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Physics NEET MCQ
Quiz 4
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Q.1
Express which of the following set ups can be used to verify Ohm's law? [ IIT 2003]
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a)
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b)
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c)
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d)
Explanation
In option a Ammeter is connected in series and voltmeter parallel to resistance Answer:(a)
Q.2
The effective resistance between point P and Q of the electrical circuit shown in the figure is [ IIT 2002]
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a)
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b)
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c)
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d)
Explanation
The circuit is symmetrical about axis PQ. The circuit above the axis PQ represent the balanced Wheatstone bridge. Hence the central resistance 2R is ineffective. Similarly in the lower part below axis PQ the central resistance 2R is ineffective Thus three horizontal rows of resistance are parallel to each other RPQ=2Rr / R+r Answer: (a)
Q.3
In the shown arrangement if the experiment of meter bridge if AC corresponds to null deflection of galvanometer is x, what would be its value if the radius of the wire is doubled? [ IIT 2003]
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a)x
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b) x/4
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c)4x
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d)2x
Explanation
At null point R1 / R2=x / (100-x) If radius of the ire is doubled, then the resistance of AC will change. But since R1 / R2 does not change so value of x will not change Answer: (a)
Q.4
The three resistance of equal value are arranged in the different combinations shown below arrange them in increasing order of power dissipation . [IIT 2003]
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a) c < b < d < a
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b) b < c < d < a
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c)a < d < c < b
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d)a < c < b < d
Explanation
Resistance of combination a=3RResistance of b combination is=(2/3)RResistance of c combination is=R/3Resistance of d combination is=1.5 R Power dissipation=I 2 R Here Power ∝ R Thus Higher the resistance higher the power Option (a) is correctAnswer: (a)
Q.5
Shown in figure is Post Office Box. In order to calculate the value of external resistance, it should be connected between [ IIT 2004]
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a)B' and C'
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b) A and D
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c)C and D
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d)B and D
Explanation
Total external resistance will be the total resistance of whole length of box. It should be connected between A and D Answer:(b)
Q.6
Six identical resistors each of resistance R, are connected as shown in figure. The equivalent resistance will be [ IIT 2004]
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a) Maximum between P and R
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b) Maximum between Q and R
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c) Maximum between P and Q
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d) All are equal
Explanation
Answer: (c)
Q.7
Find out the value of current through 2Ω resistance for the given circuit [ IIT 2005]
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a)zero
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b) 2A
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c)5A
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d)4A
Explanation
The current in 2Ω resistor will be zero because it is not a part of any closed loopAnswer: (a)
Q.8
If a steady current I is flowing through a cylindrical element ABC. Choose the correct relationship [ IIT 2006]
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a)VAB=2VBC
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b) Power across BC is 4 times the power across AB
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c)Current densities in AB and BC are equal
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d)Electric field due to current inside AB and BC are equal
Explanation
Current in element is same. Length of section AB and BC is same but radius of the section AB is 2times the section of BC thus section BC resistance=4× (resistance of AB)Option a) 4VAB=VBCOption a is incorrect Option b)Power P=I2R4PBC =PABOption b is correctOption c)Current density ∝ 1/ r2 Thus 4JAB=JBCOption d)Electric field=P.D / length Thus 4EAB=EBC Answer:(b)
Q.9
In a wheatstone's bridge, three resistance P, Q and R connected in the three arms and fourth arm is formed by two resistance S1 and S2 connected in parallel. The condition for bridge to be balance [AIEEE 2006]
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a)
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b)
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c)
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d)
Explanation
For wheatstone's bridge P/Q=R/ S But S=S1×S2 / (S1×S2) Option b is correct Answer: (b)
Q.10
A resistance of 2Ω is connected across one gap of a metrebridge ( the length of the wire is 100 cm) and an unknown resistance, greater than 2Ω, is connected across the other gap. When these resistance's are interchanged, the balance point shifts by 20cm. Neglecting any corrections, the unknown resistance is [ IIT 2007]
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a) 3Ω
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b) 4Ω
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c) 5Ω
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d) 6Ω
Explanation
Given X is greater than 2Ω when the bridge is balanced 2/l=X / 100 - l 200 -2l=Xl --eq(1) On interchanging resistance balance point shifts by 20 cm X / ( l +20)=2 / (80-l) 80X-Xl=2l + 40 --eq(2) from equation 1 and 2 80X - 200+ 2l=2l+40 80X=240 X=Ω Answer: (a)
