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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.
| 1851. |
Shown in the figure below is a meter- bridge set up will null deflection in the galvanometer. The value of the unknown resistor R isA. `220 Omega`B. `110 Omega`C. `55 Omega`D. `13.75 Omega` |
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Answer» Correct Answer - A `(55)/(R) = (20)/(100 - 20)rArr (55)/(R) = (1)/(4)` `R = 220 Omega`. |
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| 1852. |
The lead wires should haveA. Larger diameter and low resistanceB. Smaller diameter and high resistanceC. Smaller diameter and low resistanceD. Larger diameter and high resistance |
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Answer» Correct Answer - A |
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| 1853. |
An ammeter and a voltmeter of resistance `R` connected in seires to an electric cell of negligible internal resistance. Their readings are `A` and `V` respecitvely. If another resistance `R` is connected in parallel with the voltmeterA. Both `A` and `V` will increaseB. Both `A` and `V` will decreaseC. `A` will decrease and `V` will increaseD. `A` will decrease and `V` will decrease |
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Answer» Correct Answer - D (d) After connecting a resistance `R` in parallel with voltmeter, its effective resistance decreases. Hence, less voltage appears across it i.e., `V` will decrerases. Since overall resistance decreases so more current will flow i.e., `A` will increase. |
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| 1854. |
An ammeter and a voltmeter of resistance `R` connected in seires to an electric cell of negligible internal resistance. Their readings are `A` and `V` respecitvely. If another resistance `R` is connected in parallel with the voltmeterA. Both A and V will increaseB. Both A and V will decreaseC. A will decrease and V will increaseD. A will increase V will decrease |
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Answer» Correct Answer - D |
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| 1855. |
A `36 Omega` galvanometer is shunted by resistance of `4Omega`. The percentage of the total current, which passes through the galvanometer isA. `8%`B. `9%`C. `10%`D. `91%` |
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Answer» Correct Answer - C |
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| 1856. |
The measurement of voltmeter in the following circuit is A. `2.4V`B. `3.4V`C. `4.0V`D. `6.0V` |
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Answer» Correct Answer - D |
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| 1857. |
A milliammeter of range 10 mA has a coil of resistance `1 Omega`. To use it as voltmeter of range 10 volt , the resistance that must be connected in series with it, will beA. `999 Omega`B. `99 Omega`C. `1000 Omega`D. None of these |
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Answer» Correct Answer - A |
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| 1858. |
The masses of the three wires of copper are in the ratio `5:3:1` and their lengths are in the ratio `1:3:5`. The ratio of their electrical resistances isA. `15:1:125`B. `1:125:15`C. `125:1:15`D. `125:15:1` |
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Answer» Correct Answer - D Resistance in terms of length and mass is given by, `R prop (l^(2))/(m)` `:.R_(1):R_(2):R_(3)=(l_(1)^(2))/(m_(1)):(l_(2)^(2))/(m_(2)):(l_(3)^(2))/(m_(3))` `R_(1):R_(2):R_(3)=(25l_(1)^(2))/(m_(1)):(9l_(1)^(2))/(3m_(1)):(l_(1)^(2))/(5m_(1))` `=25:3:1/5` `=125:15:1`. |
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| 1859. |
In the circuit shown in figure A. power supplied by the battery is 200 WB. current flowing in the circuit is 5 AC. potential difference across `4 Omega` resistance is equal to the potential difference across `6 Omega` resistanceD. current in wire AB is zero |
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Answer» Correct Answer - A::C |
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| 1860. |
When the power delivered by a 100 volt battery is 40 watts the equivalent resistance of the circuit isA. 100ohmsB. 25ohmsC. 300ohmsD. 250ohms |
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Answer» Correct Answer - 4 Power `=(V^(2))/(R)` `4 0=((100)^(2))/(R)` or `R=(100xx100)/(40)Omega` or `R= 250Omega` |
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| 1861. |
Two cell with the same e.m.f. E and difference internal resistance `r_(1) and r_(2)` are connected in series to an external resistance `R` a value of `R` be selected such that the potential difference as the flow cell E should be zero when A. `R = r_(1)+r_(2)`B. `R = r_(1)- r_(2)`C. `R = r_(1)//r_(2)`D. `R = r_(1)=r_(2)` |
