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Physics NEET MCQ
Quiz 3
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Q.1
If an ammeter is to be used in place of a voltmeter, then we must connect with the ammeter a [ AIEEE 2002]
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a)low resistance in parallel
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b) high resistance in parallel
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c)high resistance in series
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d)low resistance in series
Explanation
Voltmeters should have very high resistance. Thus very high resistance must be connected to ammeter to use it as voltmeterAnswer: (c)
Q.2
A wire when connected to 220V mains supply has power dissipation PNow the wire is cut into two equal pieces which are connected in parallel to the same supply. Power dissipation in this case is PThen P2 :P1 is [ AIEEE 2002]
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a) 1
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b) 4
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c)2
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d)3
Explanation
In both he case resistance are connected to same Potential difference thus P ∝ (1/R)When wire is cut in two equal pieces and connected parallel the effective resistance=R/4Thus in second case power ∝ (4/R) Rhus P1 : P2=4 : 1Answer: (b)
Q.3
If current is passed through a spring then the spring will [ AIEEE 2002]
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a) Expand
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b) Compress
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c)Remain same
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d)none of these
Explanation
When current passes through a spring then current flows parallel in the adjacent turnsWhen current in two wire is parallel to each other they exerts a attractive force. Therefore spring will compress Answer:(b)
Q.4
A battery of 10V and internal resistance 0.5Ω is connected across a variable resistance R. The value of R for which the power delivered is maximum is equal to [ CBSE PMT 2001]
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a) 0.25 Ω
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b) 0.5 Ω
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c) 1.0 Ω
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d) 2.0 Ω
Explanation
Power is maximum when r=R, R=r=0.5Ω Answer: (b)
Q.5
In a Wheatstone's bridge shown in the adjoining figure , the conventional current between B and D:
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a)is from B to D
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b) is from D to B
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c)is zero
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d)is indeterminable
Explanation
Answer: (a)
Q.6
The potential difference between the terminals of a cell in an open circuit is 2.2V. When a resistor of 5Ω is connected across the terminals of the cell, the potential difference between the terminals of the cell is found to be 1.8V. The internal resistance of the cell is [ CBSE PMT 2002]
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a) (7/12) Ω
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b) (10/9) Ω
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c)( 9/10) Ω
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d)(12/7) Ω
Explanation
Open circuit potential=E=2.2 V E=V + Ir 2.2=1.8 + ( 1.8/5) × r r=(10/9) ΩAnswer: (b)
Q.7
In a Whetstone bridge all the four arms have equal resistance R. If the resistance of the galvanometer arm is also R, the equivalent resistance of the combination as seen by the battery is [ CBSE PMT 2003]
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a)2R
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b)R/4
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c)R/2
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d)R
Explanation
Since Whetstone's bridge is balanced , the resistance of galvanometer will be unaffectingThus two resistance are connected in erie=2R and combination of 2R resistance are connected in parallel . effective resistance=R Answer:(d)
Q.8
The electric resistance of certain wire of iron is R. If its length and radius are both doubled, then [ CBSE PMT 2004]
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a) the resistance and the specific resistance will both remain unchanged
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b) the resistance will be doubled and specific resistance will be halved
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c) the resistance will be halved and specific resistance will remain unchanged
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d) the resistance will be halved and the specific resistance will be doubled
Explanation
Due to change in geometry of resistance specific resistance do no change new length is 2times of initial and area is 4 times of initial as radius is doubled R'=ρ (2l/4A)=(1/2) ρ ( l/A)=(1/2)R Answer: (c)
Q.9
Resistance n, each of r Ω , when connected in parallel give an equivalent resistance of R Ω. If these resistance's were connected in series, the combination would have a resistance in ohms, equal to [ CBSE-PMT 2004]
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a)nR
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b) n2R
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d)R / n
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c)R / n²
Explanation
R=r/ n r=nRwhen connected in series R'=nr=n(nR)=n2RAnswer: (b)
Q.10
Five equal resistance's each of resistance r are connected as shown in figure. A battery of V volts is connected between A and B. The current flowing in AFCEB will be [ CBSE-PMT 2004]
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a) 2V/r
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b) 3V/r
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c)V/r
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d)V/2r
Explanation
A balanced Whetstone's bridge exists between A and B ∴ Req=r Current through circuit=V/r Current through AFCEB=V/2rAnswer: (d)
Q.11
