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Physics NEET MCQ
Quiz 8
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Q.1
Increase in which property of free electrons causes increase in the resistance of a conductor with rise in temperature?
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a)number density
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b) relaxation times
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c)mass
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d)none of above
Explanation
Answer: (d)
Q.2
A steady current is set up in a metallic wire of non-uniform cross-section. How is the rate flow of electrons (R) related to the area of cross-section A?
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a) R is independent of A
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b) R ∝ A-1
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c)R ∝ A
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d)R ∝ A2
Explanation
Answer: (a)
Q.3
Two unequal resistor are connected in series with a cell. Which of the following statement is true?
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a) Potential drop across either resistor is same
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b) Potential drop across smaller resistor is more
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c)Potential drop across larger resistor is more
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d)Any one of the above can be true depends on the emf of the cell
Explanation
Answer:(c)
Q.4
A potential difference of 10V is applied across a conductance 4 S. The current in the conductor is
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a) 40 A
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b) 2.5 A
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c) 2.4 A
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d) none of the above
Explanation
Answer: (a)
Q.5
The resistivity of a material is inversely proportional to
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a)number density of electrons as well as relaxation time
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b) number density of electrons and directly proportional to relaxation time
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c)relaxation time and directly proportional to the number density of electrons
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d)neither relaxation time nor number density of electrons
Explanation
Answer: (a)
Q.6
How will the reading in ammeter A affected if another identical bulb Q is connected in parallel to P as shown in figure . The voltage of the mains is maintained at a constant value
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a) the reading will be reduced to one half
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b) the reading will not affected
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c)the reading will be double of previous one
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d)the reading will be increased to four fold
Explanation
Answer: (c)
Q.7
Wire of resistance 0.5 Ω/m is bent into a circle of radius 1 m. The same wire is connected across a diameter AB as shown in figure. The equivalent resistance is
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a)π ohm
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b)π/ (π+2) ohm
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c)π/ (π+ 4) ohm
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d)(π+1) ohm
Explanation
R1=R2=(πr) ×0.5=0.5πΩ R3=(2r)× 0.5=1 Ω All three resistance are in parallel Answer:(c)
Q.8
A galvanometer together with an unknown resistance in series is connected across two identical cells each of 1.5V. When the batteries are connected in series, the galvanometer records a current of 1A and when the batteries are connected in parallel, the current is 0.6A. The internal resistance of the battery is
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a) 0.33 Ω
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b) 1 Ω
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c) 0.66Ω
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d) 0.5Ω
Explanation
Butteries connected in series Total potential 3V , R=Resistance of galvanometer , r is internal resistance then Current i=3 / (R+2r) ,as i=1 R+2r=3 --eq(1) Batteries sin parallel Potential 1.5 R resistance of galvanometer Total resistance R + r/2 Current 0.6=1.5 / (R +r/2) R + r/2=2.5 -- eq(2) From equation (1) and (2) we get r=0.33Ω Answer: (a)
Q.9
In figure point x and y represent the terminals of an unknown emf. On moving jockey J from A towards B it is observed that deflection of galvanometer remains in the same direction but increases. Which one of the following conclusions is correct
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a)the emf E is less than the unknown emf
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b) the emf E is more than the unknown emf
