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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.
| 1751. |
Ideal monoatomic gas is taken through a process `dQ = 2dU`. Find the molar heat capacity (in terms of `R)` for the process? (where `dQ` is heat supplied and `dU` is change in internla energy)A. 5 RB. 3 RC. RD. None of these |
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Answer» Correct Answer - B |
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| 1752. |
A battery of e.m.f. `10 V` and internal resistance `0.5 ohm` is connected across a variable resistance `R`. The value of `R` for which the power delivered in it is maximum is given byA. `2 Omega`B. `0.25 Omega`C. `1 Omega`D. `0.5 Omega` |
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Answer» Correct Answer - D For maximum power, ` therefore " " R=0.5 Omega` |
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| 1753. |
A battery of e.m.f. `10 V` and internal resistance `0.5 ohm` is connected across a variable resistance `R`. The value of `R` for which the power delivered in it is maximum is given byA. `2.0 ohm`B. `0.25 ohm`C. `1.0 ohm`D. `0.5 ohm` |
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Answer» Correct Answer - D (d) For maximum power `r = R` |
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| 1754. |
Two bulbs one of `200` volts, `60` watts & the other of `200` volts, `100` watts are connected in series to a `200` volts supply. The power consumed will beA. `37.5` wattB. `160` wattC. `62.5` wattD. `110` watt |
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Answer» Correct Answer - A `1/P=1/(P_(1))+1/(P_(2))` `P=(P_(1)P_(2))/(P_(1)+P_(2))=(300xx600)/900=200 W` |
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| 1755. |
For the circuit shown in the figure, find the charge stored on capacitor in steady state. A. `(RC)/(R+R_(0))E`B. `(RC)/(R_(0))(E-E_(0))`C. ZeroD. `(RC)/(R+R_(0))(E-E_(0))` |
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Answer» Correct Answer - B |
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| 1756. |
Consider the circuit shown in the figure. Find the charge on capacitor `C` between `A` and `D` in steady state. A. `C epsilon`B. `C epsilon // 2`C. `C epsilon // 3`D. zero |
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Answer» Correct Answer - D (d) In steady state, there will be no current in any branch. So `A` and `D` will be at same potential. Hence charge on capacitor betwork `A` to `D` will be zero |
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| 1757. |
The resistance of the network between the terminals A and B isA. `30Omega`B. `20Omega`C. `50Omega`D. `60Omega` |
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Answer» Correct Answer - B End resistors are not considered |
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| 1758. |
Most of the times we connected remote speaker to play music another room along with the buit-in speakers. These speakers are connected in parallel with the music system. At the instant represented in the picture, the a.c. voltage across the speakers is `6.00 V`. the main speaker resistance is `8.00 Omega` and the remote speaker has `4.00 Omega` resistance. The equivalent resistance of the speaker is: |
| Answer» Correct Answer - C | |
| 1759. |
If each of the resistance in the network in figure `R`, the equivalent resistance between terminals `A` and `B` is A. RB. 5RC. 3RD. 6R |
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Answer» Correct Answer - A |
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| 1760. |
A network of resistor is connected to a 16 V battery with internal resistance of `1 Omega`, a shown in fig. Obtain the current in each resistor. |
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Answer» The total current I in the circuit is `I = (epsilon)/(R + r) = (16 v)/((7 + 1)Omega) = 2A` Consider the resistor between A and B. if `I_(1)` is the current in one of the `4 Omega` resistors and `I_(2)` the current in the other. `I_(1) xx 4 = I_(2) xx 4` That is `I_(1) = I_(2)`, which is otherwise obvious from the symmetry of the two arms. But `I_(1) + I_(2) = I = 2A`. Thus `I_(1) = I_(2) = IA` That is, current in each `4 Omega` resistor is 1A, Current in `1 Omega` resistor between B and C would be 2 A. Now, consider the resistances between C and D. If `I_(3)` is the current in the `12 Omega` resistor, and `I_(4)` in the `6 Omega` resistor, `I_(3) xx 12 = I_(4) xx 6`, i.e., `I_(4) = 2I_(3)` But, `I_(3) + I_(4) = I = 2A` Thus, `I_(3) = ((2)/(3)) A, I_(4) = ((4)/(3)) A` That is,the current in the `12 Omega` resistor is (2/3)A, while the current in the `6 Omega` resistor is (4/3) A. |
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| 1761. |
