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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.
| 151. |
A potentiometer wire has resistance `40Omega` and its length is 10m. It is connected by a resistance of `760Omega` in series if emf of battery is 2V then potential gradient is:-A. `0.5xx10^(-6) V//m`B. `1xx10^(-6)V//m`C. `1xx10^(-2)V//m`D. `2xx10^(-6)V//m` |
| Answer» Correct Answer - 3 | |
| 152. |
A potentiometer wire of 10 m length and 20 ohm resistance is connected in series with a resistance R ohms and a battery of emf 2V, negligible internal resistance, potential gradient on the wire is 0.16 millivolt/centimetre then R is ohm….A. `50Omega`B. `60Omega`C. `230Omega`D. `46Omega` |
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Answer» Correct Answer - C `v=irhol` |
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| 153. |
The `n` rows each containinig `m` cells in series are joined parallel. Maximum current is taken from this combination across jan external resistance of `3 Omega` resistance. If the total number of cells used are 24 and internal resistance of each cell is `0.5 Omega` thenA. `m = 8, n = 3`B. `m = 6, n = 4`C. `m = 12, n = 2`D. `m = 2, n = 12` |
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Answer» Correct Answer - C `n` : number of rows `m` : total no. of cells in a row . `R = 3 Omega, r = 0.5 Omega` `mn = 24`…(i) From maximum current in `R` `R = (mr)/(n) rArr 3 = (m)/(n) xx 0.5` …(ii) `m = 6 n` `mn = 24 rArr 6n^2 = 24 rArr n = 2, m = 12`. |
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| 154. |
A potential difference is applied across the filament of a bulb at `t = 0` and it is maintained at a constant value while the filament gets heated to its equilibrium temperature. We find that final current in the filament is one fifth the current drawn at `t = 0`. If the temperature of filament at `t = 0` is `20^(@)C` and the temperature coefficient of resistivity at `20^(@)C` is `0.004(.^(@)C)^-1`, find the final temperature of the filament. |
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Answer» At constant potential different, current becomes `1//5^th` i.e., resistance is five times of initial resistance. `R_theta = 5 R_20` `R_20[1 + alpha(theta - 20)]= 5 R_20` `alpha(theta - 20) = 4` `theta - 20 = (4)/(alpha) = (4)/(0.004) = 1000` `theta = 1020.^@C`. |
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| 155. |
How much current is drawn by the motor of 0.5 hp from 220 volt supply? |
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Answer» Here, `P= 0.5hp = 0.5 xx 746` watt `I=P/V= 0.5xx746/220 = 1.7 A` |
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| 156. |
A and B are two identical bulbs of `40 W` connected to a `V = 12` volt cell. Switch S is closed to connect a third bulb C in the circuit. What happens to brightness of bulb A? Answer for two cases: (i) Bulb C is a very high wattage bulb. (ii) Bulb C is a very low wattage bulb. All the three bulbs have rated voltage of 12 volt. |
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Answer» Correct Answer - (i) A glows near its full brightness (ii) A becomes slightly more brighter |
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| 157. |
Assertion : Some electric appliance have three pins, even though if we remove the top pin, it will continue working. Reason: The third pin is used only as a safety device.A. If both assertion and reason are true and reason is the correct explantion of assertion.B. If both assertion and reason are true and reason is not the correct explantion of assertion.C. If assertion is false but reason is trueD. If assertion is true but reason is false. |
| Answer» Correct Answer - A | |
| 158. |
Assertion : A domestic electrical appliance working on a three pin continue working even if the top pin is removed Reason : The third pin is used only as safety device.A. both assertion and reason are true and the reason is Assertion and Reason explanation of the AssertionB. both assertion and reason are true and the reason is Assertion and the correct explanation of the AssertionC. Assertion is true , but the reason is falseD. both assertion and reason are false |
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Answer» Correct Answer - a Both assertion and reason are correct and reason is the correct explanation of Assertion |
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| 159. |
