Chandigarh TGT Science - Non Medical Mock Test - 7 - SSA Chandigarh MCQ

Long term scheduling
- Determines when new processes are admitted to the system
- Also known as admission scheduling or job scheduling
- Determines which processes should be brought into the system from the job pool or job queue
- Decides when a new process can be created and executed in the system
- Mainly focuses on managing the resources required by the processes and ensuring that the system does not become overloaded
Medium term scheduling
- Involves swapping processes between main memory and secondary memory (e.g., hard disk)
- Decides which processes should be swapped out of main memory and which should be brought back in
- It is responsible for managing the degree of multiprogramming or the number of processes in the ready state that are kept in main memory
Short term scheduling
- Also known as CPU scheduling
- Determines which process should be executed next by the CPU
- Determines the order in which ready processes are allocated the CPU for execution
- Focuses on improving the performance of the system by minimizing the waiting time and maximizing the CPU utilization
- Examples of short-term scheduling algorithms include First-Come, First-Served (FCFS), Round Robin, and Shortest Job Next (SJN)
None of these
- This option is incorrect as one of the above options is the correct answer

The size of an organization chart object
The size of an organization chart object refers to its dimensions and can be adjusted in PowerPoint.
Options:
- Option A: Is determined by the presentation design and cannot be changed.
- Option B: Is determined by the presentation design but can be changed in PowerPoint.
- Option C: Is dependent on the amount of text within the organization chart.
- Option D: Both (b) and (c).

The correct answer is Option D, which means that the size of an organization chart object can be changed in PowerPoint and is also dependent on the amount of text within the organization chart. Here's why:
- The size of an organization chart object is not solely determined by the presentation design. While the initial size may be influenced by the design template, it can be adjusted manually in PowerPoint.
- PowerPoint allows users to resize and scale objects, including organization chart objects. This flexibility enables users to customize the size of the object according to their specific needs and preferences.
- Additionally, the size of the organization chart object can also be influenced by the amount of text within it. If the chart contains a large amount of text, it may require a larger size to accommodate all the information effectively.
- Conversely, if the chart has minimal text or fewer levels, it may be possible to reduce the size of the object without compromising readability or clarity.
In summary, the size of an organization chart object can be changed in PowerPoint, and it can also vary depending on the amount of text within the chart.

Answer: B: might Explanation: - The sentence is about a possibility or uncertainty, which is best expressed using the modal verb "might." - "Mustn't" implies a prohibition or something that is not allowed. - "Have to" indicates an obligation or necessity. - "Wouldn't" is used to express a refusal or a hypothetical situation in the past.

1st term = 95
2nd term = 95 + 20.5 = 115.5
3rd term = 115.5 + 22.5 = 138
4th term = 138 + 24.5 = 162.5
5th term = 162.5 + 26.5 = 189

Thus, the missing term is 4th term = 162.5

Learning disabilities may occur due to all of the following except cultural factors as they do not affect learning disability.

For Vygotsky, although biological factors constitute the necessary pre-requisite for elementary processes to emerge, socio-cultural factors are indispensable for elementary natural processes to develop. He argues for the uniqueness of the social milieu and regards socio-cultural settings as the primary and determining factor in the development of higher forms of human mental activity such as voluntary attention, intentional memory, logical thought, planning, and problem solving.

Each pupil is different in terms of learning ability, academic standards, classroom learning and academic performance, and each has his own pace in learning. A successfully implemented remedial teaching strategy points the student towards the path which leads to goals and objectives of teaching. Hence, it is considered as a sign of motivated teaching.

IQ is a score obtained by conducting an Intelligence Test. Intelligence involves different mental activities and abilities which include logic, reasoning, planning, and problem-solving. So IQ between 90 to 110 is considered as normal. This proves that this is normal and people having IQ between 90 to 110 can understand everything, able to do planning, reasoning, etc.
Hence, the correct answer is "Normal".

