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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.
| 651. |
In the circuit shown, the currents `i_(1)` and `i_(2)` are A. `i_(1)=1.5A, i_(2)=0.5A`B. `i_(1)=0.5A, i_(2)=1.5A`C. `i_(1)=1A. i_(2)=3A`D. `i_(1)=3A, i_(2)=1A` |
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Answer» Correct Answer - B |
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| 652. |
Nine identical conducting rods are arranged as shown in the figure. The ends A and B are maintained at temperatues `T_(1)` and `T_(2)`. Calculate the ratio of rate of flow of heat across AD to that of EB in the given arrangement in steady state. |
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Answer» Correct Answer - A |
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| 653. |
In the circuit shown in figure ,find the ratio of currents `i_(1)//i_(2).` |
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Answer» Correct Answer - 8 |
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| 654. |
Find the current ( in ampere) through wire `XY` of the circuit shown in figure. |
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Answer» Correct Answer - 2 |
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| 655. |
The figure shows the tungsten filament with constant diameter except a piece of it which has half the diameter of the resy of the wire. Assume the the temperature is constant with in each part and charges suddenly between the parts. If the temperature of thick part is 2000 K, the temperature of the thin part of the filament is |
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Answer» Correct Answer - `8^(1//4).2000K` |
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| 656. |
Can the potential difference across a battery be greater than its emf? |
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Answer» Correct Answer - A |
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| 657. |
The resistance of a wire of length 40 m and radius o.25 mm is `10 Omega`. The conductivity of the material of the wire isA. `1xx10^( -7)` mho/mB. `2xx10^(-7)` mho/mC. `2xx10^(7)` mho/mD. `1xx10^(-7)` mho/m |
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Answer» Correct Answer - C `sigma = 1/rho=(l)/(RA)=(40)/(10xx3.14xx625xx10^(-10))` `=2xx10^(7)` mho/m |
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| 658. |
The mean free path of electrons in a metal is `4 xx 10^(-8)m` The electric field which can give on an average `2eV` energy to an electron in the metal will be in the units `V//m`A. `5 xx 10^(-11)`B. `8 xx 10^(-11)`C. `5 xx 10^(7)`D. `8 xx 10^(7)` |
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Answer» Correct Answer - c Average energy = workdone `= q E lambda` `:. 2eV = eE xx 4 xx 10^(-8)` or `E = (2)/(4 xx 10^(-8)m) = 8 xx 10^(7)Vm^(-1)` |
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| 659. |
Calculate the conductance and conductivity of a wire of radius `0.01 Omega` area of cross -section `10^(-4)m^(2)` length `0.1m` |
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Answer» Correct Answer - `100S ,10^(5)S m^(-1)` Conductance, `G = (1)/(R)= (1)/(0.01) = 100s` conductivity, `sigma = (1)/(rho) = (1)/(RA) = (0.1)/(0.01 xx 10^(-4))` `= 10^(5)S m^(-1)` |
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| 660. |
Currrent flows through a constriction conductor. The diameter and current density in the left of connection are `2.0mm and 1.3 xx 10^(6)Am^(-2)` (i) How much current flow through the constriction ? (ii) If the current density is tripled as it emerges from the right side of the constriction, what is the diameter of the hand is the diameter. of the right hand side of constriction? |
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Answer» Correct Answer - (i)`4.09 A`(ii)1.15 mm` Let `D_(1) and D_(2)` be the diameter on the left hand side right hand side of combination. Then `D_(1) = 2.0 mm = 2 xx 10^(3)m, J_(1) = 1.3 xx 10^(6) Am^(-2)` (i) current flowing through the constriction `I_(1) = J_(1) A_(1) = J_(1) (pi D_(1)^(2))/(4) = (1.3 xx 10^(6)) xx (22)/(7) xx ((2 xx 10^(-3))^(2))/(4)` ` = 4.09 A` (ii) For a steady flow of current, `I_(1) = I_(2)` or `J_(1)A_(1) = J_(2)A_(2)` or `J_(1) (pi D_(1)^(2))/(4) = J_(2) (pi D_(2)^(2))/(4) = 3 J_(1) (pi D_(2)^(2))/(4)` or `D_(2) = (D_(1))/(sqrt(3)) = (2.0 "mm")/(sqrt(3)) = 1.15 "mm"` |
