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Physics NEET MCQ
Quiz 7
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Q.1
The force experienced by a pole of strength 100 A-m at a distance of 0.2m from a short magnet of length 5cm and pole strength of 200A-m on its axial line will be
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a) 2.5 × 10⁻² N
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b) 2.5 × 10⁻³ N
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c)5.0 × 10⁻² N
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d)5.0 × 10⁻³ N
Explanation
Using formula for axial point find magnetic field Force=F × m Answer:(a)
Q.2
A magnet 10 cm long has a pole strength of 12 A-m. Find the magnitude of magnetic field strength B at a point on its axis at a distance of 20 cm from it. What would be the value of B, if the point were to lie at the same distance on equatorial of magnet
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a) 3.4 × 10⁻⁵ T
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b) 1.4 × 10⁻⁵ T
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c) 1.7 × 10⁻⁵ T
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d) 0.85 × 10⁻⁵ T
Explanation
since d is not very large compared to l we should use following formula for B at axial point We know that Magnetic field at equatorial point is approximately half of axial point Answer: (c)
Q.3
A magnet of moment M is lying in a magnetic field of induction B. W1 is the work done in turning it from 0° to 60 ° and W2 is the work done in turning it from 30° to 90 ° . Then
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c) W2=2W1
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a)W₂=W₁
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b) W₂=W₁/2
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d) W₂=√3 W₁
Explanation
Answer: (d)
Q.4
A bar magnet of magnetic moment 4.0 A-m2 is free to rotate about a vertices axis through its centre. The magnet is released from rest from east-west position. Kinetic energy of the magnet in north-south position will be [ H=25µT)
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a) 10-2 J
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b) 10-4 J
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c)10-6 J
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d)0
Explanation
Answer: (b)
Q.5
The length of bar magnet is 10cm and its pole strength is 10-3 Weber. It is placed in a magnetic field 4 π×10-3 T . in the direction making an angle 30° with electric field direction. The value of torque acting on the magnet will be
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a) 2π × 10-7 N-m
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b) 2π × 10-5 N-m
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c)0.5 × 102 N-m
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d)none of these
Explanation
Answer:(a)
Q.6
A bar magnet with its poles 25 cm apart and pole strength 24.0 A-m rests with its centre on a frictionless pivot. A force F is applied on the magnet at a distance of 12cm from pivot so that it is held in equilibrium at an angle of 30° with respect to a magnetic field of induction 0.25T. The value of force F is
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a) 5.62 N
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b) 2.56 N
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c)6.52 N
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d)6.25 N
Explanation
USe formula F d=m×l×Bsinθ 0.12F=24×0.25×0.25sin30 F=0.75/0.12=6.25 NAnswer: (d)
Q.7
A current of 1 mp. is flowing in a coil of 10 turns and with radius 10 cm. Its magnetic moment will be
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a) 0.314 A-m2
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b) 3140 A-m2
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c)100 A-m2
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d)µ0 A-m2
Explanation
Magnetic moment=INA where A is the cross-sectional area of coil Answer:(a)
Q.8
If the radius of circular coil is doubled and the current flowing through in it is halved then new magnetic moment will be if its initial magnetic moment is 4 units
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a) 8 units
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b) 4 units
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c) 2 units
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d) zero
Explanation
Use M=NIA for both the cases Answer: (a)
Q.9
A short bar magnet is placed with its north pole pointing south. the neutral point is 10 cm away from the centre of magnet. If H=0.4 gauss, calculate magnetic moment of the magnet
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a)2 Am2
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b) 1 Am2
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c)0.1 Am2
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d)0.2 Am2
Explanation
Answer: (d)
Q.10
At any place on earth, the horizontal component of earth's magnetic field is √3 times the vertical component. The angle of dip at that place will be
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a) 60 °
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b) 45°
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c)90°
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d)30°
Explanation
Answer: (d)
Q.11
The tangent law applied when
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a) magnet is suspended in a uniform magnetic field
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b) horizontal component of earth's field is present
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c) there are two magnetic filed's
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d) there are two uniform magnetic field acting perpendicular to each other
Explanation
Answer: (d)
Q.12
The period of oscillation of a freely suspended bar magnet is 4 second. If it is cut into equal parts length wise then the time period of each part will be
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a)4 sec
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b) 2 sec
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c)0.5 sec
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d)0.25 sec
Explanation
Note Magnet is cut along the length, thus length has remained same. Moment of inertia reduced to half and magnetic moment is reduced to half will cancel each others effect Answer: (a)
Q.13
A thin magnetic needle oscillates in a horizontal plane with a period T. It is broken into n equal parts. the time period of each part will be
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a) T
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c)Tn2
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d)T/n
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b) T/n²
Explanation
Answer: (d)
Q.14
The time period of a small magnet in a horizontal plane is T. Another magnet B oscillates at the same place in a similar manner. The size of two magnet is same but the magnetic moment of B is four times that of A. The time period of B will be
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a) T/4
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b) T/2
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c)2T
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d)4T
Explanation
Answer:(b)
Q.15
A magnet makes 10 oscillations per minute at a place where the horizontal component of earth's magnetic field (H) is 0.33 oersted. The time period of the magnet at a place where the value of H is 0.62 oersted will be
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a) 4.38 sec
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b) 0.38 sec
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c) 2.38 sec
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d) 8.38 sec
Explanation
Answer: (a)
Q.16
A magnet makes 10 oscillations per minute at one place and takes 5 seconds to complete one oscillations at another place. Compare the values of horizontal components of earth's field at two places
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a)25/36
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b) 36/25
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c)5/6
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d)6/5
Explanation
Answer: (a)
Q.17
The time period of vibration of two magnets is some position is 3 sec. When polarity of weaker magnet is reversed, the combination makes 12 oscillations per minute. compare the magnetic moments of to magnets
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a) 4
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b) 17/8
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c)8/17
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d)1/4
Explanation
Answer: (b)
Q.18
The magnetic susceptibility of paramagnetic substance is 3 ×10-It is placed in a magnetizing field of 4 × 103 amp/m. The intensity of magnetization will be ..