Find potential difference between points A and B of the network shown in fig. and distribution of given main current through different resistors. |
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Answer» Between point A and B resistors of `4Omega,6Omega` and `8Omega` resistances are in series and these are in parallel to `9Omega` resistor. ,brgt Equivalent resistance of series combination is `R_(1)=(4+6+8)ohm=18` If equivalent resistance between A and B is `R=(9xx18)/((9+18))ohm=6Omega` Potential difference between A and B is `V=IR=2.7xx6V=16.2V` Current through `9Omega` resistor `=(16.2)/(9)=1.8A` Current through `4Omega,6Omega` and `8Omega` resistors `=2.7-1.8=0.9A` |
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| 1762. |
A network of resistor is connected to a 16 V battery with internal resistance of `1 Omega`, a shown in fig. (a) complete the equivalent resistance of the network. |
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Answer» The network is a simple series and parallel combination of resistors. First the two `4 Omega` resistors in parallel are equivalent to a resistor. `= [(4 xx 4)//(4 + 4)]Omega = 2 Omega` In the same, way, the `12 Omega` and `6 Omega` resistors in parallel are equivalent to a resistor of `[(12 xx 6)//(12 + 6)] Omega = 4 Omega` The equivalent resistance R of the network is obtained by combining these resistors (`2 Omega` and `4 Omega` with `1 Omega` in series, that is `R = 2 Omega + 4 Omega + 1 Omega = 7 Omega` |
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| 1763. |
Find equivalent resistance of the network in fig. Between points (i) A and B and (ii) A and C. |
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Answer» (i) The `10Omega` and `30Omega` resistors are connected in parallel between points A and B. The equivalent resistance between A and B is `R_(1)=(10xx30)/(10+30)ohm=7.5Omega` (ii) The resistance `R_(1)` is connected in series with resistor of `7.5Omega` hence the equivalent resistance between point A and C is `R_(2)=(R_(1)+7.5)ohm=(7.5+7.5)ohm=15Omega`. |
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| 1764. |
In the network shown in fig. each resistance is R. The equivalent resistance between A and B is |
| Answer» Correct Answer - C | |
| 1765. |
If the balance point is obtained at the 35th cm in a metre bridge the resistances in the left and right gaps are in the ratio ofA. `7 :13`B. `13:7`C. `9:11`D. `11:9` |
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Answer» Correct Answer - A |
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| 1766. |
In the network shown in fig. each resistance is R. The equivalent resistance between A and B is |
| Answer» Correct Answer - A | |
| 1767. |
Two resistances are connected in two gaps of a metre bridge. The balance point is 20cm from the zero end. A resistance of`15Omega` is connected in series with the smaller of the two. The null point shifts to 40cm. Then value of the smaller resistance Is:A. `3Omega`B. `6Omega`C. `9Omega`D. `12Omega` |
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Answer» Correct Answer - B |
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| 1768. |
Potentiometer is an ideal voltmeter as voltmeter draws some current through the circuit while potentiometer needs no current to work. Potentiometer works on the principle of e.m.f comparison. In working conditions, a constant current flows through out the wire of potentiometer using standard cell of e.m.f. `e_(1)`. The wire of potentiometer is made of uniform material and cross sectional area and it has uniform resistance per unit length. the potential gradient, depends upon the current in the wire. A potentiometer with a cell of e.m.f `2 V` and internal resistance `0.4 Omega` is used across the wire `AB`. A standard cadmium cell of e.m.f. `1.02 V` gives a balance point at `66.3 cm` length of wire. The standard cell is the balance point found similarly turns out to be `82.3 cm` length of the wire. The length of potentiometer wire `AB` is `1m`. The reading of potentiometer if `4V` battery is used instead of `e`, is |
| Answer» Correct Answer - D | |
| 1769. |
Potentiometer is an ideal voltmeter as voltmeter draws some current through the circuit while potentiometer needs no current to work. Potentiometer works on the principle of e.m.f comparison. In working conditions, a constant current flows through out the wire of potentiometer using standard cell of e.m.f. `e_(1)`. The wire of potentiometer is made of uniform material and cross sectional area and it has uniform resistance per unit length. the potential gradient, depends upon the current in the wire. A potentiometer with a cell of e.m.f `2 V` and internal resistance `0.4 Omega` is used across the wire `AB`. A standard cadmium cell of e.m.f. `1.02 V` gives a balance point at `66.3 cm` length of wire. The standard cell is the balance point found similarly turns out to be `82.3 cm` length of the wire. The length of potentiometer wire `AB` is `1m`. The value of `e` is : |