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| 661. |
Calculate the conductivity of a wire of length 2 m, area of cross-sectionl `2 cm^(2)` and resistance `10^(-4) Omega`. |
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Answer» Correct Answer - `10^(8) Sm^(-1)` |
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| 662. |
A uniform wire of length `1` and radius `r` has resistance `100 Omega`. it is recast into a thin wire of (i) length `2l` (ii) radius `r//2`. Calculate the resistance of new wire in each case. |
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Answer» Correct Answer - (i) `400 Omega`,(ii)`1600 ohm` Let A, `l` be the original area of cross section and length of the given wire and `A_(1),l_(1)` be the respectively new volume when the wire is drawn to twice its original length. Then `l_(1) = 2l`. Using the relation `A l = A_(1)l_(1)`, we have `A_(1)=Al//l_(1)=Al//2l=A//2` Therefore, `R_(1) =rhol_(1)//A_(1) = rho 2 //(A//2)` `= 4 rho l//A = 4R = 4 xx 100 = 400 Omega` (ii) when wire is drawn in half its radius new area becomes `1//4th` length would become 4 times, `:. R_(2)= 16R = 16 xx 100 = 1600` ohm |
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| 663. |
A metal wire is subjected to a constant potential difference When the temperaturre of the metal in it wire increase the drift velocity of the electrons in it .A. increase, thermal velocity of the electrons decreaseB. decrease, thermal velocity of the electrons decreaseC. increase, thermal velocity of the electrons increaseD. decrease, thermal velocity of the electrons increase |
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Answer» Correct Answer - d With increase in temperature relaxation line time `t` decrease hence drift velocity `v_(d)` which is directly proportional in relaxation time `(v_(d) = tau)` decrease but thermal velocity increases |
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| 664. |
A wire `50 cm` long and `0.12mm` diameter has a resistance of `4.0 Omega` find the resistance of another wire of the same material whose length is `1.5m` and diameter is `0.15 mm` |
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Answer» Correct Answer - `1.92 Omega` Here, `l = 0.50 m, D = 0.12 xx 10^(-3)m, R = 4.0 Omega` `l_(1) = 1.5m,D_(1) = 0.15 xx 10^(-3)m , R_(1) = ?` `R = (rhol)/(pi D^(2)//4) or R prop (l)/(D^(2))` `:. (R_(1))/(R) = (l_(1))/(l) xx (D^(2))/(D_(1)^(2)) = (1.5)/(0.5) xx ((0.2 xx 10^(-3))^(2))/((0.15 xx 10^(-3))^(2)) = 1.92 Omega` |
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| 665. |
Two bulbs `X` and `Y` having same voltage rating and of power 40 watt and 60 watt respectively are connected in series across a potential difference of 300 volt, then A. X will glow brighterB. Resistance of `Y` is greater than `X`C. Heat produced in `Y` will be greater than `X`D. Voltage drop in `X` will be greater than `Y` |
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Answer» Correct Answer - A (a) Resistance `prop (1)/(power)`. Thus, `40 W` bulb has a high resistance. Because of which there will be more Potential drop across `40 W` bulb. Thus `40 W` bulb will glow brighter. |
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| 666. |
Two wire `A` and `B` of same meterial and mass have their lengths in the rato:. On connecting them to the same source, rate of heat dissiapted in `B` is found to be `5 W`. The rete of the heat dissipated is `A` isA. `10 W`B. `5 W`C. `20 W`D. None fo these |
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Answer» Correct Answer - C (c ) `(H)/(t) = P = (V^(2))/(R ) implies P prop (1)/(R )` also `R prop (l)/(A) prop (l^(2) rho)/(A l rho)` also `implies R prop (i^(2))/(m) implies R prop l^(2)` (for same mass) So `(P_(A))/(P_(B)) = (l_(B)^(2))/(l_(A)^(2)) = (4)/(1) implies P_(A) = 20 W` |
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| 667. |
The current I shown in the circuit is A. `1.33 A`B. zeroC. `2A`D. `1A` |
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Answer» Correct Answer - A |
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| 668. |
A DC ammeter has resistance `0.1 Omega` and its current ranges `0-100 A`. If the range is to be extended to `0-500`, then the following shunt resistance will be requiredA. `0.010 Omega`B. `0.011 Omega`C. `0.025 Omega`D. `0.25 Omega` |