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Answer» Correct Answer - b Current is circuit, `I= (E+E)/(r_(1)+ r_(2) +R) = (22E)/(r_(1)+ r_(2) +R) `….(i) The potential difference first cell is zero, Then `E - I r_(1) = 0 or I= E//r_(1)`…..(ii) From (i) and (ii) `(2E)/(r_(1) +r_(2)+R) = (E )/(r_(1))` On solving `R = (r_(1)- r_(2))` |
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| 1862. |
Twelve cells each having the same e.m.f are connected in series and are kept to a closed box. Some of the cell are connected in reverse order .The battery is connected in series with an ammeter an external resistance `R` and two cells of the same type as an in the battery .The current when they and support each other is `3` ampere and current is `2` ampare when the two oppose each other. How many cells are connected in servese order ?A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - a Let `epsilon` be the emf of cell and `R` be the total resistance of the circuit Let `n` cell be wrongly connected Net emf of battery `= (12- 2n)epsilon` when `2` other identical cells are connecting so that these aid the battery net emf `= (12 - 2n) epsilon + 2 and I = 3A` so `((12 - 2n) epsilon + 2 epsilon)/(R) = 3`.....(i) where`2` cell oppose the battery net emf `(12 - 2n) epsilon + 2 epsilon` As per the question Current `= ((12 - 2n) epsilon + 2 epsilon)/(R) = 2`....(ii) Dividing (i) and (ii) we get `n = 1` |
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| 1863. |
In a cyclic process, a gas is taken from state A and B via path -I as shown in the indicator diagram and taken back to state A from state B via path-II. In the complete cycle A. work is done on the gasB. heat is given to the gasC. work is done by the gasD. heat is taken from the gas |
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Answer» Correct Answer - B::C |
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| 1864. |
Assertion : If a current flows through a wire of non-uniform cross-section, potential difference per unit length of wire is same throughout the length of wire. Reason : Current through the wire is same at all cross-sections.A. If both Assertion and Reason are true and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are true and Reason is not correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true. |
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Answer» Correct Answer - D |
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| 1865. |
`20 mu A` current flows for 30 seconds in a wire, transfer of charge will beA. `2 xx 10^(-4)C`B. `4 xx 10^(-4)C`C. `6 xx 10^(-4)C`D. `8 xx 10^(-4)C` |
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Answer» Correct Answer - C |
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| 1866. |
Two rods of same material and length have their electric resistance in ratio `1 : 2`. When both rods are dipped in water, the correct statement will beA. `A` has more loss of weightB. `B` has more loss of weightC. Both have same loss of weightD. Loss of weight will be in the ratio `1 : 2` |
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Answer» Correct Answer - A (a) `R = rho (l)/(A) implies (R_(1))/(R_(2)) = (A_(2))/(A_(1)) (rho, L "constant") implies (A_(1))/(A_(2)) = (R_(2))/(R_(1)) = 2` Now, when a body dipped in water, loss of weigh `V sigma_(L) g Al sigma_(L) g` So, `(("Loss of weigth")_(1))/(("Loss of wight")_(2)) = (A_(1))/(A_(2)) = 2`, so `A` has more loss of weight. |
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| 1867. |
Two rods of same material and length have their electric resistance in ratio `1 : 2`. When both rods are dipped in water, the correct statement will beA. A has more loss of weightB. B has more loss of weightC. Both have same loss of weightD. Loss of weight will be in the ratio `1:2` |
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Answer» Correct Answer - A |
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| 1868. |
Two wires `A` and `B` of same material and same mass have radius `2r` and `r`. If resistance of wire `A` is `34 Omega`, then resistance of `B` will beA. `544 Omega`B. `372 Omega`C. `68 Omega`D. `17 Omega` |
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Answer» Correct Answer - A (a) `R = rho (1)/(A)` and mass `m` = volume `(V) xx` density `(d) = (A l) d` Since wires have same material so `rho` and `d` is same for both. Aslo they have same `implies Al =` consatnt `implies prop (1)/(A)` `implies (R_(1))/(R_(2)) = (l_(1))/(l_(2)) xx (A_(2))/(A_(1)) = ((A_(2))/(A_(1)))^(2) = ((r_(2))/(R_(1)))^(4)` `implies (34)/(R_(2)) = ((r)/(2r))^(4) = R_(2) = 544 Omega` |
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| 1869. |
Two wires `A` and `B` of same material and same mass have radius `2r` and `r`. If resistance of wire `A` is `34 Omega`, then resistance of `B` will beA. `544 Omega`B. `272 Omega`C. `68 Omega`D. `17 Omega` |
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Answer» Correct Answer - A |