| Answer» Correct Answer - A | |
| 1770. |
The equivalent resistance between x and y in the circuit shown is A. `10 Omega`B. `40 Omega`C. `20 Omega`D. `(5)/(2) Omega` |
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Answer» Correct Answer - A |
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| 1771. |
Practical diagram for meter bridge is as shown in figure. (a) G deflects zero, when `l_(1)=20cm` and `R=100Omega` then find X. (b) If R and X are interchanged then find `l_(1)` for zero deflection. |
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Answer» (a) `l_(1)=20cm, l_(2)=100-l_(1)=100-20=80cm,R=100Omega` `(P)/(Q)=(R)/(S)rArr(l_(1))/(l_(2))=(R)/(X)rArr(20)/(80)=(100)/(X)rArrX=400Omega` (b) `(l_(1))/(100-l_(1))=(X)/(R)=(400)/(100)rArrl_(1)=80cm` or R and X interchane then `l_(1),l_(2)` interchanged. |
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| 1772. |
Find the equivalent resistance across `AB` : A. `1 Omega`B. `2 Omega`C. `3 Omega`D. `4 Omega` |
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Answer» Correct Answer - A |
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| 1773. |
Two wires a and b, each of length 40 m and area of cross-section `10^(-7) m^(2)` , are connected in series and a potenital difference of 60 V is applied between the ends of this combined wire. Their resistances are respectively 40 and `20 Omega`. Determine for each wire (i) specific resistance, (ii) electric-field and (iii) current-dentsity. |
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Answer» Correct Answer - `(i) 1.0xx10^(-7) Omega m, 5.0xx10^(-8) Omega m (ii) 1.0 Vm^(-1), 0.5 Vm^(-1) (iii) 1.0xx10^(7) Am^(-2), 1.0xx10^(7) Am^(-2)` |
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| 1774. |
A 6 volt battery of negligible internal resistance is connected across a potentiometer wire is AB of length 100 cm and uniform area of cross-section. The posistive terminal of another battery of emf 4 V and internal resistance `1 Omega` is joined to the point A as shown in figure. If we take potential at B to be zero, (a) What are the potentials at points A and C. (b) At which point D of the potentiometer C? ( C) If the points C and D are connected by a wire, What will be the current through itgt (d) if the 4 V battery is replaced by 7.5 V battery, what would be the answers of parts (a) and (b)? |
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Answer» (a)Potential difference between A and B = 6 V. As potential at B = 0, So the potential at A = 6V Since the potential difference between points A and C = 4 V. Therefore, potential at C = 6 - 4 = 2 V (b) Let the potential at point D be the same as at C. Then potential difference between A and D = potential between A and C = 4V. The length AD of potentimeter wire `=100/6 xx 4 = 66.7 cm` (c ) IF points C and D are connceted by a wire, then current through this wire is zero because potential at C is equal to potential at D. (d) If 4V battery is replaced by 7.5 V battery, then potential difference between A and C = 7.5 V. As point A is at 6V, the point C will be at - 1.5 V. Since the voltage of the battery to be balanced on potentiometer wire is more than fall of potential across the potentiometer wire, no balance point like D would exist on wire. |
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| 1775. |
A set of n identical resistors, each of resistance `R Omega`, when connected in series have an effective resistance `X Omega` and when the resistors are connected in parallel, their effective resistance is `Y Omega`. Find the relation between R, X and Y. |
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Answer» When n identical resistors are connected in series `X=R+R+R+ ..............+R=nR` and r esistors are connected in parallel `(1)/(Y)=(1)/(R)+(1)/(R)+............+(1)/(R)=(1)/(R)=(n)/(R)rArrY=(R)/(n)` `rArrX=nR=((R)/(Y))R=(R^(2))/(Y)therefore XY=R^(2)` |
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| 1776. |
Three resistors are connected to form the sides of a triangle ABC , the resistance of the sides AB, BC and CA are 40 ohms , 60 ohms and 100 ohms respectively. The effective resistance between the points A and B in ohms will beA. 32B. 64C. 50D. 200 |
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Answer» Correct Answer - A |
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| 1777. |
For the given circuit, the potenital difference across P and Q will be nearest to A. 9.6 VB. 6.6 VC. 4 VD. 3.2 V |
| Answer» Correct Answer - D | |
| 1778. |
In the circuit, the potential difference across PQ will be nearest to A. 9.6 VB. 6.6 VC. 4.8 VD. 3.2 V |
