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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.
| 601. |
A boy has two spare light bulbs in his drawer. One is market 220V and 100W and the other is market 240V and 60W. He tries to decide which of the following assertions are correct?A. The 60W light bulb has more resistance and therefore burns less brightlyB. The 60W light bulb has less resistance and therefore burns less brightlyC. The 100W light bulb has more resistance and therefore burns more brightlyD. The 100W light bulb has less resistance and therefore burns more brightly |
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Answer» Correct Answer - A When the same potential difference, that is the voltage, is applied as in houses. `"Power"=VI=(V^(2))/(R)` The smallest resistance consume greater power. Here 100 W bulb has less resistance. It should glow more birghtly. This 60W bulb has more resistance and therefore statement (a) is correct. |
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| 602. |
Figure (a) and figure (b) both are showing the variation of resistivity (p) with temperature (T) for some materials. Identify the types of these materials. A. Conductor and semiconductorB. Conductor and InsulatorC. Insulator and semiconductorD. Both are conductor |
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Answer» Correct Answer - A In conductors due to increasein temperature the resistivity increases and in semiconductors if the decreases exponentially. |
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| 603. |
With increase in temperature of the conductivity ofA. metals increases and of semiconductor decreases.B. semiconductors increases and of metals decreases.C. in both metals and semiconductors increasesD. in both metals and semiconductor decreases |
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Answer» Correct Answer - B Semiconductors having negative temperature coefficient of resistivity whereas metals are having positive temperature coefficient of resistivity with increase in temperature the resistivity of metal increases whereas resistivity of semiconductor decreases. |
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| 604. |
Assertion : Bending a wire does not effect electrical resistance. Reason : The resistance of wire is proportional to the resistivity of material.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is falseD. If the assertion and reason both are false |
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Answer» Correct Answer - A |
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| 605. |
Assertion : Bending a wire does not effect electrical resistance. Reason : The resistance of wire is proportional to the resistivity of material. |
| Answer» Correct Answer - B | |
| 606. |
Assertion: When the radius of a copper wire is doubled, its specific resistance gets increased. Reason: Specific resistance is independent of cross-section of material usedA. Both (A) and (R) are true and (R) is the correct explanantion of A.B. Both (A) and (R) are true but (R) is not the correct explanation of A.C. (A) is true but (R) is falseD. (A) is false but (R) is true. |
| Answer» Correct Answer - D | |
| 607. |
A copper wire of length l and r radius r is nickel plated till its final radius is 2r. If the resistivity of the copper and nickel are `rho_(c)` and `rho_(n)`, then find the equivalent resistance of the wire. |
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Answer» Let `R_(c)` and `R_(n)` be the resistance of copper wire and nickel plated portion of wire respectively. Then `R_(c) = (rho_(c) l)/(pi r^(2)) , and R_(n) = ( rho_(n) l)/(pi (2 r)^(2) - pi r^(2))` Here, `R_(C)` and `R_(n)` are in parallel, the equivalent resistance of the wire is `R = (R_(c) R_(n))/(R_(c) +R_(n)) = (rho_(c) l //(pi r^(2)) xx rho_(n)l//[pi(2r)^(2) - pi r^(2)])/((rho_(c) l)/(pi r^(2)) + (rho_(n)l)/([pi(2r)^(2) - pir^(2)]))` `(rho_(c ) rho_(n) l)/(pi r^(2) (rho_(n) + 3 rho_(c )))` |
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| 608. |
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 assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is falseD. If the assertion and reason both are false |
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Answer» Correct Answer - C |
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| 609. |
Assertion : In metre bridge experiment, a high resistance is always connected in series with a galvanometer. Reason : As resistance increases, current through the circuit increases,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 ) The resistance of the galvanometer is fixed. In meter bridge experiments, to protect the galvanometer from a high current, high resistance is connected to the galvanometer in order to protect it from damage. |