A 6 volt battery is connected to the terminals of the three metre long wire of uniform thickness and the resistance of 100 Ω. The difference of potential between two points on the wire separated by a distance of 50Cm will be [ CBSE-PMT 2004]
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a) 1.5 V
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b) 3.0 V
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c)3.5 V
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d)1.0V
Explanation
R ∝ l For 300 cm R=100Ω For 50 cm R'=50/3 ΩIR=6 IR'=(6/R)×R'=(6/100) × (50/3)=1 V Answer:(d)
Q.12
When a wire of uniform cross-section a, length l and resistance R is bent into a complete circle, resistance between any two of diametrically opposite points will be [ CBSE-PMT 2005]
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a) R/4
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b) 4R
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c) R/8
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d) R/2
Explanation
Answer: (a)
Q.13
For a network shown in figure the value of the current i is [ CBSE-PMT 2005]
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a)9V/35
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b) 18V/5
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c)5V/9
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d)5V/18
Explanation
It is balanced Whetstone bridge. Hence resistance 4Ω can be eliminated. ∴ Req=6×9 / (6+9)=18/5 i=V/Req=5V/18Answer: (d)
Q.14
Two batteries, one of emf 18V and internal resistance 2Ω and the other of emf 12V and internal resistance 1Ω, are connected as shown. The voltmeter V will record a reading of ..[ CBSE-PMT 2005]
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a) 30 V
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b) 18 V
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c)15 V
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d)14 V
Explanation
Since cells are parallelAnswer: (d)
Q.15
In the circuit shown, if a conducting wire is connected between points A and B, the current in this wire will..[ CBSE-PMT 2006]
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a) flow in the direction which will be decided by the value of V
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b) be zero
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c)flow from B and A
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d)flow from A and B
Explanation
Potential drop over the resistance CA will br more due to higher value of resistance. S potential at A will be less as compared with at B. Hence current will flow from B to A Answer:(c)
Q.16
Two cells, having the same e.m.f are connected in series through an external resistance R. Cells have internal resistance r1 and r2 ( r1 > r2) respectively. When the circuit is closed, the potential difference across the first cell is zero. the value of R is [ CBSE-PMT 2006]
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a) (r₁ + r₂) / 2
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b) (r₁ - r₂) / 2
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c) r₁ + r₂
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d) r₁ - r₂
Explanation
Current in the circuit P.D. across first cell=E - Ir1Given P.D. across first cell=0∴ 0=E - Ir1E=Ir1 Answer: (d)
Q.17
The resistance of an ammeter is 13 Ω and its scale is graduated for a current upto 100 amp. After an additional shunt has been connected to this ammeter it becomes possible to measure current upto 750 amp by this meter. The value of shunt resistance is [ CBSE-PMT 2007]
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a)2Ω
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b) 0.2 Ω
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c)2kΩ
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d)20Ω
Explanation
From the formula Answer: (a)
Q.18
A cell can be balanced against 110cm and 100cm of potentiometer wire, respectively with and without being short circuited through a resistance of 10Ω. Its internal resistance is [ CBSE-PMT 2008]
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a) 1.0 Ω
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b) 0.5 Ω
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c)2.0 Ω
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d)zero
Explanation
Here E > ER / ( E + r) , has the length 110 cm and 100 cm are interchanged.Without being short-circuited through R, only the battery E is balancedE=(V/L)×l1 E=(V/L)×110 --eq(1)When R is connected across E RI=(V/L) ×l2 R[E/(R+r)]=(V/R) ×100 --eq(2) Dividing (1) by (2) we get R+r / R=110 /100 100R + 100r=110R as R=10Ωr=1 ΩAnswer: (a)
Q.19
For the electric circuit which equations is correct for it? [CBSE-PMT 2009]
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b)-E2 - (i1+ i2)R + i2 R2=0
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c)E1 - (i1+ i2)R + i1 r1=0
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d) E2 - (i1+ i2)R - i1 r1=0
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a)E₂ - i2r2 - E₁ - i1r1=0
Explanation
Applying Kirchoff rule Answer:(d)
Q.20
A wire of resistance 12 Ω per meter is bent to form a complete circle of radius 10cm. The resistance between its two diametrically opposite points,A and B is [ CBSE-PMT 2009]
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a) 3Ω
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b) 6πΩ
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c) 6Ω
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d) 0.6πΩ
Explanation
Total resistance of wire = 12Ω×2π×10 -1 Resistance of each half = 1.2πΩ Resistance between its two diametrically opposite points,A and B is required, it imples both the parts are connected in parallel ∴ Req = 1.2π /2 = 0.6πΩ Answer: (a)
Q.21
A student measures the terminal potential difference (V) of a cell ( of emf E and internal resistance r) as a function of the current (I) flowing through it. The slope and intercept of the graph between V and I, then respectively equal [ CBSE-PMT 2009]
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a)-r and E
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b) r and -E
