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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.
| 751. |
Two heaters are marked 200V, 300W and 200 V, 600 W. If the heaters are combined in series and the combination connected to a 200 V dc supply, which heater will produce more heat ? |
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Answer» Resistance of the two heaters are `R_(1) = V^(2)/P_(1) = (200 xx 200)/300 = 400/3 Omega` `R_(2) = V^(2)/P_(2) = (200 xx 200)/600 = 200/3 Omega` For series combination, the effective resistance is `R = R_(1) + R_(2) = 400/3 + 200/3 = 600/3 = 200 Omega` Current, `I = V/R = 200/200 = 1A` Power dissipated in 1st heater, `P_(1) = I^(2) R_(1) = 1^(2) xx 400/3 = 400/3 W` Power dissipated in second heater, `P_(2) = I^(2)R_(2) = 1^(2) xx 200/3 = 200/3 W` `:. P_(1) = 2 P_(2) or P_(1) gt P_(2)` |
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| 752. |
A resistance coil is made by joining in parllel two resistances each of `10 Omega`. An emf of 1.0V is applied between the two ends of the coil for 5 minutes. Calculate the heat produced in calories. (Given 1 cal. = 4.2 J) |
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Answer» Here, `R_(1) = 10 Omega, R_(2)= 10 Omega`, `V = 1.0 V, t = 5 xx 60s` Effective resistance in parallel combination will be `R_(p) = (R_(1) R_(2))/(R_(1) + R_(2)) = ( 10 xx 10)/( 10 +10) = 5 Omega`, Heat produced `= V^(2)/R_(p) t = 1^(2)/5 xx 5 xx 60 J` `= 60/4.2 cal = 14.3` cal. |
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| 753. |
A 10 V storage battery of negilgible internal resistance is connected across a 200V battery and a resistance of `50 Omega` resistor made of alloy manganin. How much heat energy is produced in the resistor in 1 h? What is the source of this energy? |
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Answer» Here, `V = 10 V, R = 50 Omega`, `t = 1 h = 60 xx 60 s`, Heat energy produced in 1 h is `H = (V^(2)t)/R = ((10)^(2) xx ( 60 xx 60))/50 = 7200 J` The source of this energy is the chemcal energy stored in the cell. |
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| 754. |
A uniform wire which connected in parallel with the 2 m long wire, will give a resistance of `2.0 Omega`. |
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Answer» Correct Answer - 20 m |
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| 755. |
The resistance of two conductors in series is `18 Omega` and the resistance becomes `4 Omega` when connected in parallel. Find the resistance of individual conductors. |
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Answer» Correct Answer - `12 Omega, 6 Omega` |
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| 756. |
Is the formula `V= IR` true for non ohmic device also? |
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Answer» Correct Answer - D |
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| 757. |
Two wires having resistance R and 2 R are connected in parallel, the ratio of heat generated in 2 R and R isA. ` 1 :2`B. `2 : 1`C. `1 :4`D. `4 : 1` |
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Answer» Correct Answer - B In parallel circuit potential across each resistor is same. `therefore " " P_(1) = " power across " R=(V^(2))/(R )` `P_(2) = " power across " 2R =(V^(2))/(2R)` `(P_(1))/(P_(2))=2 :1` |
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| 758. |
Calculate the current flowing through a heater rated at `2 kW` when connected to a `300V` d.c. supply. |
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Answer» Correct Answer - ` 6.67A` `I = (P)/(V) = (2kW)/(300V) = (2000W)/(300V) = 6.67A` |
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| 759. |
Internal resistance of a cell depends onA. The distance between the platesB. The area of the plantes immersed C. The concentration of the electrolyteD. All of the above |
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Answer» Correct Answer - D |
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| 760. |
Internal resistance of a cell depends onA. The distance between the platesB. The area of the plantes immersedC. The concentration of the electrolyteD. All of the above |
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Answer» Correct Answer - D |
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| 761. |
When cells are connected in parallel thenA. The current decreasesB. The curretn increasesC. The e.m.f. increasesD. The e.m.f decreases |
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Answer» Correct Answer - B |
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| 762. |
The reading on a high resistance voltmeter. When a cell is connected across it is `2.2 V`. When the terminals of the cell are connected to a resistance of `5 Omega` the voltmeter reading drop to `1.8 V`. Find the internal resistance of the cell. A. `1.2 Omega`B. `1.3 Omega`C. `1.1 Omega`D. `1.4 Omega` |