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| 610. |
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 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 - 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 to conductivity fo material. |
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| 611. |
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 true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is falseD. If the assertion and reason both are false |
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Answer» Correct Answer - B |
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| 612. |
Assertion : In metre bridge experiment, a high resistance is always connected in series with a galvanometer. Reason : As resistance increases, current through the circuit increases,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 - C (c ) The resistance of galvanometer is fixed. In metre bridge experiment, to protect is fixed. In metre bridge experiments, to protect the galvanometer from a high current, high resistance is connected to the galvanometer in order to prodect if from damage. |
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| 613. |
Assertion : There is no current in the metals in the absence of electric field. Reason : Motion of free electron are randomly.A. If both assertion and reason are ture and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason both are false. |
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Answer» Correct Answer - A (a) It is clear that electrons move in all directions haphazardly in metals. When an electric field field is applied, each free electron acquire a drift velcity. Thee is a net flow of charge, which consitute current. In the absence of electric field this is impossible and hence, there is no currrent. |
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| 614. |
Assertion : Electric field outside the conducting wire accurate measurement. Reason : The potential gradient for a potentiometer of accurate measurement. Reason : Net charge on conducting wire is zero.A. If both assertion and reason are ture and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason both are false. |
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Answer» Correct Answer - A (a) When current flows through a conductor it always remains uncharged, hence no electric field is produced outside it. |
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| 615. |
Find out the magnitude of resistance X in the circuit shown in figure, When no current flows through the `5 Omega` resistance |
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Answer» Since wheatstone bridge is balanced so `(x)/(18)=(2)/(6)` or `x=(18xx2)/(6)=6Omega` |
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| 616. |
In the given circuit, the potential of the point E is A. ZeroB. `-8V`C. `-4//3V`D. `4//3V` |
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Answer» Correct Answer - C |
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| 617. |
In the given circuit as shown below calculate the magnitude and direction of the current |
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Answer» Effective resistance of the circuit is `R_(eff)=2+2+1=5Omega` `therefore` total current in the circuit is `i=(V_(1)-V_(2))/(R_(eff))` `i=(10-15)/(5)=1A` Since the cell of larger emf decides the direction of low of current the direction of current in the circuit is from A to B through e |
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| 618. |
In the figure shown, the total resistance between A and B is A. 50 AB. 2AC. `0.5A`D. `(10)/(9)A` |
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Answer» Correct Answer - B |
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| 619. |
A moving coil galvanometer has 150 equal divisions. Its current sensitivity is 10-divisions per milliampere and voltage sensitivity is 2 divisions per millivolt. In order that each division reads 1 volt, the resistance in ohms needed to be connected in series with the coil will be -A. 99995B. 9995C. `10^(3)`D. `10^(5)` |
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Answer» Correct Answer - B (b) `I_(max) = (150)/(10) = 15 mA, V_(max) = (150)/(2) = 75 mV` resistance of galvanometer: `G = (V_(max))/(I_(max)) = (75)/(15) = 5 Omega` Now range of voltmeter `= 150 xx 1 = 150 V` `150 = (5 + R) I_(max) implies R = 9995 Omega` |
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| 620. |
The current from the battery in circuit diagram shown is A. 1AB. 2AC. 1.5 AD. 3A |
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Answer» Correct Answer - A |
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| 621. |
