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1. HTTP – Hypertext Transfer Protocol

2. HTTPS – Hypertext Transfer Protocol Secure

3. FTP – File Transfer Protocol

MMS is the combination of Voice, Video, Data

ARPANET was first recognized in 1969

Answer is (b) II, III

There are 2 phases of 5G

Secure Socket Layer

The protocols for Ethernet in datalink layer is 802.3 

(c) Both a & b

1. web

2. Protocol Converter

The term Li-Fi was first used by Harald Haas

(a) Light Fidelity

4G generation of mobile network is considered as a research stage

Sending e-mail to a friend is an example for Internet

(a) I-True, II-False

(i) Multiprotocol Label Switching

(ii) Over the Air

RFID used for radio wares to read and capture information stored on a tag attached to an object

International Telecommunication Union

(c) Hyper Text Transfer Protocol

There are 7 OSI layers

OSI model was found in 1934

(c) Induction coupling

(а) Active RFID System – larger distances

(a) File Transfer Protocol

Open System Interconnection

Answer is (b) RFID

The generations of mobile networks are as follows.

1. First Generation(lG) 1981- NMT launch

2. Second Generation(2G) 1991-GSM Launch

3. Second to Third Generation Bridge (2.5)2000 – GPRS launch ,

4. Third Generation( 3G) 2003- UK 3G launch

5. Fourth Generation (4G) 2007

6. Fifth Generation (5G) 2019+

EXTRANET: 

It is a private network that uses Internet technology and the public telecommunication system to securely share business’s information with suppliers, vendors, partners, customers, or other businesses.

There are 4 different layers of TCP/IP.

(i) User Daragram Protocol

(ii) Domian Name System.

Transmission Control Protocol/Internet Protocol

Answer is (a) I, II

Merits of Mobile Networks:

1. Higher efficiency.

2. Increased ability to communicate in and out of the workspace.

3. Greater access to modem apps and services.

4. Improved networking capabilities.

5. Quality and flexibility of services.

6. Rapid developments in cloud technologies.

Demerits of Mobile Networks:

1. Cost 

2. Vulnerable to security risks. 

3. Additional training is needed to use new technology.

4. Cyber crime

Development:

The generations of mobile networks are as follows.

1. First Generation(lG) 1981- NMT launch

2. Second Generation(2G) 1991- GSM Launch

3. Second to Third Generation Bridge (2.5)2000 – GPRS launch

4. Third Generation! 3G) 2003 - first UK 3G launch

5. Fourth Generation (4G) 2007

6. Fifth Generation (5G) 2019+

1. First Generation (1G) 1981 – NMT launch: 

1. During the initial periods the mobile systems were based on analog trAnswer:mission.

2. NMT stands for Nordic Mobile Telephone communication.

3. And a very poor voice quality.

2. Second Generation(2G) 1991-GSM Launch: 

1. Later the second generation of mobile systems were placed on digital trAnswer:mission with GSM.

2. GSM stands for (Global System for Mobile communication) was most popular standard which is used in second generation, using 900MHz and 1800MHz for the frequency bands.

3. The transfer mission used as TMDA stands for (Time Division Multiple Access) and CDMA One stands for (Code Division Multiple’Access) method to increase the amount of information trAnswer:ported on the network

3. Second to Third Generation Bridge (2.5)2000 – GPRS launch:

1. GPRS was introduced here GPRS stands for (General Packet Radio Service). GPRS is a data service which enables mobile devices to send and receive messages, picture messages and emails.

2. GSM data transfer mission rates typically reached 9.6kbit/s.

4. Third Generation( 3G) 2003- first UK 3G launch:

1. This generation of mobile systems merges different mobile technology standards, and uses higher frequency bands for transfer mission and Code Division Multiple Access to delivery data rates of up to 2Mbit/s supports multimedia services (MMS: voice, video and data).

2Mbit/s supports multimedia services (MMS: voice, video and data).

2. Data transfer mission used a WCDMA. WCDMA stands for (Wideband Code Division Multiple Access).

3. Few 3G suppliers use ATM (Asynchronous Transfer Mode) for their ‘over the air’ network with in MPLS (Multiprotocol Label Switching) or IP for theirs backbone network.

5. Fourth Generation (4G) 2007: 

1. 4G is at the research stage. 4G was based on an adhoc networking model where there was no need for a fixed infrastructure operation.

2. Adhoc networking requires global mobility features (e.g. Mobile IP) and connectivity to a global IPv6 network to support an IP address for each mobile device.

3. Logically roaming in assorted IP networks (for example: 802.11 WLAN, GPRS and UMTS) were be possible with higher data rates, from 2Mbit/s to 10- 100Mbit/s, offering reduced delays and newly services.

6. Fifth Generation (5G) 2019+:

1. 5G is the stage succeeds the 4G (LTE/ WiMAx), 3G(units) and 2G(GSM) systems.

2. 5G targets to performance the high data rate, reduced latency, energy saving, cost reduction, higher system, capacity, and massive device connectivity.

3. The ITU IMT – 2020 provides speeds up to 20 gigabits per second it has been demonstrated with millimeter waves of 15 gigahertz and higher frequency.

