Test: Physics Minor Mock Test- 4 (March 25) - NEET MCQ

According to recent researches the magnetic field of earth is considered due to a large bar magnet situated in earth's core.
It is considered that the north pole of this large magnet is situated at the geographical south of earth and vice versa and as the magnetic field due to a bar magnet is from north pole to south pole of the maget thus the earth's magnetic field is considered from geographical south to geographical north which are respectively north and south poles of the bar magnet.

μ = μ₀μ_r. μ is the permeability of medium, μ₀ is permeability of free space, and μr is relative permeability of the medium.

Work required to turn the dipole is given by
W=MB[cosθ_i- cosθ_f]
Here θi is the initial angle made by a dipole with a magnetic field and   is the final angle made by a dipole with a magnetic field.
magnetic moment is normal to the field direction.
so, θ_i=0 and θ_f=90
W = 1.5 × 0.22 [ cos0° - cos90° ]
W = 0.33J
 
magnetic moment is opposite to the field direction.
so, θ_i=0 and θ_f=180
now, W = 1.5 × 0.22 [ cos0° - cos180°]
= 0.33 [ 1 - (-1) ]
= 0.66 J

The earth's magnetic field is Be​ and its horizontal and vertical components are H_e​ and H_v​
cosθ= H_e​​/B_e​
∴cos60^o= (​0.26×10^−4​/ B_e )T
⇒B_e​=(​0.26×10⁻⁴)/ (½)​=0.52×10⁻⁴T=0.52G

Explanation:As magnetic monopole does not exist. If we split the bar magnet into two pieces  each part will have its own north and south pole.

All substances show some kind of magnetic behaviour. After all, they are made up of charged particles: electrons and protons. It is the way in which electron clouds arrange themselves in atoms and how groups of these atoms behave that determines the magnetic properties of the material. The atom (or group of atoms) in effect becomes a magnetic dipole or a mini bar magnet that can align according to the magnetic field applied. The net effect of all these dipoles determines the magnetic properties of the Magnetic Materials.

Correct Answer :- C

Explanation : Magnetic Moment = Magnetization × Volume

Side of cube = 20 cm

= 0.2 m

= 500 × 0.2 × 0.2 × 0.2

= 4Am²

 =1.5x0.22
 =15x22x10-3
 =330x10-3
Τ=0.33J
 

Explanation:for electromagnetic induction magnet and coil both may move but relative motion between must be present

Explanation : The coil of galvanometer is wound on a light metal frame. When the coil and frame rotate in the field of the permanent magnet, the eddy current set up in the frame oppose the motion so that the coil returns to zero quickly.

Explanation:A will become positive with respect to B because current  indused is in clockwise direction

e = NBAω = 20 x 3 x 10^-2 x (3.14 x 0.08 x 0.08) x 50 = 0.603C

average power loss due to joule heating

e = NBAω = NBA²πn = 8 x 0.5 x 0.09 x 2 x 3.14 x 60 = 136V

M is mutual inductance

l = 1 m

= 314V

Explanation:flux will be changed when current in C₂ is changing. Hence current will be induced in coil C₁.

Explanation:The standard voltage and frequency of alternating current supply in India is set to 220 V and 50Hz respectively by the government of India becausewhen power has to be transmitted from a power plant, the biggest challenge is to cut the transmission losses. For this purpose, the current value should be small and potential difference(Voltage) should be more. Also losses are minimal at 50 Hz/60 Hz frequency.

Resonance occurs when X_L = X_C and the imaginary part of the transfer function is zero. At resonance the impedance of the circuit is equal to the resistance value as Z = R.

I₀=250μA, v₀=3.6v .  v=1.6x10⁶ Hz
Here,
(Reactance of inductance) X_L=ωL
X_L=2πv X L
v₀/I₀=2πv x L
3.6/2.5x10^-4=2πx1.6x10^-6 x L
0.14x10^4-6=L
L=0.14x10^-2H
Now for v=16.0x10⁶Hz
XL=2πv X L
=2πx16x10⁶x14x10^-4
X_L=1407x10²Ω
Now,
v0=I0 x XL
3.6/1407x10²=I₀   [∵v0=kept constant.]
I₀=0.00256x10^-2
I₀=25.6μA

Since the current flowing through a series resonance circuit is the product of voltage divided by impedance, at resonance the impedance, Z is at its minimum value, ( =R ). Therefore, the circuit current at this frequency will be at its maximum value of V/R

Angular frequency of the ac signal w=2πν
∴ w=2π(50)=100π
Capacitive reactance X_c​=1/wC​
∴ X_c​=1/(100π×786×10⁻⁶)​=4Ω
Inductive reactance X_L​=w_L
∴ X_L​=100π×(25.48×10−3)=8Ω
Impedance of the circuit Z=√[R²+(X_L​−X_c​)²​]
∴ Z=√[3²+(8−4)²​]=5Ω
 Phase difference ϕ=tan⁻¹[(XL​−Xc​​)R]
Or ϕ=tan⁻¹((8−4​)/3)=tan⁻¹(4/3​)
⟹ ϕ=53.13o
Power factor cosϕ=cos53.13o=0.6
Power dissipated in the circuit
P=I_v²R
Now, Iv=I₀/√2=E₀/√2Z=283/(1.414×5)=40A
∴P=I_v²R=(40)²×3=4800 watt

Radio waves have frequencies as high as 300 gigahertz (GHz) to as low as 30 hertz (Hz). Like all other electromagnetic waves, radio waves travel at the speed of light in vacuum (and close to the speed of light in the Earth's atmosphere, which acts as the transmission media for the vast majority of terrestrial use).
 

The electromagnetic wave travels in vacuum along z-direction. Then, the electric field vector and magnetic field vector will be in the XY plane.
Frequency of the wave = 30 MHz
Converting into Hz, v = 30× 10⁶ Hz.
speed of light, c = 3 × 10⁸ m/s
Wavelength of the wave can be calculated using the following relation:
λ = c/v
⇒ λ = 3 × 10⁸/30 × 10⁶
⇒λ = 10 m

The frequency of radiation arising from the two close energy levels in hydrogen known as Lamb shift i.e. 1057 MHz is radio waves as it belongs to the short wavelength end of the electromagnetic spectrum.

The frequency of radiation arising from the two close energy levels in hydrogen known as Lamb shift i.e. 1057 MHz is radio waves as it belongs to the short wavelength end of the electromagnetic spectrum. this wavelength correspond to microwave region of electromagnetic waves.

Long distance radio broadcasts use short-wave bands because they can be reflected by the ionosphere of the earth's atmosphere and thus can be send to longer distances.