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Physics NEET MCQ
Quiz 12
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Q.1
A galvanometer of 50 Ω resistance has 25 divisions. A current of 4×10⁻⁴ ampere gives a deflection of one per division. To convert this galvanometer into voltmeter having a range of 25 volts, it should be connected with a resistance of ... [ CBSE-PMT 2004]
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a) 2450 Ω in series
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b) 2500 Ω
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c) 245 Ω
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d) 2550 Ω
Explanation
Current capacity of galvanometer Ig=25 ×4×10⁻⁴=10-2 Resistance of galvanometer G=50 Ω Range of voltmeter=25 V From formula resistance to be connected in series R=(V/Ig - G R=2450 Ω Answer: (a)
Q.2
Two straight parallel wires both carrying 1 amp current in the same direction attracts each other with a force of 1×10⁻³ N. If both the currents are doubled, the force of attraction will be [ MPPMT 1994]
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a)1×10⁻³ N
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b) 2×10⁻³ N
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c)4×10⁻³ N
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d)0.25×10⁻³ N
Explanation
We know that Force per unit length F ∝ i1 I2 if distance is constantThus by making current double in each wire force will be four times of initialThus force will be 4×10⁻³ NAnswer: (c)
Q.3
A particle with 10-11 coulomb charge and 10-7 kg mass is moving with velocity of 108 m/s along the y-axis. A uniform static magnetic field B=0.5 Tesla is acting along x-direction. The force on the particle is [ MPPMT 1997]
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b) 5×103 N along k
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a)5×10⁻¹¹ N along i
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c)5×10⁻¹¹ N along -j
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d)5×10⁻⁴ N along -k
Explanation
F=q(v×B) F=10-11(108j×0.5i) F=5×10⁻⁴(-k)Answer: (d)
Q.4
A coil of 100 turns and area 5 square centimetre is placed in a magnetic field B=0.2T. The normal to the plane of the coil makes angle of 60° with the direction of the magnetic field. The magnetic flux linked with the coil is [ MPPMT 1997]
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c) 10-2 Wb
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d) 10-4 Wb
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a) 5×10⁻³ Wb
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b) 5×10⁻⁵ Wb
Explanation
Φ=NBAcosθ Φ=100×0.2×(5×10⁻⁴) ×cos60Φ=5×10⁻³ Wb Answer: (a)
Q.5
A proton enters a magnetic field of flux density 1.5 weber/metre2 with a velocity of 2×104 metre / sec at an angle of 30° with the field. The force on the proton will be [ MPPMT 1994]
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a) 2.4×10⁻¹² N
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b) 0.24×10⁻¹² N
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c) 24×10⁻¹² N
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d) 0.024×10⁻¹² N
Explanation
F=qvB sinθ F=1.6×10⁻¹⁶ ×2×104 ×1.5×sin30 F=2.4×10⁻¹² Answer: (a)
Q.6
In the hydrogen atom, the electron revolves in circular orbit of radius 0.53×10⁻¹⁰ metre and makes 6.6×1015 r.p.s. Then the magnetic dipole moment is approximately [ MPPMT 1999]
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a)10-29 amp×metre2
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b) 10-27 amp×metre2
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c)10-23 amp×metre2
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d)10-19 amp×metre2
Explanation
Current due to revolution=Q/T=e/Tperiodic time T=1/ number of revolution per second1 /T=number of revolution per second Area of circular loop=π r2A=0.88×10⁻²⁰Now magnetic dipole moment µ=iA Answer: (c)
Q.7
A conductor in the form of right angle ABC, with AB=3 cm and BC=4 cm, carries a current of 10A. There is uniform magnetic field of 5T, perpendicular to the plane of the conductor. The force on the conductor AC will be [ MPPMT 1997]
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a) 1.5 N
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b) 2.0 N
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c) 2.5 N
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d) 3.5 N
Explanation
Given AB=3 cm and BC=4 cm ΔABC is right angled triangle thus AC=5cm Force on AC=Bil=5×5×10⁻²=2.5 N Answer: (c)
Q.8
A current of 5 ampere is passed through a straight wire of length 6cms then the magnetic induction at a point 5cms from either end of the wire is
