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


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

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Test: Physics Minor Mock Test- 4 (March 25) - Question 1

Which of the following statements about earth's magnetism is correct

Detailed Solution for Test: Physics Minor Mock Test- 4 (March 25) - Question 1

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.

Test: Physics Minor Mock Test- 4 (March 25) - Question 2

magnetic permeability of the substance μ is

Detailed Solution for Test: Physics Minor Mock Test- 4 (March 25) - Question 2

μ = μ0μr. μ is the permeability of medium, μ0 is permeability of free space, and μr is relative permeability of the medium.

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Test: Physics Minor Mock Test- 4 (March 25) - Question 3

A bar magnet of magnetic moment 1.5 J/T lies aligned with the direction of a uniform magnetic field of 0.22 T. What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment opposite to the field direction?

Detailed Solution for Test: Physics Minor Mock Test- 4 (March 25) - Question 3

Work required to turn the dipole is given by
W=MB[cosθi- cosθf]
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, θi=0 and θf=90
W = 1.5 × 0.22 [ cos0° - cos90° ]
W = 0.33J
 
magnetic moment is opposite to the field direction.
so, θi=0 and θf=180
now, W = 1.5 × 0.22 [ cos0° - cos180°]
= 0.33 [ 1 - (-1) ]
= 0.66 J

Test: Physics Minor Mock Test- 4 (March 25) - Question 4

In the magnetic meridian of a certain place, the horizontal component of the earth’s magnetic field is 0.26G and the dip angle is 60o. What is the magnetic field of the earth at this location

Detailed Solution for Test: Physics Minor Mock Test- 4 (March 25) - Question 4

The earth's magnetic field is Be​ and its horizontal and vertical components are He​ and Hv​
cosθ= He​​/Be
∴cos60o= (​0.26×10−4​/ Be )T
⇒Be​=(​0.26×10−4)/ (½)​=0.52×10−4T=0.52G

Test: Physics Minor Mock Test- 4 (March 25) - Question 5

Which of the following statements about bar magnets is correct ?

Detailed Solution for Test: Physics Minor Mock Test- 4 (March 25) - Question 5

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.

Test: Physics Minor Mock Test- 4 (March 25) - Question 6

Magnetic material differences are explained by

Detailed Solution for Test: Physics Minor Mock Test- 4 (March 25) - Question 6

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.

Test: Physics Minor Mock Test- 4 (March 25) - Question 7

A cube-shaped permanent magnet is made of a ferromagnetic material with a magnetization 500M of about The side length is 20 cm. Magnetic dipole moment of the magnet is.

Detailed Solution for Test: Physics Minor Mock Test- 4 (March 25) - Question 7

Correct Answer :- C

Explanation : Magnetic Moment = Magnetization × Volume

Side of cube = 20 cm

= 0.2 m

= 500 × 0.2 × 0.2 × 0.2

= 4Am2

Test: Physics Minor Mock Test- 4 (March 25) - Question 8

A bar magnet of magnetic moment 1.5 J/T lies perpendicular to the direction of a uniform magnetic field of 0.22 T. What is the torque acting on it?

Detailed Solution for Test: Physics Minor Mock Test- 4 (March 25) - Question 8


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

Test: Physics Minor Mock Test- 4 (March 25) - Question 9

electromagnetic induction i.e currents can be induced in coils (Select the best)

Detailed Solution for Test: Physics Minor Mock Test- 4 (March 25) - Question 9

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

Test: Physics Minor Mock Test- 4 (March 25) - Question 10

Damping in galvanometers is based oUpdate Questionn

Detailed Solution for Test: Physics Minor Mock Test- 4 (March 25) - Question 10

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.

Test: Physics Minor Mock Test- 4 (March 25) - Question 11

Predict the polarity of the capacitor in the situation described by fig.

Detailed Solution for Test: Physics Minor Mock Test- 4 (March 25) - Question 11

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

Test: Physics Minor Mock Test- 4 (March 25) - Question 12

A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s-1 in a uniform horizontal magnetic field of magnitude 3×10-2 T. If the coil resistance is 10Ω, maximum emf induced in the coil,maximum value of current in the coil and average power loss due to Joule heating are

Detailed Solution for Test: Physics Minor Mock Test- 4 (March 25) - Question 12

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

Test: Physics Minor Mock Test- 4 (March 25) - Question 13

An ac generator consists of 8 turns of wire, each of area A=0.0900m2 , and the total resistance of the wire is 12.0Ω. The loop rotates in a 0.500-T magnetic field at a constant frequency of 60.0 Hz. Maximum induced emf is

Detailed Solution for Test: Physics Minor Mock Test- 4 (March 25) - Question 13

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

Test: Physics Minor Mock Test- 4 (March 25) - Question 14

There are two stationary coils near each other. If current in coil-1 is changing with time, relationship of current in coil-2 to it's emf is described by

Detailed Solution for Test: Physics Minor Mock Test- 4 (March 25) - Question 14

M is mutual inductance

Test: Physics Minor Mock Test- 4 (March 25) - Question 15

A metallic rod of 1 m length is rotated with a frequency of 50 rev/s about an axis passing through the centre point O. Other end of the metallic rod slides on a Metallic ring . A constant and uniform magnetic field of 2 T parallel to the axis is present everywhere. What is the emf between the centre and the metallic ring?