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c)-E and r
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d)E and -r
Explanation
The terminal potential difference of a cell is given by V=E-IrdV/dI=-r [ E is constant]Also for i=0 then V=E ∴ slope=-r , intercept=EAnswer: (a)
Q.22
A potentiometer circuit is set up as shown. The potential gradient, across the potentiometer wire, is k V/cm and the ammeter, present in the circuit reads 1.0A when two way key between the terminals (i) 1 and 2 (ii) 1 and 3, is plugged in, are found to be at lengths l1 cm and l2 cm respectively. The magnitudes of the resistors R and X in ohms, are then, equal, restively to [ CBSE - PMT 2010]
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a) k( l2 - l1) and kl2
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b) kl1 and k(l2 - l1)
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c)k(l2 - l1) and kl1
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d)kl1 and kl2
Explanation
(i) When key between the terminals 1 and 2 is plugged inP.D. across R=IR=kl1I as I=1Amp R=kl1(ii) When key between terminals 1 and 3 plugged in P.D. across (X + R )=kl2 I X + R=kl2 ∴ X=kl2 -RX=kl2 - kl1X=k(l2 - l1 ) Answer: (b)
Q.23
Consider the following two statements:1) Kirchoff's junction law follows from the conservation of charge2) Kirchoff's loop law follows from the conservation of energy Which of the following is correct [ CBSE-PMT 210]
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a) Both 1 and 2 are wrong
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b) 1 is correct and 2 is wrong
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c)1 is wrong and 2 is correct
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d)Both 1 and 2 are correct
Explanation
Answer:(d)
Q.24
A current of 2A flows through a 2Ω resistor when connected across a battery. The same battery supplies a current of 0.5A when connected across a 9Ω resistor. The internal resistance of the battery is [ CBSE-PMT 2011]
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a) 0.5 Ω
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b) (1/3) Ω
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c) (1/4) Ω
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d) 1 Ω
Explanation
Let the internal resistance of the battrey be r. Then the current flowing through the circuit is given by I=E / ( R + r) In first case 2=E / ( 2+r) --eq(1) In second case 0.5=E / ( 9 + r) --eq(2) from 1 and 2 4 + 2r=4.5 + 0.5r 1.5r=0.5 r=(1/3) Ω Answer: (b)
Q.25
In the circuit shown, if potential at point A is taken zero, the potential at point B is [ CBSE PMT 2011]
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a)-1V
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b) +2V
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c)-2V
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d)+1V
Explanation
Current from D to C=1A ∴ VD - VC=2 ×1=2V VA=0 ∴ VC=1 V ∴ VD - VC =2 VD - 1=2 VD=3 VD - VB=2 3 - VB=2 ∴ VB=1 V Answer: (d)
Q.26
A battery of internal resistance 4 Ω is connected to the network of resistance as shown. In order that the maximum power can be delivered to the network, the value of R in Ω should be [ IIT 1995]
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a) 4/9
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b) 2
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c)8/3
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d)18
Explanation
The equivalent circuit are as shown in the figureThe circuit represents Wheatstone Bridge. Hence 6RΩ resistance is infective.Req=2R For maximum power External resistance=Internal resistance 2R=4 ∴ R=2 ΩAnswer: (b)
Q.27
A steady current flows in a metallic conductor of non-uniform cross-section. The quantity/quantities constant along the length of the conductor is /are [ IIT 1997]
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a) current, electric field and drift velocity
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b) drift speed only
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c)current and drift speed
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d)current only
Explanation
As the conductor is of non-uniform cross section drift speed an electric field will non uniform as they depends on cross sectional areaHence only current remains constant along the length of the conductor. Answer:(d)
Q.28
In the circuit P ≠ R, the reading of the galvanometer is same with switch S open or closed. Then [ IIT 1999]
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a) IR=IG
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b) IP=IG
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c) IQ=IG
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d) IQ=IR
Explanation
Since the opening or closing the switch does not affect the current through G, it means that in both the cases there is no current flowing through switch. Or Wheatstone bridge is balanced. Thus current through R must be equal to current through G Answer: (a)
Q.29
In the given circuit, with steady current, the potential drop across the capacitor must be [ IIT 2001]
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a)V
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b) V/2
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c)V/3
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d)2V/3
Explanation
No current will flow through capacitor during steady conditionOn applying Kirchoff's law We gait I=V/(3R) Now voltage across capacitor=voltage across R Vcap=(V/3R)×R=V/3Answer: (c)
Q.30
In the given circuit, it is observed that the current I is independent of the value of the resistance RThen the resistance value must satisfy [ IIT 2001]
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a) R1R2R5=R3R4R6
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b)
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c)R1R4=R2R3
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d)R1R3=R2R4=R5R6
Explanation
Since current I is independent of R6, it follows that the resistance R1, R2, R3 and R4 must form a balanced Wheatstone bridge ∴ R1R4=R2R3Answer: (c)
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