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| 1870. |
The resistivity of alloys `=R_(alloy)`, the resistivity of constituent metals `R_("metal")`. Then, usuallyA. `R_("alloy") = R_("metal")`B. `R_("alloy") lt R_("metal")`C. There is no simple relation between `R_("alloy")` and `R_("metal")`D. `R_("alloy") gt R_("metal")` |
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Answer» Correct Answer - D |
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| 1871. |
Alloys of metals have greater resistivity than that of their constituent metals. Why? |
| Answer» In an alloy like nichrome (made of nickel and chromium), the free electron finds a disordered arrangment of nickel ions and cromium ions. Due to it, the electron is scattered by them randomly and very frequently. As a result of it, the value of relaxation time of electorn decreases and hence resistivity increases because `rho prop 1//tau`. | |
| 1872. |
There are two concentric spheres of radius a and b respectively. If the space between them is filled with medium of resistivity `rho`, then the resistance of the intergap between the two spheres will be (Assume `bgta`) |
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Answer» Consider a concentric spherical shell of radius `x` and thickness dx its resistance is `dR,dR=(rhodx)/(4pix^(2))` Total resistance `R=int_(a)^(b)dR=R=((rho)/(4pi))int_(a)^(b)(dx)/(x^(2))=(rho)/(4pi)[(1)/(a)-(1)/(b)]` |
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| 1873. |
Space between tow concentric spheres of radii `r_(1) and r_(2)` such that `r_(1) lt r_(2)` is filled with a material of resistivity `rho`. Find the resistance between inner and outer surface of the material A. `(r_(1))/(r_(2))(p)/(2)`B. `(r_(2)-r_(1))/(r_(1)r_(2))(p)/(4pi)`C. `(r_(1)r_(2))/(r_(2)-r_(1))(p)/(4pi)`D. none of these. |
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Answer» Correct Answer - B Since, `R=rho(l)/(a) therefore R=rho=(dl)/(4pil^(2))` (where l is any radius and dl is small element). `R=(rho)/(4pi) underset(r_(1))overset(r_(2))(int)(dl)/(l^(2))=(rho)/(4pi)[-(1)/(l)]_(r_(1))^(r_(2))=(rho)/(4pi)[(1)/(r_(1))-(1)/(r_(2))]` `R=[(r_(2)-r_(1))/(r_(1)r_(2))](rho)/(4pi)` |
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| 1874. |
A hollow cylinder of length `l` and of radii a and b is filled with a material of resistivity `rho` and is connected to a battery of emf `E` through an ammeter. Find the current through ammeter |
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Answer» Correct Answer - `p log_(e)((b)/(a)) Consider an imaginary cylinder of radius x and thickness dx of the given cylinder. Its resistance is `dR = (rho dx)/(2pi x l)` Total resistance, `R=underset(a)overset(b)(int)(rho dx)/(2pixxl)=(rho log_(e)((b)/(a)))/(2pil)` Current, `I=(E)/(R)=(Exx2pil)/(rho log_(e)((b)/(a)))`. |
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| 1875. |
The external diameter of a 5 metre long hollow tube is 10 cm and the thickness of its wall is 5 mm. If the specific resistance of copper be `1.7xx10^(5)` ohm-metre, then determine its resistance.A. `5.7xx10^(-5)Omega`B. `2.7xx10^(-5)Omega`C. `2xx10^(-5)Omega`D. `5xx10^( -5)Omega` |
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Answer» Correct Answer - A Area of cross section of tube, `A= pi (r_(2)^(2)-r_(1)^(2))` `=3.14[(0.05)^(2)-(0.045)^(2)]` `=14.91xx10^(-4)m^(2)` Thus, `R=(rho l)/(A)=(1.7xx10^(-8)xx5)/(14.91xx10^(-4))=5.7xx10^(-5)Omega` |
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| 1876. |
The resistance in the two arms of the meter bridge are `5 Omega` and `R Omega`, respectively. When the resistance `R` is shunted with an equal resistance, the new balance point is `1.6 l_(1)`. The resistance `R` is A. `10 Omega`B. `15 Omega`C. `20 Omega`D. `25 Omega` |
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Answer» Correct Answer - B (b) Initially, `(5)/(l_(1)) = (R )/(100 - l_(1))` Finally, `(5)/(1.6 l_(1)) = (R )/(2 (100 - 1.6l_(1))` `implies (R )/(1.6 (100 - l_(1))) = (R )/(2 (100 = 1.6 l_(2)))` `implies 160 - 1.6 l_(1) = 200 - 3.2 l_(1)` `implies 1.6 l_(1) = 40` `implies l_(1) = 25` From Equaiton (i), `(5)/(25) = (R )/(75) implies R = 15 Omega` |
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| 1877. |
Length of a hollow tube is `5 m`, its outer diameter is `10 cm` and thickness of its wall is 5 mm. If resistivity of the material of the tube is `1.7 xx 10^(-8) Omega xx m` then resistance of tube will beA. `5.6 xx 10^(–5) Omega`B. `2 xx 10^(-5) Omega`C. `4 xx 10^(–5) Omega`D. None of these |
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Answer» Correct Answer - A |
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| 1878. |
Two electric bulbs of 50 W and 100 W are given. When they are (i) connected in series (ii) connected in parallel, which bulb will glow more ? |