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Answer» Correct Answer - C |
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| 763. |
You are given `n` resistors each of resistance `r` .These are first connected to get minimum resistance .In the secoind case these are again connected differently to get maximum possible resistance. Compute the ratio between the minimum and maximum value of resistance so obtained. |
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Answer» Correct Answer - `(1)/(n^(2))` To get minimum posibale resistance, the resistors are to be connected in parallel .Then the connected minimum resistance is given by `(1)/(R_(min)) = (1)/(r ) + (1)/(r ) + … n "times" = (n)/(r ) or R_(min) = r//n` For maximum position resistance, the resistors are to be conneted in series. Then the combined maximum resistance is given by `R_(min) = r+ r + .... "times" =nr` `:.(R_(min))/(R_(min)) = (r//n)/(nr) = (1)/(n^(2))` |
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| 764. |
The reading on a high resistance voltmeter. When a cell is connected across it is `2.2 V`. When the terminals of the cell are connected to a resistance of `5 Omega` the voltmeter reading drop to `1.8 V`. Find the internal resistance of the cell. |
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Answer» Correct Answer - ` (10)/(9) Omega` Here, `epsilon = 2.2 V, V = 1.8 V, R = 5 Omega , r = ?` As `r = (( epsilon - V)/(V)) R = ((2.2 - 1.8))/(1.8) xx 5 = (10)/(5) Omega` |
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| 765. |
Figure shows a potentiometer with a cell of 2.0 V and internal resistance `0.40 Omega` maintain a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents upto a few mA) gives a balances point at 63.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of `600 Omega` is put is series with it, which is shortedd close to the balance point. The standard cell is then replaced by a cell of unknown emf `epsilon` and the balance point found similarly, turns out to be at 82.3 cm length of the wire. What purpose does the high resistance of 600 k `Omega` have? |
| Answer» The purpose of using high resistance of `600 k Omega` is to allow very small current through the galvanometer when the movalbe contact is far from the balance point. | |
| 766. |
Figure shows a potentiometer with a cell of 2.0 V and internal resistance `0.40 Omega` maintain a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents upto a few mA) gives a balances point at 63.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of `600 Omega` is put is series with it, which is shortedd close to the balance point. The standard cell is then replaced by a cell of unknown emf `epsilon` and the balance point found similarly, turns out to be at 82.3 cm length of the wire. What is the value `epsilon` ? |
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Answer» Here `epsilon = 1.02 V, L_(1) = 67.3cm, epsilon_(2) = epsilon = ?, L_(2) = 82.3 cm` Since `(epsilon_(2))/(epsilon_(1)) = (L_(2))/(L_(1)) :. Epsilon = (L_(2))/(L_(1)) xx epsilon_(1) = (82.3)/(67.3) xx 1.02 = 1.247 V` |
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| 767. |
Determine the equivalent resistance of networks shown in Fig. |
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Answer» (a) The given network is a series combination of 4 equal units. Each unit has 4 resistances in which 2 resistance (`1Omega` each in series) are in parallel with 2 other resistances (`2 Omega` each in series) `:.` Effective resistance of two resistance (each of `1 Omega`) in series `= 1 + 1 = 2 Omega` Effective Resistance of two resistance (each of `2 Omega`) in series `= 2 + 2 = 4 Omega` If R is the resistance of one unit of resistance them `(1)/(R_(P)) = (1)/(2) + (1)/(4) = (3)/(4)` or `R_(P) = (4)/(3) Omega`. `:.` Equivalent resistance in network `= 4 R_(P) = 4 xx (4 Omega)/(3) = (16 Omega)/(3)` (b) Total resistances each of the value of R are connected in series. Their effective resistance = 5R |
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| 768. |
If a constant potential difference is applied across a bulb, the current slightly decreases as time passes and then becomes constant. Explain. |
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Answer» Correct Answer - C |
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| 769. |
A proton beem is going from west. Is there an electric current ? If yes, in what direction? |
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Answer» Correct Answer - A |
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| 770. |
Pressure versus temperature graph of an ideal gas at constant volume V is shown by the straight line A. Now mass of the has is doubled and the volume is halved, then the corresponding pressure versus temperature graph will be shown by the line A. AB. BC. CD. None of these |