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c)the potential drop of potential across potentiometer wire is less than the unknown emf
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d)the unknown emf is wrongly connected
Explanation
Answer: (d)
Q.10
In figure the potentiometer wire AB has a resistance of 5Ω and length 10 m. The balancing length AJ for the emf of 0.4V is
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a) 8 m
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b) 0.8 m
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c)4 m
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d)0.4 m
Explanation
Total resistance in circuit=45+5=50Ω Current through wire=5/50=0.1 AResistance between AJ R=0.4/0.1=4ΩResistance per meter of wire=5/10=0.5 Ω Thus 4Ω resistance will have length=4/0.5=8 mAnswer: (a)
Q.11
Circuit whose resistance is R is connected to n similar cells. If the current in the circuit is the same whether the cells are connected in series or in parallel, then the internal resistance r of each cell is given by
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a) r=R/n
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b) r=nR
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c)r=R
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d)r=1/R
Explanation
Current in series connection of battery I=nE / (R +rn) Current in parallel connection of battery I'=E / ( R+ r/n) from above question r=R Answer:(c)
Q.12
In the following figure, the reading of an ideal voltmeter V is zero. Then the relation between R, r1 and r2 is
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a) r=r₂ - r₁
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b) R=r₁ - r₂
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c) R=r₁ + r₂
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d) R=r1r2 / (r₁ + r₂ )
Explanation
Current I=2E / ( r1 + r2R) V1=E - Ir1=0 ( given)Or E=Ir1 E=2Er1 /(r1 + r2R) r1 + r2R=2r1 R=r1 -r2 Answer: (b)
Q.13
A primary cell has emf 2 volts. When short circuited, it gives current of 4 amp. Its internal resistance in ohms will be
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a)0.5
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b) 2
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c)5
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d)8
Explanation
Answer: (a)
Q.14
A cell of e.m.f E volts and internal resistance r ohm is supplying a current of i ampere to external circuit, 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: (b)
Q.15
In the following figure, the value of resistor to be connected between C and D so that the resistance of the entire circuit between A and B does not change with the number of elementary sets used is( each resistance in circuit is of value R :
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a)R
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b) R(√3 -1)
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c)3R
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d)R(√3 + 1)
Explanation
the ladder is infinite. So the effective circuit is Answer:(b)
Q.16
Twelve equal resistors, each of resistance r, are connected to form a skeleton cube. Then the equivalent resistance taken between tow diagonally opposite corners is
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a) r
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b) 12r
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c)5r/6
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d)(7/12)r
Explanation
Follow any path between A and D let R be the equivalent resistance Then 6IR=5Ir R=(5/6)rAnswer: (c)
Q.17
In the above question, the equivalent resistance between the adjacent corners of any one face of the cube is
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a) r
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b) 12r
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c)5r/6
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d)(7/12)r
Explanation
Applying kirchoff's II law to mesh ABCDA zr + 2zr +zr - (x-z)r=0 x=5z Again applying kirchoff's II law to mesh A'B'C'D'A' xr + (x-z)r + xr - yr=0 3x - z=y 3(5z) - z=y y=14z VA'D'=(2x+y) rA'D'=yr rA'D'=yr / (2x+ y) y=14z and and x=5z from above rA'D'=14z / (10z+14z )=14zr/ 24z=7r/12 Answer:(d)
Q.18
In the following figure the resistance of galvanometer G is 50Ω. Of the following alternatives in which case are the currents arranged strictly in the order of decreasing magnitudes with the larger coming earlier
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a) i, i2, ig, i1
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b) i , ig, i1, i2
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c) i, i2, i1, ig
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d) ig, i1, i2, i
Explanation
Bridge is balanced ig=0. Answer: (c)
Q.19
In Wheatstone's bridge, P=9 ohms, Q=11 ohms, R=4 ohms and S=6 ohms. How much resistance must be put in parallel to the resistance S to balance bridge?