Detailed Solution for Test: Physics Minor Mock Test- 4 (March 25) - Question 15

l = 1 m

= 314V

Test: Physics Minor Mock Test- 4 (March 25) - Question 16

Coils C1 and C2 are stationary. C1 connected in series with a galvanometer while C2 is connected to a battery through a tapping key. If the tapping key is pressed and released the galvaometer

Detailed Solution for Test: Physics Minor Mock Test- 4 (March 25) - Question 16

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

Test: Physics Minor Mock Test- 4 (March 25) - Question 17

Domestic power supply in India is

Detailed Solution for Test: Physics Minor Mock Test- 4 (March 25) - Question 17

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.

Test: Physics Minor Mock Test- 4 (March 25) - Question 18

For a series LCR circuit the input impedance at resonance

Detailed Solution for Test: Physics Minor Mock Test- 4 (March 25) - Question 18

Resonance occurs when XL = XC 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.

Test: Physics Minor Mock Test- 4 (March 25) - Question 19

A resistor of 100 Ω and a capacitor of 10μF are connected in series to a 220 V 50 Hz ac source. Voltage across the capacitance and resistor are

Test: Physics Minor Mock Test- 4 (March 25) - Question 20

The current amplitude in a pure inductor in a radio receiver is to be 250 μA when the voltage amplitude is 3.60 V at a frequency of 1.60 MHz (at the upper end of the AM broadcast band). If the voltage amplitude is kept constant, what will be the current amplitude through this inductor at 16.0 MHz?

Detailed Solution for Test: Physics Minor Mock Test- 4 (March 25) - Question 20

I0=250μA, v0=3.6v .  v=1.6x106 Hz
Here,
(Reactance of inductance) XL=ωL
XL=2πv X L
v0/I0=2πv x L
3.6/2.5x10-4=2πx1.6x10-6 x L
0.14x104-6=L
L=0.14x10-2H
Now for v=16.0x106Hz
XL=2πv X L
=2πx16x106x14x10-4
XL=1407x102Ω
Now,
v0=I0 x XL
3.6/1407x102=I0   [∵v0=kept constant.]
I0=0.00256x10-2
I0=25.6μA

Test: Physics Minor Mock Test- 4 (March 25) - Question 21

A 0.180-H inductor is connected in series with a 90.0 Ω resistor and an ac source. The voltage across the inductor is (-12.0 V)sin(480 rad/s)t What is VR at t = 2ms ?

Detailed Solution for Test: Physics Minor Mock Test- 4 (March 25) - Question 21

 

Test: Physics Minor Mock Test- 4 (March 25) - Question 22

At resonance the current in an LCR circuit

Detailed Solution for Test: Physics Minor Mock Test- 4 (March 25) - Question 22

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

Test: Physics Minor Mock Test- 4 (March 25) - Question 23

A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 Ω, L = 25.48 mH, and C = 796 μF. Power dissipated in the circuit; and the power factor are

Detailed Solution for Test: Physics Minor Mock Test- 4 (March 25) - Question 23

Angular frequency of the ac signal w=2πν
∴ w=2π(50)=100π
Capacitive reactance Xc​=1/wC​
∴ Xc​=1/(100π×786×10−6)​=4Ω
Inductive reactance XL​=wL
∴ XL​=100π×(25.48×10−3)=8Ω
Impedance of the circuit Z=√[R2+(XL​−Xc​)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=Iv2R
Now, Iv=I0/√2=E0/√2Z=283/(1.414×5)=40A
∴P=Iv2R=(40)2×3=4800 watt

Test: Physics Minor Mock Test- 4 (March 25) - Question 24

A resistor is connected in series with a capacitor. The voltage across the resistor is vR =  (1.20 V) cos(2500 rad/s)t . Capacitive reactance is

Test: Physics Minor Mock Test- 4 (March 25) - Question 25

According to Maxwell’s equations

*Multiple options can be correct
Test: Physics Minor Mock Test- 4 (March 25) - Question 26

Electromagnetic waves for TV and radio have a frequency about

Detailed Solution for Test: Physics Minor Mock Test- 4 (March 25) - Question 26

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).
 

Test: Physics Minor Mock Test- 4 (March 25) - Question 27

Plane electromagnetic wave travels in vacuum along z-direction. If the frequency of the wave is 30 MHz, its wavelength is

Detailed Solution for Test: Physics Minor Mock Test- 4 (March 25) - Question 27

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

Test: Physics Minor Mock Test- 4 (March 25) - Question 28

Electromagnetic waves propagate

Detailed Solution for Test: Physics Minor Mock Test- 4 (March 25) - Question 28

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.

Test: Physics Minor Mock Test- 4 (March 25) - Question 29

State the part of the electromagnetic spectrum to which 1057 MHz (frequency of radiation arising from two close energy levels in hydrogen; known as Lamb shift)..Belongs

Detailed Solution for Test: Physics Minor Mock Test- 4 (March 25) - Question 29

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.

Test: Physics Minor Mock Test- 4 (March 25) - Question 30

Long distance radio broadcasts use short-wave bands because

Detailed Solution for Test: Physics Minor Mock Test- 4 (March 25) - Question 30

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.

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