'आपबीती' अव्ययीभाव समास का उदाहरण नही है। अतः 'आपबीती' सही उत्तर है।
आपबीती - अपने पर बीती हुई - तत्पुरुष समास
Key Points

• बेखटके - बिना खटके के - अव्ययीभाव समास
• प्रतिदिन - प्रत्येक दिन - अव्ययीभाव समास
• 'पुश्तानुपुश्त' :- पुश्तो के बाद पुश्ते - अव्ययी भाव समास
• यथासम्भव - जैसे भी सम्भव हो - अव्ययीभाव समास

Additional Information

विकल्प 1, ‘निशापति - दिवाकर’ यह युग्म सही नहीं है।

Key Points

• "निशापति - दिवाकर" यह युग्म समानार्थी शब्दों को नहीं दर्शाता।
• निशापति का अर्थ -
  • चंद्रमा; राकेश।
• दिवाकर का अर्थ -
  • दिनकर, सूर्य।
• अन्य सभी विकल्प सही सुमेलित हैं। वे सभी चन्द्रमा के समानार्थी शब्द हैं।

Additional Information

• समानार्थी का अर्थ होता हैं (समान+अर्थ) अर्थात किसी शब्द का समान अर्थ वाले दूसरे शब्द या उसी के सामान कोई दूसरा नाम (वस्तु)। सामान्यत: हिन्दी में एक ही वस्तु के अनेक समान अर्थ वाले शब्द है।
• जब किसी बात (विचार), वस्तु या भाव का केवल अर्थ बताना हो और उसके अर्थ को बताने के लिए जिस शब्द का प्रयोग करें, वह शब्द समानार्थी शब्द कहलाएगा। जबकि किसी एक ही बात (विचार), वस्तु, भाव को किन-किन नामों से या किन किन शब्दों से जाना जाता है यह पूछा जाए तो बताने के लिए प्रयुक्त शब्दों के समूह में सभी पर्यायवाची होंगे।
  • जैसे:-, 'अश्व' का समानार्थी 'घोड़ा' होगा। इसके विपरीत यदि यहाँ यह कहा जाये कि अश्व को और किन-किन शब्दों से संबोधित किया जाता है तब उत्तर में घोटक, बाजी, हय, तुरंग, दधिका, सैंधव, घोड़ा, रविपुत्र, सर्ता आदि बताए जायेंगे और ये सभी शब्द 'अश्व' के पर्यायवाची होंगे।

F=-eE .
we know that,
F= ma ,
then
a=F/m.
a=-eE/m.

1^st case[series]
I=E+E/r+r+1-----1
2^nd case[parallel]
I=E[E will be same in parallel]/(1/r)+(1/r)+(1/1)----2
Current in both cases are the same.
1=2
2E/2r+1=E/(2/r)+1
R=1 Ω

If a hollow sphere is cut into parts and the outer surface of the cut part is painted. Then it becomes a mirror with its inner surface as the reflecting surface This kind of mirror is known as concave mirror. Light converges at a point when it strikes and reflects back from the reflecting surface of concave mirror. Hence, it is also known as converging mirror.  When concave mirror is placed very close to the object, a magnified and virtual image is obtained. But if we increase the distance between the object and the mirror then the size of the image reduces and a real image is formed. So the image formed by concave mirror can be small or large and it can also be real or virtual.

Here, f = 6cm, μ = 1.5, R₁ = R, R₂ = −2R
According to lens maker's formula

1. Current: This is a general term for the flow of electric charge. It can refer to any type of charge flow, not specifically electrons.

2. Electronic current: This specifically refers to the flow of electrons through a conductor.

3. Conventional current: This is the flow of positive charge and is considered to flow from the positive side to the negative side of a power source. Historically, this was defined before the discovery of electrons, so it is opposite to the direction of electron flow.

4. Photonic current: This is not a standard term in electrical engineering or physics. It might refer to currents related to photons in some contexts, but it is not related to the flow of electrons.

So, if you are referring to the flow of electrons specifically, electronic current is the accurate term.

