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Quiz 6
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Q.1
in 1g of solid, there are 5×1021 atoms. If one electron is removed from every one of 0.01% of atoms of the solid, the charge gained by the solid is
0%
a) +0.08 C
0%
b) 0.8C
0%
c)-0.08C
0%
d)-0.8C
Explanation
Number atoms from which one electron is removed=5×1021× (0.01/100)=5×1017Since one electron is removed from atom charge on atom=1.6×10-19CTotal charge on the sphere=5×1017×1.6×10-19=8.0×10-2=0.08CAnswer: (a)
Q.2
You are travelling in a car during a thunder storm. In order to protect yourself from lightning would you prefer to
0%
a) remain in the car
0%
b) take shelter under a tree
0%
c)get out and lie flat on the ground
0%
d)touch the nearest electric pole
Explanation
Answer:(a)
Q.3
Two point charges placed at a distance of 20cm in air repel each other with certain force. When a dielectic slab of thickness 8cm and dielectric constant K is intoduced between these point charges, the force of interaction becomes half of its previous value. Then K is approximately
0%
a) 2
0%
b) 4
0%
c) √2
0%
d) 1
Explanation
Dilectric slab is of thickness 8cm its equivalent distance in air=√k (8) Let F be the force without dielectric and F' be the force with dielectric, given F=2F' Answer: (b)
Q.4
Capacitor C, 2C, 4C, ...∞ all are connected in parallel, their effective capacitance will be
0%
a)C
0%
b) C/2
0%
c) ∞
0%
d)0
Explanation
Answer: (c)
Q.5
A one micro ampere beam of protons with a cross-sectional area of 0.5 sq.mm is moving with velocity of 3×104 m/sec. Then the charge density of the beam is
0%
a)6.6 ×10-9 coulomb per metre3
0%
b) 6.6 ×10-7 coulomb per metre3
0%
c)6.6 ×10-6 coulomb per metre3
0%
d)6.6 ×10-5 coulomb per metre3
Explanation
change in one second=I=10-6Volume per second=vlocity × cross sectional area=3×104 × 0.5 × 10-6=1.5×-2 ∴ chrge density=charge / volume=10-6 /1.5×-2=6.66×10-5 Answer: (d)
Q.6
A metal sphere is held fixed on a smooth horizontal insulated plate and another small metal sphere is placed at some distance away. If the fixed sphere is given a charge, then
0%
a) the another sphere will be attracted towards first
0%
b) the another sphere will be repelled by the first
0%
c)there will be no effect at all
0%
d)the another sphere will move towards the first sphere and after contact it will be repelled
Explanation
Answer: (d)
Q.7
In the following diagram a particle with small charge -q is free to move up or down but not sideways near a larger fixed charge Q. The small charge is in equilibrium because in the positions shown shown the electrical upward force is equal to the weight of the particle. Which statement is true
0%
a)In fiure A, -q is in stable equilibrium
0%
b) In figure A , -q is in neutral equilibrium
0%
c)in figure B , -q is in stable equilibrium
0%
d)neither in figure A nor in B, q is in stable equilibrium
Explanation
Answer:(c)
Q.8
In the electric field of a point charge q, a certain charge is carried from point A to B, C, D, and E. Then work done :
0%
a) is least along the path AB
0%
b) is least along the path AD
0%
c) is zero along any one of the path AB, AC, AD, AE
0%
d) is least along AE
Explanation
Point A, B, C, D are equipotentail Answer: (c)
Q.9
The variation of potential with distance R from a fixed point is as shown below. The electric field at R=5m is
0%
a)2.5 volts/m
0%
b) -2.5 volts/m
0%
c)2/5 volts/m
0%
d)-2/5 volts/m
Explanation
E=- dv/dr=- slope of curve.At point R=5m slope=(0-5) / (6-4)=-2.5 V/m∴ E=+2.5 V/mAnswer: (a)
Q.10
A table tennis ball, which has been covered with a conducting paint is suspended by a silk thread so that it hangs between two metal plates. One plate is earthed, while the other plate is attached to the heigh voltage generator the ball
0%
a) is attracted to the high voltage plates and stay there
0%
b) hangs without moving
0%
c)swings backward and forward hitting each plate in turn
0%
d)is repelled by the earthed plate and stays there
Explanation