Q.11
A circuit is connected as shown in figure with the switch S open. When the switch is closed, the total amount of charge that flows from Y to X is [ IIT 2007]
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a)0
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b) 54µC
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c)27µC
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d)81µC
Explanation
When steady state is reached Current from battery=1 AThus P.D. across 3Ω resistance=3V P.D. across 6Ω resistance is=6V Potential across 3µF=3VCharge on 3µF=Q1=3×3=9µFPotential across 6µF=Q2=9V Charge on 6µF=Q2=6×6=36µF∴ Charge (-Q1) is shifted from positive plate of 6µF capacitor. The remaining charge on the positive plate of 6µF capacitor is shifted through the switch.∴ Charge passing the switch=36 - 9=27µCAnswer: (c)
Q.12
If in the circuit, power dissipation is 150W, then R is ..[ AIEEE 2002]
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a) 2 Ω
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b) 6 Ω
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c)5 Ω
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d)4 Ω
Explanation
Req=2R/ (R + 2) Power P=V2 /Req Answer: (b)
Q.13
The length of a wire of potentiometer is 100 cm and the e.m.f. of its standard cell is E volt. It is employed to measure the e.m.f of a battery whose internal resistance is 0.5Ω. If the balance point is obtained at l=30cm from the positive end, the emf of the battery is [ AIEEE 2003]
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a) 30E / 100.5
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b) 30E / (100 -0.5)
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c)30(E-0.5i) / 100
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d)30E / 100
Explanation
In this arrangement , the internal resistance of the battery E does not play any role as current is not passing through the battery.From the principle of potentiometer, V ∝ l V / E=l / LV=emf of battery . E=emf of standard cellL=potentiometer wire V=El/ L=30E/100 Answer:(d)
Q.14
An ammeter reads up to 1 ampere. Its internal resistance is 0.81Ω. To increase the range of 10A the value of the required shunt is [ AIEEE 2003]
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a) 0.03Ω
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b) 0.3Ω
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c) 0.9Ω
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d) 0.09Ω
Explanation
Answer: (d)
Q.15
The circuit shown in the figure consists of eight resistors, each with resistance R, and a battery with terminal voltage V and negligible internal resistance. What is the current flowing through the battery?
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a) V/3R
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b) 2V/3R
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c)V/6R
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d)3V/2R
Explanation
nodes the horizontal resistors connect to each have the same voltage. it doesn't matter if the horizontal resistors are there or not so they can be removed without changing the solution Effective resistance in each brachis 2R and such three series combinations are parallel thus effective resistance is Reff= 2R/3 NowI = V/Reff I = 3V/2R Answer: (d)
Q.16
The total current supplied to the circuit by the battery is .. [ AIEEE 2004]
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a)4 A
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b) 2A
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c)1 A
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d)6A
Explanation
From the diagram 2Ω and 6Ω resistance's are parallel to each other thus effective resistance is=6/4=1.5 ΩAbove resistance is in series with 1.5Ω resistance to give 3ΩThis 3Ω resistance is in parallel with 3Ω resistance in circuit thus resultant resistance is 1.5ΩNow V=IR I=V/R=6 /1.5=4AAnswer: (a)
Q.17
The resistance of the series combination of two resistance's is S, when they are joined in parallel the total resistance is P. IF S=nP the minimum possible value of n is [ AIEEE 2004]
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a) 2
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b) 3
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c)4
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d)1
Explanation
Let R1 and R2 be the two resistance S=R1 + R2P=R1 × R2 / (R1 + R2)given S=nPR1 + R2=n [R1 × R2 / (R1 + R2)] (R1 + R2)2=n(R1 + R2) Minimum value of n=4 For that value (R1 + R2)2=4(R1 ×R2)2)⇒ ((R1 - R2)2=0 Answer:(c)
Q.18
An electric current is passed through a circuit containing two wires of the same material, connected in parallel. If the lengths and radii are in ratio of (4/3) and (2/3), then the ratio of the current passing through the wire will be [ AIEEE 2004]
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a) 8/9
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b) 1/3
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c) 3
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d) 2
Explanation
Resistance of wire is given by R=ρ ( l/A) But A=πr2 ∴ R=ρ[ l /(πr2)] Potential across both the conductor is same (given) Answer: (b)
Q.19
In a meter bridge experiment null point is obtained at 20cm from one end of the wire when resistance X is balanced against another resistance Y. If X< Y, then where will be the new position of the null point from the same end, if one decides to balance a resistance of 4X against Y [ AIEEE 2004]
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a)40 cm
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b) 80 cm
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c)50 cm
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d)70 cm
Explanation
R1 / R2=l1 / l2 [ Here l2=100 -l1]In first case X/Y=20/80 --eq(1)In second case 4X/Y=l/ (100-l) --eq(2) from equation (1) and (2) ⇒ l=50 Answer: (c)
Q.20
In the circuit, the galvanometer G shows zero deflection. If the batteries A and B have negligible internal resistance, the value of the resistor R will be [ AIEEE 2005]
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a) 100 Ω