Assertion : In lower combination of electric bulbs the bulb of lower power emit more light than that of highest power bulb Reason : The power power bulb in series gets more of higest power bulbA. both assertion and reason are true and the reason is Assertion and Reason explanation of the AssertionB. both assertion and reason are true and the reason is Assertion and the correct explanation of the AssertionC. Assertion is true , but the reason is falseD. both assertion and reason are false |
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Answer» Correct Answer - c In series, power `prop` resistance of a bulb of law is higher that of high power bulb . |
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| 160. |
A potentiometer wire of length 1m and resistance `10 Omega` is connected in series with a cell of emf 2V with internal resistance `1 Omega` and a resistance box including a resistance R. If potential difference between the ends of the wire is 1 mV, the value of R isA. `20000 Omega`B. `19989 Omega`C. `10000 Omega`D. `9989 Omega` |
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Answer» Correct Answer - B |
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| 161. |
The emf of a battery is 2V and its internal resistance is `0.5Omega` the maximum power which it can deliver to any external circuit will beA. 8wattB. 4wattC. 2wattD. none of the above |
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Answer» Correct Answer - 3 Power `P=(V^(2))/(r)=( (2)^(2))/(0.5)=(4)/(0.5)=8` watt |
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| 162. |
Power `P` is to be delivered to a device via transmission cables having resistance `R_(c)`. If `V` is the voltage across `R` and `I` the current through it , find the power wasted and how can it be reduced. |
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Answer» Power delivered, P = voltage `xx` current `=VI` or `I = P//V`. If `R_(c)` is the resistance of transmission line then power wasted, `P_(C) = l^(2)R_(C) = P^(2)/v^(2) R_(C)` In order to reduce `P__(C), V` should be large. Therefore the power should be transmitted at high voltage. |
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| 163. |
Assertion : In parallel combination of electrical appliances, total power combination is equal to the sum of the powers of the individual appliances, Reason : In parallel combination , the voltage across each appliance is the same as required for the proper working of of eletrical applianceA. both assertion and reason are true and the reason is correct explanation of AssertionB. both assertion and reason are true and the reason is not correct explanation of AssertionC. Assertion is true , but the reason is falseD. both assertion and reason are false |
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Answer» Correct Answer - a In parallel combination of electrical appliances `(1)/(R ) = (1)/(R_(1)) + (1)/(R_(2)) + (1)/(R_(3))` As voltage `(V)` across each appliance is same so `(V^(2))/(R )= (V^(2))/(R_(1)) + (V^(2))/(R_(2)) + (V^(2))/(R_(3))orP=P_(1)+P_(2)+P_(3)` |
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| 164. |
A battery has six cells in series. Each has an emf 1.5 V and internal resistance 1 ohm. If an external load of `24Omega` is connected to it. The potential drop across the load isA. 7.2 VB. 0.3VC. 6.8VD. 0.4 V |
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Answer» Correct Answer - A `V=((nE)/(R+nr))R` |
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| 165. |
Long distance power transmission is carried on high voltage line. Why ? |
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Answer» When current `I` is transmitted through a power line of resistance R, Power loss `= I^(2)R`. If the power P is transmitted at voltage V, then `P= VI or I = (P//V) :.` Power loss `= P^(2)/V^(2) R` For a given power and given is line, P and R are constant. Hence, power loss `prop (1//V^(2))` It means if power is transmitted at high voltage, power loss will be small and vice-versa. |
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| 166. |
A voltmeter reads the potential difference across the terminals of an old battery as `1.40 V`, while a potentiometer reads its voltage to be `1.55 V`. The voltmeter resistance is `280 Omega`.A. the emf of the battery is 1.4 VB. the emf of the battery is 1.55 VC. the internal resistance r of the battery is `30 Omega`D. the internal resistance r of the battery is `5 Omega` |