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Answer» Resistance of a bulb `R = V^(2)//P or R prop1//P`. Thus the resistance of 50 W bulb is more than thtat of 100 W bulb. As glow `prop` power of a bulb. (i) In series connection, the current is same, so power `=I^(2)R`, i.e., `P prop R`. `:.` 50 W bulb will glow more than 100 W bulb. (ii) In parallel connection, power `(P) = V^(2)//R`, i.e., `P prop 1//R`. Thus 100 W bulb will glow more than that of 50 W bulb. |
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| 1879. |
Of the bulbs in a house, one glows brighter than the other, which of the two has a large resistance. |
| Answer» Power of dim bulb is less than the power of bright bulb, i.e., `P_(dim) lt P_("bright")` . At constant voltage, `P prop 1//R` :so `R_(dim) gt R_("bright")` | |
| 1880. |
In the circuit shown in Fig.6.2, the heat produced in `5Omega` resistor, due to the current flowing through it is 10 calorie per second. Find the heat produced in `4Omega` resistor.A. `1cal/s`B. `2cal/s`C. `3cal/s`D. `4 cal/s` |
| Answer» Correct Answer - B | |
| 1881. |
A heater joined in series with a ` 50 W` bulb is connected to the mains . If the `50 W` bulb is replaced by a `100 W` bulb , then will the heater now give more heat , less heat, or same heat ? Why ? |
| Answer» Resistance of 50 watt bulb is more than that of 100 watt bulb. When 100 watt bulb is replaced by 50 watt bulb, connected in series with heater, the resistance of circuit increases and hence current decrease. As `H prop I^(2)`, therefore the heater will now give less heat. | |
| 1882. |
Seven resistance are connected as shown in the firgure. The equivalent resistance between `A` and `B` is A. `3 Omega`B. `4 Omega`C. `4.5 Omega`D. `5 Omega` |
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Answer» Correct Answer - B |
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| 1883. |
In the circuit shown here, the readings of the ammeter and voltmeter are A. `6A, 60 V`B. `0.6A, 6V`C. `6//1A, 60//11A`D. `11//6A, 11//60V` |
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Answer» Correct Answer - C |
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| 1884. |
A 50 W bulb is in series with a room heater and the combination is conneted across the mains. To get max, heater output, the 50 W bulb should be replaced by :A. 25 WB. 10 WC. 100 WD. 200 W |
| Answer» Correct Answer - D | |
| 1885. |
A Steady current flows in a metalic conductor of non uniform cross section. The quantity/quantities which remain constant along the length of the conductor is/areA. Current, electric field and drift speedB. Drift speed onlyC. Current and drift speedD. Current only |
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Answer» Correct Answer - D (d) If `E` electric field, then current denstiy `j = sigma E` Also we know that current density `j = (i)/(A)` Hence `j` is different for different area of cross-section. When `j` is different, then `E` is also different. Thus `E` is not constant for different `j` value. Hence, only current `i` will be constant. |
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| 1886. |
The equivalent resistance between the points A and B is: A. `36/7Omega`B. `10Omega`C. `85/7Omega`D. |
| Answer» Correct Answer - C | |
| 1887. |
A battery of internal resistance `4Omega` is connected to the network of resistance as shown . In order that the maximum power can be delivered to the network, the value of R in `Omega` should be .A. `4//9`B. `2`C. `8/3`D. `18` |
| Answer» Correct Answer - B | |
| 1888. |
A battery of internal resistance `4Omega` is connected to the network of resistance as shown . In order that the maximum power can be delivered to the network, the value of R in `Omega` should be .A. `4//9`B. `8//9`C. 2D. 18 |
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Answer» Correct Answer - C |
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| 1889. |
In the figure, the value of resistors to be connected between C and D so that the resistance of the entire circuit between A and B does not change with the number of elementary sets used is A. RB. `R (sqrt(R)-1)`C. 3RD. `R(sqrt(3)-1)` |
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Answer» Correct Answer - B |
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| 1890. |
In the given circuit, it is observed that the current I is independent of the value of the resistance `R_6`. Then the resistance values must satisfy .A. `R_1R_2R_5 = R_3R_4R_6`B. `1/R_5 + 1/R_6 = 1/(R_1+R_2) + 1/(R_3+R_4)`C. `R_1R_4 = R_2R_3`D. `R_1R_3 = R_2R_4 = R_5R_6` |
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Answer» Correct Answer - C (c) Since current I is independent of `R_6`, it follows that the resistance `R_1, R_2, R_3 and R_4` must from the balanced Wheatstone bridge. `:. R_1R_4 = R_2R_3`. |
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| 1891. |