An infinite sequence of resistance is shown in the figure. The resultant between `A` and `B` will be, when `R_(1) = ohm` and `R_(2) = 2 ohm` A. InfinityB. `1 Omega`C. `2 Omega`D. `1.5 Omega` |
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Answer» Correct Answer - C |
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| 622. |
In the figure shown, the total resistance between `A` and `B` is A. `12 Omega`B. `4 Omega`C. `6 Omega`D. `8 Omega` |
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Answer» Correct Answer - D |
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| 623. |
A moving coil galvanometer has 150 equal divisions. Its current sensitivity is 10-divisions per milliampere and voltage sensitivity is 2 divisions per millivolt. In order that each division reads 1 volt, the resistance in ohms needed to be connected in series with the coil will be -A. 99995B. 9995C. `10^(3)`D. `10^(3)` |
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Answer» Correct Answer - B (b) Voltage sensitivity `= ("Current sensitivity")/("Resistance of galvaonometer G")` `G = (10)/(2) Omega` Here `i_(g) =` Full scale deflection current `= (150)/(10) = 15 mA`. `V =` Voltage to be measued `= 150 xx 1 = 150 V`. Hence `R = (V)/(i_(g)) - G = (150)/(15 xx 10^(-3)) 5 = 99995 Omega` |
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| 624. |
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. `19 Omega`C. `20 Omega`D. `25 Omega` |
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Answer» Correct Answer - b Initially `(5)/(R )= (l_(1))/((100 -l_(1))`…(i) Finally `(5)/(R//2)= (1.6l_(1))/((100 -16l_(1)))` or `(5)/(R )= (1.6l_(1))/(2(100 -16l_(1))`….(ii) `:.(l_(1))/((100 -l_(1))) = (16l_(1))/(2(100 -16l_(1)))` On solving we get `l_(1) = 25 cm` From(i) , `(5)/(R )=(25)/((100 - 25)) = (1)/(3) or R = 15 Omega` |
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| 625. |
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. `15 Omega`B. `20 Omega`C. `25 Omega`D. `10 Omega` |
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Answer» Correct Answer - A Initial condition, `(5)/(l_1) = ( R)/(100 - l_1)` …(i) Final condition, `(5)/(1.6 l_1) = (R//2)/(100 - 1.6 l_1)` …(ii) (i)//(ii) `rArr 1.6 = (2(100 - 1.6 l_1))/(100 - l_1)` `160 - 1.6 l_1 = 200 - 3.2 l_1` `1.6 l_1 = 40 rArr l_1 = (400)/(16) = 25` From (i) `(5)/(25) = (R)/(100 - 25) rArr R = 15 Omega`. |
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| 626. |
Across a metallic conductor of non-uniform cross-section a constant potential difference is applied. The quantityA. Current densityB. Current flows form `E` to `F`C. Drift velocityD. Electric field |
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Answer» Correct Answer - B (b) Across a metallic conductor of non-uniform cross section the rate of flow of charge through every cross section is constant hence . Current is constant. |
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| 627. |
In a ammeter `0.2%` of main current passes through the galvanometer. If resistance of galvanometer is `G`, the resistance of ammeter will beA. `(499)/(500) G`B. `(1)/(500) G`C. `(500)/(499) G`D. `(1)/(499) G` |
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Answer» Correct Answer - B `I_g G = (I - I_g) S` `0.002 I G = (I - 0.002 I) S` `S = (G)/(499)` Resistance of ammeter `R_A = (GS)/(G + S) = (G.(G)/(499))/(G +(G)/(499)) = (G)/(500)`. |
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| 628. |
Two wires A and B of the same material have their lengths in the ratio `5: 3` and diameter in the radius `2:3`. If the resistance of wire A is `15 Omega`. find the resistance of wire B |
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Answer» Correct Answer - `4 Omega` `R = (4 rhol)/(pi D^(2)) or R = (l)/(D^(2))` `:.(R_(2))/(R_(1)) = (l_(2))/(l_(1)) xx (D_(1)^(2))/(D_(2)^(2)) = (3)/(5) xx ((2)/(3))^(2) = (4)/(15)` or `R_(2) = (4)/(15) R_(1) = (4)/(15) xx 15 = 4 Omega` |
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| 629. |
A cylindrical metallic wire is stretched to increase its length by 5%. Calculate the percentage change in resistance. |
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Answer» Given % change in length, `(dl)/(l) = 5%` Resistance of wire `R = (rho l^(2))/(V)` % change in Resistance of wire, `(dR)/(R ) = 2 (dl)(l) = 2 xx 5% = 10 %` |
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| 630. |
Calculate the mass of copper required to draw a wire `5 km` long having resistance of `5 Omega` .The density of copper is `8.9 xx 10^(3) kg m^(-3)` and resistance of copper is `1.7xx 10^(-8) Omega m` |
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Answer» Correct Answer - `756.5 kg` `R = (rhol)/(A) or A = (rhol)/(R ) = ((1.7 xx10^(-8)) xx (5000))/(5)` `= 1.7 xx 10^(-5)m^(2)` Mass of copper required, m = volume `xx` density `= A l xx d` `= (1.7 xx 10^(-5)) xx 5000 xx (8.9 xx 10^(3)) = 756.5 kg` |
