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Physics NEET MCQ
Quiz 10
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Q.1
If θ1 and θ2 be the apparent angles of dip observed in two vertical planes angles to each other then the true angle of dip θ is given by .. [ NEET 2017]
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a) cot2 θ= cot2 θ1 + cot2 θ2
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b) tan2 θ= tan2 θ1 + tan2 θ2
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c) cot2 θ= cot2 θ1 - cot2 θ2
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d) tan2 θ= tan2 θ1 - tan2 θ2
Explanation
Let one plane makes angle of δ1 with magnetic meridian as shown in figure B’ is the magnetic field in plane. Now vertical component of both magnetic meridian and plane is B’ . B’sinθ1= Bsinθ … (i) Now Bcosθcosδ1= B’cos θ1 …(ii) Substituting value of B’ from (i) in (ii) Similarly for other plane which is perpendicular Since δ1 + δ2 = π/2 cos δ2 = cos (π/2 – δ1) = sin δ1 Squaring (iii) and (iv) and adding we get Answer:(a)
Q.2
A bar magnet having a magnetic moment of 2×104 JT-1 is free to rotate in a horizontal plane. A horizontal magnetic field B=6×10⁻¹ T exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction 60° from the field is .. [ CBSE-PMT 200]
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a) 12J
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b) 6J
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c)2J
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d)0.6J
Explanation
Work done=MB(cos θ1 - cos θ2W=MB( cos 0 -cos60°)W=MB/2 On substituting values of B and M we getW=6 JAnswer: (b)
Q.3
A short magnet of moment 6.75 Am2 produces a neutral point on its axis. If the horizontal component of earth's magnetic field is 5×10⁻⁵ wb/m2, then distance of the neutral point should be [ SCRA 1994]
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a) 10 cm
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b) 20 cm
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c)30 cm
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d)40 cm
Explanation
At null point magnetic field due to bar magnet=Earth's horizontal component 03 m=30 cm Answer:(c)
Q.4
A bar magnet has a magnetic moment equal to 5×10⁻⁵ weber×metre. It is suspended in a magnetic field which has a magnetic induction (B) equal to 8π×10-4 tesla. The magnet vibrates with period of vibration equal to 15 seconds. the moment of inertia of the magnet is [ MPPMT 1993]
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a) 22.5 kg×m²
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b) 11.25 kg×m²
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c) 5.62 kg×m²
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d) 7.16×10⁻⁷ kg×m²
Explanation
From the formula for time period solve equation of I Answer: (d)
Q.5
The value of horizontal component of earth's magnetic field at a place is 0.367 × 10⁻⁴ weber per metreIf at this place, the angle of dip is 60°, the value of vertical component of earth's magnetic field in weber per metre2 is about [ MP MPT 1985]
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a)0.12 × 10⁻⁴
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b) 0.24 × 10⁻⁴
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c)0.40 × 10⁻⁴
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d)0.62 × 10⁻⁴
Explanation
Answer: (d)
Q.6
A bar magnet of magnetic moment 104 J/T is free to rotate in a horizontal plane. the work done in rotating the magnet slowly from direction parallel to a horizontal magnetic field of 4 ×10⁻⁴ T to a direction 60° from the field will be [ MPPMT 1995]
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a)0.2J
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b) 2.0 J
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c)4.18 J
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d)2 × 102 J
Explanation
W=MB ( 1 - cosθ) Answer: (b)
Q.7
The value of horizontal component of earth's magnetic field at a place is 0.36 × 10⁻⁴ weber per metreIf at that place, the angle of dip is 60°, the value of vertical component of earth's magnetic field in weber per metre2 is about
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a) 0.12 ×10⁻⁴ Wb/m²
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b) 0.24 ×10⁻⁴ Wb/m²
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c)0.4 ×10⁻⁴ Wb/m²
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d)0.62 ×10⁻⁴ Wb/m²
Explanation
Answer:(d)
Q.8
A bar magnet 8cm long is placed in the magnetic meridian with the N-pole pointing towards geographical north. Two neutral points separated by a distance of 6cms are obtained on the equatorial axis of the magnet If H=3.2 ×10⁻⁵ tesla then the pole strength of the magnet is
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a) 5 ab-amp × cm
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b) 10 ab-amp × cm
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c)2.5 ab-amp × cm
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d)20 ab-amp × cm
Explanation
The magnet is in tanB position. LEt P and Q be the neutral points, thenH=m×8 / (5)3 m=5 ab-amp × cm Answer:(a)