Answer: (c)
Q.11
A small circular ring has a uniform charge distribution. On a far-off axial point distant x from the centre of the ring, the electric field is proportional to
0%
a) x-1
0%
b) x-3/2
0%
c)x-2
0%
d)x5/4
Explanation
Answer:(c)
Q.12
In figure A is a point on the axis of an electric dipole. The electric field at the location of A is E, If distance of of point A is doubled from O, the field will be ( assume x > a)
0%
a) E/4
0%
b) E/8
0%
c) E/2
0%
d) E
Explanation
Electric field due to dipole ∝ 1/r3 Answer: (b)
Q.13
Two insulated charged sphere of radii 20cm and 25cm respectively and having an equal charge Q are connected by a copper wire and then they are separated
0%
a)Both the spheres will have the same charge Q
0%
b) Charge on the 20cm sphere will be greater than that on the 25cm sphere
0%
c) Charge on the 25cm sphere will be greater than that on the 20cm sphere
0%
d)Charge on each sphere will be 2Q
Explanation
Answer: (c)
Q.14
Six identical capacitors are joined in parallel, charged to a potential difference of 10V, separated and then connected in series. Then the potential difference between the free plates is
0%
a) 10 V
0%
b) 30 V
0%
c)60 V
0%
d)10/6 V
Explanation
Answer: (c)
Q.15
A parallel plate capacitor is made by taking n equally spaced plates connected alternate. If the capacitance between ant two successive plates is C, then the capacitance of equivalent system is
0%
a) C
0%
b) nC
0%
c)(n-1)C
0%
d)(n+1)C
Explanation
Answer:(c)
Q.16
Two spherical conductors A1 and A2 of radii r1 and r2 ( r2 > r1 ) are placed concentrically in air. A1 is given a charge +Q while A2 is earthed. Then the equivalent capacitance of the system is
0%
a)
0%
b)
0%
c) 4πεor2
0%
d) 4πεor1
Explanation
Answer: (a)
Q.17
A charged conductor of radius R is connected momentarily to another uncharged spherical conductor of radius r by means of thin conducting wire, then the ratio of the surface charge density of the first to the second conductor is
0%
a)R2 / r2
0%
b) R/r
0%
c)r/R
0%
d)1/1
Explanation
Potential becomes equal therefore σ1R/ εo=σ2r/ εoσ1/σ2=r/R Answer: (c)
Q.18
A parallel plate air capacitor has a capacitance of 100µµF. The plates are at a distance d apart. A slab of thickness (t < d) and dielectric constant 5 is introduced between the parallel plates. Then the capacitance can be
0%
a) 50 µµF
0%
b) 100 µµF
0%
c)200 µµF
0%
d)500 µµF
Explanation
since, whole of the space between the plates is filed then capacitance will be 500µµF but whole space is not filled thus capacitance will be more than 100µµF but less than 500µµF Answer should be 200µµFAnswer: (c)
Q.19
A parallel plate capacitor has circular plates of 10 cms radius separated by an air gap of 1mm. It is charged by connecting the plates to a 100V battery. Then the change in energy stored in the capacitor when the plates are moved to a distance of 1cm. When the plates are maintained in connection with battery is
0%
a) loss of 12.5 erg
0%
b) gain of 12.5 ergs
0%
c)gain of 125 ergs
0%
d)loss of 125 ergs
Explanation
∴ There is loss in energy=12.5 erg Answer:(a)
Q.20
A foil of aluminum of negligible thickness is inserted in between the space of a parallel plate condenser. If the foil is electrically insulated, the capacity of the condenser will
0%
a) increase
0%
b) decrease
0%
c) remain unchanged
0%
d) become zero
Explanation
Answer: (c)
Q.21
A capacitor C1=4µF is connected in series with another capacitor C2=1µF. The combination is connected across a d.c. source of voltage 200 V. The ratio of potential across C2 and C1 is
0%
a)1:4
0%
b) 4:1
0%
c)1:2
0%
d)2:1
Explanation
Charge through each capacitors Q is same Potential across C1 is V1=Q/C1=Q/4Potential across C2 is V2=Q/C2=Q/1 V2 / V1=4:1Answer: (b)
Q.22
A parallel plate capacitor has a capacity C. The separation between plates is doubled and a dielectric medium is inserted between plates. The new capacity is 3C. The dielectric constant of medium is
0%
a) 1.5
0%
b) 3.0
0%
c)6.0
0%
d)12.0
Explanation
C=εoA/d 3C=KεoA / 2d∴ K=6Answer: (c)
