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Physics NEET MCQ
Quiz 6
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Q.1
If two conducting spheres are separately charged and then brought in contact ---
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a) the total energy of the two spheres is conserved
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b) the total charge on the two spheres is conserved
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c) both the total energy and charge are conserved
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d) the final potential is always the mean of the potential of the two spheres
Explanation
Answer: (b)
Q.2
A parallel plate capacitor is charged and then disconnected from the battery. If the plates of the capacitor are then moved away from each other by the use of insulated handel, then there is increase in
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a)the charge on either plate
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b) the capacitance of the capacitor
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c)voltage across the capacitor
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d)all the above three
Explanation
Answer: (c)
Q.3
Two identical air filled parallel capacitors are charged to the same potential in the manner shown in figure by closing the switch S. If now the switch is opened and the space between the plates if filled with a dielectric of relative permitivity εr then
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a) the p.d across A remains constant and the charge on B remains unchanged
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b) the p.d. across B remains constant while the charge on A remains unchanged
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c)the p.d. as well as charge on each capacitor goes up by a factor εr
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d)the p.d as well as the charge on each capacitor down up by a factor εr
Explanation
Answer: (a)
Q.4
What physical quantity may X and Y represent? ( Y represent first mentioned quantity)
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a)pressure v/s temperature of given gas at constant volume
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b) kinetic energy v/s velocity of particle
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c)capacitance v/s charge to given constant potential
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d)potential v/s capacitance to given constant charge
Explanation
Answer:(d)
Q.5
A number of capacitors, each of capacitance 1µF and each one of which gets punctured if a potential difference just exceeding 500V is applied, are provided. Then an arrangement suitable for giving a capacitor of capacitance 3µF across which 2000V may be applied requires at least
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a)4 component capacitor
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b) 12 component capacitor
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c)48 component capacitor
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d)3 component capacitor
Explanation
We can connect a four group of capacitors in series such that potential drop across each group is 500 V Let C be the capacitance of group of capacitor Then 1/3=1/C + 1/C + 1/C + 1/C=4/C C=12µF Let each group contain 'n' capacitor connected in parallel Thus 12 µF=2×1¯ n=12Thus number of capacitors required=12×4=48 Answer: (c)
Q.6
Three capacitors, with capacitance of 1µF, 2µF, 3µF, are connected in series. Each capacitor gets punctured if potential difference just exceeding 100 volt is applied. If the group is connected across a 300 Volt circuit then the capacitor most likely to puncture first is
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a) of capacitance 1µF
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b) of capacitance 2µF
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c)of capacitance 3µF
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d)of capacitance either 1µF, 2µF, 3µF
Explanation
Let C' be the resultant capacitance Charge through the circuit=Q=C'V Now potential across capacitor=C'V/C From above equation it is clear that smaller is the capacitance greater is the voltage drop thus 1µF capacitor will puncture firstAnswer: (a)
Q.7
A slab X is placed between the two parallel isolated charged plates, as shown. IF Ep and Eq denotes the intensity of electric field at P and Q
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a) Ep is reduced by the presence of X if X is metallic
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b) Eq is increased by the presence of X if X is dielectric
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c)Eq is in opposite sense to Ep if X is a dielectric
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d)Eq is zero if X is metallic
Explanation
Answer:(c,d)
Q.8
Two spherical conductors of capacitances 3.0pF and 5.0pF are charged to potentials of 300V and 500V. The two are connected resulting in redistribution of charges. Then the final potential is
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a) 300 V
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b) 500 V
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c) 425 V
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d) 400 V
Explanation
Answer: (c)
Q.9
A battery of e.m.f V volts, resistors R1 and R2, a condenser C and switches S1 and S2 are connected in a circuit as shown in the figure below. The condenser will get fully charged to V volts when
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a)S1 and S2 are both closed
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b) S1 and S2 are both open
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c)S1 is open and S2 is closed
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d)S1 is closed and S2 is open
Explanation
Answer: (d)
Q.10
The amount of work done in increasing the voltage across the plates of a capacitor from 5V to 10V is W. The work done in increasing it from 10V to 15V will be
