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Quiz 10
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Q.1
In figure, AB is potentiometer wire of length 10 m and resistance 1Ω. The balancing length for 2 volt p.d. is 8m. The emf of the battery E is
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a) 2.75 V
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b) 13.75 V
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c) 27.5 V
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d) 5.5 V
Explanation
Potential across 8 cm is 2 V hence potential difference across 10 m=2.5 Ω Resistance of wire 1Ω is 2.5 V Current I through wire=2.5/1=2.5 A Total resistance in circuit=10+ 1=11 Ω Potential E=I×R=2.5×11=27.5 V Answer: (c)
Q.2
64 cells, each of emf 2volt and internal resistance 2Ω are connected to supply a maximum current through an external resistance of 8×10-3 Ω. Then the cells must be connected in
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a)series only
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b) parallel only
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c)mixed series and parallel arrangement
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d)any of the above three combination
Explanation
Answer: (c)
Q.3
Four identical cells each having emf E, internal resistance zero are connected as shown in figure. The potential difference between A and B is :
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a) 4E
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b) 2E
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c)E
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d)zero
Explanation
Two cells are connected in series Thus I=2E/2r , here r is internal resistance I=E/r potential drop across any cell when giving current is V=E -Ir V=E - (E/r) ×r=0Hence, potential difference across each cell is zero.So across ABAnswer: (d)
Q.4
In a potentiometer experiment, it is found that no current flows through the galvanometer when the terminals of the cells are connected across 52 cm of the potentiometer wire. If cell is shunted by a resistance of 5Ω a balance is found when the cell is connected across 40 cms of the wire. the internal resistance of the cell is
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a) 5 Ω
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b) (200/52) Ω
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c) (52/8) Ω
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d) 1.5 Ω
Explanation
r=[(l1 / l2) - 1] ×R r=[(52/40) - 1]×5=1.5 Ω Answer:(d)
Q.5
A short piece of iron wire glows red hot when placed across the terminals of a2 volts accumulator ( storage battery), but the same piece of wire does not get worm when connected across the terminals of 9 volt 'dry' battery. This is because
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a) the dry battery is smaller
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b) the dry battery does not contain enough electric charges
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c) the wire makes better contact with accumulator
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d) the dry battery has a high internal resistance
Explanation
Answer: (d)
Q.6
In the circuit show below R1=10Ω, R2=20Ω, R3=30Ω and the potentials of points A, B and C are 10V, 6V and 5V respectively. The current through resistance R1 is :
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a)0.1 A
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b) 0.2 A
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c)0.3 A
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d)0.4 A
Explanation
Answer: (b)
Q.7
In an electrolyte 3.2×1018 bivalent positive ions drift to the right per second while 3.6×1018 monovalent negative ions drift to left per second. Then the current is
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a) 1.6 amp. to the left
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b) 1.6 amp to the right
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c)0.45 amp. to the left
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d)0.45 amp to the right
Explanation
Answer: (b)
Q.8
A cell of emf E and internal resistance r is connected in series with an external resistance nr. Then the ratio of the terminal potential difference to emf is
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a) 1/n
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b) 1 (n+1)
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c)n/(n+1)
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d)(n+1)/ n
Explanation
Terminal potential difference V=E - Ir but I=E / (nr+r) V=E - [ E / (nr+r)]rV=E - E/(n+1)V=E (n/n+1)V/E=n / (n+1) Answer:(c)
Q.9
In comparing emf's of two cells with the help of a potentiometer, at the balance point the current flowing through the wire is taken from
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a) one of these cells
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b) both of these cells
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c) the battery in the main circuit
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d) none of above
Explanation
Answer: (c)
Q.10
A number of cells are to be grouped so as to obtain maximum current. The internal resistance of each cell is r and the total external resistance is R. Which of the following is wrong?