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Answer» Correct Answer - B::C |
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| 167. |
Prove that in series combination of electrical appliances, the reciprocal of total power consumption is equal to the sum of the reciprocal of the powers of the individual appliances. |
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Answer» Let `P_(1),P_(2),P_(3)` be the powers of electrical appliances at V volt. If `R_(1), R_(2), R_(3)` are their resistance, then `R_(1)=V^(2)/P_(1), R_(2) = V^(2)/P_(2) and R_(3) = V^(2)/P_(3)` When the electrical appliances are connected in series, then effective resistance, R is `R = R_(1) + R_(2) +R_(2)` or `R/V^(2) = R_(1)/V^(2) + R_(2)/V^(2) + R_(3)/V^(2)` or `1/(V^(2)//R) = 1/(V^(2)//R_(1)) + 1/(V^(2)//R_(2)) + 1/(V^(2)//R_(3))` or `1/P = 1/P_(1) + 1/P_(2) + 1/P_(3)` |
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| 168. |
For what value of load resistance, the power transfer is maximum when two identical cells each of emf E and internal resistance r are connected (a) in series (b) in parallel. |
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Answer» (a) When cells are connecteed in series to the external load resistance R, then effective emf `= E + E = 2E`. Total resistance of the circuit `= R + r + r` `= R + 2r` Current in load, `I= (2E)/(R + 2r)` Power delivered to the external load is, `P = I^(2)R = ((2E)/(R + 2r))^(2) R` Power deliverd to the external load will be maximum if external resistance = internal resistance of cells, i.e., `R = 2r`. Then maximum power, `P_(max) = ((2E)/(R + 2r))^(2) xx 2r = E^(2)/(2r)` (b) When cells are connected in parallel to the external load resistance R, then effective emf = E. Total resistance of the circuit `R + (r xx r)/(r+r) = (R + r/2)` Current in load, `I = E/((R + r//2))` Power delivered to the external load, `P = I^(2)R = (E/(R + r//2))^(2)R` Power is maximum when `R = r//2`. Then `P_(max) = ((E)/(r//2 + r//2))^(2) r//2 = E^(2)/(2r)` Note that in both the cases, maximum power deliverd to the loas is the same. |
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| 169. |
(a) A voltmeter with resistance `R_v` is connected across the terminals of a battery of emf `E` and internal resistance `r`. Find the potential difference measured by the voltmeter. (b) If` E = 7.50 V` and `r = 0.45 Omega`, find the minimum value of the voltmeter resistance `R_v` so that the voltmeter reading is within `1.0%` of the emf of the battery. (c) Explain why your answer in part (b) represents a minimum value. |
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Answer» Correct Answer - A::B::C::D a. `V=E-ir` `=E-(E/(r+R_V))r` `V=((ER_V)/(r+R_V))`………….i b. `V=E/100` substituting in eq. i we get `R_V=4.5xx10^-3Omega` c. `V=E(1/(1+r/R_V))` If `R_V` is increased from this value `V` will increased |
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| 170. |
A battery has an emf E and internal resistance r. A variable resistance R is connected across the terminals of the battery. Find the value of R such that (a) the current in the circuit id maximum (b) the potential difference across the terminals is maximum. |
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Answer» Current in the circuit, `I = E/(R + r)` Power deliverd to the resistance R is, `P = I^(2) xx R = (E^(2)R)/(R +r)^(2)` Power delivered to the load will be maximum when `R/(R +r)^(2)` is maximum, i.e., `(d)/(dR) [R (R + r)^(-2)]=0` or `[(R +r)^(-2) + R(-2) (R + r)^(-3)]=0` or `(R + r)^(-3) [r +R]=0` As `(R + r)^(-3)!= 0` for finite value of R, So, r-R =0 or R=r When R = r, the power transferred to the load is `= E^(2)r/(r+r)^(2) = E_(2)/4r = max`. |
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| 171. |
A galvanometer with `50` divisions on the scale has a resistance of `25 Omega`. A current of `2 xx 10^-4` A gives a deflection of one scale division. The additional series required to convert it into a voltmeter reading up to `25 V` is.A. `1200 Omega`B. `1225 Omega`C. `2475 Omega`D. `2500 Omega` |
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Answer» Correct Answer - C `i_g = 50 xx 2 xx 10^-4 = 10^-2 A, G = 25 Omega` `V = i_g(R + G)` `25 = 10^-2(R + 25) rArr R = 2475 Omega`. |
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| 172. |
On a six fold increase in external resistance of a circuit the voltage across the terminals of the battery has increased from 5 V to 10 V. The emf of battery isA. 15 VB. 18 VC. 12.5 VD. 11 V |