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a) 0.6W
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b) W
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c)1.25W
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d)1.67W
Explanation
change in energy stored in capacitor=work done W=(1/2) C[102 - 52]=(1/2) C(75)W'=(1/2)C[152 - 102]=(1/2) C(125)W'/W=125/75=5/3W'=(5/3)WW'=1.67W Answer:(d)
Q.11
A parallel plate capacitor is charged to 100volts and then connected to an identical capacitor in aparallel. The second capacitor has some dielectric between its plates. If the common potential is 20V then the dielectric constant of the dielectric is
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a) 2.5
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b) 4
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c) 5
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d) 8
Explanation
let dielectric constant be K Let capacitance of first capcitor be C capacitance of second capacitor=CK since charge is conserved charge before connecting another capacitor=sum of charges on the both capacitor CV=CV' +CK(V') V=V' + KV' 100=20( 1 + K) K=4Answer: (b)
Q.12
The electric field midway between two charges 0.1µC and 0.4µC separated by a distance of 60cm is
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a) 5 ×103 N/C
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b) 9 ×104 N/C
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c)5 ×104 N/C
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d)3 ×104 N/C
Explanation
Net electric field between mid way=|E|=|E1| -| E2| Let 2r be the distance between the charges Answer:(d)
Q.13
Two identical parallel plate capacitors are connected in series and connected to a constant voltage source of Vo Volt. If one of the capacitors is completely immersed in a liquid of dielectric constant K, the potential difference between the plates of the other capacitor change to
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a)
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b)
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c)
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d)
Explanation
When one capacitor is immersed in a liquid of dielectric constant K, then the capacity of this condenser becomes KC. The new capacity of system is CR=C(KC) / (KC+C)=KC /(K+1) hence the potential difference across the plates of other capacitor becomesV'=q/C=CRVo / C=kVo / (K+1)Answer: (b)
Q.14
A parallel plate capacitor is filled by a dielectric whose permitivity varies with applied voltage according to relation εr=αV where α=1 volt-The same capacitor containing no dielectric charged to a voltage of 72 V is connected parallel to the first non liner uncharged capacitor. The final voltage across capacitor is
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a) 19V
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b) 6V
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c)36V
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d)8V
Explanation
Capacity of non linear capacitor C'=εrClet V be the common potential then leaving negative root, we get V=8 VoltsAnswer: (d)
Q.15
In a circuit diagram potential difference between points A and B is 200 volts, the potential difference between a and b when the switch S is open is
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a) 100 V
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b) (200/3) V
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c)(100/3)V
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d)50 V
Explanation
Net capacity of both the branches is equal=[ (3×6) / (6+3)]=2µF ∴ charge on each capacitor is=CV=2×200=400 µC Now potential difference across A and B is potential across 3µF capacitor=Q/C=400/3 Potential difference across A and B is potential across 6µF capacitor=Q/C=400/6 (VA - Va ) - (VA - Vb=Va - Vb=400/3 -400/6=400/6=200/3 Volts Answer:(b)
Q.16
Five identical capacitor plates, each of area A , are arranged such that the adjacent plates are at a distance d apart. The plates are connected to battery of e.m.f E volts as shown
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a)
0%
b)
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c)
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d)
Explanation
The arrange meant is equivalent to four capacitors each capacitor of capacitance C=εoA / d Charge on each capacitor is Q=CE=εoAE/d Now plate 1 is common only to one capacitor and is connected to positive terminal of battery therefore charge on itis εoAE/d the plate 4 is common to to capacitors therefore it has a charge -2εoAE/d Answer: (a)
Q.17
Three uncharged capacitors of capacitance C1, C2 and C3 are connected as shown in figure to one another and to point A, B, and D at potentials VA , VB , VD . then the potential at O will be
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a)
0%
b)
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c)
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d)
Explanation
Potential drop across C1=VA - VO q1 / C1=VA - VO q1=(VA - VO)C1similarly q2=(VB - VO)C2q3=(VD - VO)C3From the law of conservation of charges q1 + q2 + q3=0 (VA - VO)C1 + (VB - VO)C2 + (VD - VO)C3=0 on rearranging turns we getAnswer: (b)
Q.18
The Gaussian surface for calculating the electric field due to a charge distribution is
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a) any closed surface around the charge distribution
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b) any surface near the charge distribution
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c)a spherical surface
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d)a symmetrical closed surface at every point of which electric field has a single fixed value
Explanation
Answer: (d)
Q.19
Three capacitors of capacitance 4µF, 6µF and 12µF are connected first in series and then in parallel. What is the ratio of equivalent capacitance in the two cases?