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a)Series combination for R >r and parallel for R < r
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b) Series combination for R
r
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c)Mixed combination for R=r
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d)None of the above
Explanation
Answer: (a)
Q.11
Two dry cells P and Q have same emf. The dimensions of cells P are more than that of Q. r1 and r2 denotes internal resistance of P and Q respectively. Then
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a) r1 and r 2
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b) r1 > r 2
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c)r1 < r 2
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d)r1 and r 2 are unrelated to one another
Explanation
Answer: (b)
Q.12
A battery is connected in series with an external resistance. The current in the circuit is 1A and 0.7 A when external resistance equals 5Ω and 8Ω respectively. The internal resistance of battery is
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a) 0.2Ω
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b) 0.5Ω
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c)2Ω
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d)0.6Ω
Explanation
Use IR=E -Ir Answer:(c)
Q.13
In the circuit shown in the figure a voltmeter reads 30V when it is connected across 400Ω resistance. Calculate what the same voltmeter will read if it is connected across 300Ω resistance
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a) 30 V
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b) 60 V
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c) 22.5 V
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d) zero
Explanation
when voltmeter is connected across 400Ω resistance Let G be the resistance of voltmeter . Total resistance in circuit is Current I through circuit I=60/R Effective resistance of voltmeter and 400Ω resistance is Reff=400G / (400+G) Calculation of G Potential difference 30=V Now voltmeter is connected across 300Ω resistance, the total resistance in circuit is R' Current through circuit I'=60/640 Reading in voltmeter Resistance of voltmeter and 300Ω combination R"=300×1200 / (300+1200)=240 Reading in voltmeter=I ×R" V=(60/64)×240=22.5V Answer: (c)
Q.14
Figure represents a part of closed circuit. The potential difference between A and B i.e. VA - VB
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a)24 V
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b) 0 V
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c)6 V
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d)18 V
Explanation
Applying kirchoff's law VA + 3-(1×3)- (6×3)=VB VA - 18=VB VA - VB=18 Answer: (d)
Q.15
A conductor contains 8×1022 free electrons per cubic metre. The conductor carries a current of 1A and has a length of 10cm. the are of cross section of the conductor is 7.81×the time taken by an electron to move from one end of the conductor to the other end is
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a) 10-2s
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b) 10-4 s
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c)105 s
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d)106
Explanation
Use formula I=neAvd and t=l/vd Answer: (a)
Q.16
In figure current in 3Ω and 6Ω resistance is
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a)7.33A ; 3.67A
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b) 3.67A; 7.33A
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c)10.0A ; 15.0A
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d)3.37A ; 6.74A
Explanation
USe formula I1=IR2 / (R1 + R2)and I2=IR1 / (R1 + R2) Answer:(a)
Q.17
In a potentiometer experiment, the balancing length is 8m when two cells E1 and E2 are joined in series. When the two cells connected in opposite the balancing length is 4 m. The ratio of the emf of two cells ( E1 / E2=is
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a)1:2
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b) 2:1
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c)1:3
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d)3:1
Explanation
(E1 + E2) ∝ 8and (E1 + E2) ∝ solve above equationsAnswer: (d)
Q.18
The resistance in the left and right gap of a meter bridge are 3 Ω and 5Ω respectively. When resistance in the left gap increases by 10 percent, the balance point shifts by
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a) 2.2 cm left
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b) 22 cm left
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c) 22 cm right
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d) 2.2 cm right
Explanation
We know that (3/5) = E/S but R ∝ l and S ∝(100-l) ∴ 3/5 = kl / (100 - l)k l = 37.5 When the resistance in the left gap is increased by 10% then P = 3.3Ω HEnce 3.3/5 = l' / ( 100-l') l' = 39.7 cm. Hence the shift of balance point = (39.7 - 37.5) = 2.2 cm to the right Answer: (d)
Q.19
When resistance P and Q are kept in the left gap of a metre bridge the balance point is obtained at 40cm and 60 cm. respectively.When P and Q are connected in series in the same gap the balance point will be nearly at
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a) 68.4 cm from left
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b) 64.4 cm from right
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c) 64.6 cm from left
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d) 64.6 cm from right
Explanation
P/x = 40/60 = 2/3 P = 2x/3 and Q/x = 60/40 Q = 3x/2 P+Q = (2x/3) + (3x/2) = 13x/6 Now [(P+Q)/ x ] = kl/ k(100-l) ∴ 13/6 = l / (100 - l) l = 68.5 cm from left Answer:(a)
Q.20
Potential gradient is defined as:
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a) Fall of potential per unit length of the wire
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b) Fall of potential per unit area of the wire
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c) Fall of potential between two ends of the wire
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d) Potential at any one ends of the wire
Explanation
Answer: (a)
Q.21
Calculate the resistance between points P and Q in the circuit shown in the figure