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Answer» Correct Answer - C (E-V)/(V)R=r implies (E-5)/(5) R=r` `(E-10)/(10)6R=r` ` Equating Eqs. (i) and (ii), we get `(E-5)/(5)R=(E-10)/(10)6R implies E-5=3E-30` `2E=25 implies E=12.5V` |
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| 173. |
When a `12 Omega` resistor is connected with a moving coil galvanometer, then its deflection reduces form 50 divions to 10 divisions. The ressitance of the galvanometer isA. `24 Omega`B. `36 Omega`C. `48 Omega`D. `60 Omega` |
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Answer» Correct Answer - C (c ) `i_(g) = (iS)/(S + G) implies 10 = (50 xx 12)/(12 + G) implies 12 + G = 60` `implies G = 48 Omega` |
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| 174. |
For a cell the graph between the p.d. (v) across the terminals of the cells and the current (I) drawn from the cell as shown. The emf and internal resistance isA. `(3)/(2)Omega`B. `(1)/(3)Omega`C. `3Omega`D. `(2)/(3)Omega` |
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Answer» Correct Answer - D `i=(E)/(r)` |
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| 175. |
For a cell, a graph is plotted between the potential difference V across the terminals of the cell and the current I drawn the cell. The emf and the internal resistance of the cell are E and r, respectively. Then A. `2 V, 0.5 Omega`B. `2 V, 0.4 Omega`C. `gt 2 V, 0.5 Omega`D. `gt 2 V, 0.4 Omega` |
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Answer» Correct Answer - B `P.d` across call `V = E - Ir` When `I = 0, V = E = 2 V` When `V = 0, 0 = E - 5 r rArr 0 = 2 - 5 r` `r = 0.4 Omega`. |
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| 176. |
The emf of a daniel cell is 1.08 V. When the terminals of the cells are connected to a ressitance of `3Omega`, the potential difference across the terminals is found to be 0.6 V. Then the internal resistance of the cell isA. `1.8Omega`B. `2.4Omega`C. `3.24Omega`D. `0.2Omega` |
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Answer» Correct Answer - B `r=((E-V)/(V))R` |
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| 177. |
for a cell, the graph between the p.d. (V) across the terminals of the cell ad te current I drawn from the cell is shown in the fig. the emf and the internal resistance of the cell is E and r respectively. A. `E=2V,r=0.5Omega`B. `E=2V,r=0.4Omega`C. `Egt2,r=0.5Omega`D. `Egt2V,r=0.4Omega` |
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Answer» Correct Answer - B `V=E-ir` `i=0,V=E=2V` `V=0,r=(E)/(i)=0.4Omega` |
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| 178. |
The potential difference across the terminals of a battery is 10 V when there is a current of 3 A in the battery from the negative to the positive terminal. When the current is 2A in the reverse direction, the potential difference becomes 15 V. The internal resistance of the battery isA. `2.5 Omega`B. `5.0 Omega`C. `2.83 Omega`D. `1 Omega` |
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Answer» Correct Answer - D |
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| 179. |
The potential difference across te terminals of a battery is 10 V when there is a current of 3 A in the battery from the negative to the positive terminal. When the current is 2A in the reverse direction, the potential difference becomes 15 V. The internal resistance of the battery isA. `1Omega`B. `0.4Omega`C. `0.6Omega`D. `0.8Omega` |
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Answer» Correct Answer - A `E-3r=10` …(i) `E+2r=15` ..(ii) solving (i) & (ii) `r=1` |
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| 180. |
The potential difference across terminals of a battry is `9.0 V`, when a current of `3.5A` flows through it from its negative terminal to the positive terminal .When a current of `2A` flows through in the opposite direction, the terminal potential difference is `12V`. Find the internal resistance and emf of the battery |
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Answer» Correct Answer - `2 Omega, 16 V` When current flow through the battery from its the negative to positive terminal (i.e , current is draw from battery ) then `V = epsilon - 1 r or 9.0 = epsilon - 3.5 r `…..(i) When current flow through the battrey from its the positive negative to terminal (i.e , battery is charged) then `V = epsilon 1 r or 12 = epsilon - 2 r `…..(ii) On solving (i) and (ii) we get `r = 2 Omega and epsilon = 16 V` |
