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Physics NEET MCQ
Quiz 2
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Q.1
The electron drift speed is small and the charge of the electron is also small but still, we obtain large current in a conductor. This is due to [AFMC 2008]
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a) the conducting property of the conductor
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b) the resistance of the conductor is small
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c) the electron number density of the conductor is small
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d) the electron number density of the conductor is enormous
Explanation
Electron number density is very large. Therefore as soon as one electron enters in conductor from the battery one electron leaves conductorAnswer: (d)
Q.2
A 30 V, 90 W lamp is to be operated on a 120V DC line. For proper glow, a resistor of ..... Ω should be connected in series with the lamp. [AFMC 2008]
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a) 40
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b) 10
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c) 20
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d) 30
Explanation
If we want to operate lower voltage rated bulb on higher voltage then we have to connect resistance to control current. Hence first we will find current rating of bulbP=VI hence I=P/V , Here Power P=90W , Voltage V=30I=3ANow we will find resistance of bulb P=V2/R or R=V2/P R=10 Ω If bulb is to operate on 120V then required total resistance R1 to maintain current of 3A can be calculatedR1=V’/I Here V’ is new potential 120VR1=120/3=40 ΩR1=R + R’ Here R’ is additional resistance, R is the resistance of bulbThus R’=40-10=30 ΩAnswer : (d)
Q.3
he resistance of a wire at 20°C is 20 ohms and at 500°C is 60 ohms. At which temperature its resistance will be 25 ohm? [AFMC 2010]
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a) 80oC
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b) 70oC
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c) 60oC
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d) 50oC
Explanation
According to formulaRt=R0 [1 + α (t-t0)]By substituting values we get 60=20[1+ α ( 500-20)] 3=1+ α ( 480) α=2/480, Substituting value of α and resistance at unknown temperature we get25=20 [1+ 1/240 ( t-20)]After solving equation for T we gett=80 oC Ans : (a)
Q.4
A material 'B' has twice the specific resistance of 'A'. A circular wire made of 'B' has twice the diameter of wire made of 'A' then for the two wires to have the same resistance, the ratio of lB / lA of their respective length must be [ AIEEE 2006]
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a) 1
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b) 1/2
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c)1/4
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d)2
Explanation
Answer:(d)
Q.5
Calculate the current through the battery and through a resistance of 6Ω in the given circuit: [AFMC 2010]
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a) 2.66amp, 2amp
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b) 1.33amp, 1amp
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c) 0.66amp, 0.5amp
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d) None of these
Explanation
Resistance 7Ω, 1 Ω and 10 Ω are in series hence R’=18R’ connected parallel with 6 Ω hence R”=R’x6 / (R’+6)=18x6 /(18+6)=4.5 ΩR” is in series with 2 Ω, 0.5 Ω and 8 Ω resistance. Hence total resistance in the circuit R=4.5+2+0.5+8=15ΩTotal current through battery=20/15=4/3=1.33 ampSince R’=18 Ω connected parallel with 6 Ω circuit changes to as shown in figure Current of I=1.33 amp will get divided at junction A if I1 is the current through 6 Ω If R1 and R2 are connected in parallel then current through R1 is I1 given byNow I=1.33 amp; R1=6 Ω ; R2=R'=18 Ω On substituting the values in above equation we get current through 6 Ω=I1=1 ampAnswer: ( b)
Q.6
A battery is charged by supply of 100V as shown in figure. The charging current is 1.0A The value of R is .. [AFMC 2010]
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a) 88 立
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b) 68 立
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c) 44 立
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d) None of these
Explanation
Supply potential=Voltage drop across battery + Voltage drop across Resistance100=12 +VR Here VR is the potential drop across resistanceVR=88 V Now current is 1 amp givenhence R=88 ΩAnswer : (a)
Q.7
If the conduction-electron density in copper wire is n, the area of cross-section of wire is A and the average drift speed of electron Vd, then the net number of electrons crossing a cross-section of the wire in a unit time interval ∆t will be[AFMC 2010]
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a) nVdA∆t