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a) 2:3
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b) 1:11
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c)11:1
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d)1:3
Explanation
Answer:(b)
Q.20
A parallel plate capacitor has a capacitance of 50pf in air and 105pf when immersed in oil. The dielectric constant of the oil is
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a) 50/105
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b) 1
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c) 2.1
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d) ∞
Explanation
C'=kC Answer: (c)
Q.21
When a capacitor is connected to a battery
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a)an alternating current flows in the circuit
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b) no current flows at all
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c)a current flows for some time and finally decreases to zero
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d)current keeps on increasing and reaches maximum after some time
Explanation
Answer: (c)
Q.22
When a charge q is moved once round a circle of radius R with charge Q at the centre of the circle, work done is
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a) zero
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b) Qq / 4πεoR
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c)Qq / 4πεoR2
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d)Qq /R
Explanation
Answer: (a)
Q.23
A work of 100 J is performed in carrying a charge of -5C from infinity to a particular point in an electrostatic field. The potential of this point is
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a) 100V
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b) 5V
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c)-20V
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d)20V
Explanation
Potential at point=Work/charge V=100/(-5)=-20V Answer:(c)
Q.24
Two copper spheres A and B of same size are charged to some potential. A is hollow and B is solid. Which of the two holds more charge>
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a) Solid sphere cannot hold any charge
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b) Hollow sphere can not hold any charge
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c) Both have zero charge
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d) Both have same charge
Explanation
Answer: (d)
Q.25
An electron initially at rest is accelerated through a P.D. of one volt. The energy acquired by electrons is
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a) 1 J
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b) 1.6×10⁻¹⁹ J
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c)10-19 J
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d)1.6×10⁻¹⁹ erg
Explanation
Answer: (b)
Q.26
The S.I. unit of surface integral of electric field is
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a) NC-1
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b) Nm2C
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c)J/C
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d)V/m
Explanation
Answer:(c)
Q.27
A bird sitting on a high power line
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a) gets killed instantly
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b) gets a mild shock
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c) is not affected practically
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d) gets a fatal shock
Explanation
Answer: (c)
Q.28
The mutual interaction between two charged bodies is best understood in terms of
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a) the action at a distance view
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b) the field concept
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c)a material connection between the charges
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d)medium in between the charges
Explanation
Answer: (b)
Q.29
Torque on a dipole in an electric field is maximum when angle between p and E is
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a) 0°
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b) 90°
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c)45°
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d)180°
Explanation
Answer:(b)
Q.30
When an electric dipole is placed in a uniform electric field, it experiences
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a) a force as well as torque
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b) a torque but not force
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c) a force but not torque
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d) neither for nor torque
Explanation
Answer: (b)
Q.31
The value of a charge q at the centre of two equal and like charges Q so that the three charges are in equilibrium [ kerala P.M.T 2004]
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a)-Q/4
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b) +Q/4
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c)Q
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d)Q/2
Explanation
resultant for on Q should be zeroLet distance between two Q charges be 2r kqQ/r2 =kQQ/(2r)2 q=Q/4 since Q charges are like q should be negative ∴ q=-Q/4Answer: (a)
Q.32
A charged spherical shell does not produce an electric field at any [ kerala P.M.T. 2004]
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a) interior point
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b) outer point
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c)beyond 2 meters
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d)beyond 10 meters
Explanation
Answer: (a)
Q.33
The charge on 4µF capacitor in the given circuit is
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a)12 µF
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b) 24 µF
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c)36 µF
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d)32 µF
Explanation
1µF and 5µF condensers are in parallelCp=1+5=6µF Now 4µF and 6µF capacitors are in series∴ (1/Cs)=(1/4) + (1/6)=(5/12) Cs=12/5=2.4 µF Charge on each capacitor 4µF and 6µF capacitor=Cs V=2.4×10=24µC Answer:(b)
Q.34
Two equal metal balls are charged to 10 and -20 units of electricity. Then they are brought in contact with each other and then are brought in to original distance. The ratio of magnitudes of force between the two balls before and after contact is [ Kerala P.M.T 2004]
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a) 8:1
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b) 1:8
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c) 1:2
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d) 2:1
Explanation
Here q1=10 units and q2=-20 units On touching, charges are redistributed charge on each sphere is q=(-20+10) /2=-5 units Now F1 / F2=q12 / q=200/25=8:1 Answer: (a)
Q.35
A conductor having a cavity is given a positive charge. Then field strengths EA, EB and EC at points A ( with in cavity) at B (within conductor) and C ( outside conductor) will e [ kerala P.M.T 2004]
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a)EA=0, EB=0, EC=0
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b) EA=0, EB=0, EC ≠=0
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c)EA ≠=0, EB=0, EC≠=0
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d)EA ≠=0, EB≠=0 , EC≠=0
