Bihar STET Paper 2 Physics Mock Test - 3 - Bihar PGT/TGT/PRT MCQ

• Eddy currents are the currents which are induced in a conductor whenever the amount of linked magnetic flux with the conductor changes.
• These were discovered by Foucault in the year 1895 and hence they are also called Foucault currents.

Eddy currents:

The Critical Angle
a light ray is in the more dense medium and approaching the less dense medium.
The angle of incidence for the light ray is greater than the so-called critical angle.
When the angle of incidence in water reaches a certain critical value, the refracted ray lies along the boundary, having an angle of refraction of 90-degrees. This angle of incidence is known as the critical angle; it is the largest angle of incidence for which refraction can still occur. For any angle of incidence greater than the critical angle, light will undergo total internal reflection.

Option A: True. The velocity vector of a particle at any point on its path is always tangent to the path at that point, as velocity represents the direction of motion.

Option B: True. For a particle in uniform circular motion, the centripetal acceleration points towards the center. Over one complete cycle, the acceleration vectors cancel out, resulting in an average acceleration of zero (null vector).

Option C: True. The net acceleration in uniform circular motion is centripetal acceleration, which always points towards the center of the circle along the radius.

Option D: Not true. This statement is incorrect because it implies that the net acceleration in all types of circular motion is always radial. While in uniform circular motion, the acceleration is purely radial, in non-uniform circular motion, there is an additional tangential component of acceleration due to the change in speed. Thus, the net acceleration in such cases is not always directed towards the center.

The heat supplied under these conditions is the change in internal energy Q = ΔU
The heat supplied Q = i²RT = 1 × 1 × 100 × 5 × 60 = 30,000 J = 30 kJ

Resultant force is zero, as three forces on the particle can be represented in magnitude and direction by three sides of a       triangle in same order. Hence, by Newton's 2nd law 

Particle velocity  will be same

A projectile can have the same range if the angle of projection are complementary ie, θ and (90⁰ - θ). Thus, in both cases:

Decrease in GPE of B₁ = Increase in Gravitational PE of B₂ + Increase in elastic PE of spring. 

Again applying law of conservation of mechanical energy at the mean position we get 

Time period, T=2π√l/g
=>T∝√l
=>ΔT/T=1/2 Δl/l
Δl/l=1%
Therefore, ΔT/T=1/2x1%=0.5%

When Rutherford saw the results of the experiment by Geiger and Marsden, he said:
“It was quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you.”
Rutherford used the results of this experiment to develop a new model for the atom. This model proposed a central nucleus with a positive charge. It was this positively charged nucleus that was responsible for the strong backward deflection of the positively charged alpha particles.
The model also proposed that negatively charged electrons surrounded this nucleus. However, as most of the alpha particles passed through the gold foil with no deflection at all, Rutherford realised that most of the atom was empty space. So, his model placed the electrons at some distance from the nucleus.
 

On equatorial line, the direction of electric field is reversed to that of angle of axial line. Therefore, the angle between dipole moment and electric field is 180°.

The restoring force in a simple harmonic motion is maximum in magnitude when the particle is instantaneously at rest because in SHM object’s tendency is to return to mean position and here particle is instantaneously at rest after that instant restoring force will be max to bring particle to mean position.

As work is a path function rather than a state function, we can easily say that work can often be graphically represented as the area under the PV graph. And as cyclic processes are represented as closed shapes on PV graph it is obvious that they have non zero area and thus work done is non zero.

Width of the magnetic field region where 'R' is its radius of curvature inside magnetic field,

Difference in masses = 20.17 – 20.15 = 0.02 g
In subtraction, the final result should retain as many decimal places as there are in the number with the least decimal places.

Work done = potential*charge by definition. We know that the potential of a point is the amount of work done to bring a unit charge from infinity to a certain point. Therefore, work done W=q*V=4*10^-3*200J=0.8J. The work done is positive in this case.

We know: Vrms​=​Vo​​/√2
 
Substituting values (Here RMS Voltage is 220 V)
We get: Maximum V (or amplitude) =311V
 

3. 1.5

Solution: Let the radius of curvature of the convex surface of the plano-convex lens be RRR. The focal length of a concave mirror is related to the radius of curvature by f=R2f = \frac{R}{2}f=2R​.

1. When the plane side is silvered, the system behaves as a concave mirror with focal length 30 cm. The effective focal length fefff_{\text{eff}}feff​ of this system can be found using the mirror formula and lens formula combined, but given the scenario, we focus on this system.

2. When the convex side is silvered, the system behaves like a concave mirror with focal length 10 cm.

By using the relations for combined focal length and refractive index of the plano-convex lens, and applying these boundary conditions, the refractive index nnn of the material of the lens is determined to be 1.5.