Difference Between Potential Difference And Emf |
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Answer» (i) emf not depends on current and internal resistance of cell while terminal potential difference depends. (ii) emf is the potential difference when no current related to cell, while terminal potential difference is the potential difference when some current relates to cell. (iii) emf is constant for a given cell while terminal potential difference depends on circuit connected to cell. it may different for a given cell. |
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| 1892. |
Can we produce high voltage on the human body without getting a shock ? |
| Answer» Yes we can. For this person must stand on a highly insulating platform, therefore with high voltage on the body, no charge will flow to the ground through the body and person will not get any shock. | |
| 1893. |
When high power heater is used at homes all bulbs being used become dim. Why? |
| Answer» All appliances are connected in parallel in house. When high power heater is used, very high current passes through it. So potential difference on mains wires increases and potential drop at mainswitch board of house decreases (V=E-ir, here r is resistance of main line) which results in decrease in intensity of bulbs. | |
| 1894. |
Assertion : The drift velocity of electrons in a metallic wire will decrease, if the temperature of the wire is increased. Reason : On increasing temperature, conductivity of metallic wire decreases.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 - B (b) On increasing temperature of wire the kinetic energy of free electrons increase and so they collide more rapidly with each other and hence their drift velocity decreases. Also when temperature increases, resistivity increase and resistivity is inversely proportional ot conductivity of material. |
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| 1895. |
Assertion : The electric bulbs glows immediately when switch is on. Reason : The drift velocity of electrons in a metallic wire is vary high.A. If both the assertion and reason are true and reason is a true explanation of the assertion.B. If both the assertion and reason are true but the reason is not the correct explanation of assertion.C. If the assertion and reason are falseD. If both the assertion and reason are false. |
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Answer» Correct Answer - C (c ) As the conductor is foil of electrons, with slight drift anywhere the queue starts moving. As a reusult bulb starts glowing instantly. Drift velocity of electrons in a good conductor is very low. |
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| 1896. |
A copper wire of resistance `R_(0)` is strerched till its length is increased to n times of its original length. What will be its new resistance? |
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Answer» Original resistnace, `R_(0) = rho l_(0)//A_(0)` New resistance, `R = rho l//A`. As the volume of wire remains constant, so `l_(0)A_(0)=lA= (n l_(0)) A or A= A_(0)//n` Now `R=(rho l)/(A)= rho (n l_(0))/(A_(0)//n)=n^(2)((rho l)/(A))=n^(2) R_(0)` |
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| 1897. |
Assertion : Bending a wire does not effect electrical resistance. Reason : The resistance of wire is proportional to the resistivity of material.A. If both the assertion and reason are true and reason is a true explanation of the assertion.B. If both the assertion and reason are true but the reason is not the correct explanation of assertion.C. If the assertion and reason are falseD. If both the assertion and reason are false. |
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Answer» Correct Answer - A (a) Resistance wire `R = rho (l)/(A)` where `rho` is resistivity of material which does not depend on the geometry of wire. Since when wire is banded, resistivity, length and area of cross-section do not change, therefore resistance of wire also remain same. |
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| 1898. |
Explain why bending a wire does not affect its electrical resistance? |
| Answer» Free electrons in a wire have small value of drift velocity and hence low value of inertia of motion. Due to it, they are able to go around the bends easily. That is why, the electrical resistance is not affected on bending the wire till the area of cross-section remains the same at the bend. | |
| 1899. |
At absolute zero silver wire behaves asA. Super conductorB. semi conductorsC. perfect insulatorD. semi insulator |
| Answer» Correct Answer - A | |
| 1900. |
Fuse wire is a wire of :A. low melting point and low value of `alpha`B. high melting pointand high value of `alpha`C. high melting point and low value of `alpha`D. low melting point and high value of `alpha` |
| Answer» Correct Answer - D | |