As a very few α-particles were deflected by sharp angles, it was concluded that the positive charge is concentrated in a very small volume of the atom.

Metal oxides react with acid to form salt and water because most of the metal oxides are basic or amphoteric in nature.

NH₄NO₃ (s) → N₂O (g) + 2H₂O ;     ΔH = -37.0 KJ/mol from the above data, we can show that ΔH is for 1 mole of NH₄NO₃.
For 2.5 g of NH₄NO₃, we have 2.5/80 = 0.03125 moles
Therefore, heat released = (-37)×(0.03125) = -1.16 kJ

Oxidation Number Transition in the Conversion of Br₂ to BrO₃

• The oxidation number of an element in its free or uncombined state is zero. This rule applies to bromine in Br₂, so the initial oxidation state of Br in Br₂ is zero.
• In BrO₃^-, the combined state of Bromine, Oxygen has an oxidation state of -2. However, the overall charge of the ion is -1. Since we have three Oxygen atoms, the total contribution of Oxygen is -6. To balance this, the Bromine must have an oxidation state of +5.
• Therefore, in the conversion of Br₂ to BrO₃^-, the oxidation state of Bromine changes from 0 in Br₂ to +5 in BrO₃^-
• Hence, the correct answer is Option A: zero to +5.
   

Energy of one photon
Energy supplied by one mole photon in KJ/mole = 3.11×6×10⁻¹⁹×6×10²³×10⁻³=297KJ/mole
% of energy converted to K.E.= 

Create ((X+B) + (A-B)) in numerator and then apply Sin(a+b) formula then you will be able to solve it.

Probability that first and second student can solve
=0.7×0.5=0.35
Probability that first can solve and second cannot solve
=0.7×0.5=0.35
Probability that first cannot solve and Amisha can solve
=0.3×0.5=0.15
Therefore, probability that at least one of them will solve
=0.35+0.35+0.15 = 0.85
=> 85/100
= 17/20

In the question, it should be f’(2) instead of f”(2) 
Explanation:- f(x) = ∫(0 to x) log(1+x²)
f’(x) = 2xdx/(1+x²)
f’(2) = 2(2)/(1+(2)²)
= 4/5

Correct answer is C.

r = -2i + 3j + 7k + λ( -i - 2j - 3k)
xi + yj + zk = (-2-λ)i + (3-2λ)j + (7-3λ)k
Equating the terms, we get
x = -2-λ     y = 3-2λ   z = 7-3λ
(x+2)/(-1) = λ,   (y-3)/(-2) = λ,    (z-7)/(-3) = λ
(x+2)/(1) =  (y-3)/(2) =  (z-7)/(3) 

As, 867=255 × 3 +102
255 = 102 × 2 + 51
102 = 51 × 2 + 0
So, HCF (867,255) = 51

Let the other nos. be a and b
then (x+y+1+2+6)/5 = 4.4
x + y = 13 ---------------------(1)
Variance = 8.24


41.2 = 19.88 + (x2 + 19.36 – 8.8x) + (y2 + 19.36 – 8.8y) 
21.32 = x2 + y2 + 38.72 – 8.8(x + y) 
x2 + y2 + 38.72 – 8.8(13) – 21.32 = 0 
(using equation (1)) 
x2 + y2 – 97 = 0 …(2) 
Squaring equation (1) both the sides, 
we get (x + y)2 = (13)^2 
x2 + y2 + 2xy = 169 
97 + 2xy = 169 
(using equation (2)) 
xy = 36 or x = 36/y (1)
⇒ 36/y + y = 13 
y2 + 36 = 13y 
y2 – 13y + 36 = 0 
(y – 4)(y – 9) = 0 
Either (y – 4) = 0 or (y – 9) = 0 
⇒ y = 4 or y = 9 
For y = 4 x = 36/y 
= 36/4 = 12 
For y = 9 
x = 36/9 
x = 4 
Thus, remaining two observations are 4 and 9.

U = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A = {2, 4, 6, 8}
A’ = U – A
A’ = {1, 3, 5, 7, 9}