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Answer» Correct Answer - B |
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| 771. |
Gas at a pressure `P_(0)` in contained as a vessel. If the masses of all the molecules are halved and their speeds are doubles. The resulting pressure P will be equal toA. `4p_(0)`B. `2p_(0)`C. `p_(0)`D. `(p_(0))/(2)` |
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Answer» Correct Answer - B |
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| 772. |
By a cell a current of 0.9 A flows through 2 ohm resistor and `0.3 A` through 7 ohm resistor. The internal resistance of the cell isA. `0.5 Omega`B. `1.0 Omega`C. `1.2 Omega`D. `2.0 Omega` |
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Answer» Correct Answer - A |
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| 773. |
Current is flowing through a wire of non-uniform cross section. Cross section of wire at A is les than the cross section of wire at B. Then match the following |
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Answer» Correct Answer - `(A)S, (B)Q,(C)Q,(D)Q` |
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| 774. |
Pressure `P`, Volume `V` and temperature `T` of a certain material are related by the `P=(alphaT^(2))/(V)`. Here `alpha` is constant. Work done by the material when temparature changes from `T_(0)` to `2T_(0)` while pressure remains constant is :A. `6alphaT_(0)^(3)`B. `(3)/(2)alphaT_(0)^(2)`C. `2alphaT_(0)^(2)`D. `3alphaT_(0)^(2)` |
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Answer» Correct Answer - D |
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| 775. |
In the circuit shown in figure, match the match the following |
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Answer» Correct Answer - `(A)Q, (B)P,(C)R,(D)Q` |
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| 776. |
In the circuit shown in figure , `R_(1)=R_(2)=R_(3)=R.` Match the following |
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Answer» Correct Answer - `(A)P, (B)P,(C)P,` |
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| 777. |
Four resistance are connected to a DC battery as shown in figure. Maximum power will be developed across ......... Ohm resistance. A. 3B. 6C. 12D. 36 |
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Answer» Correct Answer - C |
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| 778. |
A voltmeter and an ammeter are connected in the circuit as shown. Resistance of ammeter is say `(R)/(10)` and that of voltmeter is 10R. Then A. percentage error in the reading of ammeter (compared to that measured, if both ammeter and voltmeter were ideal) is - 1.0%B. percentage error in the reading of voltmeter is - 10.0%C. both (a) and (b) are correctD. both (a) and (b) wrong |
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Answer» Correct Answer - C |
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| 779. |
A cell of emf E having an internal resistance R varies with R as shown in figure by the curve A. AB. BC. CD. D |
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Answer» Correct Answer - B |
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| 780. |
A wire of resistance `2.20 Omega` has a length `2m`. Calculate the length of the similar wire which connected in parallel with `2m` length wire will give a resistance of `2.0 Omega` |
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Answer» Correct Answer - `20m` Resistance per unit length of wire `= 2.2//2 = 1.1 Omegam^(-1)`, Let `R_(1)` be the resistance connected in parallel to `R(=2.20 Omega)` so that effective `R_(P) = 2.0 Omega`. Then `(1)/(R_(1)) = (1)/(R_(P)) - (1)/(R ) = (1)/(2) - (1)/(2.2) = (0.1)/(2.2) = (1)/(22)` or `R_(1) = 22 Omega` `:. ` Length of wire neended `= 22//1 = 20m` |
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| 781. |
Find the value of colour coded resistance shoen in fig.A. `520+-10%`B. `5200+-1%`C. `52000+-10%`D. `52000+-1%` |
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Answer» Correct Answer - C `R=52xx10^(3)+-10%` |
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| 782. |
The resistance of a wire is `2Omega` if it is drawn in such a way that it experiences a longitudinal strain `200%` its new resistance isA. `4Omega`B. `8Omega`C. `16Omega`D. `18Omega` |
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Answer» Correct Answer - D `Rpropl^(2)` |
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| 783. |
Two wires of same dimension but resistivities `rho_1` and `rho_2` are connected in series. The equivalent resistivity of the combination isA. `rho_(1)+rho_(2)`B. `1//2 (rho_(1)+rho_(2))`C. `sqrt(rho_(1)rho_(2))`D. `2(rho_(1)+rho_(2))` |
| Answer» Correct Answer - B | |
| 784. |