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| 631. |
A wire stretched to increase its length by`5%`. Calculate percentage charge in its resistance. |
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Answer» Correct Answer - `10.25 %` When a wire is stretched, its volume remain constant hence `l_(1)A_(1) = l_(2)A_(2) = V`, the volume Now `R_(1) = (rhol_(1))/(A_(1)) = (rhol_(1) xx l_(1))/(l_(1)A_(1)) = (rhol_(1)^(2))/(V)` , i.e. `R_(1) prop l_(1)^(2)` Hence or `R_(2) = 1.1025 R_(1)` `:. %` change in resistance `= ((R_(2) - R_(1))/(R_(1))) xx 100= ((R_(2))/(R_(1)) - 1) xx 100` `= (1.1025 - 1) xx 100 = 10.25 %` |
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| 632. |
A theostat has `1000` turn a wire radius `0.4 mm` having resistivity `49xx 10^(-8) Omega m`. The diameter of each turn is `4 cm. What are the maximum value of conductance and conductivity of rheostat wire. |
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Answer» Correct Answer - `24.5 Omega` Here length of wire used , `l = 100 pi D` `R = (rhol)/(pi r^(2)) = (rho 100 pi D)/(pi r^(2)) = (100 rho D)/(r^(2))` `= (100 xx (4.9 xx 10^(-8)) xx (5.0xx 10^(-2)))/((0.1 xx 10^(-3))^(2)) = 24.5 Omega` |
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| 633. |
In the circuit shown, the reading of the voltmeter and the ammeter areA. `4V.0.2A`B. `2V,0.4A`C. `3V,0.6A`D. `4V,0.04A` |
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Answer» Correct Answer - D combination of resistors |
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| 634. |
The resistance of a wire of 100 cmlength is `10Omega`. Now, it is cut into 10 equal parts and all of them are twisted to form a single bundle. Its resistances isA. `1Omega`B. `0.5Omega`C. `5Omega`D. `0.1Omega` |
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Answer» Correct Answer - D `underset(eff)(R)=(R)/(n^(2))` |
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| 635. |
Resistivity of the material of a conductor of uniform cross-section varies along its length as `rho=rho_(0)(1+alpha x)`. Find its resistance if its length is L and area of cross-section is A. |
| Answer» `dR=rho(dx)/(A)=rho_(0)(1+alphax)(dx)/(A),thereforeR=int_(0)^(L)dR` | |
| 636. |
A wire with 15 Q resistance is stretched by one tenth of its original length and volume of wire is kept constant. Then its resistance will beA. `15.18Omega`B. `81.15Omega`C. `51.18Omega`D. `18.15Omega` |
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Answer» Correct Answer - D If the wire is stretched by `(1//10)^(th)` of its original length then the new length of wire become `l_(2)=l+(l)/(10)=(11l)/(10)....(i)` As the volume of wire remains constant then `pir_(1)^(2)l=pir_(2)^(2)l_(1)=pir_(2)^(2)l((11l)/(10)) ("using" (i))` `Rightarrow r_(2)^(2)=(10)/(11)r_(1)^(2)....(ii)` Now the resistance of stretched wire. `R_(2)=(p((11)/(10)l))/(pir_(2)^(2))=(((11)/(10)rhol))/(pixx(10)/(11)r_(1)^(2))xx(rhol)/(pir_(1)^(2))` `(therefore R_(1)=(rhol)/(pir_(1)^(2))=15Omega)` `therefore R_(2)=((11)/(10))^(2)xx15=18.15Omega` |
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| 637. |
A cell has emf of 2.2 V, when connected to a resistance of `5 Omega`, the potential difference between the terminals of the cell becomes 2.1 V, the internal resistance for the cell isA. `0.12 Omega`B. `0.48 Omega`C. `0.24 Omega`D. `0.50 Omega` |
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Answer» Correct Answer - C `E=2.2 V, R=5 Omega, V=2.1 V`, `E=lR+ir implies (E-V)/(V)R=r` `(2.2-2.1)/(2.1)xx5=r implies (0.1)/(2.1)xx5=r` `r=0.24 Omega` |
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| 638. |
For the circuit shown in figure. The point F is grounded. Which of the following is wrong statement ? A. D is at 5 VB. E is at zero potentialC. The current in the circuit will be 0.5 AD. None of the above |
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Answer» Correct Answer - A Effective emf of circuit =10-3=7 V Total resistance of circuit `=2+5+3+4=14 Omega` Current `l=7//14=0.5 A` Potential difference between A and D `=0.5xx10=5A` Potential at D=10-5=5 V Potential at E=5-3=2 V Hence, E cannot be at zero potential, as there is a potential drop at E. |
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| 639. |
A negligbly small current is passed through a wire of length 15 cm and uniform cross-section `6.0xx10^(-7) m^(2)` and its resistance is measured to be `5.0 Omega`. What is the resistivity of the material at the temperature of the experiment ? |
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Answer» Correct Answer - `2xx10^(-7) Omega m` |
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| 640. |
A metal wire of specific resistance `64xx10^(-6) Omega` m and length 1.98 cm has a resistance of `7 Omega`. Find its radius. |
