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Physics NEET MCQ
Quiz 11
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Q.1
A non conducting ring of radius 0.5 m carries a total charge of 1.1×10⁻¹⁰C distributed non-uniformly on its circumference producing an electric field E everywhere in space. The value of the integral ( l=0 being center of ring) in volts is [ IIT 1997]
0%
a) +2
0%
b) -1
0%
c)-2
0%
d)zero
Explanation
potential at centre Vo=kq/R and potential infinity=0Answer: (a)
Q.2
An oil drop carrying a charge of 4 electrons has mass of 3.2×10⁻¹⁷ kg. It is falling freely in air with terminal speed. The electric field required to make the drop move upwards with the same speed is ( g=10 ms-2)
0%
a) 2×103 V/m
0%
b) 1×103 V/m
0%
c)3×103 V/m
0%
d)8×103 V/m
Explanation
In first case mg=kv In second case, motion up with velocity v so resulatant force is kv upkv=Eq -kv Eq=2kvEq=2mgE=2mg /q On subtituting values we get E=103 V/m Answer: (b)
Q.3
A hollow metallic sphere of radius 10cm is given a charge of 3.2×10⁻⁹ C. The electric intensity at a point 4cm from the centre is
0%
b) 288 V
0%
c)2.88 V
0%
d)zero
0%
a) 9×10⁻⁹ V
Explanation
Answer: (d)
Q.4
How many electrons must be added to spherical conductor of radius 10cm to produce a field of 2×10⁻³ N/C just above the surface
0%
a) 1.39×104
0%
b) 1.6×105
0%
c)3×104
0%
d)9.1×104
Explanation
The entire charge on the spherical conductor behaves as if the entire charge is concentrated at the centre of the sphere, with respect to an external pointE=KQ/r2=K×ne/r2 Answer:(a)
Q.5
A ball of mass 100 gm and having a charge 4.9×10⁻⁵C is released from rest in vertical region where a horizontal electric field 2×104 N/C exists. The resultant force acting on the ball is
0%
a) 1.4 N in the direction of g
0%
b) 1.4 N making an angle of 45° with the horizontal
0%
c) 0.98 N in the direction of electric field
0%
d) 1.96 N along vertical
Explanation
angle between Electric and gravitational force is 90° Answer: (b)
Q.6
A positively charge oil droplet remains stationary in the electric field between two horizontal plates separated by a distance of 1cm. If the charge on the drop is 9.6×10⁻¹⁶ C and the mass of the droplet is 10-11g. The potential difference between the plates is
0%
a)2.0 V
0%
b) 3.1 V
0%
c)1.02 V
0%
d)10.2V
Explanation
Drop is stationary resultant force is zero mg=qE E=mg/q potential=E×d potential=(mg/q) ×d here d=1 cmon solving we get V=1.02 VAnswer: (c)
Q.7
The charge on the drop of water is 3×10⁻⁸C. If its surface potential is 500V, its radius must be equal to
0%
a)81 cm
0%
b) 54 cm
0%
c)27 cm
0%
d)108cm
Explanation
use formula v=kQ/r Answer: (b)
Q.8
An electric dipole consists of two opposite charge each of magnitude 1×10⁻⁶ C separated by a distance 2cm. The dipole is placed in electric field of 10×105 N/C. The maximum torque on the dipole is
0%
a) 0.2× 10⁻³ N-m
0%
b) 1.0×10⁻³ N-m
0%
c)2×10⁻³ N-m
0%
d)4×10⁻³ N-m
Explanation
Answer:(c)
Q.9
A pendulum bob of mass 80mg and carrying a charge of 2×10⁻⁸ C is at rest in a horizontal uniform electric field of 20,000 N/m. Find the tension in the thread of pendulum
0%
a) 8.8 × 10⁻² N
0%
b) 8.8 × 10⁻³ N
0%
c) 8.8 × 10⁻⁴ N
0%
d) 8.8 × 10⁻⁵ N
Explanation
Because of horizontal electric field and down direction gravitational field boob will make an angle with horizontal as shown in figure From diagram Tcosθ=mg and Tsinθ=q E solve above two equations to get T Answer: (c)
Q.10
An electron is situated 3×10⁻⁹ m from one α-particle and 4×10⁻⁹m from another α particle. The magnitude of force on electron, when two α particles are 5 ×10⁻⁹ m apart is
0%
a)5.64×10⁻¹¹ N
0%
b) 56.4×10⁻¹¹ N
0%
c)0.564×10⁻¹¹N
0%
d)564×10⁻¹¹ N
Explanation
locations of all particle must be as show in figureFind magnitude of force between the electron and alpha particle. Then calculate resultant forceAnswer: (a)
Q.11
Two metal pieces having a potential difference of 800V are 0.02 m apart horizontally. A particle of mass 1.96×10⁻¹⁵ kg is suspended in equilibrium between the plates. If e is elementary charge, then charge on the particle is
0%
a) e
0%
b) 3e
0%
c) 6e
0%
d) 8e
Explanation
Answer: (b)
Q.12
A non-conducting ring of radius 0.5m carries a total charge of 1.11×10⁻¹C distributed non-uniform;y on its circumference producing electric field everywhere in space. The value of the line integration of -E.dl form infinity to l=0 centre of ring is in volt is..