Consider a current carrying wire (current I) in the shape of a circle. Note that as the current progresses along the wire, the direction of `vecJ` (current density) changes in an exact manner, while the current I remains unaffected. The agent that is essentially responsible for isA. source of e.mf.B. electric field produced by charges accumulated on the surface of wire.C. the charges just behind a given segment of wire which push them just right way by repulsionD. the charges just behind a given segment of wire which push them just right way by repulsion |
| Answer» Correct Answer - B | |
| 785. |
Consider a current carrying wire (current I) in the shape of a circle. Note that as the current progresses along the wire, the direction of `vecJ` (current density) changes in an exact manner, while the current I remains unaffected. The agent that is essentially responsible for isA. source of emfB. electric field produced by charges accumulated on the surface of wireC. the charges just behind a give segment of wire which push them just the right way by repulsionD. the charges ahead |
| Answer» Current per unit area (taken normal to the current) `(I)/(A)` is called current density and is denoted by j. The SI units of the current density are `(A)/(m^(2))`. The current density is also directed along E and is also a vector and the relation ship is give by (`J=SE`) The j changes due to electric field produced by charges accumulated on the surface of wire. | |
| 786. |
Consider a current carrying wire (current I) in the shape of a circle. Note that as the current progresses along the wire, the direction of `vecJ` (current density) changes in an exact manner, while the current I remains unaffected. The agent that is essentially responsible for isA. source of emfB. electric field produced by charges accumulated on the surface of wireC. the charges just behind a given segment of wire which push them just the right way by repulsionD. the charges ahead |
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Answer» Correct Answer - B |
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| 787. |
A wire of resistance `12 Omega m^(-1)` is bent to from a complete circle of radius `10 cm`. The resistance between its two diametrically opposite points, `A` and `B` as shown in the figure, is A. `0.6piOmega`B. `3Omega`C. `6piOmega`D. `6Omega` |
| Answer» Correct Answer - A | |
| 788. |
A wire of resistance `12 Omega m^(-1)` is bent to from a complete circle of radius `10 cm`. The resistance between its two diametrically opposite points, `A` and `B` as shown in the figure, is A. `0.6 pi Omega`B. `3 pi Omega`C. `6 pi Omega`D. `6 pi Omega` |
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Answer» Correct Answer - A (a) Circumference of circule `= 2 xx pi (10)/(100) = (2 pi)/(10) = (pi)/(5)` Resistance of wire `= 12 xx (pi)/(5) = (12 pi)/(5)` Resistance of each section `= (12 pi)/(10) Omega` `:.` Equivalent resistance `R_(eq) = ((12 pi)/(10) xx (12pi)/(10))/((12pi)/(10) + (12 pi)/(10)) = (6 pi)/(10) = 0.6 pi Omega` |
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| 789. |
When a wire of uniform cross-section `a`, length `I` and resistance `R` is bent into a complete circle, resistance between two of diametrically opposite points will beA. `(R)/(4)`B. `(R)/(8)`C. `4R`D. `(R)/(2)` |
| Answer» Correct Answer - A | |
| 790. |
For the network shown in the figure the value of the current i is A. `(9V)/(35)`B. `(5V)/(18)`C. `(5V)/(9)`D. `(18V)/(5)` |
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Answer» Correct Answer - B |
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| 791. |
The emf of a cell E is 15 V as shown in the figure with an internal resistance of `0.5Omega`. Then the value of the current drawn from the cell isA. `3A`B. `2A`C. `5A`D. `1A` |
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Answer» Correct Answer - D |
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| 792. |
Copper and carbon wires are connected in series and the combined resistor is kept at `0^(@)C`. Assuming the combined resistance does not vary with temperature the ratio of the resistances of carbon and copper wires at `0^(@)C` is (Temperature coefficient of resistivity of copper and carbon respectively are `4xx(10^(-3))/(``^(@)C)` and `-0.5xx(10^(-3))/(``^(@)C)`A. 2B. 4C. 8D. 6 |
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Answer» Correct Answer - C `R_(1)alpha_(1)=R_(2)alpha_(2)` |
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| 793. |
Copper and carbon wires are connected in series and the combined resistor is kept at `0^(@)C`. Assuming the combined resistance does not vary with temperature the ratio of the resistances of carbon and copper wires at `0^(@)C` is (Temperature coefficient of resistivity of copper and carbon respectively are `4xx(10^(-3))/(``^(@)C)` and `-0.5xx(10^(-3))/(``^(@)C)`A. 4B. 8C. 6D. 2 |
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Answer» Correct Answer - B |