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Answer» Correct Answer - D potential different across PQ. Ie. Potentail difference across the resistance of 20 `Omega ` ,which is V=lxx20 and `i=(48)/((100+100+80+20))` `0.16 A `v=0.16xx20=3.2 V` |
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| 1779. |
A 3 volt battery with negligible internal resistance is connected in a circuit as shown in the figure. The current I, in the circuit will be A. `1//3 A`B. `1A`C. `1.5A`D. `2A` |
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Answer» Correct Answer - C |
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| 1780. |
Two resistance of `400 Omega` and `800 Omega` are connected in series with 6 volt battery of negligible internal resistance. A voltmeter of resistance `10,000 Omega` is used to measure the potential difference across `400 Omega`. The error in measurement of potential difference in volts approximatley isA. `0.01`B. `0.02`C. `0.03`D. `0.05` |
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Answer» Correct Answer - D |
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| 1781. |
In the circuit, the potential difference across PQ will be nearest to A. `9.6 V`B. `6.6 V`C. `4.8V`D. `3.2 V` |
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Answer» Correct Answer - D |
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| 1782. |
A 3 volt battery with negligible internal resistance is connected in a circuit as shown in the figure. The current I, in the circuit will be A. `1 A`B. `1.5 A`C. `2 A`D. `1/3 A` |
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Answer» Correct Answer - B `i=3/(((6xx3)/(6+3)))=3/2=1.5 A` |
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| 1783. |
Two resistance wires on joining in parallel the resultant resistance is ohms `(6)/(5)` ohms. One of the wire breaks, the effective resistance is 2 ohms . The resistance of the broken wire isA. `(3)/(5)` ohmB. 2 ohmC. `(6)/(5)` ohmD. 3 ohm |
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Answer» Correct Answer - D |
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| 1784. |
The length of a given cylindrical wire is increased by `100%`. Due to the consequent decrease in diameter the change in the resistance of the wire will beA. `200 %`B. `100 %`C. `50%`D. `300%` |
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Answer» Correct Answer - D `R=rho l/A=rho(l^(2))/Vimplies R prop l^(2)` `as l to2l` `impliesR to 4R` `implies %` change in `R=300 %` |
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| 1785. |
The length of a given cylindrical wire is increased by `100%`. Due to the consequent decrease in diameter the change in the resistance of the wire will beA. `300 %`B. `200%`C. `100 %`D. `50%` |
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Answer» Correct Answer - A (a) If suppose initial length `l_(1) = 100` then `l_(2) = 100 + 100 = 200` `(R_(1))/(R_(2)) = ((l_(1))/(l_(2)))^(2) = ((100)/(200))^(2) implies R_(2) = 4 R_(1)` `(Delta R)/(R ) xx 100 = (R_(2) - R_(1))/(R_(1)) xx 100 = (4 R_(1) - R_(2))/(R_(1)) xx 100 = 300 %` |
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| 1786. |
The length of a given cylindrical wire is increased by `100%`. Due to the consequent decrease in diameter the change in the resistance of the wire will beA. `300%`B. `200%`C. `100%`D. `50%` |
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Answer» Correct Answer - A (a) If suppose initinal length `l_(1) = 100` then `l_(2) = 100 + 100 = 200` `(R_(1))/(R_(2)) = ((l_(1))/(l_(2)))^(2) = ((100)/(200))^(2) implies R_(2) = 4 R_(1)` `(Delta R)/(R ) xx 100 = (R_(2) - R_(2))/(R_(1)) xx (4 R_(1) - R_(1))/(R_(1)) xx 100 = 300%` |
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| 1787. |
Two resistance wires on joining in parallel the resultant resistance is ohms `(6)/(5)` ohms. One of the wire breaks, the effective resistance is 2 ohms . The resistance of the broken wire isA. `(3)/(5)Omega`B. `2Omega`C. `(6)/(5)Omega`D. `3Omega` |
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Answer» Correct Answer - D |
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| 1788. |
The specific resistance of a wire is `rho`, its volume is `3 m^(3)` and its resistance is `3 ohms`, then its length will beA. `sqrt((1)/(rho))`B. `(1)/(sqrtrho)`C. `(1)/(rho) sqrt3`D. `rho sqrt((1)/(3))` |
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Answer» Correct Answer - B (b) Volume `= Al = 3 implies A = (3)/(l)` Now `R = rho (l)/(A) implies 3 = (rho xx l)/(3//l) = (rho l^(2))/(3) implies l^(2) = (9)/(rho) = (3)/(sqrt(rho))` |
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| 1789. |
The length of a given cylindrical wire is increased by `100%`. Due to the consequent decrease in diameter the change in the resistance of the wire will beA. 3B. 2C. 1D. 0.5 |
| Answer» Correct Answer - 0.01 | |