Q.9
A magnet 10 cm long and having a pole strength 2 amp-m is deflected through 30° from the magnetic meridian is 0.32 ×10⁻⁴ Tesla. The value of deflecting couple is
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a)16 × 10⁻⁷ N-m
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b) 32 × 10⁻⁷ N-m
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c)45 × 10⁻⁷ N-m
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d)64 × 10⁻⁷ N-m
Explanation
Use formula τ=MH sin θ M=m × l τ=m × l ×H sin θ Answer: (b)
Q.10
The angle of dip at place i 40.6° and the intensity of the vertical component of the earth's magnetic field V=6 × 10⁻⁵ tesla. The total intensity of earth's magnetic field at this place is
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a) 7 × 10⁻⁵ tesla
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b) 6 × 10⁻⁵ tesla
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c) 5 × 10⁻⁵ tesla
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d) 9.2 × 10⁻⁵ tesla
Explanation
Answer: (d)
Q.11
A magnet having a magnetic moment of 1.0 × 104 J/T is free to rotate in horizontal plane where a magnetic field 4 × 10⁻⁵ T exists. Find the work done in rotating the magnet slowly done in rotating the magnet slowly from a direction parallel to the field to a direction 60° from the field
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a) 0.4 J
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b) 2 J
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c) 0.2 J
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d) 1 J
Explanation
Answer: (c)
Q.12
A magnetic needle of magnetic moment 60 amp-m2 experience a torque of 1.2 ×10⁻³ N-m directed in geographical north. If the horizontal intensity of earth's magnetic field at that place is 40 Wb/m2, then the angle of direction will be
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a)30°
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b) 45°
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c)60°
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d)90°
Explanation
Answer: (a)
Q.13
The horizontal component of earth's magnetic field at any place is 0.36×10⁻⁴ Weber/mIf the angle of dip at that place is 60° then the value of vertical component of earth's magnetic field will be ( in Wb/m2)
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a) 0.12 × 10⁻⁴
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b) 0.24 ×10⁻⁴
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c)0.40 ×10⁻⁴
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d)0.62 ×10⁻⁴
Explanation
Answer:(d)
Q.14
The radius of the coil of tangent galvanometer is 16 cm. How many turns of the wire should be used if a current of 40 mA is to be produced a deflection of 45°, given horizontal component of earth's field is 0.36 ×10⁻⁴ T.
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a) 458
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b) 229
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c) 200
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d) 115
Explanation
Use formula Answer: (b)
Q.15
A short bar magnet place with axis at 30°, with uniform external magnetic field of 0.25 T experiences a torque of 4.5 × 10⁻² N-m. Magnetic moment of the magnet is ...
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a) 0.36 J/T
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b) 0.72 J/T
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c)0.18 J/T
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d)zero
Explanation
Answer: (a)
Q.16
The magnetic moment of a magnet is 0.1 amp × mIt is suspended in a magnetic field of intensity 3×10⁻⁴ weber/mThe couple acting upon it when deflected by 30° from the magnetic field is
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a) 1 × 100-5 Nm
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b) 1.5 × 100-5 Nm
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c)2 × 100-5 Nm
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d)2.5 × 100-5 Nm
Explanation
Answer: (b)
Q.17
Force acting on a magnetic pole of 7.5 ×10⁻² A-m is 1.5 N. Magnetic field at the point is
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c)112.5 T
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d)2.0 T
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a) 20 Wb/m²
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b) 50 Wb/m²
Explanation
Answer:(a)
Q.18
A bar magnet of magnetic moment 104 J/T is free to rotate in a horizontal plane. the work done in rotating the magnet slowly from a direction parallel to a horizontal magnetic field of 4×10⁻⁵ T to a direction 60° from the field will be [ MPPMT 1995]
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a) 0.2 J
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b) 2.0 J
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c)4.18 J
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d)2×102 J
Explanation
W=BM ( 1 - cosθ) W=104×4×10⁻⁵ ( 1 -cosθ) W=0.2JAnswer: (a)
Q.19
If a dip needle is suspended at an angle of 30° to the magnetic meridian, it makes an angle of 45° with the horizontal. The real dip is [ CPMT 1998]