0%
a) +2
0%
b) -1
0%
c)-2
0%
d)zero
Explanation
Use formula for potential at centre V=KQ/RAnswer: (a)
Q.13
Two points P and Q are maintained at the potential of 10V and -4V, respectively. The work done in moving 100 electrons from P to Q is [ AIEEE 2009]
0%
a) 9.6×10⁻¹⁷ J
0%
b) -2.24×10⁻¹⁶ J
0%
c) 2.24×10⁻¹⁶J
0%
d) -9.6×10⁻¹⁷ J
Explanation
WPQ=q( VQ - VP) WPQ=(-100×1.6×10⁻¹⁹) ( -4-10) WPQ=+2.24×10⁻¹⁶J Answer: (c)
Q.14
Each capacitor in figure has capcity 5µF. The voltmeter, connected in parallel reads 100V. The charge on each plate of cpacirtor is
0%
a) 0.05 mC
0%
b) 0.5 mC
0%
c) 5 mC
0%
d) 1 mC
Explanation
As volymeter is paraellel to capacitor voltage across capacitor is 100 q=CV=5×10⁻⁶×100=0.5 mC Answer: (b)
Q.15
Four capacitors each of 25 μF are connected as shown in diagram. The DC voltmeter reads 200 volt. The charge on each plate of capacitor will be [AFMC1997]
0%
a) 5×10⁻²C
0%
b)2× 10⁻²C
0%
c)5× 10⁻³ C
0%
d) 2× 10⁻³
Explanation
From the figure two capacitors are connected in parallel and such two combinations are connected in series.Now Voltmeter is connected parallel to one of the parallel combination of capacitor. Hence Potential difference across each capacitor in a parallel combination will be same i.e 200 V.Now C=Q/V Therefore Q=CV Q=25×10⁻⁶ x 200 here Q=25×10⁻⁶ and V=200 V givenQ=5 × 10⁻³ C. Since Another combination is similar to voltmeter combination and all the capacitors equal. Charge across each capacitor will be same. Answer is (c)
Q.16
The mean free path of electrons in metal is 4×10⁻⁸ m. The electric field which can give on an average 2eV energy to an electron in the metal will be in units of V/m [ CBSE-PMT 2009]
0%
c) 5×107
0%
d) 8×107
0%
a) 5×10⁻¹¹
0%
b) 8×10⁻¹¹
Explanation
E=V/d E=2/ (4×10⁻⁸) E=5×107 Vm-1Answer: (c)
Q.17
The electric intensity due to a dipole of length 10 cm and having a charge of 500µC, at a point on the axis at a distance 20cm from one of the charge in air, is [ CBSE-PMT 2001]
0%
a) 6.25×107 N/C
0%
b) 9.28×107N/C
0%
c) 13.1×1011 N/C
0%
d) 20.5×107 N/C
Explanation
Asked to find electric field at a axial pointGiven length of dipole 2a=10cm ∴ a=5cm=5×10⁻² Dipole moment P=q(2a)=500×10⁻⁶ ×0.1=5×10⁻⁵ distance of point from mid-point of dipole, r= 5cm +20 cm = 25cm Electric field intensity due to dipole at equatorial point is Answer: (a)
Q.18
The mean free path of electrons in a metal is 4×10⁻⁸ m. The electric field which can give on an average 2eV energy to an electron in metal will be in units of V/m [ CBSE-PMT 2009]
0%
c)5×107
0%
d)8×107
0%
a) 5×10⁻¹¹
0%
b) 8×10⁻¹¹
Explanation
Energy =2eV Hnece potential fifference is 2V E=V/d E=2 / (4×10⁻⁸)W=5×107 Vm-1Answer: (c)
Q.19
An electric dipole , consisting of two opposite charges of 2×10⁻⁶C each separated by a distance 3cm is placed in an electric field of 2×105 N/C. Torque acting on the dipole is .. [ CBSE-PMT 1995]
0%
a)12×10⁻¹N-m
0%
b) 12×10⁻² N-m
0%
c)12×10⁻³ N-m
0%
d)12×10⁻⁴ N-m
Explanation
torque τ=qEdBy substituting values τ=2×10⁻⁶×2×105×3×10⁻² τ=12×10⁻³ N-mAnswer: (c)
Q.20
If the potential of a capacitor having capacity 6µF is increased from 10V to 20V, then increase in its energy will be [ CBSE-PMT 1995]
0%
a)4×10⁻⁴ J
0%
b) 6×10⁻⁴ J
0%
c)9×10⁻⁴
0%
d)12×10⁻⁶ J