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Answer» Correct Answer - C |
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| 669. |
A galvanometer of resistance `50 Omega` is connected to a battery of 8 V along with a resistance of `3950 Omega` in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 15 divisions, the resistance in series should be .... `Omega`.A. 1950B. 7900C. 2000D. 7950 |
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Answer» Correct Answer - D |
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| 670. |
A metal plate weighting 750g is to be electroplated wiwth 0.05% of its weight of silver. If a current of 0.8A is used. Find the time needed for depositing the required weight of silver. `"E.C.E. of silver is" 11.18xx10^(-7)kgC^(-1)`.A. 5 min 32 sB. 6 min 37 sC. 4 min 16 sD. 6 min 10 s |
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Answer» Correct Answer - B |
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| 671. |
A ring is made of a wire having a resistance `R_(0) = 12 Omega`. Find the point `A` and `B`, as shown in the figure, at which a current carrying conductor should be connected so that the resistance `R` of the subcircuit between these points is equal to `(8)/(3) Omega` A. `(l_(1))/(l_(2)) = (1)/(2)`B. `(l_(1))/(l_(2)) = (5)/(8)`C. `(l_(1))/(l_(2)) = (1)/(3)`D. `(l_(1))/(l_(2)) = (3)/(8)` |
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Answer» Correct Answer - A (a) `((A)/(rho l_(1)) + (A)/(rho l_(2))) = (3)/(8)` `(rho (l_(2) + l_(2)))/(A) = 12` Solving `(l_(1))/(l_(2)) = (1)/(2)` |
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| 672. |
A current of `3 A` flows through the `2 Omega` resistor as shown in the circuit. The power dissipated in the `5 Omega` resistor is A. `4 W`B. `2 W`C. `1 W`D. `5 W` |
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Answer» Correct Answer - D (d) Voltage across `2 Omega` is the same as voltage across arm containing `1 Omega` and `5 Omega` resistance. Voltage across `2 Omega` resistance, `V_(1) = 6 V` Current across `5 Omega, I = (6)/(1 + 5) = 1 A` Thus, power across `5 Omega`, `P = I^(2) R = (1)^(2) xx 5 = 5 W` |
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| 673. |
In the determination of the internal resistance of a cell using a potentiometer, when the cell is shunted by a resistance R and connected in the secondary circuit, the balance length is found to be `L_(1)`. On doubling the shunt resistance, the balance length is found to increase to `L_(2)`. The value of the internal resistance isA. `(2R(L_(2)L_(1)))/((L_(1)-2L_(2)))`B. `(2R(L_(2)-L_(1)))/((2L_(1)-L_(2)))`C. `(R(L_(2)-L_(1)))/((L_(1)-2L_(2)))`D. `(R(L_(2)-L_(1)))/((2L_(1)-L_(2)))` |
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Answer» Correct Answer - B `((E)/(R+r))R=K.l` |
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| 674. |
Figure shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.8 cm. when a resistor of `9.5 Omega` is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell. |
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Answer» Here `l_(1) = 76.3 cm, l_(2) = 64.8 cm` `r = ? R = 9.5 Omega` Now, `r = ((l_(1) - l_(2))/(l_(2))) R = ((76.3 - 64.8)/(6.8)) 9.5 = 1.68 Omega` |
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| 675. |
Figure `6.13` shows a `2.0 V` potentiometer used for the determination of internal resistance of a `1.5 V` cell. The balance point of the cell in open circuit is `76.3 cm`. Whan a resistor of `9.5 Omega` is used in the external circuit of the cell, the balance point shifts to `64.8 cm`, length of the potentiometer. Dentermine the internal resistance of the cell. |
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Answer» Here, `l_(1) =76.3cm , l_(2) = 64.8cm,` `r=?, R=9.5Omega` (br) Now, `r=((l_(2)-l_(2))/l_(2))R = ((76.3-64.8)/(64.8))xx9.5 =1.68Omega` |
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| 676. |
Figure shows a potentiometer circuit for comparison of tworesistances. The balance point with a standard resistor R = 10.0 Ωis found to be 58.3 cm, while that with the unknown resistance X is68.5 cm. Determine the value of X. What might you do if you failedto find a balance point with the given cell of emf ε ? |