Explanation
Electric field inside the cavity and inside metallic conductor is zero, and out side the conductor electric field is not zeroAnswer: (b)
Q.36
The potential at point P, which is forming a corner of a square of side 93mm with charges Q1=33 nC, Q2 -51 nC, Q3=47nC located at the other three corners is nearly [ kerala P.E.T 2004]
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a) 16 kV
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b) 4 kV
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c)400 V
0%
d)160 V
Explanation
Sides are 93nm thus diagonal QR=93√2 Potential at P is scalar addition of potential at P due to individual charges Answer: (b)
Q.37
The plates of a parallel plate capacitor are charged up to 100V. A 2mm thick plate is inserted between the plates, then to maintain the same potential difference between capacitor plates, the distance between the capacitor plate is increased by 1.6 mm. The dielectric constant of the plate is
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a)5
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b) 1.25
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c)4
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d)2.5
Explanation
To maintain same potential capacitance in both the cases must be same thus εA/d=εA / d' Thus d=d' let t be the thickness of dielectric slabbut d'=d - t +t/k+1.6 d=d -t +t/k+1.6 t -t/k-1.6=0t( 1- (1/k) -1.6=02( 1- (1/k)=1.61 - (1/k)=0.8 1/k=1-0.8k=5Answer: (a)
Q.38
If one penetrates a uniformly charged spherical shell, the electric field strength E
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a) increases
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b) decreases
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c)remains same as surface
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d)is zero at all points
Explanation
Answer: (d)
Q.39
Two small spheres, each carrying a charge q, placed r meters apart, repel each other with a force F. If one of the spheres is taken around the other one in a circular path of radius r, the work done will be
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a) F × r
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b) F × 2πr
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c)F / 2πr
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d)zero
Explanation
Answer:(d)
Q.40
An electric dipole consists of two opposite charges each of magnitude 1.0µC separated by a distance of 2.0cm. The dipole is placed in an external field of 1.0×105 N/C. The max. torque on the dipole is
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a) 0.2×10⁻³ N-m
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b) 1.0×10⁻³ N-m
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c) 2.0×10⁻³ N-m
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d) 4.0×10⁻³ N-m
Explanation
p=q(2a) τ=pEsinθ Answer: (c)
Q.41
A positively charged sphere hangs from a silk thread. We put a positive test charge qo at a point and measure F/qo then it can be predicted that the electric field strength E is
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a) > F/qo
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b)=F/qo
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c)< F/qo
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d)can not be estimated
Explanation
Here, the electric field due to test charge +qo opposes the field of positive charge being measured. therefore, the measured value is less than the actual value (F/qo)Answer: (a)
Q.42
Hollow metal sphere of radius 5 cm is charged such that the potential on its surface is 10V. The potential at a distance of 2cm from the centre of the sphere is
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a) zero
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b) 10 V
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c) 4 V
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d) 10/3 V
Explanation
Potential inside the hollow sphere=potential at its surface Answer: (b)
Q.43
Particle A has charge +Q and particle B has charge +4Q with each of them having the same mass. When allowed to fall from rest through same electrical potential difference, ratio of their speed VA / VB will become
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a)2:1
0%
b) 1:2
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c)1:4
0%
d)4:1
Explanation
Potential of A=qV and potential energy of B=4qV Thus ratio of kinetic energy=ratio of squares of velocities as mass of both is same vA / vB=1/2Answer: (b)
Q.44
Which of the following is a volt
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a) erg per cm
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b) joule per coulomb
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c) erg per ampere
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d) newton/(coulomb×metre2)
Explanation
Answer: (b)
Q.45
The capacitance of an isolated conducting sphere of radius R is proportional to
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a) R2
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b)1 / R2
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c)1 /R
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d)R
Explanation
Answer:(d)
Q.46
A condenser is connected across another charged condenser. The energy in the two condenser will
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a) Be equal to the energy in the initial condenser
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b) Be less than that in the initial condenser
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c) Be more than that in the initial condenser
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d) BE more or less depending on the relative capacitances of the two condenser
Explanation
Answer: (b)
Q.47
Work done in moving a unit positive charge through a distance of x meter on an equipotential surface is [ Panjab C.E.T 1997]
0%
a)x joule
0%
b) (1/x) joule
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c)zero
0%
d)x2 joule
Explanation
Answer: (c)
Q.48
If a positive charge is shifted from lower potential to region of higher potential region, the electric potential energy
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a) decreases
0%
b) increases
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c)remains same
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d)may increase / decrease
Explanation
Answer: (b)
Q.49
If the flux of electric field through a closed surface is zero
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a) electric field must be zero everywhere on the surface
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b) electric field may be zero everywhere on the surface
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c)charge in the vicinity of the surface must be zero
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d)charge inside the surface must be zero
Explanation
Answer:(b, d )
Q.50
The electric potential V at any point x, y, z ( all in meters) in space is given by V=4x2 volt. The electric field at the point (1m, 0, 2m) in volt/meter is
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a) 8 V/m along negative X -axis
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b) 8 V/m along positive X -axis
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c) 16 V/m along negative X axis
0%
d) 16 V/m along positive Z axis
Explanation
Thus 8V/m along negative X-axis Answer: (a)
0 h : 0 m : 1 s
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