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a)7.5 Ω
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b) 12 Ω
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c)50 Ω
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d)20 Ω
Explanation
Loop TSRU is whetstone bridge hence no current flows through 40Ω resistance . solve furtherAnswer: (b)
Q.22
Amperes hour is the unit of
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a) quantity of charge
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b) strength of current
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c)power
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d)energy
Explanation
Answer: (a)
Q.23
In circuit shown, the cell E1 and E2 have emf of 4V and 8V. Internal resistance of 0.5Ω and 1Ω respectively. Then the potential difference across E1 and E2 will be
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a) 4.25 V; 4.25V
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b) 3.75V; 3.75V
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c)4.25V; 7.5V
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d)3.75V; 7.5V
Explanation
Effective emf=of the circuit=8-4=4VTotal resistance of the circuit=1+0.5+4.5 + (3×6/3+6)=8Ω∴ current in the circuit I=4/8=0.5ATerminal potential difference across E2 ( cell is discharging)=8 -0.5×1==7.5 VTerminal potential difference across E1 ( cell is charging)=4 +0.5×0.5=4.25 Answer:(c)
Q.24
In the circuit, when key K1 is closed, the ammeter reads Io whether K2 is open or closed. But when K1 is open, the ammeter reads Io/2 when K2 closed. Assuming that the ammeter resistance is much less than R2, find value of r and R1( in ohms)
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a) 100, 50
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b) 50, 100
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c) 0, 100
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d) 0, 50
Explanation
Case I) When K1 is closed, and K2 is open ,the resistance R1 is ineffective. I0=E / (100+r) CaseII) When K1 is closed, and K2 is closed ,the resistance R1 is ineffective.Total resistance in circuit=50+r and current through circuit I=E / (50+r) current through ammeter will be half of total thus Io=E / 2(50+r)from above equations E / (100+r)=E / 2(50+r) Therefore r=0 Case II) When k1 and K2 R1 becomes effective Total resistance in the circuit=R1 + r+ 50 But r=0 ∴ total resistance=R1 + 50 Current through circuit I=E/(R1 + 50) Current trough Ammeter will be half of total current=I/2 Current through ammeter=E/2(R1 + 50) Given that current through ammeter is Io/2 Thus Io/2=E/2(R1 + 50) Now from above I0=E / (100+r) but r=0 ∴ I0=E / (100) Thus E / 2(100)=E/2(R1 + 50) R1=50Ω Answer: (d)
Q.25
The terminal potential difference of a battery exceeds its emf when it is connected ( Pb.C.E.T. 1997)
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a)in parallel with battery of higher emf
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b) in series with a battery of higher emf
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c)in series with battery of lower emf
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d)in parallel with a battery of lower emf
Explanation
Answer: (a)
Q.26
The current I in a conductor varies with time t as I=2t + 3t2, where I is in ampere and t in seconds. Electric charge flowing through a section of the conductor during t=2s to t=3s is ( Orissa JEE 2003]
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a) 10 C
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b) 24C
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c)33C
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d)44C
Explanation
Answer: (b)
Q.27
According to Joule's law if potential difference across a conductor having a material of specific resistance ρ, remains constant, then heat produced in the conductor is directly proportional to
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a) 1 / √ρ
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b) 1/ρ
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c)ρ
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d)ρ2
Explanation
When potential is constant Heat ∝ 1/R But R ∝ρThus H ∝ 1/ρ Answer:(b)
Q.28
If in the circuit. The internal resistance of the battery is 1.5Ω and VP and VQ are potentials at P and Q respectively. What is the potential difference between the points P and Q?
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a) Zero
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b) 4 volts ( Vp > VQ)
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c) 4 volts (VQ > VP)
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d) 2.5 volts (VQ > VP)
Explanation
Total current in circuit=20/ (1.5+2.5)=5.0 A Current get equally divided in both arms=2.5 Total potential across each arm=5×2.5=12.5 V Potential across 3Ω resistance=3×2.5=7.5 Thus potential at P=12.5 - 7.5=5.0 V Potential across 2Ω resistance=2×2.5=5.0 V Potential at Q=12.5 - 5.0=7.5 VQ - VO=7.5-5.0=2.5 Answer: (d)
Q.29
In the circuit shown, the ammeter reads 2A. The resistance of ammeter is negligible. The value of R is
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a)2 Ω
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b) 4 Ω
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c)6 Ω
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d)8 Ω
Explanation
R, 2R and 3R are parallel and this combination equivalent resistance=6R/11 is is in series with 5R/11 resistance thus equivalent resistance=6R/11 + 5R/11=R Now V=IR 12=2 ×R ∴ R=6ΩAnswer: (c)
Q.30
All resistance shown un circuit are 2Ω each. The current in the resistance between D and E is
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a) 5 A
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b) 2.5 A
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c)1 A
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d)7.5 A
Explanation
Resistance in arm Ab has no significance because Resistance in arm Ab and Fg forms a balanced Whetstone bridgeResistance of Whetstone bridge=R switch is parallel to resistance in arm DGEquivalent resistance=R/2=1Ω∴ I=10/1=10AThus Current through arm DE using current division=10/4=2.5Answer: (b)
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