Explanation
The increase in energy ΔU =½ C ( V22 -V12)=½ ×( 6×10⁻⁶)×(202 - 102 ) =9×10⁻⁴ J Answer: (c)
Q.21
Two metal plates having a potential difference of 800V are 2cm apart. It is found that a particle of mass 1.96×10⁻¹⁵ kg remain suspended in the region between the plates. The charge on the particle must be ( e=elementary charge) : [ PET 1999]
0%
a) 3e
0%
b) 4e
0%
c)6e
0%
d)8e
Explanation
V=800, d=2×10⁻² mE=V/d=800 / (2×10⁻² )=4×104 V/mFor equilibrium Eq=mg q=mg / E Answer:(a)
Q.22
An electric dipole consists of two opposite charges each of magnitude 1.0µC separated by distance of 2.0cm. The dipole is placed in an external field of 1.0×105N/C. The maximum torque on the dipole is [ CPMT 1990]
0%
a) 0.2×10⁻³ N-m
0%
b) 1.0×10⁻³ N-m
0%
c)2.0×10⁻³ N-m
0%
d)4.0×10⁻³ N-m
Explanation
τmax=pE=10-6 ×2×10⁻²×1.0×105=2×10⁻³ N-mAnswer: (c)
Q.23
Two point charges 100µC and 5µC are placed at point A and B respectively, with AB=40cm. The work done by the external force in displacing the charge 5µC from B to C where BC=30cm, angle ABC=π/2 and (1/4πεo )=9×109 [ MPPMT 1997]
0%
a) 9 J
0%
b) (81/20)J
0%
c) (9/25)J
0%
d) (-9/4)J
Explanation
Charge 5µC is moved in electric field of 100µC change Potential at B=VB Potential at C=VB WBC=(VC - VB)q WBC=(-2.5×10+ +1.8×106)(5×10⁻⁶=-9/4 JAnswer: (d)
Q.24
Capacitance (in F0 of a spherical conductor with radius 1 m is [ AIEEE 2002]
0%
a) 1.1×10⁻¹⁰
0%
b) 10-6
0%
c)9×10⁻⁹
0%
d)10-3
Explanation
fro an isolated sphere capacitance C C=4πεor=1/9×109=1.1×10⁻¹⁰ FAnswer: (a)
Q.25
A 4 µF conductor is charged to 400V and then its plates are joined through a resistance of 1kΩ. The heat produced in the resistance is [ CBSE 1994]
0%
a) 0.61 J
0%
b) 1.28J
0%
c) 0.64J
0%
d) 0.32J
Explanation
Energy stored in cpaitor=energy lost in resistance energy=½ CV2 energy=½ 4×10⁻⁶ ×(400)2 energy=0.32J Answer: (d)
Q.26
Two parallel plates, separated by a distance of 5mm, are kept at a potential difference of 50V. A particle of mass 10-15kg and charge 10-11 C enters in it with a velocity 107 m/s. The acceleration of the particle will be [ MPPMT 1997]
0%
a) 108 m/s²
0%
b) 5 ×105 m/s²
0%
c) 105 m/s²
0%
d) 2 × 103 m/s²
Explanation
E=V/d=50/5×10⁻³=104 V/m a=F/m=Eq/m=104 ×10⁻¹¹ / 10-15 a=108 m/s2 Answer: (a)
Q.27
Condenser A has a capacity of 15µF when it is filled with a medium of dielectric constantAnother condener B has a capacity 1µF with air between the plates. Both are charged separately by battery of 100V. After charging, both are connected in parallel with the and dielectric material being removed. The common potential now is [ MNR 1994]
0%
a)400 V
0%
b) 800 V
0%
c)1200 V
0%
d)1600 V
Explanation
Charge on the first capacitor filled with dielectric medium=15×10⁻⁶ ×100=15×10⁻⁴Charge on second capacitor=1×10⁻⁶ ×100=1×10⁻⁴ Total charge on both the capacitor=16×10⁻⁴ Capacity of first capacitor before filling dielectric=C/k=15µF/15=1µFCharge on second capacitor after connecting in parallel=Now V Q2 / C2V=8×10⁻⁴ / (1×10⁻⁶=800VAnswer: (b)
Q.28
Due to a charge inside a cube the electric field is Ex=600x1/2 , Ey=Ez=The charge inside the cube is
0%
a) 600µC
0%
b) 60µC
0%
c) 7µµC
0%
d) 6µµC
Explanation