Q.23
ABC is an equilateral triangle of side 1m. Charges are placed at its corners as shown in figure. O is the mid point of side BC. The potential at point is
0%
a)2.7×103 V
0%
b) 1.52×105 V
0%
c)1.3×103 V
0%
d)-1.52×105 V
Explanation
Answer: (d)
Q.24
A charge particle of charge qo is moved around a charge +q along the circular path from A to B . the work done is :
0%
a)
0%
b)
0%
c)
0%
d)zero
Explanation
Answer: (d)
Q.25
two balls with equal charges are in vessel with ice at -10°C at a distance of 25cm from each other. On forming water at 0°C, the balls are brought nearer to 5cm from the interaction between them to be same. If the dielectric constant of water at 0°C is 80, the dielectric constant of ice at -10°C is
0%
a) 40
0%
b) 3.2
0%
c)20
0%
d)6.4
Explanation
In ice In water Answer:(b)
Q.26
A free proton and a free α-particle initially at a separation of 1Å are released, the kinetic energy of proton and that of α particle when at infinite distance, bears the ratio
0%
a) 1 :1
0%
b) 1:2
0%
c) 14:4
0%
d) 4:1
Explanation
HEre asked to calculate the ratio of enrgy. No need to calculate potential at 1Å Potential energy at nfinity is zero Now according to law of conservation of momentum mpvp + mαvα=0 vp / vα=mα / mp mα=4×mp vp / vα=4 Answer: (d)
Q.27
In the given figure the resultant capacity will be
0%
a) n ( n+1)C
0%
b) n (n2+1)C
0%
c)nC / (n2 + 1)
0%
d)n2C /(n2 + 1)
Explanation
Answer: (c)
Q.28
A charge +q is carried from a point A(r,135°) to point B(r, 45°) following a path which is quadrant of circle of radius r. If the dipole moment is P. the ork done by the external agent is
0%
a)zero
0%
b)
0%
c)
0%
d)
Explanation
Potential at a point P ling on a line passing through centre of the dipole and making an angle θ with the axis of dipole is Answer: (c)
Q.29
A metallic plate of thickness t and face area of one side A is inserted between the plates of parallel plate air capacitor with a separation d and face area A. Then the equivalent capacitance is
0%
a) εoA/d
0%
b) εoA/ (d×A)
0%
c)εoA/ (d-t)
0%
d)εoA/ (d+t)
Explanation
Answer: (c)
Q.30
A parallel plate capacitor is charged and then isolated. What is the effect of increasing the plate separation
0%
a) Charge remains constant, Potential remains constant, Capacitance decreases
0%
b) Charge increases, Potential increases , Capacitance decreases
0%
c)Charge remains constant, Potential decreases, Capacitance increases
0%
d)Charge remains constant, Potential remains increases, Capacitance decreases
Explanation
Answer:(d)
Q.31
If two conducting spheres are separately charged and then brought in contact ---
0%
a) the total energy of the two spheres is conserved
0%
b) the total charge on the two spheres is conserved
0%
c) both the total energy and charge are conserved
0%
d) the final potential is always the mean of the potential of the two spheres
Explanation
Answer: (b)
Q.32
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
0%
a)the charge on either plate
0%
b) the capacitance of the capacitor
0%
c)voltage across the capacitor
0%
d)all the above three
Explanation
Answer: (c)
Q.33
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
0%
a) the p.d across A remains constant and the charge on B remains unchanged
0%
b) the p.d. across B remains constant while the charge on A remains unchanged
0%
c)the p.d. as well as charge on each capacitor goes up by a factor εr
0%
d)the p.d as well as the charge on each capacitor down up by a factor εr
Explanation
Answer: (a)
Q.34
What physical quantity may X and Y represent? ( Y represent first mentioned quantity)
0%
a)pressure v/s temperature of given gas at constant volume
0%
b) kinetic energy v/s velocity of particle
0%
c)capacitance v/s charge to given constant potential
0%
d)potential v/s capacitance to given constant charge
Explanation
Answer:(d)
Q.35
The two condensers of capacitance 2 and 3µF are in series. The outer plate of the first condenser is at 1000 V and the outer plate of the second condenser is earthed. The potential of the inner plate of each condenser is