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Answer» Resistance of the standard resistor, `R = 10.0 Omega` Balance point for this resistance, `l_(1)=58.3` cm Current in the potentiometer wire `= l` Hence, potential drio across `R, E_(1)=iR` Resistance of the unknown resistor = X Balance point for this resistor, `I_(2)= 68.5` cm Hence, potential drop across `X, E_(2)= iX` The relation connecting emf and balance point is, `(E_(1))/(E_(2))=(l_(1))/(l_(2))` `(iR)/(iX) = (l_(1))/(l_(2))` `(iR)/(iX)=(l_(1))/(l_(2))` `X = (l_(1))/(l_(2))xxR` `= (68.5)/(58.3) xx 10 = 11.789 Omega` Therefore, the value of the unknown resistance, X, is `11.75 Omega` If we fall to find a balance point with the given cell emf `epsilon`, then the potential drop across R and X must be reduced by putting a resistance in series with it. Only if the potential drop across R or X is smaller than the potential drop across the potentiometer wire AB, a balance point is obtained. |
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| 677. |
A 2.0V potentiometer is used to determine the internal resistance ofa I.SY cell. The balance point of the cell in the open circuit is obtained at 75cm. When· a resistor of `10Omega` is connected across the cell, the balance point shifts to 60cm. The internal resistance of the cell is:A. `1.5Omega`B. `2.5Omega`C. `3.5Omega`D. `4.5Omega` |
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Answer» Correct Answer - B |
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| 678. |
In a potentiometer, potential difference across the potentiometer wire is directly proportional to itsA. LengthB. areaC. resistanceD. all of these |
| Answer» Correct Answer - A | |
| 679. |
The e.m.f. of two cells can be compared byA. potentiometerB. ammeterC. luxmeterD. speedometer |
| Answer» Correct Answer - A | |
| 680. |
If the length of potentiometer wire is increased, then the accuracy in the determination of null pointA. decreasesB. increasesC. remains unaffectedD. none of these |
| Answer» Correct Answer - B | |
| 681. |
Figure `6.12` shows a potentiometer circular for comparison of two resistances. The balance point with a standard resistor `R = 10.0 Omeag ` is found to be `58.3 cm`, while that with the unknows resistance `X` is `68.5 cm`. Determine the value of `X`. What would you do if you fail to find a balance point with the given cell `E`? |
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Answer» Here, `l_(1) = 58.3cm , l_(2) = 68.5cm ,R=10Omega ,X=?` Let `I` be the current in the potentiometer wire and `Epsilon_(1)` and `Epsilon_(2)` be the potential drops across `R` and `X` respectively. When connected in circuit by closing respective key. Then `epsilon_(2)/epsilon_(1) = (IX)/(IR) =X/R or X = epsilon_(2)/epsilon_(1)R` ...(i) But `epsilon_(2)/epsilon_(1) = l_(2)/l_(1)` `:.` From (i), `X=l_(2)/l_(1)R = 68.3/58.3xx10.0 = 11.75Omega` If there is no balance point with given cell of e.m.f. `epsilon`, it means potential drop across `R` or `X` is greater than the potential drop across the potentiometer wire `AB`. In order to obtain the balance point, the potential drops across `R` and `X` are to be reduced, Which is possible be reducing the current in `R` and `X`. For that, either a suitable resistance should be put in series with `R` and `X` or a cell of smaller e.m.f. `epsilon` should be used. Another possible way is to increase the potential drop across the potentiometer wire by increasing the voltage of driver cell. |
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| 682. |
Ten cells each of emf `1 V` and internal resistance `1Omega` are connected inseries. In this arrangement, polarity of two cells is reversed and the system is connected to an external resistance of `2Omega`. Find the current in the circuit. |
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Answer» Correct Answer - A::B Net emf `=(n-2m)E` `=(10-2xx2)(1)` `=6V` `i=("Net emf")/("Net resistance")` `=6/(10+2)=0.5A` |
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| 683. |
A resistance of `4 Omega ` is connected across a cell Then it is replaced by another resistance of `1 Omega` it is found the power dissipeted in resistance in both the is `16` walt ThenA. internal resistance of the cell is `2 Omega`B. emf of the celll is `12V`C. maximum power the can be dissipated in the external resistance is `18` waltD. short circuit current from the cell is infinite |
