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Physics NEET MCQ
Quiz 1
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Q.1
Which of the following represents Biot-Savart's law correctly? [ CBSE 1996]
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a)
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b)
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c)
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d)
Explanation
Answer: (c)
Q.2
An ideal transformer has primary power of 10kW. The secondary current when the transformer is on load is 25A. If the primary and secondary turns are in ratio is 8:1 then the potential difference applied in the primary coil is ...[ AFMC 1998]
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a)
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b)
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c)
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d)
Explanation
For ideal transformer power in primary coil=power in secondary coil V1×I1=V2×I2 10000=V2×25V2=10000 / 25 Now V1 / V2=n1×n2 V1=V2( n1 / n2 Answer: (c)
Q.3
One tesla is equal to [ AFMC 1998]
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a) 104 gauss
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b) 10-8 gauss
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c)10-7 gauss
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d)10-4 gauss
Explanation
Answer: (a)
Q.4
Cyclotron is a device to .. [ AFMC 1998]
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a) accelerate electron
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b) accelerate proton
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c)measure voltage
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d)measure charge
Explanation
Answer:(b)
Q.5
A current carrying wire in the neighborhood produces ...[AFMC 1999]
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a)electric and magnetic fields
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b) magnetic field only
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c)no field
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d)electric field
Explanation
Answer: (a)
Q.6
Two parallel wire in free space are 10 cm apart and each carries a current of 10A in the same direction. The force exerted by one wire on other per metre of length of the wire is..[ AFMC 1999]
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a) 2×10⁻⁶N
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b) 2×10⁻⁴N
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c)2×10⁻³N
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d)2×10⁻²N
Explanation
Force of interaction between two current carrying conductor is given by formula value of µ0=4π10-7Substituting value of current and perpendicular distance 'r' in above equation we getAnswer: (c)
Q.7
A long hollow copper pipe carries a current, then magnetic field produced is ..... [ AFMC 1999]
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a) both inside and outside the pipe
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b) neither inside nor outside the pipe
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c)outside the pipe only
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d)inside the pipe only
Explanation
As shown in diagram direct of current through copper pipe's opposite faces is same hence magnetic field produced at centre will be zero.whereas magnetic field will be possible only out side the copper pipe carrying current Answer:(c)
Q.8
A galvanometer of 100Ω resistance gives full scale deflection with 0.01A current. How much resistance should be connected convert it into an ammeter of range 10A? [ AFMC 2002]
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a)0.1 Ω in parallel
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b) 0.1Ω in series
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c)0.2Ω in parallel
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d)0.2 Ω in series
Explanation
Ig=0.01A Range I=10A, Resistance of Galvanometer G=100Ó Formula for shunt S=Ig G /( I - Ig) S=(0.001)(100) / ( 10-.01)=0.1 Ω To convert Galvanometer to Ammeter we have connect resistance parallel to galvanometerAnswer: (a)
Q.9
To convert a galvanometer into voltmeter we must connect a ...[ AFMC 2002]
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a) high resistance in series
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b) low resistance in parallel
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c)high resistance in parallel
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d)low resistance in series
Explanation
Voltmeter is connected parallel to the resistance to measure potential drop , hence minimum value of current should passed through galvanometer thus we must connect high value of resistance across galvanometer, whose value is given by R=(V/Ig) - GHere V is the voltage to be measured, Ig and G is the resistance of galvanometer Answer:(a)
Q.10
Two wires carry current in different direction they will ...[ AFMC 2003]
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a) attract each other
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b) repel each other
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c) create gravitational field
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d) any of the above
Explanation
Answer: (b)
Q.11
To convert a galvanometer into an ammeter, one needs to connect a [ CBSE-PMT 1992]
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a)low resistance in parallel
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b) high resistance in parallel
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c)low resistance in series
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d)high resistance in series