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a) tan⁻¹ (√3/2)
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b) tan⁻¹ (√3)
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c)tan⁻¹ √(3/2)
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d)tan⁻¹ (2 /√3)
Explanation
Let Φ be the actual deep in magnetic meridian tanΦ=V / H V=H tanΦ --(1) Vertical components are same for both the plane but horizontal component will be different tan45=V/ H' V=H'But plane of needle makes an angle with magnetic meridian of 30° thus Horizontal component in plane have component H'=H cos30 in magnetic meridian V=H cos30 --(2) from equation (1) and (2) we get H tanΦ=H cos30 tanΦ=√3 / 2 Φ=tan⁻¹( √3 / 2) General equation can be derived as follows: Φ = is real deep in magnetic meridian= to calculate θ = is apparent or deep in geographical meridian δ = angle of declination or angle between magnetic and geographical meridian For magentic and geographical meridian verticle componant of B is same But hoorizonatl componant of B are different Thus for real deep δ V/H = tanδ Now Horizontal componant of B in geographical meridian is H' and angle betwwen magentic and geographical meridian is δ thus H' = H cosδ ---- (1) Apperent angle of deep in geographical meridian = θ V/H' = tan θ or H' = V/tanθ ---- (2) From (1) and (2) Or tan(real deep) = tan(apperent deep) cos(angle of declination) Answer:(a)
Q.20
A uniform magnetic needle is suspended from its centre by a thread. Its upper end is now loaded with a mass of 50 mg when the needle becomes horizontal. If the strength of each pole is 98.1 ab-amp × cm and g is 981 cm/sec2, then the vertical componenet of the earth's magnetic induction is
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a)0.50 gauss
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b) 0.25 gauss
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c)0.05 gauss
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d)0.005 gauss
Explanation
At equilibrium, algebric sum of moment of forece about centreO must be zeromgl- MBl - MBl=0B=mg/2M B=(50 × 10⁻³ × 981) / (2×98.1)B=0.25 gaussAnswer: (b)
Q.21
In hydrogen atom, an electron revolves with frequency of 6.8 × 109 megahertz in an orbit of diameter 1.06 Å. The equivalent magnetic moment is
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a) 7.9 × 10⁻²⁴ Am2
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b) 9.7 × 10⁻²⁴ Am2
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c)9.7 × 1024 Am2
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d)7.9 × 1024 Am2
Explanation
Current I=charge × frequency I=1.6×10⁻¹⁹ ×6.8 × 1015 I=10.88 ×10⁻⁴ amp Magnetic moment=IA Magnetic moment=10.88 ×10⁻⁴ × π × (0.59 × 10⁻¹⁰ Magnetic moment=9.7 × 10⁻²⁴ Am2 Answer:(b)
Q.22
There are 2.0 × 1024 molecular dipoles in a paramagnetic salt. Each has dipole moment 1.5 ×10⁻²³ A-mFind the maximum ( saturation) magnetization in the specimen [ GUJCET 2008]
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a)30 A-m2
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b) 20 A-m2
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c)50 A-m2
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d)200 A-m2
Explanation
Maximum dipole moment per unit volume=NM/V MAximum dipole moment per unit volume=nM=2.0×1024×1.5×10⁻²Answer: (a)
Q.23
A solenoid is 1.5 m long and its inner diameter is 4.0cm. It has 3 layers of windings of 1000 turns each and carries a current of 2.0 A. The magnetic flux for a cross-section of the solenoid is nearly.....
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a) 4.1 ×10⁻⁵ Wb
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b) 5.2 ×10⁻⁵ Wb
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c)6.31 ×10⁻³ Wb
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d)2.5 ×10⁻⁷ Wb
Explanation
Number of turns per unit length=(3 × 100) / 1.5=2000 B=µo nI B=4π × 10-7 × 2000 × 2=16π ×1-4and Φ=NBA Area of coil=πr2=π(2×10⁻²)2=4π×10-4therefore magnetic flux Answer: (c)
Q.24
Two short bar magnets of length 1 cm each have magnetic moments 1.20 Am2 and 1.00 Am2 respectively. They are placed on a horizontal table parallel to each other with their N poles pointing towards the South. They have a common magnetic equator and are separated by a distance of 20.0 cm. The value of the resultant horizontal magnetic induction at the mid - point O of the line joining their centre is close to (Horizontal component of earth′s magnetic induction is 3.6 × 10⁻⁵ Wb / m2)
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a) 3.6 × 10⁻⁵ Wb/m²
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b) 2.56 × 10⁻⁴ Wb/m²
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c) 3.50 × 10⁻⁴ Wb/m²
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d) 5.80 × 10⁻⁴ Wb/m²
Explanation
Both the magnetic will produce magnetic filed in same direction from S to N can be calculated using formula Field produced by both the magnet Earth magnetic filed is also from S to N Thus total = 2.2×10⁻⁴ + 0.36×10⁻⁴ = 2.56×10⁻⁴ Wb/m2 Answer:(b)
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