R and alpha = (1)/(2)` |
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Answer» Correct Answer - (a) `(alpha)/((1-alpha)((alphaR)/(R_(0))+1)+alpha)` (b) `alpha` (c) `(4)/(9)` |
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| 1280. |
The balancing lengths for the cells of e.m.f. `E_(1)` and `E_(2)` are `l_(1)` and `l_(2)` respectively. If they are connected separately thenA. `(E_(1))/(E_(2))=(l_(1))/(l_(2))`B. `(E_(1))/(E_(2))=(l_(1)+l_(2))/(l_(1)-l_(2))`C. `(E_(1))/(E_(2))=(l_(1)+l_(2))/(l_(2))`D. `(E_(1))/(E_(2))=(l_(1)-l_(2))/(l_(2))` |
| Answer» Correct Answer - A | |
| 1281. |
Instead of voltmeter, potentiometer is always used to measure e.m.f. of cell, because at the null point, the potentiometerA. does not draw current from balanced cellB. draw current from driven cellC. ends its current through driven cellD. none of these |
| Answer» Correct Answer - A | |
| 1282. |
Potentiometer is used toA. measure e.m.f. of cellB. compare e.m.f. of two cellsC. determine internal resistance of cellD. all of these |
| Answer» Correct Answer - D | |
| 1283. |
The equation, `( E_(1))/(E_(2))= (l_(1)+l_(2))/(l_(1)-l_(2))`, is used to compare e.m.f. of two cells byA. individual methodB. sum and difference methodC. null deflection methodD. equal deflection method |
| Answer» Correct Answer - B | |
| 1284. |
The internal resistance of the cell can be determined by,A. ohm meterB. galvanometerC. voltmeterD. potentiometer |
| Answer» Correct Answer - D | |
| 1285. |
The unit of potential gradient isA. volt cmB. ohm cmC. volt/cmD. volt/ampere |
| Answer» Correct Answer - C | |
| 1286. |
The potentiometer wire of resistance R is connected in series with a cell of e.m.f. E and of resistance `R_(h)`. The current flowing through the potentiometer wire isA. `I=(E)/(R+r+R_(h))`B. `I=(E)/(R+r)`C. `I=(RE)/(R_(h)+r)`D. `I=(E)/(R)` |
| Answer» Correct Answer - A | |
| 1287. |
Statement-1: Two bulbs of `25 W` and `100 W` rated at `200 V` are connected in series across `200 V` supply. Ratio of powers of both the bulb in series is `2:1` Statement-2: In series current in both bulbs is same, therefore power depends on resistance of bulb.A. Statement-1 is True,Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True,Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1C. Statement -1 is True, Statement-2 is false.D. Statement-1 is False, Statement-2 is True. |
| Answer» Correct Answer - D | |
| 1288. |
Statement-1: For zero value of `R` in circuit power transfer in external resistance will be maximum. Statement-2: Since `R_(1)gtr` in the given circuit, So, power transfer in external resistance will be maximum when `R=0`,A. Statement-1 is True,Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True,Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1C. Statement -1 is True, Statement-2 is false.D. Statement-1 is False, Statement-2 is True. |
| Answer» Correct Answer - C | |
| 1289. |
Assertion : When current through a bulb is increased by `2%` power increases by `4%` Reason : Current passing through the bulb is `prop (1)/(Resistance)`.A. If both assertion and reason are ture and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason both are false. |
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Answer» Correct Answer - D (b) `P = sum^(2) R` `100 % xx ((dP)/(P)) = 2 ((dI)/(I)) 100 %` |
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| 1290. |
Find equivalent resistance between A and B A. RB. `(3R)/(94)`C. `(R)/(2)`D. `2R` |
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Answer» Correct Answer - C |
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| 1291. |
Statement I: Internal resistance of a battery is drawn parallel to a battery in electical circuit. Statement II: Heat generated in a battery is due to internal resistance.A. If both assertion and reason are ture and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason both are false. |
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Answer» Correct Answer - D (d) Internal resistance is drawn in series with battery. |
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| 1292. |
What are thermistors? Explain their use in brief. |