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| 181. |
The following combination are concerned with experiments of the characterization and use of a moving coil galvanometer. The series combination of variab resistance R, one `100 Omega` resistor and moving coil galvanometer is connected to a mobile phone charger having negligible internal resistance. The zero of the galvanometer lies the centre and the pointer can move 30 division full scale on either side depending on the directin of current. The reading of the galvanometer is 10 divisions and the voltages across the galvanometer and `100 Omega` resistor are respectively 12mV and 16 mV. The figure os merit of the galvanometer is microampere per division is :A. 16B. 20C. 32D. 10 |
| Answer» Correct Answer - A | |
| 182. |
The following combination are concerned with experiments of the characterization and use of a moving coil galvanometer. The series combination of variab resistance R, one `100 Omega` resistor and moving coil galvanometer is connected to a mobile phone charger having negligible internal resistance. The zero of the galvanometer lies the centre and the pointer can move 30 division full scale on either side depending on the directin of current. The reading of the galvanometer is 10 divisions and the voltages across the galvanometer and `100 Omega` resistor are respectively 12mV and 16 mV. A `24 Omega` resistcnce is conneted to a 5 V battery with internal resistance of `1Omega`. A `25 Omega` resistance is conneted in series with the galvanometer and this combination is used to measur the voltage across the `24 Omega` resistance. The number of divisons shown in the galvanometer isA. 6B. 8C. 10D. 12 |
| Answer» Correct Answer - D | |
| 183. |
A resistance of `2Omega` is connected across one gap of a meter bridge (the length of the wire is `100 cm`) and an unknown resistance, greater than `2Omega`is conneted across the other gap. When these resistances are interchanged, the balance point shifts by `20 cm`. Neglecting any corrections,the unknown resistance isA. `3 Omega`B. `4 Omega`C. `5 Omega`D. `6 Omega` |
| Answer» Correct Answer - A | |
| 184. |
Why is a meter bridge also called a slide wire bridge ? |
| Answer» This is because during experiment, the jockey is to be slided over the bridge wire. | |
| 185. |
When a resistor of `5 Omega` is connected across a cell, its terminal potential differnce is balanced by 140 cm of potentiometer wire and when a resistance of `8 Omega` is connected across the cell, the terminal potential difference is balanced by 160 cm of the potentiometer wire. Find the internal resistance of the cell. |
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Answer» Here, `R_(1) = 5 Omega , l_(1) = 140 cm` , `R_(2) = 8 Omega , l_(2) = 160 cm` Let emf of the cell be balanced by length l of the potnetiometer wire. Then as per question, In the first case, `r = R_(1)((l_(1)-l_(1))/(l_(1))) or (r l_(1))/R_(1) = l-l_(1)` In the second case, `r = R_(2)((l_(1)-l_(2))/(l_(2))) or (r l_(2))/R_(2) = l-l_(2)` Subtracting (ii) from (i), we get `r[(l_(1))/(R_(1)) - l_(2)/R_(2)]=(l - l_(1)) - (l - l_(2))= l_(2) - l_(1)` or `r=(l_(2) - l_(1))/(l_(1)/R_(1) - l_(2)/R_(2)) = (160-140)/(140/5 - 160/8) = 2.5Omega` |
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| 186. |
In the circuit shown, potential difference between X and Y will be A. ZeroB. 20VC. 60VD. 120V |
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Answer» Correct Answer - D |
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| 187. |
In an electrolyte `3.2 xx 10^(18)` bivalent positive ions drift to the right per second while `3.6 xx 10^(18)` monovalent negative ions drift to the per second. Then the current isA. 1.6 amp to the leftB. 1.6 amp to the rightC. `0.45` amp to the rightD. `0.45` amp to the left |
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Answer» Correct Answer - B |
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| 188. |
What is terminal potential differnce of a cell? Can its value be greater than the emf of a cell? Explian.A. being dischargedB. open circuitC. being chargedD. being either changed or discharged |
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Answer» Correct Answer - C |
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| 189. |