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a) 0.25 gauss
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b) 0.12 gauss
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c)0.15 gauss
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d)0.30 gauss
Explanation
sinθ=3/5 and r=4 cm=4×10⁻² From formula Answer:(c)
Q.9
A circular coil of 50 turns carries a current of 0.2A. The earth's magnetic field is 6×10⁻⁵ T. the net field at the centre of coil is zero. The radius of the coil is
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a)1.047 cm
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b) 10.47 cm
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c)1.047 m
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d)10.47 m
Explanation
use formula B=(µon i ) / 2R given value of B=6×10⁻⁵ TAnswer: (b)
Q.10
A circular coil of radius 4 cm having 5 turns carries a current of 2A. It is placed in uniform magnetic field of intensity 0.1 weber/mThe work done to rotate the coil from the equilibrium position through 180° is
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a) 0.1 J
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b) 0.2 J
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c)0.4 J
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d)0.8 J
Explanation
Magnetic moment of coil=NIAB=50×2×π(4×10⁻²)2)=0.5024Work=MB(1 - cos θ) Work=(0.5024) ( 0.1) ( 1- cos180) Work=0.1 JAnswer: (a)
Q.11
The plane of a rectangular loop of wire with sides 0.05 m and 0.08 m is parallel to a uniform magnetic field of induction 1.5×10⁻² tesla. A current of 10.0 amp. flows through the loop. If the side of length 0.08 m in normal and the side of length 0.05 m is parallel to the line of induction, then torque acting on it is
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a) 6000 Newton × metre
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b) Zero
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c) 1.2 × 10⁻² newton × metre
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d) 6 × 10⁻⁴ newton × metre
Explanation
Equal and opposite force will be on side of length 0.08m τ=BINA sin90° τ=1.5×10⁻²×10×1×0.08×0.05 τ=6×10⁻⁴ N-m Answer: (d)
Q.12
A beam of ion with velocity 2 ×105 m/s, enters perpendicularly to a uniform magnetic field of 4 ×10⁻² Tesla. If the specific charge of the ion is 5 ×10⁻⁷ C/ kg, the radius of the circular path will be ....
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a)0.1 m
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b) 0.16 m
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c)0.2 m
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d)0.25 m
Explanation
use r=mv/eB, given e/m=5 ×10⁻⁷ C/ kg,Answer: (a)
Q.13
A long starlight wire carrying a current of 30A is placed in an external uniform magnetic field of induction 4 × 10⁻⁴ T. The magnetic field is acting parallel to the direction of current. The magnitude of the resultant magnetic induction in tesla at a point 2cm away from wire is .....
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a)10-4
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b) 3 × 10⁻⁴
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c)5 × 10⁻⁴
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d)6 × 10⁻⁴
Explanation
MAgnetic induction because of wire B=µoI/ 2πr=3 × 10⁻⁴ T External magnetic field is perpendicular to magnetic field produced by wire Resultant magnetic field Answer: (c)
Q.14
A long solenoid has 1000 turns. When a current of 4A flows through it, the magnetic flux linked with each turn of the solenoid is 4×10⁻³Wb. The self inductance of the solenoid is
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a) 4H
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b) 3H
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c) 2H
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d) 1H
Explanation
Flux linked with each turn = 4×10⁻³Wb ∴ Total flux linked Φ = 1000(4×10⁻³) Wb = 4Wb Φ = LI 4= L (4) ⇒ L = 1 H Answer:(d)
Q.15
A 250-Turn rectangular coil of length 2.1 cm andwidth 1.25 cm carries a current of 85 μA and subjected to a magnetic field of strength 0.85 T.Work done for rotating the coil by 180° against the torque is [ NEET 207]
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a) 9.1 µJ
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b) 4.55 µJ
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c) 2.3 µJ
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d) 1.15 µJ
Explanation
W = MB [cosθ1 – cosθ2] As θ1 = 0 and θ2 = 180° W = 2MB Now M = nIA W= 2×250×85×10⁻⁶×( 2.1×1.25×10⁻⁴)×0.85= 9.48×10⁻⁶ W = 9.1 µJ Answer:(a)
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