| Answer» A thermisor is a heat sensitive device whose resistivity changes very rapidally with change of temperature. The temperature coefficient of resistivity of a thermistor is very high, which may be positive or negative. A thermistor can have a resistance in the range of `0.1 Omega` to `10^(7) Omega` Thermistors are used (i) to detect small temperature changes (ii) to safe guard the filament of picture tube of a television set against the variation of current (iii) in temperature contorl units of industry (iv) for voltage stabilisation and remote sensing. | |
| 1293. |
Four resistances, each of `10 Omega`, are connected to form a square as shown in (Fig. 3.52), find the equivalent resistance between the opposite corners `A and C`. .A. 10 ohmB. 40 ohmC. 20 ohmD. `10//4` ohm |
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Answer» Correct Answer - A |
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| 1294. |
Assertion: Terminal potential difference of a cell is always less than its emf. Reason: Potential drop on internal resistance of cell increases terminal potential difference.A. If both Assertion `&` Reason are True `&` the Reason is a correct explanantion of the AssertionB. If both Assertion `&` Reason True but Reason is not a correct explanation of the AssertionC. If Assertion is True but the Reason is FalseD. If both Assertion `&` Reason are False. |
| Answer» Correct Answer - 4 | |
| 1295. |
A storage battery with emf `2.6 V` loaded with external resistance produces a current `1 A`. In this case, the potential difference between the terminals of the storage battery equals `2 V`. Find the thermal power generated in the battery and the net power supplied by the battery for external circuit. |
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Answer» Correct Answer - B `V=E-ir` `:. r=(E-V)/i=(2.6-2)/1` `=0.6Omega` Now, power generated in the bttery `P=i^2r` `=(1)^2(0.6)=0.6W` Power supplied by the battery `=Ei` `=2.6W` `:.` Net supplied for external circuit `=2.6-0.6=2.0W` |
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| 1296. |
The resistors of resistances `2 Omega, 4 Omega` and `8 Omega` are connected in parallel, then the equivalent resistance of the combination will beA. `(8)/(7) Omega`B. `(7)/(8)Omega`C. `(7)/(4)Omega`D. `(4)/(9)Omega` |
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Answer» Correct Answer - A |
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| 1297. |
Assertion : The e.m.f. of the drivercell in the potentiometer experiment should be greater than the e.m.f. of the cell to determined.A. If both assertion and reason are ture and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason both are false. |
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Answer» Correct Answer - A (a) If either the e.m.f. of the driver cell or potential difference across whole potentiometer wire is lesser than the e.m.f of the experimental cell, then balance point will not obainted. |
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| 1298. |
Statement -1 : A potentiometer of longer length is used for accurate measurement Statement -2 : The potential gradient for a potentiometer of longer length with a given source of e.m.f becomes smallA. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is falseD. If the assertion and reason both are false |
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Answer» Correct Answer - A |
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| 1299. |
Thermistors differ ordinary resistors. Explain. |
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Answer» A thermistor differs from an ordinary resistance in the following ways. (i)The resistivity and hence the resistance of thermistor changes very rapidly with change of temperaure. (ii) The temperature coefficient of resistivity of thermistor is very high. (iii) The temperature coefficient of resistivity of thermistor can be both, positive and negative. |
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| 1300. |
Assertion : In the following circuit e.m.f. is `2 V` internal resistance of the cell is `1 Omega` and `R = 1 Omega` the reading of the voltmeter is `1 V`. Reason : `V = E - ir`, where `E = 2 V, i= 1 A` and `R = 1 Omega`A. If both assertion and reason are ture and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason both are false. |
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Answer» Correct Answer - A (a) Here `E = 2 V, I = (2)/(2) = 1 amp` and `r = 1 Omega` `:. V = E - ir = 2 - (1)(1) = 1` volt |
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