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Answer» Correct Answer - a,b,c `((epsilon)/(4+r))^(2) xx 4 = ((epsilon)/(1+r))^(2) xx1` `r = 2 Omega` Current in circuit,`I = (epsilon)/(4+2)` `16 = ((epsilon)/(4+2))^(2) xx 4or epsilon = 12 V` power dissipated is maximum when exertnal resistance is equal to internal resistance i.e. `R = r = 2 Omega` `P_(max) = ((12)/(2+2))^(2) xx 2 = 18W` |
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| 684. |
12 cells each having same emf are connected in series with some cells wrongly connected. The arrangement is connected in series with an ammeter and two cells which are in series. Current is 3 A when cells and battery aid each other and is 2 A when cells and battery oppose each other. The number of cells wrongly connected isA. 4B. 1C. 3D. 2 |
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Answer» Correct Answer - B |
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| 685. |
In the adjoining circuit diagram each resistance is of `10 Omega`. The current in the arm AD will be A. `(2i)/(5)`B. `(3i)/(5)`C. `(4i)/(5)`D. `(i)/(5)` |
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Answer» Correct Answer - A |
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| 686. |
In the circuit the galvanometer `G` shows zero deflection. If the batteries A and B have negligible internal resistance find the value of the resitor `R` |
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Answer» Correct Answer - `100 Omega` Since gavanometer shows no deflection potential difference across `R = 2V` .Therefore potential difference across `500 Omega = 12 - 2 = 10V` Current through `500 Omega = (10)/(500) = (1)/(50)A` As same current flows through `R` `:. R = ("pot. diff")/("current") = (2)/((1//50)) = 100 Omega` |
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| 687. |
Calculate the energy stored in the condenser the given circuit |
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Answer» Correct Answer - `4 xx 10^(-4)J` Total resistance of arm `ABCD` `R = 2+ 3 + 5 = 10 Omega` Potential difference across `A and D` `V = 1 xx R = 2 xx 10 = 20V` Energy stored in capctor `= (1)/(2) CV^(2)` `= (1)/(2) xx (2 xx 10^(-6)) xx (20)^(2) = 4 xx 10^(-4)J` |
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| 688. |
A coil of enamelled copper wire of resistance `50 Omega` is embedded in a block of ice and a potential difference of 210 V applied across it. Calculate, how much ice will melt in half minute. Latent heat of ice is 180 cal per gram. |
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Answer» Here, `R = 50 Omega, V = 210, t = 30s, L = 80 cal//g` Heat produced, `H = (V^(2)t)/4.2R = ((210)^(2)xx30)/(4.2 xx 50) = 210 xx 30 cal` Let m gram of ice be melted in half minute, then `mL = H or m = H/L = (210 xx 30)/80 = 78.75g` |
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| 689. |
A parallel plate capacitor with plates of length l is included in a circuit as shown in figure-3.164. The EMF of the source is `epsilon`, its internal resistance is r and the distance between the plates is d. An electron with a velocity u files into the capacitor, parallel to the plates . What resistance R should be connected in parallel with the capacitor so that the electron files out of the capacitor at an angle of `37^(@)` to the plates? Assume that circuit is in steady state. Given values of parameters as `l=91cm epsilon=3V,r=2Omega,d=(1//3)mm,u=4xx10^(7)m//s m_(e)=9.1xx10^(-31)kg` and `e=1.6xx10^(-19)C`. |
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Answer» Correct Answer - 3 ohm |
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| 690. |
An electric motor runs on a d.c. source of emf `epsilon` and internal resistance r. show that the power output of the source is maximum when the current drawn by the motor is `epsilon//2r`. |
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Answer» Let `I` be the current drawn by motor. Then power output of the source, `P = epsilon I - I^(2) r` P will be maximum, when `(dP)/(dI) = 0 or (d)/(dt) (epsilon I - I^(2)r) =0` or `epsilon - 2 Ir = 0 or I = (epsilon)/(2r)` |
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| 691. |
Four identical cells of emf `epsilon` and internal resistance r are to be connected in series. Suppose, if one of the cell is connected wrongly, then the equivalent emf and effective internal resistance of the combination isA. 2E and 4rB. 4E and 4rC. 2E and 2rD. 4E and 2r |