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b) neVdA∆t
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c) neVdA
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d) nVd∆t
Explanation
Current I in copper wire I=neAVd Here N is the number of electron crossing the cross section in time ∆t N=nAVd∆tAnswer: (a)
Q.8
A potentiometer is connected between A and B and the balance point is obtained at 203.6cm. When the end of the potentiometer connected to B is shifted to C, then the balance point is obtained at 24.6cm. If now the potentiometer be connected between B and C, the balance point will be… [AFMC 2011]
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a) 179.0cm
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b) 197.2cm
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c) 212.0cm
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d) 228.0cm
Explanation
When potential meter connected between A and B it measure the potential across Battery 1; E1 = 203.6 V When potentiometer is connected across A and C it measure the potential across two batteries since they are opposing. Let Potential of other battery be E2 E = E2-E2 24.6 = 203.6-E2 E2= 203.6-24.6 = 179.0 V This potential will be measured by potential meter hence balance point will be at 178.0 cm Answer: (a)
Q.9
Five resistance each of 5 Ω, are connected as shown in figure. Find the equivalent resistance between points (1) A and B (2) A and C , all resistance are of 5 Ω [AFMC 2011]
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a) (1) 7.5 Ω (2) 2.25 Ω
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b) (1) 5 Ω (2) 2.5 Ω
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c) (1) 2.5 Ω (2) 3.1 Ω
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d) (1) 3 Ω (2) 2.5 Ω
Explanation
Resistance between A and B: from figure It is clear resistance R1 and R2 are in series let effective resistance br R' = 10 and resistance R3 and R4 are in series let effective resistance be R"=10 Ω . Now R' , R" and R5 are connected in parallel by using formula for equivalent resistance of parallel connection we get R = 2.5 Ω Resistance between A and C : Resistance R1 and R2 are series R’ = 10 Ω R’ is parallel to R5therefore by substituting R'=10 Ω in above equation we get R" = 10/3 Ω = 3.33 Ω Now R” is in series with R3 therefore R’” = 3.33 +5 = 8.33 Ω R”’ is parallel with R4 by substituting the value of R" in above equation we gat R'" = 3.12 Ω Answer: (c)
Q.10
A 60 volt battery would spend 1200Watt power across a resistance of .. [AFMC 2011]
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a) 12 Ω
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b) 10 Ω
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c) 60 Ω
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d) 3 Ω
Explanation
Substituting value of P = 1200 w and Potential V = 60 in above equation we get R = 3 Ω Answer : (d)
Q.11
What is the resistance between point A and B in the circuit given below, all resistance are of 1 Ω
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a)(√3+1)
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b)(√3-1)
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c)(1/√3 )
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d)(1- √3)
Explanation
Let the required equivalent resistance be X. The network is infinite. Therefore, adding one more stage to the network should not change the value of X. Figure given below shows one such addition Here X and 1 Ω are in parallel The equivalent resistance between A and B is A resistance between P and Q of value 2R is found in series The equivalent resistance between P and Q ( which should be X) Answer: (a)
Q.12
Find equivalent resistance between A and B [ CBSE PMT 1988]
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a)13/11Ω
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b) 39/29Ω
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c)6/4Ω
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d) 2Ω
Explanation
To solve Unbalanced Wheatstone bridge problem we can use Kirchhoff's first law We will consider potential difference across A and B is 10V So we can consider potential at A as 10V and potential across B as 0 Now algibric sum of current at a and y points is zero according to Kirchhoff's first law For point y 3 y- x = 10 ..(1) For point x 5x -2y = 10 ... (2) On simplify two equations we get x=50/13 and y = 60/13 Now I1 = y-0/1 = 60/13A I2 = x-0/2 = 50/13A Total current = 60/39+50/13 =110/13 Now V = I Req Req = V/I = 13/11Ω Answer: (a)
Q.13
n equal resistors are first connected in series and then parallel. What is the ratio of the maximum to minimum resistance ? [ CBSE PMT 1989]
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a) n
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c)n2
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d)1/n
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b) 1 / n²
Explanation
