SSA Chandigarh TGT Math Mock Test -10 - SSA Chandigarh MCQ

Alcohol = 40% of 120 = 48

Hence, Option C is correct.

Hence, Option B is correct.

Evolution of Ethernet from a Coaxial Cable Bus to Hub Managed, Twisted Pair Network
The first step in the evolution of Ethernet from a coaxial cable bus to a hub managed, twisted pair network was the introduction of the Star LAN. Here is a detailed explanation of this evolution:
1. Coaxial Cable Bus:
- Initially, Ethernet networks were based on a coaxial cable bus topology.
- In this setup, all devices were connected to a single cable, forming a bus-like structure.
- Data was transmitted over the cable, and each device could listen to the data being transmitted.
2. Limitations of Coaxial Cable Bus:
- The coaxial cable bus had limitations in terms of distance, speed, and scalability.
- It had a limited maximum distance for data transmission.
- As the network grew, the signal quality degraded, leading to performance issues.
- The bus topology was also susceptible to collisions, resulting in data collisions and network congestion.
3. Introduction of Star LAN:
- To overcome the limitations of the coaxial cable bus, the Star LAN topology was introduced.
- In the Star LAN, each device is connected to a central device called a hub or switch using twisted pair cables.
- The hub or switch acts as a central point for data transmission and management.
4. Advantages of Star LAN:
- The Star LAN topology addressed the limitations of the coaxial cable bus.
- It allowed for longer cable distances, as the hub or switch could regenerate and amplify the signal.
- Each device had its own dedicated connection to the hub or switch, eliminating the signal degradation issues of the bus topology.
- The star topology also reduced the likelihood of collisions, improving network performance and reliability.
5. Further Evolution:
- After the introduction of the Star LAN, Ethernet continued to evolve.
- Ring LAN and Mesh LAN are other network topologies that were developed, but they did not directly lead to the hub managed, twisted pair network evolution.
- Ring LAN uses a ring topology, and Mesh LAN employs multiple direct connections between devices.
In conclusion, the first step in the evolution of Ethernet from a coaxial cable bus to a hub managed, twisted pair network was the introduction of the Star LAN topology. This topology addressed the limitations of the coaxial cable bus and provided improved performance, scalability, and reliability for Ethernet networks.

HTTPS:
HTTPS stands for Hypertext Transfer Protocol Secure. It is the secure version of HTTP, which is the protocol used for transmitting data over the internet.
What does the 'S' in HTTPS stand for?
The 'S' in HTTPS stands for:
- Secured: The 'S' in HTTPS stands for 'Secured'. It indicates that the communication between the web browser and the web server is encrypted and secure.
How does HTTPS work?
HTTPS works by using SSL/TLS (Secure Sockets Layer/Transport Layer Security) protocols to encrypt the data transmitted between the web browser and the web server. Here is a brief overview of how HTTPS works:
1. Handshake: The web browser sends a request to the web server to establish a secure connection.
2. Server authentication: The web server sends its SSL/TLS certificate to the web browser to prove its identity.
3. Encryption: The web browser and the web server negotiate an encryption algorithm and exchange encryption keys.
4. Data transmission: The web browser and the web server encrypt and decrypt the data using the shared encryption keys, ensuring that the data cannot be intercepted or tampered with during transmission.
Why is HTTPS important?
HTTPS is important for several reasons:
- Data security: HTTPS ensures that the data transmitted between the web browser and the web server is encrypted and secure, protecting it from unauthorized access.
- Authentication: HTTPS verifies the identity of the web server, ensuring that the user is connecting to the intended website and not a malicious imposter.
- Trust and credibility: Websites that use HTTPS are often seen as more trustworthy and credible by users, as they prioritize data security.
Conclusion:
In summary, the 'S' in HTTPS stands for 'Secured'. HTTPS is crucial for ensuring the security and integrity of data transmitted over the internet, providing encryption, authentication, and trustworthiness.

Microsoft and OpenAI are jointly embarking on a $100 billion project to create “Stargate,” an AI supercomputer. Scheduled for launch in 2028, it follows OpenAI’s AI upgrade in 2022. The project aims to surpass existing data centers’ capabilities by 100 times. Divided into five phases, with Stargate as the fifth, a smaller fourth-phase supercomputer is slated for release in 2026. Significant costs will be allocated to acquiring essential AI chips.

Deductive Reasoning is also known as deductive logic. It is a process of reasoning that involves moving from one or more statements to a logically certain conclusion or moves from general to particular. For eg- All women are geniuses. Rhea is a woman. Thus, we can conclude that Rhea is a genius. 
Hence, the correct answer is, 'Reasoning from general to particular.'

If any teacher is involved in gender discrimination, then it may lead to diminished effort or performance of the students. Many times it happens that a teacher is biased towards a particular gender which includes he/she is giving more marks to either girls or boys. Not only this, even boys are getting more harsh punishments than girls even if they have done the same type of mistakes. All these decrease the morale of the other students. So, a teacher should not be partial on any grounds. 
Hence, the correct answer is, 'may lead to diminished effort or performance of the students.'

