OSSSC TGT Science PCM Mock Test - 9 - OTET MCQ

A documentary provides actual facts related to the topic. Thus, a documentary on water pollution is the most suitable way for teaching the topic.

When dealing with the students at the primary level, the best method to enable a child to express his views on public issues is using wall magazines. Wall magazine is a means by which students can share their views with each other in the form of charts, articles, drawings, poems, etc.

A teacher collects and reads the work of the class, then plans and adjusts the next lesson to meet the student's needs. He/She is doing an Assessment for learning. Assessment for learning is also known as diagnostic and formative assessments. It is conducted for the purpose of determining the skills and abilities of the students or their progress. So, teachers can adjust their classrooms as per the individual needs of the students. In this way, students are provided with their feedback also.
Hence, the correct answer is, 'Assessment for learning.'

First generation learners are defined as students whose parent(s)/legal guardian(s) has/have not completed any formal schooling. This means that they are the first in their family to attend school. Generally, there is no one at home to encourage such children for self-studying. So, option 2 is the correct answer.

a car battery of 12V provides a current of about 200A, which is the current needed to operate a starter motor. So a battery has a very small internal resistance.
also,The cause of non-correctable battery problems and.A weak cell holds less capacity and is discharged more quickly.

The resistance of a material at temperature T is given by 
R=R₀​[1+α(T−T₀​)]
Where R₀​= resistance at reference temperature T​₀=0^oC and α is the temperature coefficient of resistance. 
From the above equation, the unit of the temperature coefficient is per degree Celsius (/^oC). So, dimension will be [C⁻¹]

Long distance radio broadcasts use shortwave bands because only these bands can be refracted by the ionosphere.

Fringe width β=λD/d​
On d′=d/2​, D′=2D
New fringe width β′= λD′​/d′=4β

The principle of reversibility of light states that light follows the same path if the direction of the travel of light is reversed.
The principle of reversibility presumes that the attenuation of a light ray during its passage through an optical medium because of reflection, refraction, and absorption is not affected by a reversal of the direction of the ray. In its simplest explanation, the principle of reversibility is a deduction of Snell’s law of the refraction of light. The principle of reversibility presumes that the attenuation of a light ray during its entrance through an optical medium due to reflection, refraction, and absorption is not changed by a reversal of the direction of the ray.

Lenz’s law is used for determining the direction of induced current.

Lenz’s law of electromagnetic induction states that the direction of induced current in a given magnetic field is such that it opposes the induced change by changing the magnetic field.

Following is the formula of Lenz’s law:

ϵ=−N (∂ϕ_B/∂t)

Where,

• ε is the induced emf

• ∂Φ_B is the change in magnetic flux

• N is the number of turns in the coil

Lenz’s law finds application in electromagnetic braking and in electric generators

The process of changing solid directly into gases without changing into liquid is called sublimation. In the substance given above only Iodine is a sublime substance.

In the year 1829, Johann Wolfgang Dobereiner,a German scientist, was the first to classify elements into groups based on John Dalton's assertions. He grouped the elements with similar chemical properties into clusters of three called 'Triads'. The distinctive feature of a triad was the atomic mass of the middle element.When elements were arranged inorder of their increasing atomic mass,the atomic mass of the middle element was approximately the arithmetic mean of the other two elements of the triad.

The greater the valency of the element, the greater is its unreactivity. Also, the valency in a period increases from the left to the middle, then again decreases from middle to the right. So we can say that the unreactive elements are located in the center of the period. Since phosphorus and Silicon are located at the center, they have the highest number of Valence electrons, and hence are the most unreactive in the period.

Changes of state are physical changes in matter. They are reversible changes that do not involve changes in matter’s chemical makeup or chemical properties. Common changes of state include melting, freezing, sublimation, deposition, condensation, and vaporization.

- Acids and Bases: Blue litmus turns red in the presence of an acid.
- Identifying the Acid:
- Mg(OH)₂: Magnesium hydroxide is a base.
- NaOH: Sodium hydroxide is a strong base.
- NH₄OH: Ammonium hydroxide is a weak base.
- CH₃COOH: Acetic acid is a weak acid.
- Correct Answer: Acetic acid (CH₃COOH) can turn blue litmus red because it is an acid.

y = x is tangent to the parabola
y=ax²+c
if a= then c=?
y′ =2ax
y’ = 2(7/2)x  =1
x = 1/7
1/7 = 2(1/7)² + c
c = 1/7 * 2/49
c = 7/2

Mean deviation of a data set is the average of the absolute deviations from a central point (Average value).

f(x) = ln(1+x) - x ≤ 0
f(x) = ln(1+x) - x
f’(x) = 1/(1+x) - 1
= (1 - 1 - x)/(1+x) 
= -x/(1+x)
f’(x) ≤ 0
-x/(1+x) ≤ 0
0 ≤ x /(1+x)
for(x = 2) (-2)/(1-2)
= 2 > 0, therefore x > - 1

cos^-1[(1-x²)/(1+x²)]
Put x = tanθ
y = cos^-1(1 - tan²θ)/(1 + tan²θ)
y = cos^-1(cos²θ)
y = 2θ……………….(1)
Put θ = tan^-1 x in eq(1)
y = 2(tan^-1 x)
dy/dx = 2(1/(1+x²))
dy/dx = 2/(1+x²)

Four circles of radius 5 cm are cut out.
Area cut = 4*(π5²) = 100π
Total area of circular sheet = π(20)² = 400π
Area uncut = 300 π
The ratio of uncut to cut portion is 3 : 1

Given, y = (sinx+cosx)/(sinx−cosx)
∴ dydx = [(sinx−cosx)(cosx−sinx)−(sinx+cosx)(cosx+sinx)]/(sinx−cosx)² 
[by quotient rule]
= [(−sinx−cosx)² − (sinx+cosx)²]/(sinx−cosx)²
− [(sinx−cosx)² − (sinx+cosx)²]/(sinx−cosx)²
= − [(sin²x + cos²x − 2sinxcosx + sin²x + cos²x + 2sinxcosx)]/(sinx−cosx)²
= −2/(sinx−cosx)²
= -2/(sin²x + cos²x - 2sinxcosx)
= - 2/(1 - sin2x)

f(x) = x^x     f(x) = x^(-x)
f’(x) = x^x(1+ln x)          f’(x) = -x^(-x)(1+ln x)
1+ln x = 0                   x = 1/e
x = 1/e
Min value = (1/e)^(1/e)        f(1/e) = e^(1/e)
Product = (e^-1/e)(e)^1/e = 1

Distance from origin = 3p 

|a + b|² = |a|² + |b|² + 2|a||b|.cosθ
|a|² + |b|² = |a|² + |b|² + 2|a| + |b|  ∵ −1 ⩽ cosθ ⩽ 1
⇒ 2|a||b|.cosθ ⩽ 2|a||b|
So, |a + b|² ⩽ (|a| + |b|)²
⇒ |a + b| ≤ |a| + |b|
This is also known as Triangle Inequality of vectors.