effective resistance in series=nr (maximum resistance) effective resistance when connected in parallel=r/n (minimum resistance)Ratio=maximum / minimum=n2Answer: (c)
Q.14
Forty electric bulbs are connected in series across 220V supply. After one bulb is fuse the remaining 39 are connected again in series across the same potential. the illumination will be [ CBSE PMT 1989]
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a) more with 40 bulbs than with 39 bulbs
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b) more with 39 bulbs then with 40 bulbs
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c)equal in both cases
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d)in ratio 402 : 392
Explanation
The voltage terminals same for both the casesH ∝ (1/R). the combination with 39 bulbs have low resistance than 40 bulb. Hence 39 bulb combination will glow more Answer:(b)
Q.15
A current of 2A passing through a conductor produces 80J of heat in 10 seconds. The resistance of the conductor in Ohm is [ CBSE PMT 1989]
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a) 0.5
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b) 2
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c) 4
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d) 20
Explanation
H=I2Rt R=H / (I2t) R=80 /(4×10)=2 Ω Answer: (b)
Q.16
A cell of e.m.f E volt, internal resistance r ohm is being charged with current of i ampere, then the terminal potential difference is :
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a) E
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b) E - ir
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c) E + ir
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d) E ± ir
Explanation
Answer: (c)
Q.17
You are given several identical resistance each of value R=10Ω and each capable of carrying a maximum current of one ampere. It is required to make a suitable combination of these resistance of 5Ω which can carry a current of 4 ampere. The minimum number of resistance's of the type R that will be required for this job is [ CBSE PMT 1990]
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a)4
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b) 10
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c)8
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d)20
Explanation
Let m resistance's of resistance 'r' are connected in series and such n combinations are connected in parallel thus total resistance is R=(m/n)r (m/n)r=5 given (m/n)10=5 m/n=1/2 or 2m=n --eq(1) since each resistance can carry one ampere and combination can carry 4 amp Thus n=4 from eq91) m=n/2=4/2=2 Total resistance=m×n=4×2=8Answer: (c)
Q.18
In the network shown in figure each resistance is 1 Ω. The effective resistance between A and B is [ CBSE PMT 1990]
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a) (4/3) Ω
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b) (3/2) Ω
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c)7 Ω
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d)(8/7) Ω
Explanation
Due to symmetry current will distribute as shown in figure from figure a it is clear that Current In AO and OB is same thus we can separate triangular arrangement of resistance and calculate the resistance as shown n figure bRrangular arrangement with O=two resistance in series as one parallel to series combinationR'=2×1 / ( 2 +1)=2/3This R' is in series with two 1Ω resistance R"=8/3above combination is parallel with bottom two 1 Ω resistance effective resistance R=8/7 ΩAnswer: (d)
Q.19
Three resistance's each of 4Ω are connected to form a triangle. The resistance between any two terminal is [CBSE PMT 193]
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a) 12 Ω
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b) 2 Ω
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c)6 Ω
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d)(8/3) Ω
Explanation
In triangle two are connected in series=8 Ωand one para ell to this 8Ω resistance=8/3Ω Answer:(d)
Q.20
The current through 3Ω resistor is 0.8 amp., then potential drop through 4 Ω resister is[ CBSE PMT 1993]
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a) 9.6 V
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b) 2.6V
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c) 4.8 V
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d) 1.2V
Explanation
For a parallel combination of resistance I1=IR2 / ( R1 + R2) 0.8 (6+3)=I(6) I=1.2 A Thus current through 4Ω resistance is 1.2A , so potential drop=4×1.2=4.8 V Answer: (c)
Q.21
In the circuit shown in figure the current in 4Ω resistance is 1.2A. What is the potential difference between B and C [ CBSE PMT 1994]
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a)3.6 V
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b) 6.3V
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c)1.8 V
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d)2.4 V
Explanation