Answer is (a) (i), (ii)

The Application layer of the TCP/IP model is similar to the Session, Presentation, and Application layers of the OSI Reference Model.

The most popular Application layer protocols are:

(i) Hypertext Transfer Protocol (HTTP): The core protocol of the World Wide Web.

(ii) File Transfer Protocol (FTP): enables a client to send and receive complete files from a server.

(iii) Telnet: connect to another computer on the Internet.

(iv) Simple Mail Transfer Protocol (SMTP): Provide e-mail services.

(v) Domain Name System (DNS): Refer to other host computers by using names rather than numbers

Network Interface Layer: 

1. It is the bottom most level layer.

2. It is comparable to that of the Open System Interconnection Physical and Data Link layers.

3. Different TCP/IP protocols are being used at this layer, Ethernet and Token Ring for local area networks and protocols such as X.25, Frame Relay, and ATM for wide area networks.

4. It is assumed to be an unreliable layer.

Gmail is not the social media

(a) Mobile devices

(d) Social media

(a) Work function,

ϕ₀ = 1.9 eV

= 1.9 x 1.6 x 10^-19 J

λ₀ = {6.6 x 10^-34 Js x 3 x 10⁸ m/s}/{1.9 x 1.6 x 10^-19 J}

= 652 nm

(b) ϕ = hv₀

v₀ = ϕ/h = {1.9 x 1.6 x 10^-19}/{6.63 x 10^-34}

= 4.598 x 14 s^-1

Accuracy: Accuracy refers to the difference between true value and that obtained from a very large number of tests.

Precision: Precision refers to degree of reproducibility of results of an experiment or method as distinct from its accuracy. The precision of a measurement depends upon the measuring device and the skill with which it is used.

For example, distance between two cities is 200 km. The distance between two cities can be measured by an odometer fitted in automobile, (car, bus, truck). If an odometer records 1 km for every kilometer passed over, it will record the distance between two cities as 200 km. If it records 0.9 km for every 1 km passed over, the distance recorded will be 180 km. Therefore, the odometer reading is not accurate.

But if the odometer records 180 km each time it is used for the trip, the odometer reading is precise. The measurement has high degree of precision between the same reading (180 km) is reproduced time after time but the accuracy is poor because the odometer is not calibrated properly and there is difference between the exact value (200 km) and the value recorded (180 km).

Determination of Empirical formula:- The determination of empirical formula of an organic compound involves the following steps:

1. From quantitative analysis, the percentage of the various elements present in the compound is found out.

2. By dividing the percentage of each element by its atomic mass, the relative number of atoms of various elements is known.

3. Divide the number obtained by the smallest of them so as to get a simple atomic ratio.

4. Multiply the simple atomic ratio of fraction by some suitable integer to get a whole number ratio.

5. The symbols of various elements are written side by side with the above whole numbers as subscripts to the lower side of each to get empirical formula of the compound.

Determination of Molecular formula:- The determination of molecular formula of an organic compound involves the following steps:

1. Calculate the empirical formula as stated above.

2. Determine the empirical formula mass by adding the atomic masses of the constituent atoms taken in the ratio in which they are present.

3. Divide the molecular mass of the compound by empirical formula mass to get the value of n.

4. Multiply the empirical formula with n so as to get the molecular formula of the compound.

Correct option: (B) different molecular weights

The compound is glucose. 

Molecular formula = C₆H₁₂O₆ 

Empirical formula = CH₂O

Correct option: (A) CO₂ 

Correct option: (C) C₆H₆ 

Empirical formula mass = CH = 12 + 1 = 13

Molecular mass = 78

r = \(\cfrac{Molecular\, mass}{Empirical\, mass}\) = \(\cfrac{78}{13}\) = 6

Molecular formula = r x Empirical formula

= 6 x CH

= C₆H₆

• Steps involved in the calculation of empirical formula : 

• Convert the mass percentage into grams: considering 100 g of the compound, the given mass percentages represent the masses of the elements in grams.
• Calculate the number of moles of each element.
• Calculate the simplest molar ratio: divide the moles obtained by the least value from amongst the values obtained for each element.
• Calculate the simplest whole number ratio: multiply all the simplest atomic ratios by a suitable integer.
• Write the empirical formula: write the symbols of each element along with their whole number ratios side by side.

• Molecular formula = empirical formula x n

n = 1,2,3,…..etc.

n =  \(\frac{molar\, mass }{empirical \,formula \,mass}\)

Correct option: (B) CH₂O₂ 

Empirical formula of an acid is CH₂O₂

(Empirical formula) x n = Molecular formula

n = whole number multiple i.e., 1,2,3,4....

If n = 1 molecular formula CH₂O₂.

(i) Both A and R are true and R is the correct explanation of A.

Plastics are non-biodegradable in nature. Over usage of plastic bags can lead to accumulation of plastic waste in garbage cans. Disposal of plastic wastes is an issue. Burning plastic wastes in air would pollute the air. Disposing plastic in water endangers aquatic life and pollutes water. Since plastic is non-biodegradable, it cannot be decomposed by natural processes. Therefore, the usage of plastics should be minimised.