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Answer» Correct Answer - 0.024 cm |
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| 641. |
Two identical cells whether connected in parallel or in series gives the same current, when connected to an external resistance `1.5Omega`. Find the value of internal resistance of each cell.A. `1Omega`B. `0.5Omega`C. ZeroD. `1.5Omega` |
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Answer» Correct Answer - D Given, n = 2, m = 2 `R = 1.5 Omega` Let n cells e in series and m in parallel gives the same current, then `(nE)/(R+nr)=(E )/(R+(r )/(m))` `rArr " " n[R+(r )/(m)]=R + nr` `rArr nRm+nr=Rm+mnr` `rArr 6+2r=3+4r` `rArr 2r=3 rArr r=1.5 Omega` |
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| 642. |
The current pasing through the ideal ammeter in the circuit given below is A. 1.25 AB. `1A`C. 0.75 AD. 0.5 A |
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Answer» Correct Answer - D Here `2Omega` and `2 Omega` are in parallel. `(1)/(R )=(1)/(2)+(1)/(2) rArr R =(2xx2)/(2+2)=1 Omega` Now, internal resistances `1Omega, 2Omega`and resistance R are in series. `therefore " " R_("net")=1Omega +2 Omega+4Omega+1Omega = 8 Omega` Hence, current `l=(V)/(R )=(4)/(8)=0.5 A` |
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| 643. |
In the network shown the potential difference between `A` and `B` is `(R = r_(1) = r_(2) = r_(3) = 1 Omega, E_(1) = 3V, E_(2) = 2V, E_(3) = 1V`) A. 1VB. 2VC. 3VD. 4V |
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Answer» Correct Answer - B At junction point E we have, `I_(1)+I_(2)+I_(3)=0` ...(i) Apply KVL for the loop `E E_(1) F E` we have, `I_(1)-3+2-I_(2)=0` `I_(1)-I_(2)=1` `:.I_(2)=I_(1)-I` ...(ii) Apply KVL for the loop `E E_(1) FE_(3)E` we have, `I_(1)-I_(3)=2` `:.I_(3)=I_(1)-2` ...(iii) From equation (i), (ii) and (iii) we have, `I_(1)=1A, I_(2)=0` and `I_(3)=-1A` Now, potential difference between point A and B = potential difference between E and F `:.V_(EF)=E_(2)-I_(2)r_(2)=2-0xx1=2V` |
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| 644. |
In the network shown the potential difference between `A` and `B` is `(R = r_(1) = r_(2) = r_(3) = 1 Omega, E_(1) = 3V, E_(2) = 2V, E_(3) = 1V`) A. `1 V`B. `3 V`C. `3 V`D. `4 V` |
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Answer» Correct Answer - B (b) No current through `R`, so potential difference across `AB` is `V = ((E_(1))/(r_(1)) + (E_(2))/(r_(2)) + (E_(3))/(r_(3)))/((1)/(r_(1)) + (1)/(r_(2)) + (1)/(r_(3))) = ((3)/(1) + (2)/(1) + (1)/(1))/((1)/(1) + (1)/(1) + (1)/(1)) = 2 V` |
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| 645. |
For the same potential difference, a potentiometer wire is replaced by another one of a high specific resistance. The potential gradient then (`r=R_h=0`)A. decreasesB. remains sameC. increasesD. data is incomplete |
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Answer» Correct Answer - B `x=V/L` = V and L are constant |
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| 646. |
Can you express the potential gradient in terms of specific resistance of the wire ? If yes, find the relation. |
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Answer» Yes. Potential gradient, `K=("potential difference applied (V)")/("length of wire"(l)),`, i.e., `K = V/l = ((IR))/l = I/l(rho l//A) = (I rho)/A` |
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| 647. |
What is the unit of potential-gradient ? If the potential gradient along the potentiometer wire be decreased, will the zero-deflection position be obtained at longer length or shorter length ? |
| Answer» Unit of potential gradient is `V cm^(-1)` or `V m^(-1)`. Since fall of potetential, `V = K l` where K is the potential gradient, therefore, for the given value of V, if K is decreased, l will increase. | |
| 648. |
One mole of an ideal monotomic gas undergoes a linear process from A to B in which its pressure p and its volume V change as shown in figure. The maximum temperature of the gas during this process isA. `(p_(0)V_(0))/(2R)`B. `(p_(0)V_(0))/(4R)`C. `(3p_(0)V_(0))/(4R)`D. `(3p_(0)V_(0))/(2R)` |
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Answer» Correct Answer - B |
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| 649. |
In the network shown, the equivalent resistance between A and B is A. `(4)/(3)Omega`B. `(3)/(4)Omega`C. `(24)/(17)Omega`D. `(17)/(24)Omega` |
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Answer» Correct Answer - A |
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| 650. |
In the circuit shown in figure ,find the ratio of currents `i_(1)//i_(2).` A. 2B. 8C. `0.5`D. 4 |
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Answer» Correct Answer - B |
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