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a)24 ohms
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b) (44/9) ohms
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c)26.4 ohms
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d)18.7 ohms
Explanation
P/Q=R/SS=RQ/P by substituting values except S we get S=44/9 But actual is 6 let x be the resistance connected in parallel 9/44 \\ 1/6 + 1/x X=26.4 ohmsAnswer: (c)
Q.20
In figure to make the bulb of rating 5W, 2V glow at normal intensity, the emf E is its internal resistance is 0.46Ω
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a) 2 V
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b) 2.76 V
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c)3.75 V
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d)5 V
Explanation
Rating of bulb is 5 W and 2 Vresistance of bulb=r=V2/ P=4/5=0.8 ΩAnd current i=V/r=2/0.8=2.5A Thus current of 2.5 A should flow through bulb Let I current flow through 1.6 ohm resistance i/I=R/r 2.5/I=1.6/0.8 I=1.25A Thus battery should provide total current of 2.5+1.25=3.75 A Total resistance of the circuit R=[(0.8)(1.6) / (0.8+1.6)]+ 0.46=1 Potential of battery=Current × resistance E=3.75 × 1=3.75 V Answer: (c)
Q.21
A battery of emf having internal resistance 1Ω is connected to a external resistor of resistance 4Ω . the rate of energy dissipation in battery is
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a) 2W
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b) 4W
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c)6W
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d)8W
Explanation
total resistance in circuit=4+1=5 Ω current through battery=V/R=10/5=2 APower dissipation in battery=I2×r=4×1=4W Answer:(b)
Q.22
A current of 2A flows in a system of conductors as shown in the following figure. The potential difference VA - VB will be ( in volts)
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a) +2
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b) +1
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c) -1
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d) -2
Explanation
resistance between DAC=DBC=5 Ω Current will get equally divided in two branches=1A VD - VA=2×1=2VVD - VB=3×1=3VVD - VB -(VD - VA)=3-2VA - VB=1 VAnswer: (b)
Q.23
When a current is divided between two resistors according to Kirchhoff's law, then the heat produced is
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a)zero
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b) negligible
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c)minimum
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d)maximum
Explanation
Answer: (c)
Q.24
An electric cell is
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a) a source of charge
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b) a source of energy
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c)an energy converter
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d)both a source of charge and of energy
Explanation
Answer: (c)
Q.25
10 cells each of emf 1V and internal resistance 0.1Ω are to send maximum current through an external resistance of 100Ω, the cell be connected in
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a) series
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b) parallel
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c)mixed grouping
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d)can not say
Explanation
Answer:(a)
Q.26
The potential difference V across a filament lamp is related to the current I by V=2I + 8IThe lamp is connected in one arm of a wheatstone bridge and a resistance of 4Ω each is connected in other arms. The potential difference that must be applied to the bridge, to obtain balance is
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a) 1 V
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b) 2 V
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c) 4 V
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d) 8 V
Explanation
Let I be the current through the battery when bridge is balanced Since all other resistance are same current will equally divided in two branches current through each branch is I/2 from given equation for potential across bulb V=2(I/2) + 8(I/2)2 V=I +2I2 Resistance of bulb must be 4Ω to balance bridge Potential across bulb V=IR=(I/2)×4=2I Volts Thus 2 I=I + 2I2 I=(1/2) A The net resistance of the circuit will be 4 Ω V'=(1/2)×4=2 V Answer: (b)
Q.27
To get maximum current in a resistance of 3Ω, one can use n rows connected in parallel. Each row contains m cells connected in series. If the total number of cells is 24 and the internal resistance of the cell is 0.5Ω, then
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a) m = 12, n = 2
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b) m = 8, n=3
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c) m = 2, n = 12
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d) m = 6, n =4
Explanation
Current will be maximum if the internal resistance = external resistance Total internal resistance = mr/n = 3 m/n = 3/0.5 = 6 m = 6n Also mn = 24 6m2 = 24 n = 2 and m = 12 Answer: (a)
Q.28
A star ( as shown in figure) is made of uniform wire. The resistance of the arm EL is r ohm. The resistance of the star between the terminals C and F is
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a) 1.946r
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b) 0.937r
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c) 0.62r
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d) 3.892r
Explanation
∠ BOC = 72°, therefore∠BEC = 36° ∠FEM = 18° Side FL = 2 ×FM = 2(rsin18 ) = 0.56r Consider a ΔFEL resistance of FE and LE are in series = 2r and resistance of FL is parallel to above combination Thus total resistance of ΔFEL = (2r)×'(0.56r) / 2.5r = 0.4375r Now, the resistance of each Δ KDL, ADF and BGH same as ΔFEL Hence arrangement becomes as shown in figure There is no current through HK as potential at H and K is same So effective resistance between C and F is Answer: (b)
Q.29
Five identical resistance each of 1100Ω are connected to 220 volt as shown in the following figure. The reading of an ideal ammeter A is
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a) 1 A
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b) (3/5) A
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c) (1/5) A
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d) (2/5) A
Explanation
Current through each resistor I = 220/1100= (1/5) A Ammeter carries current only due to last three lamps. Therefore the current in ammeter is (3/5)A. Answer:(b)
Q.30
In figure the current through resistor R is
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a) 3.0 A
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b) 13.0 A
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c) 6.5 A
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d) 9.0 A
Explanation
Use Kirchoff's first law for current through R (8-5)A = 3A Answer: (a)
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