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a) 3 × 108 A/am
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b) 12 × 108 A/m
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c)12 A/m
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d)24 A/m
Explanation
USe formula χ=M/H here χ is magnetic susceptibility.H is magnetizing fields and M is intensity of magnetization Answer:(c)
Q.19
A magnetizing field of 2 × 103 amp/m produces a magnetic flux density of 8π Tesla in an iron rod. the relative permeability of the rod will be
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a) 102
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b) 100
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c) 104
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d) 101
Explanation
B be the magnetic flux density H be magnetizing field µ be the permeability of iron . Then µ=B/H If µr represents relative permeability Then µr=µ / µo ∴ µ=B/(µoH) Answer: (c)
Q.20
The correct curve between χ and 1/T for paramagnetic is
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a)
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b)
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c)
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d)
Explanation
For magnetic material χ ∝ 1/T thus straight line curveAnswer: (a)
Q.21
The mass of an iron rod is 80 gm and its magnetic moment is 10 A-mIf the density of iron is 8 gm/cc then the value of intensity of magnetization will be
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a) 106 A/m
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b) 104 A/m
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c)102 A/m
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d)10 A/m
Explanation
Volume of rod=80/8=10 cc 10 cc=10-5 m3 Intensity of magnetization=10/10-5=106 A/mAnswer: (a)
Q.22
A bar magnet is cut into two equal halves by a plane parallel to the magnetic axis, of the following physical quantities the one which remains unchanged is
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a) pole strength
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b) magnetic moment
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c)intensity of magnetization
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d)moment of inertia
Explanation
Answer:(c)
Q.23
Two isolated point pole of strength 30 A m and 60 A m are placed at a distance of 0.3 m. The force of repulsion
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c) 2 × 105 N
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a) 2 × 10⁻³ N
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b) 2 × 10⁻⁴ N
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d) 2 × 10⁻⁵ N
Explanation
Answer: (a)
Q.24
Magnetic lines of force are
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a)continuous
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b) discontinuous
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c)some times continuous and some times discontinuous
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d)nothing can be said
Explanation
Answer: (a)
Q.25
Unit of magnetic moment are
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a) JT
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b) JT-1
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c)Am2
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d)Am-2
Explanation
The unit A m2 is a correct SI unit for magnetic moment, though, unless the concept of “current in a coil” needs to be emphasized in a particular context, it is perhaps better to stick to N m T-1 or JT-1 Explanation: magnitude of the magnetic moment is defined as the maximum torque experienced by the magnet when placed in unit external magnetic field. The magnitude and direction of the torque is given by the equation τ = P × B SI unit of torque = N.m, torque is measured in Joules (J) SI unit of Magnetic field = T Answer: (b)
Q.26
Units of pole strength of magnet are
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a) A m-1
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b) A m2
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c) A m-2
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d)A m
Explanation
Answer:(d)
Q.27
A tiny loop of current behaves as magnetic dipole
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a) true
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b) falls
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c) may be true or falls
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d) nothing can be said
Explanation
Answer: (a)
Q.28
Magnetic field intensity due to a dipole various as dn where n=...
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a)2
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b) -2
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c)3
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d)-3
Explanation
Let the solenoid consists of n turns per unit length. Let its length be 2l and radius a. We can evaluate the axial field at a point P, at a distance r from the centre O of the solenoid. To do this, consider a circular element of thickness dx of the solenoid at a distance x from its centre. It consists of n d x turns. Let I be the current in the solenoid. The magnetic field on the axis of a circular current loop at point P due to the circular element is The magnitude of the total field is obtained by summing over all the elements — in other words by integrating from x = – l to x = + l . Thus, Consider the far axial field of the solenoid, i.e., r >> a and r >> l . Then the denominator is approximated by Note that the magnitude of the magnetic moment of the solenoid is, m = n (2l)I(πa2) = (total number of turns × current × cross-sectional area). Thus, Answer: (d)
Q.29
The dimensions of magnetic permeability are
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a) [MLT-2A-2]
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b) [ML2T-2A-2]
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c) [ML2T-2A-1]
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d) [M-1LT-2A-1]
Explanation
Answer: (a)
Q.30
A circular coil of radius 4cm having 20 turns carries a current of 3A. It is placed in a magnetic field of intensity 0.5 weber/mThe magnetic dipole moment of the coil is
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a)0.15 amp ×m²
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b) 0.3 amp × m²
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c)0.45 amp × m²
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d)0.6 amp × m²
Explanation
USe formula magnetic moment=NIA here A is cross-sectional area of coilAnswer: (b)
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