Explanation
Answer: (a)
Q.12
Energy in a current carrying coil is stored in the form of ... [ CBSE-PMT 1989]
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a) electric field
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b) magnetic field
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c)dielectric strength
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d)heat
Explanation
Answer: (b)
Q.13
A coil carrying electric current is placed in uniform magnetic field, then... [ CBSE-PMT 1993]
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a)torque is formed
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b) e.m.f. is induced
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c)both (a) and (b) are correct
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d)none of the above
Explanation
A current carrying coil has magnetic dipole moment. Hence, a torque Pm ×B acts on it in magnetic field. Answer:(a)
Q.14
A positively charged particle moving due east enters a region of uniform magnetic field directed vertically upwards. The particle will.. [ CBSE-PMT 1997]
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a) continue to move due east
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b) move in a circular orbit with its speed unchanged
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c) move in a circular orbit with its speed increases
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d) get deflected vertically upwards
Explanation
in perpendicular magnetic field, the path of a charged particle is a circle, and the magnetic field does not cause any change in energy. Answer: (b)
Q.15
A straight wire of diameter 0.5mm carrying a current of 1A is replaced by another wire of 1mm diameter carrying same current. The strength of magnetic field far away is ... [ CBSE-PMT 1999]
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a)twice the earlier value
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b) same as earlier value
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c)one-half of the earlier value
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d)one-quarter of the earlier value
Explanation
magnetic field due to current carrying element is given by B=µoi / 2πr distance is far away thus diameter of wire don't affect the distance rcurrent is same therefore value of magnetic field will be same Answer: (b)
Q.16
If a long hollow copper pipe carries a current then magnetic field is produced.. [ CBSE-PMT 1999]
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a) inside the pipe only
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b) outside the pipe only
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c)both inside and outside the pipe
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d)no where
Explanation
Inside a hollow pipe carrying current, the magnetic field is zero, according to Ampere's law But for external points, the current behaves as if it was concentrated at the axis only, so outside the pipe magnetic field will be produced Answer: (b)
Q.17
A galvanometer can be converted into a voltmeter by connecting .. [ CBSE-PMT 2002]
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a)A high resistance in parallel
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b) A low resistance in series
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c)A high resistance in series
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d)A low resistance in parallel
Explanation
Answer:(c)
Q.18
A galvanometer acting like a voltmeter will have .. [ CBSE-PMT 2004]
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a) a low resistance in series with its coil
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b) a high resistance in parallel with its coil
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c) high resistance in series with its coil
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d) low resistance in parallel with its coil
Explanation
Answer: (c)
Q.19
A 10 eV electron is circulating in a plane at right angle to uniform field of magnetic induction 10-4 Wb/mThe orbital radius of the electron is .. [ CBSE-PMT 1996]
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a)12 cm
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b) 16 cm
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c)11 cm
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d)18 cm
Explanation
We know that for circular motion in magnetic fieldAnswer: (c)
Q.20
When a charged particle moving with velocity v is subjected to a magnetic field of induction B, the force on it is zero. This implies that... [ CBSE-=PMT 2006]
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a) angle between v and B can have any value other than 90°
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b) angle between v and B can have any value other than zero and 180°
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c)angle between v and B is either zero or 180°
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d)angle between v and B is necessarily 90°
Explanation
Magnetic force on moving charged particle is F=q(v × B )For F=0 angle between v and B must be either 0° or 180° Answer: (c)
Q.21
A particle of mass m, charge Q and kinetic energy T enters a transverse uniform magnetic field of induction B. After 3 seconds, the kinetic energy of the particle will be... [CBSE-PMT 2008]
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a)3T
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b)2T
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c)T
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d)4T
Explanation
When a charged particle enters a transverse magnetic field its follows circular path kinetic energy remains constant Answer:(c)