Since Ey=Ez=0, therefore flux is linked only with face 1 and face 2 According to Gauss's theorem E.A=q/εo q=εo E.A Angle between area vector of face 1 and electric field=0 and angle between area vector of face 2 and electric field is 180° x for face 1=0.2m and distance for face 2=0.1m q=ε[ ExA cos0 + ExA cos180] q=ε[ Ex cos0 + Ex cos180]A q=8.86×10⁻¹² [ 600(0.2)1/2 - 600(0.1)1/2] (0.1)2 q=µµCAnswer: (c)
Q.29
Abullet of mass 2g is moving with a speed of 10 m/s. If the bullet has a charge of 2µC, through what potential must it be accelerated, starting from rest to acquire the same speed?
0%
a)5 kV
0%
b) 5 V
0%
c)50 V
0%
d)50 kV
Explanation
(1/2) mV2=qV (1/2) ×0.2×10⁻³×100=2×10⁻⁶×V V=50kVAnswer: (d)
Q.30
A particle of mass 2gm and charge 1µC is held at rest on a frictionless horizontal surface at a distance of 1 m from the fixed charge of 1 mC. If the particle is released it will be repelled. the speed of the particle when it is at a distance of 10 m from the fixed charge is
0%
a) 100 m/s
0%
b) 90 m/s
0%
c)60 m/s
0%
d)45 m/s
Explanation
Initail potential energy Final potential energy By conservation of energy ΔK=ΔK (1/2)mv2=(9-0.9) (1/2)×2×10⁻³v2=8.1 v2=8100 v=90 m/s Answer:(b)
Q.31
In a certain region of space there exists a uniform electric field of 2×103 k V/m. A rectangular coil of dimensions 10cm ×20cm is placed in XY plane. The electric flux through the coil is
0%
a)zero
0%
b) 4×105
0%
c)40
0%
d)4
Explanation
Pcoil is in XY plane therefore area vector is along z axis Φ=EAcosθΦ=(2×103×200×10⁻⁴Φ=40Answer: (c)
Q.32
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.33
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.34
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.35
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.36
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.37
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.38
Potential energy of two equal negative point charges 2µC each held apart in air is
0%
a)2J
0%
b) 2eV
0%
c)4J
0%
d)0.036J
Explanation
From formula V=Kq1q2 / rV=9×109×4×10⁻¹² / 1=0.036JAnswer: (d)
Q.39
The number of lines of force that radiate outwards from one coloumb of charge is
0%
a)9×109
0%
c)infinite
0%
d)1.13×1011
0%
b) 8.85×10⁻¹⁰
Explanation
from Gauss's theorem φ=∑ (q/εo φ=1 / (8.85×10⁻¹²=1.13×1011Answer: (d)
Q.40
The electric potential at the surface of an atomic nucleus ( Z=50) of radius 9.0×10⁻¹⁵ m is
0%
a) 80V
0%
b) 8 ×106V
0%
c)9 Volts
0%
d)9 ×105 volts
Explanation
given number of protons Z=50. Charge on nucleus Q=50×1.6×10⁻¹⁹ C use formula V=kQ/r Answer:(b)
Q.41
Two capacitor 2µF and 4µF are connected in series across 120V supply. The potential across the 2µF capacitor is
0%
a) 120 V
0%
b) 40 V
0%
c)80 V
0%
d)60 V
Explanation
Total capacitance of combination=(4/3) µF charge in both the capacitor is same Charge flowing, Q=CV=160 × 10⁻⁶Potential across 2µF capacitor=160/2=80VAnswer: (c)
Q.42
A pendulum bob of mass 80mg, carrying a charge of 2×10⁻⁸ C is at rest in a horizontal uniform electric field of 20000 V/m. The tension in the thread of the pendulum and the angle makes with the vertical are
0%
a) 9×10⁻⁴ N, tan⁻¹ (1/2)
0%
b) 9×10⁻⁴ N, tan⁻¹ (2)
0%
c) 9×10⁻³ N, tan⁻¹ (1.5)
0%
d) 8×10⁻³ N, tan⁻¹ (1/2)
Explanation