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Answer» Correct Answer - A |
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| 692. |
In the series combination of n cells each cell having emf `epsilon` and internal resistance r. If three cells are wrongly connected, then total emf and internal resistance of this combination will beA. `"n"epsi,(nr-3r)`B. `("n"epsi-2r),nr`C. `("n"epsi-4epsi),nr`D. `("n"epsi-6epsi),nr` |
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Answer» Correct Answer - D Since due to wrong connection of each cell the total emf reduced to `2epsilon` then for wrong connected of three cells the total emf will reduced to `(nepsi-6epsi)` whereas the total or equivlent of cell combination will be nr. |
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| 693. |
`N` identical cells, each emf `E` and internal resistance `r` are joined in series. Out of `N` cells, `n` cells are wrongly connected i.e., their terminals are connected in reverse of the required for series connection `(n lt (N)/(2))`. Let `E_(0)` be the emf of resulting battery and `r_(0)` be its internal resistance. ThenA. `epsilon_0 = (N - n) epsilon, r_0 = (N - n) r`B. `epsilon_0 = (N - 2n) epsilon, r_0 = (N - 2n) r`C. `epsilon_0 = (N - 2n) epsilon, r_0 = Nr`D. `epsilon_0 = (N - n) epsilon, r_0 = Nr`] |
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Answer» Correct Answer - C `i=(nE)/(nr) = ( E)/( r)` `p.d` across each cell, `V = E - ir = E - (E )/( r) r = 0`. |
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| 694. |
The reading of a voltmoter when a cell is connected to it is 2.2 V. When the terminals of the cell are connected to a resistance of `4 Omega`, the voltmeter reading drops to 2 V. Find the internal resistance of the cell. |
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Answer» Correct Answer - `0.4 Omega` |
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| 695. |
The potential difference across a cell is 1.8 V when a current of 0.5 A is drawn from it. The p.d. falls to 1.6 V when a current of 1.0 A is drawn. Find the emf and the internal resistance of the cell. |
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Answer» Correct Answer - `2.0 V, 0.4 Omega` |
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| 696. |
In the circuit shown in Fig. 4.67, the resistance of the ammeter A is negligible and that of thevoltmeter V is very high. When the switch S is open, the reading of voltmeter is 1.53 V. On closing the switch S, the reading of ammeter drops to 1.03 V. Calculate : (i) emf of the cell (ii) internal resistance of the cell (iii) value of R. |
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Answer» Correct Answer - `(i) 1.53 V (ii) 0.50 Omega (iii) 1.03 Omega` |
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| 697. |
A `100W-220V` bulb is connected to 110 V source. Calculate the power consumed by the bulb. |
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Answer» Power of the bulb, `P=100W` Operating voltage `V=200V` Resistance of the bulb,`R=(V^(2))/(P)=((220)^(2))/(100)=484Omega` Actual operating voltage `V^(1)=110V` Therefore, power consumed by the bulb, `P^(1)((V^(1))^(2))/(R)=((110)^(2))/(484)=25W` |
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| 698. |
Which has greater resistance : 1 kW electroheater or a 110 W filament bulb, both marked for 220V? |
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Answer» Resistance of heater, `R_(1) = V^(2)/P_(1) = (220)^(2)/1000 Omega` Resistance of bulb, `R_(2) = V^(2)/P_(2) = (220)^(2)/100 Omega` `R_(1)/R_(2) = 100/1000 = 1/10 or R_(2) = 10R_(1)` Thus resistance of bulb is greater than heater. |
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| 699. |
What do you understand by maximum power rating of a resistor ? |
| Answer» Maximum power rating of a resistor is the maximum power which it can be tolerate in the form of heat without being melt. | |
| 700. |
A metallic wire has variable cross sectional area. Cross sectional area at cd is twice the area at ab. The wire is connected to a cell as shown. find the ratio of heat dissipated per unit volume at section cd to that at section `ab`. |
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Answer» Correct Answer - `(1)/(4)` |
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