 So the range is 88 - 198 pF

Heat extracted from source Q1= 700 cal
Heat rejected to the sink Q2= 500 cal
Temperature of source T1= 150 +273= 423K
Temperature of sink T2 to be found:
 Q2/Q1 = T2/T1
So,
T2 = (Q2/Q1) x T1
= (500 x 423)/700
= 302.02 K
= 29.02 degree C.
 

Total emf = 3 x 4 = 12 V
Total resistance = 6 + 3r
Current in the circuit = 2 A
Using ohm's law
2 = 12/ (6+3r)
2(6+3r)=12
12+6r=12
6r = 12-12 = 0
r = 0/6 = 0

NEP 2020 advocates for the use of technology as a supplement to traditional teaching methods to enhance learning experiences.

The greatest cause of failure of a new teacher is in the area of interpersonal relations. If the teacher is not able to develop a bond or relationship with the students, she will be rejected by the students and will face a lot of hardships.

The sounds th, ph, ch are phonemes. The phoneme is the unit of sound.  A phoneme combines with other phonemes to give rise to meaningful words. Phonemes distinguish one word from another. These are established by the use of minimal pairs, such as pat vs bat or tap vs tab.
Hence, the correct answer is, 'Phonemes.'

The correct answer is National Higher Education Regulatory Council (NHERC).
Important PointsThe National Education Policy (NEP) 2020 proposed the creation of a Higher Education Commission of India (HECI) as a single overarching umbrella body for higher education.

• This new regulatory body would replace the existing separate regulatory institutions in the higher education sector to improve coordination and maintain standards.
• The HECI is set to have four independent verticals, namely:

1. National Higher Education Regulatory Council (NHERC)
2. General Education Council (GEC)
3. Higher Education Grants Council (HEGC)
4. National Accreditation Council (NAC)

Each of these verticals has specific roles and responsibilities, from regulation, to setting standards, to funding, to accreditation.

• According to NEP 2020, the first vertical of the HECI will be the National Higher Education Regulatory Council (NHERC).
• This council will be the primary regulator for higher education in India, including teacher education, excluding medical and legal education.
• This implies that all other bodies, like UGC and AICTE, would cease to be regulatory bodies and would only be responsible for standards setting and providing grants.

Therefore, the correct answer would be the "National Higher Education Regulatory Council (NHERC)".

Visual Impairment: The term 'visual impairment' is used to refer to not only those who are blind but also those who have their own vision. When the person has residual vision above a certain level, he is said to have low vision or poor vision. However, if the residual vision is below that level, he is considered to be blind.

Important Points

In a classroom situation, the total number of blind students should be 10-12, so that teacher can adopt the strategies which cater individual differences in the class.

Strategies to teach blind students in the classroom:

• Teachers who teach blind children in ordinary classrooms need to provide adequate special materials and equipment. There is almost universal agreement that the special materials and equipment needed by a visually impaired student in a regular school are essential to success.
• In addition, Friendly user materials are good teaching materials for blind children, and creative materials which are provided to blind children enhance dialogue and social interaction to complete an assignment. 
• Students with visual impairments who's knowledge of concepts are strong would have difficulty keeping pace with this rate of instruction unless it is entirely verbal.
• Purposeful sampling was used to select teachers who teach blind children in ordinary classrooms. Legibility for teacher’s participation in the study was based on the following criteria:
  • experience in teaching blind children with their distance vision assessment results indicating that they are blind with (visual acuity of (3/60 or worse).
  • Teaching ordinary classrooms,
  • fully attending school at the time of the study, and
  • gender balance was desirable but not necessary.
• The teacher who is guiding the blind children in ordinary classrooms is supposed to know more than the blind pupils who are being guided, that is why very many blind children have fallen victims to wrong teaching. Guidance is needed for blind children since they can hear, it is easy for them to learn.

Hence, we can conclude that in a classroom situation, the total number of blind students should be 10-12.

Manganese is one of the important raw materials used for smelting iron ore.

Important Points

• Manganese is used for making alloys.
• It is used in manufacturing of enamel, dry cell batteries, plastics, chemicals and glasses.
• It requires almost 10 kg of manganese to manufacture one tonne of steel.
• Odisha is the largest producer of manganese in India with multiple mines located over the iron ore belt of India.
• Some of the major mines are Bonai, Kendujhar, Sundergarh, Gangpur, Koraput, Kalahandi, and Bolanger.
• After Odisha, Karnataka, Maharashtra, Telangana, Goa, and Jharkhand lies in the list of manganese producer in India.
• In 2000-2001, Odisha produced one-third of the country's total production.

Thus, we can conclude that Odisha is the largest producer of manganese.