from figure Tcosθ=mg and Tsinθ=qE tanθ=qE/mg on solving θ=tan⁻¹(1/2) Tension in string T=qE/sinθ on solving T=8.94×10⁻⁴ N Answer: (a)
Q.43
A uniform electric field of 400 V/m is directed at 45° above the x-axis as shown in the figure. The potential difference VA - VB is given given by
0%
a)0
0%
b)4V
0%
c)6.4V
0%
d)2.8V
Explanation
We know that E=-(dV/dr)dV=-Edr E=Ecos45i + Esinj E=(400/√2)i + (400/√2)jPotential at along Y axis E electric field Ey=4(400/√2)jSimilarly Potential at B along x axis VB - Vo=(-400×3×10⁻²) /&adic2 Thus VA - VB=400/&radic2×10⁻²=2.83 V Answer:(d)
Q.44
Between the plates of a parallel plate capacitor there is a metallic plate whose thickness takes up 0.6 of the capacitor gap. When the plate is absent, the capacitor has a capacity C=20 nF. The capacitor is connected to a d.c voltage source V=100V. The metallic plate is slowly extracted from the gapThe mechanical work performed in the process of extraction is
0%
a) 0.35mJ
0%
b) 0.25mJ
0%
c)0.15mJ
0%
d)0.10mJ
Explanation
In absence of plate C1=20 nF thickness of metallic plate=0.6dIn presence of plate capacitance C2=Aεo / (d - 0.6d)=Aεo /0.4d=C1 /0.4=20nF/0.4=50 nF Potential V=100 VInitial energy Ui=(1/2)C1V2 Final energy Uf=(1/2)C2V2 Change in Energy=Uf - Ui=(1/2)V2 (C2V2 - C1) ΔU=(1/2)(100)2 (50-20)×10-9 ΔU=15×10⁻³=0.15 mJAnswer: (c)
Q.45
Two point charges is air at a distance of 20 cm from each other interact with a certain force. At what distance from each other should these charges be placed in oil of relative permitivity 5 to obtain the same force of interaction
0%
d)8.94×102 m
0%
a) 8.94×10⁻² m
0%
b) 0.894×10⁻² m
0%
c)89.4×10⁻² m
Explanation
distance d in air then equivalent distance with dielectric=d /√(k)here k is dielectric constant thus d'=20/√5 cm=8.93cm ≈8.94×10⁻² mAnswer: (a)
Q.46
A conductor gets a charged of 60pC when it is connected to a battery of emf 6mV. Then the capacitance of the conductor is ...
0%
d)10×10- F
0%
a)120×10⁻⁶ F
0%
b) 1.2×10⁻⁹ F
0%
c)10×10⁻⁹ F
Explanation
Q=60pF=60×10⁻¹²C Use formula C=Q/VAnswer: (c)
Q.47
Two capacitors 3pF and 6pF are connected in series and a potential difference of 5000V is applied across the combination. They are then disconnected and reconnected in parallel. The potential between the plates is
0%
a)2250 V
0%
b) 2222 V
0%
c)2.25V×106 V
0%
d)1.1 ×106V
Explanation
Capacitance of combination=2pF charge through the combination=C×V=2pF×5000=10-8 Potential across capacitor 6pF V1=Q/C=10-8 / 6×10⁻¹² V1=104/6 Potential across capacitor 3pF V1=Q/C=10-8 / 3×10⁻¹²=104/3 Now from the formula when connected parallel Answer: (b)
Q.48
An electric dipole is placed at an angle of 30° with an electric field intensity 2×105 N/C. It experiences a torque equal to 4 Nm. The charge on the dipole ,if the dipole length is 2 cm, is … [ NEET II : 2016]
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a) 5 mC
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b) 7 μC
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c) 8 mC
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d) 2 mC
Explanation
length of dipole be l = 2cm = 2×10⁻² m τ = PE sinθ τ = ql E sin sinθ 4 = q × 2 × 10⁻² × 2 × 105 sin 30° ∴ q = 2 mC Answer:(d)
0 h : 0 m : 1 s
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