Group projects must be given to the students as these projects help the students to understand the importance of working in a team and solve different problems effectively.

Let 
E₁ and E₂ are events that a person has disease and the person has not disease.

Let A  = event that the test result is positive

 

Let, the angle be A
⇒ Its complement angle = 90° – A
According to the question
(90 – A) = A + 60°
⇒ 90 - 60 = A + A
⇒ 2A = 30°
⇒ A = 15°

As the angle increases from 270° to 360°, the sine increases from -1 to 0. As the angle increases from 270° to 360°, and the value of cos 270° = 0
and cos 360° = 1
So, the cosine increases from 0 to +1.

x = 2y = -3z     -4x = 6y = -z
x/1 = y/(½) = z(-⅓)                   x/(-¼) = y/(⅙) = z/(-1)
Cosθ = [(a1a2 + b1b2 + c1c2)/(a1 + b1 + c1)^½ * (a2 + b2 + c2)^½]
Cosθ ={[(1*(-¼)) + (½)(⅙) + (-⅓)(-1)]/[(1)² + (½)² + (-⅓)²]1/2 * [(-¼)² + (⅙)² + (-1)²]1/2}
= {[(-¼ + 1/12  - ⅓)]/[2 + 1 - ⅔]^1/2 * [ -½ + ⅓ -½]^½}
Cosθ = 0
θ = 90deg

The chord of a parabola through the focus and perpendicular to the axis is called the latus rectum.

The length of an arc of a unit circle is numerically equal to the measurement in radians of the angle that it subtends; one radian is just under 57.3 degrees.

dy/dx = 2^x(tanx)' + tanx (2^x)'
dy/dx = 2^x (secx)² + 2^x tanx log2 , {since , (a^x)' = a^x loga (x)'}
dy/dx = 2^x[(Secx)² + log2 tanx]

 (x)^½ (a - x)^½ dx
=  ∫(ax - x²)^½ dx
=  ∫{-(x² - ax)^½} dx
=  ∫{-(x² - ax + a^2/4 - a^2/4)^½} dx
=  ∫{-(x - a/2)² - a^2/4} dx
=  ∫{(a/2)² - (x - a/2)²} dx
=  ½(x - a/2) {(a/2)² - (x - a/2)²} + (a^2/4) (½ sin^-1(x - a)/a²)
= {(2x - a)/4 (ax - x²)^½} + {a^2/8 sin^-1(2x - a)/a} + c

Given: y = sin 3x . sin³

= 3 sin²x.sin 4x

Here A ∩ C = Ø ⇒ A ∩ B ∩ C = Ø
n (A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C)
 = 25 + 20 + 27 - 5 - 7 - 0 - 0
= 60

Required area :

Surface area of a cuboid = 2 × (lb + bh + hl)
So the answer = 2 × (10 × 8 + 8 × 5 + 5 × 10) = 340 cm²
Hence, Option A is correct.

The two curves parabola and the line meet where,

Therefore , the required area is :

For continuity: LHL = RHL

At x=2,
LHL: x < 2 ⇒ f(x) = 2*a
RHL: x ≥ 2 ⇒ f(x) = 4
For continuity: LHL = RHL
⇒ 2a = 4 ⇒ a = 2

At x = 0,
LHL: x < 0 ⇒ f(x) = b
RHL: x ≥ 0 ⇒ f(x) = 0 * a
For continuity: LHL = RHL
⇒ b = 0

Solve x = 9 and x² – y² = 9 equation of tangent at point of contact is S₁ = 0

Let number of rackets made = x 
And number of bats made = y 
Therefore , the above L.P.P. is given as : 
Maximise , Z = x +y , subject to the constraints : 1.5x +3y ≤ 42 and. 3x +y ≤ 24, i.e.0.5x + y ≤ 14 i.e. x +2y ≤ 28 and 3x +y ≤ 24 , x, y ≥ 0.

Here Z = 16 is maximum. i.e Maximum number of rackets = 4 and number of bats = 12.
Here , profit function is P = 20x + 10y 
Profit is maximum at x = 4 and y = 12 . 
Therefore , maximum profit = 20(4) + 10 ( 12) = 200.i.e. Rs.200.

(cos135° - cos120o)/ (cos135° + cos1200)
= cos(180° - 45°) - cos(180° - 60°)/cos(180° - 45°) + cos(180°-60°)
=  (-cos45° + cos60°)/(- cos45° - cos60°)
=  (-1/√2 + 1/2)/(- 1√2 - 1/2)
=  -2 + √2 / -2 - √2 
= (-2 + √2 / -2 - √2) * (-2 + √2)/(-2 + √2)
=  (-2 + √2)2/(-2)2 - (√2)2
= (4 + 2  - 4√2) / 2
= (6 - 4√2)/2
= 3 - 2√2

Distance of pt P from plane = 5

Gen. pt. on line is (λ + 1, 2λ - 2, -2λ + 1)
This pt line on plane
(λ + 1 + 2 2λ - 2) - 2 (-2λ + 1) = 10