| 1790. |
The specific resistance of a wire is `rho`, its volume is `3 m^(3)` and its resistance is `3 ohms`, then its length will beA. `sqrt((1)/(rho))`B. `(3)/(sqrt(rho))`C. `(1)/(rho)sqrt(3)`D. `rho sqrt((1)/(3))` |
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Answer» Correct Answer - B |
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| 1791. |
When a piece of aliminium wire of finite length is drawn through a series of dies to reduce its diameter to half its original value, its resistance will becomeA. two timesB. four timesC. eight timesD. sixteen times |
| Answer» Correct Answer - D | |
| 1792. |
When a piece of aliminium wire of finite length is drawn through a series of dies to reduce its diameter to half its original value, its resistance will becomeA. Two timesB. Four timesC. Eight timesD. Sixten times |
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Answer» Correct Answer - D |
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| 1793. |
A piece of wire of resistance 4 ohm s is bent through `180^(@)` at its mid point and the two halves are twisted together, then the resistance isA. 8 ohmsB. 1 ohmC. 2 ohmsD. 5 ohms |
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Answer» Correct Answer - B |
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| 1794. |
A piece of wire of resistance 4 ohm s is bent through `180^(@)` at its mid point and the two halves are twisted together, then the resistance isA. `8Omega`B. `1Omega`C. `2Omega`D. `5Omega` |
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Answer» Correct Answer - B |
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| 1795. |
In a meter bridge experiment, null point is obtained at `20 cm` from one end of the wire when resistance `X` is balanced against another resistance `Y`. If `X lt Y`, then the new position of the null point from the same end, if one decides to balance a resistance of `4 X` against `Y` will be at.A. 50 cmB. 80 cmC. 40 cmD. 70 cm |
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Answer» Correct Answer - A `(X)/(Y) = (20)/(100 - 20) rArr 4 X = Y` `(4 X)/(Y) = (l)/(100 -l) rArr 1 = (l)/(100 - l)` `l = 50 cm`. |
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| 1796. |
In a meter bridge experiment, null point is obtained at `20 cm` from one end of the wire when resistance `X` is balanced against another resistance `Y`. If `X lt Y`, then the new position of the null point from the same end, if one decides to balance a resistance of `4 X` against `Y` will be at.A. `50 cm`B. `80 cm`C. `40 cm`D. `70 cm` |
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Answer» Correct Answer - A `x/20=y/80 implies 4x=y` `implies` New position of null point will be at the mid point. |
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| 1797. |
In a meter bridge experiment null point is obtained at 20 cm. from one end of the wire when resistance X is balanced against another resistance Y. If XltY, then where will be the new position of the null point from the same end, if one deicdes to balance a resistance of 4 X against YA. 40 cmB. 80 cmC. 50 cmD. 70 cm |
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Answer» Correct Answer - C (c) In the first case `X/Y = 20/80 = 1/4` In the second case `(4X)/Y = l/(100-l) rArr l = 50`. |
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| 1798. |
The termistors are usually made ofA. metal oxides with high temperature coefficient of resistivityB. metals with high temperature coefficient of resistivityC. metals with low temperature coefficient of resistivityD. semiconducting materials having low temperature coefficient of resistivity |
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Answer» Correct Answer - A (a) Thermistors are usually made of metaloxides with high temperature coefficient of resistivity. |
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| 1799. |
Time taken by a 836 W heater to heat one litre of water from `10^@C to 40^@C` isA. 150 sB. 100 sC. 50 sD. 200 s |
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Answer» Correct Answer - A (a) `DeltaQ = mC xx DeltaT` `= 1 xx 4180 xx (40-10) = 4180 xx 30 ` `(:. DeltaQ = heat supplied in time t for heating 1L water from `10^@C to 40^@C`) `also Delta Q = 836 xx t rArr t = (4180 xx 30)/836 = 150 s.` |
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| 1800. |
Time taken by a 836 W heater to heat one litre of water from `10^@C to 40^@C` isA. `50s`B. `100s`C. `150s`D. `200s` |
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Answer» Correct Answer - C Let time taken in boiling the water by the heater is `t` sec. then `Q=ms DeltaTimplies(Pt)/J=msDeltaT` `836/4.2t=1xx1000(40^(@)-10^(@))` `836/4.2t=1000xx30` `t=(1000xx30xx4.2)/836` `=150 ` second |
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