Q.22
An electron moves in circular orbit with uniform speed v. It produces a magnetic field B at the centre of the circle. the radius of the circle is proportional to ... [ CBSE-PMT 2005]
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a)
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b)
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c)
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d)
Explanation
electron is revolving in circular path of radius r with velocity v, periodic time is v/2πr charge on electron is e, now by formula i= q/T current due to motion of electron = ev/2π Magnetic filed at centre of circular loop is given by Answer: (c)
Q.23
A beam of electron passes un deflected through mutually perpendicular electric field and magnetic field. If the electric field is switched off, and the same magnetic field is maintained, the electrons move ... [ CBSE-PMT 2007]
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a)in a circular orbit
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b) along a parabolic path
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c)along a straight line
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d)in an elliptical orbit
Explanation
After switching off electric field magnetic field is perpendicular to velocity hence electron will follow circular pathAnswer: (a)
Q.24
A beam of electrons is moving with constant velocity in a region having simultaneous perpendicular electric and magnetic fields of strength 20V/m and 0.5T respectively at right angles to the direction of motion of the electrons. Then the velocity of electrons must be .. [ CBSE-PMT 1996]
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a) 8 m/s
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b) 20m/s
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c)40 m/s
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d)(1/40) m/s
Explanation
qvB=qEv=E/B=20/0.5=40 m/sAnswer: (c)
Q.25
A particle having charge q moves with velocity v through a region in which both an electric field E and magnetic field B are present. The force on the particle is ... [ CBSE-PMT 2002]
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a) qE + q( v × B )
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b) qE ⋅ q( B × v )
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c)qv + q( E × B )
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d)qE × q( B × v )
Explanation
Answer:(a)
Q.26
A square current carrying loop is suspended in a uniform magnetic field acting in the plane of the loop. if the force on one arm of the loop is F, the net force on the remaining three arms of the loop is .. [ CBSE-PMT 2010]
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a)3F
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b) -F
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c)-3F
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d)F
Explanation
The force on arms parallel to magnetic field will be zero. Hence force on remaining arm will be -FAnswer: (b)
Q.27
Charge q is uniformly spread on the conductor radius R. The ring rotates about its axis with uniform frequency f Hz. the magnitude of magnetic field induction at the centre of the ring is .... [ CBSE-PMT 2011]
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a)
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b)
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c)
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d)
Explanation
Current I=q/T=qf Magnetic field at the centre of the ring is Answer: (a)
Q.28
A uniform electric field and uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along the direction of fields, then the electron ... [ CBSE-PMT 2011]
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a) Will turn towards right of direction of motion
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b) Speed will decrease
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c)Speed will increase
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d)Will turn towards left direction of motion
Explanation
v and B are in same direction so magnetic force on electron become zero, only electric force acts. But force on electron due to electric field is opposite to the direction of velocity Answer:(b)
Q.29
A charged particle having charge q is moving in a circle of radius R with uniform speed v. The associated magnetic moment µ is given by.. [ CBSE-PMT 2007]
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a) qvR2
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b) qvR2 / 2
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c) qvR
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d) qvR/2
Explanation
periodic time T=2πR /v Thus frequency f=v/ 2πR Current I=qf I=qv/ 2πR Magnetic moment µ=IA A=πR2 µ=qvR / 2 Answer: (d)
Q.30
A galvanometer having a coil resistance of 60Ω shows full scale deflection when current of 1.0 amp passes through it. It can be converted into an ammeter to read current up to 5.0amp by.. [ CBSE-PMT 2009]
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a)putting in series a resistance of 15Ω
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b) putting in series a resistance of 240Ω
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c)putting in parallel a resistance of 15Ω
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d)putting in parallel a resistance of 240Ω
Explanation
To convert galvanometer in to ammeter resistance should be connected in parallelShunt resistance formula Here Current capacity of galvanometer, Ig=1 AmpCurrent to be measure I=5 AmpResistance of Galvanometer G=60ΩOn substituting values in above formula we get S=15Ω to be connected in parallelAnswer: (c)
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