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| 794. |
V- I graph for a metallic wire at two different temperature `T_(1)` and `T_(2)` is shown in figure. Which of the two temperature is higher and why? A. `cos 2 theta`B. `sin 2 theta`C. `cot 2 theta`D. `tan 2 theta` |
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Answer» Correct Answer - c `R_(1) = tan theta = R_(0) ( 1+ alpha T_(1))` and`R_(2) = cos theta = R_(0) ( 1+ alpha T_(2))` `:. cot theta - tan theta= = R_(0) ( 1+ alpha T_(2))- R_(0) ( 1+ alpha T_(2)) ` `R_(0)alpha (T_(2) - T_(1))` or `T_(2) - T_(1) = (1)/(alpha R_(0)) (cot theta - tan theta)` `(1)/(alpha R_(0)) ((cot theta)/(sin theta) -(sin theta)/(cos theta))= (2 cos 2 theta)/(alpha R_(0) sin 2 theta)` `= (2)/(alpha R_(0)) cot 2 theta or T_(1) - T_(2) prop cot 2 theta` |
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| 795. |
The variation of resistance of a metallic conductor with temperature is shown in figure. (i) Calculate the temperature coefficient of resistance from the graph. (ii) State why the resistance of the conductor increases with the rise in temperature. |
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Answer» (i) From graph, corrensponding to point A, let the resistance be `R_(0)` and corresponding to point B, the resistance be R. When temperature changes from 0 to `theta^@C`, the resistance changes from `R_(0)` to R . Thus change in temperature `= theta - 0 = theta^@C` Change in resistance `= R - R_(0)` Temperature coefficient of resistnace, `alpha= ("change in resistance")/("original resistance" xx "change in temp.")` `(R - R_(0))/(R_(0) xx theta) = (OC - OA)/(OA xx OE)` (ii) With the rise of temperature of conductor, the resistance of a conductor increases because the frequency of collision of electrons with ions/atoms of the conductor increases, resulting decreases in relaxtion time `(tau)` of electrons. As `R prop 1// tau`, so R increases as `tau` decreases. |
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| 796. |
V- I graph for a metallic wire at two different temperature `T_(1)` and `T_(2)` is shown in figure. Which of the two temperature is higher and why? |
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Answer» We know that resistance, `R=V/l` From figure, for given V, `I_(1) gt I_(2)` , therefore, `R_(1) ltR_(2)`. As `R prop T_(1)` so `T_(1) lt T_(2)` i.e., `T_(2)` is greater than `T_(1)` . |
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| 797. |
The voltage-current variations of two metallic wire X and Y at constant temperature is shown in fig. Assuming that the wires have the same length and the same diameter, explain which of the two wires will have larger resistivity. |
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Answer» Slope of I - V line for a wire represents the coductance of the wire. From graph. Slope of I - V line for wire X gt slope of I - V line for wire Y. `:.` Conductance of wire X gt conductance of wire Y. As resistance `=1/"conductance"`, so Resistance of wire X lt resistance of wire Y. `rho X l/A lt rho Y l/A or rho X lt rho Y` Thus, wire Y has larger resistivity. |
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| 798. |
Consider a current carryiing wire (current I) in the shape of a circleA. source of emfB. electric field produced by charges accumulated on the surface of wireC. the charges just behind a given segment of wire which push them just the right way by repulsionD. The charges ahead. |
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Answer» Correct Answer - B Current per unit area (taken normal to the current( I/A, is called current density and is denoted by j. the SI units of the current density are `A//m^(2)`. The current density is also directed along E and is also a vector and the relationship is given by `j=sE` The j changes due to electric field produced by charges accumulated on the surface of wire. |
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| 799. |
In the given circuit the potential at point B is zero, the potential at points A and D will be A. `V_(A)=4V, V_(D)=9V`B. `V_(A)=3V, V_(D)=4V`C. `V_(A)=9V, V_(D)=3V`D. `V_(A)=4V, V_(D)=3V` |
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Answer» Correct Answer - D `V_(A)-V_(B)=2xx2=4V` `thereforeV_(A)-0=4VRightarrow V_(A)=4V` According to the question `V_(B)=0` Point D is connected to positive terminal of battery of emf 3V. |
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| 800. |
The equivalent resistance of resistors connected in series is alwaysA. Equal to the mean of component resistorsB. Less than the lowest of component resistorsC. In between the lowest and the highest of component resistorsD. Equal to sum of component resistors |
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Answer» Correct Answer - D |
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