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b) 200 Ω
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c)1000 Ω
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d)500 Ω
Explanation
By applying kirchoff's law for path NXYM 12-500i -2=0 i=10/500=1/50By applying Kirchoff's law for path NXPQ12 -500i -iR [ given no current flows through Ammeter )i=12 /( 500 + R) ∴ 500 +R=600 R=100ΩAnswer: (a)
Q.21
Two sources of equal emf are connected to an external resistance R. The internal resistance of the two sources are R1 and R2 ( R1 > R2). If the potential difference across the source having internal resistance R2 is zero, then [ AIEEE 2005]
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a) R=R₂ - R₁
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b) R=R₂ × (R₁ + R₂) / (R₂ - R₁)
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c) R=R1R2 / (R₂ - R₁)
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d) R=R1R2 / (R₁ - R₂)
Explanation
Now I=2E / 9 R+R1+R2) Potential difference across second cell=V=E - IR2 Which is zero given ∴ E -IR2=0 Answer: (a)
Q.22
In a potentiometer experiment the balancing with a cell is at length 240cm. On shunting the cell with resistance of 2Ω, the balancing length becomes 120cm. The internal resistance of the cell is [ AIEEE 2005]
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a)0.5 Ω
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b) 1Ω
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c)2Ω
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d)4Ω
Explanation
The internal resistance of cell r Answer: (c)
Q.23
An energy source will supply constant current into the load if its internal resistance is [ AIEEE 2005]
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a) very large as compared to the load resistance
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b) equal to the resistance of the load
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c)nonzero but less than the resistance of the load
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d)zero
Explanation
I=E / ( R + r) here r is internal resistance If r=0 then I=E/R=constantAnswer: (d)
Q.24
A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and the work done by the battery will be [ AIEEE 2007]
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a)1/2
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b) 1
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c)2
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d)1/4
Explanation
Required ratio=Energy stored in Capacitor / Work done by battery=½ CV2 / qV=½ ( CV/q) but C=q/v Ratio=1/2Answer: (a)
Q.25
The resistance of wire is 5 Ω at 50° and 6Ω at 100°C. The resistance of the wire at 0° will be [ AIEEE 2007]
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a) 3 Ω
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b) 2 Ω
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c)1 Ω
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d)4 Ω
Explanation
We know that Rt Ro ( 1 + αt)R50=Ro ( 1 +50α) 5=Ro ( 1 +50α) 5 - Ro=50×Roα --eq(1)R100=Ro ( 1 +100α) 6=Ro ( 1 +100α) 6 - Ro=100×Roα --eq(2)eq(1) / eq(2)5 - Ro / (6 - Ro)=1/210- 2Ro=6 - RoRo=4Answer: (d)
Q.26
Two uniform wires A and B are of same metal have equal masses. The radius of wire A is twice that of wire B. The total resistance of A and B when connected in parallel is [ MNR 1994]
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a)4Ω when the resistance of wire A is 4.25Ω
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b) 5Ω when the resistance of wire A is 4Ω
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c)4Ω when the resistance of wire B is 4.25Ω
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d)5Ω when the resistance of wire B is 4Ω
Explanation
given radius rA=2rAMass of both wire is same thus Thus RA / RB= When resistance connected in parallel effective resistance is (16/17)RA Onl;y option 'a' satisfy the condition Answer:(a)
Q.27
At 0 K Germanium behaves as a [ EAMCET 1989]
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a) conductor
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b) insulator
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c)super conductor
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d)ferromagnetic substance
Explanation
Answer: (b)
Q.28
Two non-idel batteries are connected in parallel. Consider the following statements(i) The equivalent e.m.f. is smaller than either of the two e.m.f.'s(ii) The equivalent internal resistance is smaller than either of the two internal resistance's [MPPMT 1999]
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a)Both (i) and (ii) are correct
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b) (i) is correct but (ii) is incorrect
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c)(ii) is correct but (i) is incorrect
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d)Both (i) and (ii) are incorrect
Explanation
Answer:(c)
Q.29
The specific resistance and cross section area of potentiometer wire is ρ and A respectively. If a current I is passed through the wire, the potential gradient of wire will be [ Raj.PMT 1996]
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a) (Iρ) /A
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b) I/(Aρ)
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c) (IA) / ρ
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d) IAρ
Explanation
V=IR V=I( ρl/A) Potential gradient V/l=Iρ / A Answer: (a)
Q.30
A box with two terminals is connected in series with a 2V battery, an ammeter and a switch. When a switch is closed the needle of ammeter moves quickly across the scale and then drop back to zero. The box contains
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a)a 20 Ω resistor
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b) a strip of copper
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c)a diode
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d)a fuse wire
Explanation
On switching on high value of current passes through the fuse which melts hence no deflection.Answer: (d)
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