Is it necessary to keep the length of the slide-bridge wire 1 meter ? Explian |
| Answer» It is not necessary to keep the length of the slide - bridge wire 1 meter. The working of silde - bridge is based on Wheatstone bridge principal. According to this when bridge is balanced `PQ//= R//S`. We can use wire of any length in this bridge and note down the resistances used in the four arms of Wheatstone bridge by knowing the resistance per unit length of the bridge wire. | |
| 190. |
What is terminal potential differnce of a cell? Can its value be greater than the emf of a cell? Explian.A. Being dischargedB. In open circuitC. Being chargedD. Being either charged or discharged |
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Answer» Correct Answer - C |
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| 191. |
All bulbs in the circuit shown in figure are identical. Which bulb glows most brightly? A. BB. AC. DD. C |
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Answer» Correct Answer - B |
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| 192. |
In an electrolyte, the positive ions move from left to right and negative ions from right to left. Is there anet current? If yes, in what direction? |
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Answer» Correct Answer - C |
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| 193. |
If the length of the wire be (i) doubled and (ii) halved, what will be effect on the position of zero deflection in a potentiometer ? Explian |
| Answer» (i) When length of the wire is doubled, the potential gradient across the potentiometer wire will decrease. Due to it, the position of zero deflection will occur at longer length. (ii) The reverse will be true when length is halved. | |
| 194. |
When the resistance connected in series with a cell is halved, the current is not exactly doubled but slightly less, why? |
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Answer» Correct Answer - A |
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| 195. |
Lights of a car become dim when the starter is opereterd. Why? |
| Answer» When the motor starter of a car is operated, it draws more current from the battery for the operation of car. Therefore, the voltage across the light bulb is lowered, hence the light of a car is dimmed. | |
| 196. |
A filament bulb `(500 W, 100 V)` is to be used in a `230 V` main supply. When a resistance `R` is connected in series, it works perfectly and the bulb consumers `500 W`. The value of `R` isA. `230 Omega`B. `46 Omega`C. `26Omega`D. `13 Omega` |
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Answer» Correct Answer - C |
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| 197. |
A potentiometer is an accurate and versatile device to make electrical measurements of `E.M.F.` because the method involvesA. cellsB. potential gradientsC. a condition of no current flow through the galvanometerD. a combination of cells, galvanometer and resistance |
| Answer» Correct Answer - C | |
| 198. |
A potentiometer is an accurate and versatile device to make electrial measurements of `E.M.F.` because the method involvesA. Potential gradientsB. A condition of no current flow through the galvanometerC. A combination of cells, galvanometer and resistanceD. Cells |
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Answer» Correct Answer - B (b) When a potentiometer is connected across a circuit, in zero deflection condition the potentionmeter draws no current hernce the reading of potentiometer is accurate. |
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| 199. |
In the circuit shown in figure, the value of R is A. `1Omega`B. `2Omega`C. `3Omega`D. `4Omega` |
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Answer» Correct Answer - B `R_(1)=1Omega, R_(2)=2 Omega, R_(3)=Romega, I_(1)=6 A, I_(2)=2A, R=?` From figure, `I_(3)=11-8=3A` Now in figure` R_(1), R_(2)` and `R_(3)` are in parallel. `:.V=I_(1)R_(1)=I_(3)R_(3)` `:.R_(3)=(I_(1)R_(1))/(I_(3))=(6xx1)/(3)` `R_(3)=R=2Omega` |
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| 200. |
To convert a `800 mV` range milli voltmeter of resistance `40 Omega` into a galvanometer of 100 mA` range, the resistance to be connected as shunt isA. `10 Omega`B. `20 Omega`C. `30 Omega`D. `40 Omega` |
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Answer» Correct Answer - A |
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