0%
a) 300 V
0%
b) 500 V
0%
c) 600 V
0%
d) 400 V
Explanation
effective capacitance C=(2×3) / (2+3)=6/5=1.2µF Charge through circuit Q=CV=1.2×10-6×1000=1.2×10-3 potential difference first capacitor (PQ) V=Q/C1=1.2×10-3/2×10-6=0.6×103=600 V Thus potential at common plate=1000 - 600=400V Answer: (d)
Q.36
The area of the plates of parallel plate condenser is 100cmThe paper ( K=2.5) of thickness 0.005 cm is pit in between the plates. If the paper can tolerate a field of 5×107 volts/m, the minimum potential difference up to which the condenser can be charged is
0%
a) 2500 Volts
0%
b) 7500 Volts
0%
c)500 Volts
0%
d)10000 Volts
Explanation
E=V/d V=Ed=5×107×0.005×10-2V=2500 volts Answer:(a)
Q.37
The capacitance of a parallel plate capacitor is 2.5µF. When it is half filled with dielectric as shown in the figure, its capacitance becomes 5µF. The dielectric constant of the dielectric is
0%
a) 7.5
0%
b) 3
0%
c) 0.33
0%
d) 4
Explanation
C=2.5×10-6=εoA/d As shown in figure two capacitors are connected in parallel whose plate area is A/2 C1=εoA/2d and C1=KεoA/2d effective capacitance CR= Answer: (b)
Q.38
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
0%
a)4 component capacitor
0%
b) 12 component capacitor
0%
c)48 component capacitor
0%
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.39
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
0%
a) of capacitance 1µF
0%
b) of capacitance 2µF
0%
c)of capacitance 3µF
0%
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.40
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
0%
a) Ep is reduced by the presence of X if X is metallic
0%
b) Eq is increased by the presence of X if X is dielectric
0%
c)Eq is in opposite sense to Ep if X is a dielectric
0%
d)Eq is zero if X is metallic
Explanation
Answer:(c,d)
Q.41
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
0%
a) 300 V
0%
b) 500 V
0%
c) 425 V
0%
d) 400 V
Explanation
Answer: (c)
Q.42
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
0%
a)S1 and S2 are both closed
0%
b) S1 and S2 are both open
0%
c)S1 is open and S2 is closed
0%
d)S1 is closed and S2 is open
Explanation
Answer: (d)
Q.43
In the circuit shown in figure the key is first inserted between points 1 andThen keey is inserted between 1 andThe heat produced in 300Ω resistance is
0%
a) 10J
0%
b) 6.25 J
0%
c)3.75 J
0%
d)7.8 J
Explanation
When the key is inserted between points 1 and 2 the kinetic energy stored between the plates of capacitor U=(1/2) CV2=(1/2)× 500×10-6×(200)2U=10 JOn inserating the key between points 1 and 3 . The capacitor discharges through the resistance 300Ω and 500Ω. Since the ciurrent in both these resistance is the same, the heat is distributed in the ratio of resistors (H ∝ R). If H1 and H2 are the heat produced in 300Ω and 500Ω resistance then,H1 /H2=3/5, and H1 + H2=10J solving we get H1=3.75JAnswer: (c)
Q.44
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
0%
a) 0.6W
0%
b) W
0%
c)1.25W
0%
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.45
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
0%
a) 2.5
0%
b) 4
0%
c) 5
0%
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.46
The electric field midway between two charges 0.1µC and 0.4µC separated by a distance of 60cm is
0%
a) 5 ×103 N/C
0%
b) 9 ×104 N/C
0%
c)5 ×104 N/C
0%
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.47
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
0%
a)
0%
b)
0%
c)
0%
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.48
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
0%
a) 19V
0%
b) 6V
0%
c)36V
0%
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.49
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
0%
a) 100 V
0%
b) (200/3) V
0%
c)(100/3)V
0%
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.50
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
0%
a)
0%
b)
0%
c)
0%
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)
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