4 Ω and 8 Ω resistance are parallel. Thus current circuit is Potential difference between B and C=2×1.8=3.6VAnswer: (a)
Q.22
If a resistance R is melted and recasted to half of its length, then the new resistance of the wire will be CBSE PMT 1995]
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a) R/4
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b) R/2
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c)R
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d)2R
Explanation
Given l2=l/2 thus A2=2A New resistance=(ρl2) / A2 New resistance=ρ (l/A) (1/4)=R/4Answer: (a)
Q.23
Two wire of same metal have same length but their cross-sections are in the ratio 3:They are joined in series. The resistance of thick wire is 10Ω. The total resistance of combination will be [ CBSE 1995]
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a) 10 Ω
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b) 20 Ω
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c)40 Ω
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d)100 Ω
Explanation
length is same Thus R ∝ ( 1/A) Thus second wire resistance is=3 ×10=30Total resistance=10+30=40Ω Answer:(c)
Q.24
What will be the equivalent resistance of circuit shown in figure between point A and D.. each resistance is 10Ω [ CBSE PMT 1999]
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a) 10Ω
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b) 20Ω
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c) 30Ω
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d) 40Ω
Explanation
Equivalent circuit Equivalent Resistance of circuit=10 + [(20×20) / ( 20+20)] + 10=30 Ω Answer: (c)
Q.25
Kirchoff's first law, i.e. Σi=0 at a junction, deals with the conservation of [ CBSE PMT 1992]
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a) charge
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b) energy
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c)momentum
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d)angular momentum
Explanation
We know from the Kirchoff's first law that the algebraic sum of the current meeting at any junction in the circuit is zero or the total charge remain constant. Therefore, Kirchoff's law at a junction deals with the conservation of chargeAnswer: (a)
Q.26
There are three copper wires of length and cross sectional area ( L, A) , (2L, ½ A), (½L, 2A). In which case is the resistance minimum? [ CSE PMT 1997]
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a) It is the same in all three cases
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b) wire of cross sectional area 2A
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c)wire of cross-sectional area A
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d)wire of cross sectional area ½ A
Explanation
Answer:(b)
Q.27
The resistance of discharge tube is [ CBSE PMT 1999]
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a) zero
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b) ohmic
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c) non-ohmic
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d) infinity
Explanation
In discharge tube the current is due to flow of positive ions and electrons. Moreover, secondary emission of electrons is also possible. So V-I curve is non-liner, hence resistance is non-ohmic Answer: (c)
Q.28
The current (I) in the given circuit is [ CBSE PMT 1999]
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a)1.6 A
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b) 2.0 A
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c)0.32 A
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d)3.2 A
Explanation
Both 6 Ω resistance are in series thus effective resistance is 12 Ω which is in parallel with 3Ω resistance effective resistance of this combination is=12/5 ΩThus current I=V/R=4.8/(12/5)=2AAnswer: (b)
Q.29
In a meter -bridge, the balancing length from the left end when standard resistance of 1Ω is in right gap is found to be 20cm. The value of unknown resistance is [ CBSE-PMT 1999]
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a) 0.25Ω
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b) 0.4Ω
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c)0.5Ω
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d)4Ω
Explanation
Let the unknown resistance be X. Then condition of Whetstone's bridge gives X/R=20r/80R. Where r is resistance of wire per cm. ∴ X=(20/80)×R=(1/4)×1=0.25ΩAnswer: (a)
Q.30
Potentiometer measures potential more accurately because [ CBSE PMT 2000]
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a) it measures potential in the open circuit
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b) it uses sensitive galvanometer for null deflection
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c)it uses high resistance potentiometer wire
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d)it measures potential in the closed circuit
Explanation
Answer:(a)
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