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11 months ago

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P
Payal Gupta

Contributor-Level 10

13.2 Atomic mass of Ne1020 neon isotope, m1 = 19.99 u ad the abundance η1 = 90.51 %

Atomic mass of Ne1021 neon isotope, m2 = 20.99 u ad the abundance η2 = 0.27 %

Atomic mass of Ne1022 neon isotope, m3 = 21.99 u ad the abundance η3 = 9.22 %

The average atomic mass of neon is given as:

m = m1η1+m2η2+m3η3η1+η2+η3 = 19.99×90.51+20.99×0.27+21.99×9.2290.51+0.27+9.22 = 2017.71100 = 20.1771 u

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11 months ago

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Vishal Baghel

Contributor-Level 10

  1. Physical appearance: All the alkali metals are silvery white, soft and light metals.
  2. Density: Because of the large size, these elements have low density which increases down the group except for potassium which is lighter than sodium (most likely due to an unexpected increase in the atomic size.).
  3. The melting and boiling points of the alkali metals are low indicating weak metallic bonding due to the presence of only a single valence electron in them.
  4. Atomic volume:The atomic volume, atomic and ionic radii rise as the group number reduces from Li to Cs.
  5. Melting and boiling points:The weak crystal lattice bonding causes low melting and boili
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11 months ago

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11 months ago

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Jagriti Shukla

Contributor-Level 8

Here are the steps to apply for Google scholarships:

  1. Research: Find open Google scholarships on their official site.
  2. Check Eligibility: Ensure you meet all requirements.
  3. Prepare Documents: Resume, transcripts, recommendation letters, essays.
  4. Apply Online: Fill form, upload documents.
  5. Submit: Review and submit before the deadline.
  6. Follow Up: Check email for updates.

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11 months ago

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Shiksha Ask & Answer
Abhishek singh

Contributor-Level 10

The minimum cutoff for a B.Tech in Computer Science and Engineering (CSE) at Indian Institutes of Technology (IITs) generally varies based on the specific IIT, the category of the candidate, and the year of admission. For top IITs like Bombay, Delhi, and Madras, the closing ranks for the General category can be below 200, while for other IITs, it may be in the range of 250 to 1000. For newer IITs, the closing ranks can range from 3000 to 6000. 

The UPSC cut off marks is 75.41 for Prelims, and 741 for Mains. The final CSE cut off marks is 947 for the UPSC CSE 2024 based on the cumulative marks obtained by candidates in the Mains and

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11 months ago

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Sanjana Dixit

Contributor-Level 10

No, candidates looking for course admission at Alliance School of Computer Science and Engineering cannot enrol for direct admissions. The institute admits students based on various national-level entrance exam scores, such as JEE Main/JEE Advance/KCET/CUET, etc., for various courses such as BTech, MCA, among others. It also considers the candidates' academic performance for final selection. The college also uses other parameters, such as scholastic and extracurricular records, for admission. The final selection of the candidate will be based on the overall performance in the Alliance admission selection process.

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11 months ago

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Nishtha Shukla

Guide-Level 15

Yes, Mohammad Ali Jauhar University offers Diploma courses through Mother Teresa Institute of Paramedical Sciences. The university offers Diploma in Optometry, Operation Theatre Technician, Physiotherapy and Nursing. Candidates must complete Class 12 in relevant subjects with at least 45% aggregate to apply for diploma. 

Besides, the university also offers Diploma in various Engineering specialisations such as Civil, Computer Science, Electrical, Automobile, etc. Candidates can apply for Diplom in Engineering after completing Class 10 from a recognised board. Besides, the university also offers Diploma in Taxation/ Finance Accounti

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11 months ago

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Satyam Fashion Institute,noida Noida

Beginner-Level 3

Candidates can join Satyam School of Journalism and Mass Communication (SSJMC) by following a straightforward admission process. Interested applicants must first meet the eligibility criteria, which typically includes passing the 10+2 examination for undergraduate programs and holding a relevant bachelor's degree for postgraduate courses. Admissions are generally merit-based, and candidates may be required to fill out an application form available on the official website or at the campus. After submitting the form along with necessary documents and the application fee, shortlisted candidates may be invited for a personal interview or c
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11 months ago

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Payal Gupta

Contributor-Level 10

14.15 A acts as two inputs of the NOR gate and Y is the output. As shown in the following figure. Hence the output of the circuit is A + A ? = A ?

The truth table for the same is given as:

A

Y = ( A ? )

0

1

1

0

This is the truth table of a NOT gate. Hence, this circuit functions as a NOT gate.

A and B are the inputs and Y is the output of the given circuit. By using the result obtained in solution (a), we can infer that the outputs of the first two NOR gates are A ? and B ?  , as shown in the following figure

Above is given the inputs for the last NOR gate.

Hence, the output for the circuit can be written as:

Y = A + B ? = A ? . B ? ?   = A.B

The truth table for the same can b

...more

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11 months ago

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A
Aryan Saini

Beginner-Level 5

With 422 marks in the 12th grade and a cutoff score of 105, your chances of getting into a good college in Chennai are good, particularly for private engineering colleges or those with lower cutoffs.Here are some colleges in Chennai - SSN College of Engineering, Chennai Institute of Technology (CIT), Vel Tech Rangarajan Dr.Sagunthala R&D Institute of Science and Technology.

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11 months ago

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Nishtha Shukla

Guide-Level 15

Mohammad Ali Jauhar University offers more than 50 courses at UG, PG, diploma, certificate and doctoral levels. The university offers programmes in various streams such as Engineering, Management, Arts, Science, etc. Admission to most of the programmes offered at the university is merit-based. The university conducts an admission entrance test to select candidates for a few programmes such as BTech, MBA, B.Ed, BPharm, etc.

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11 months ago

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Nishtha

Contributor-Level 10

The institute admits students based on various national-level entrance exam scores, such as JEE Main/JEE Advance/KCET/CUET, etc., for various courses such as BTech, MCA, among others. It also considers the candidates' academic performance for final selection. Students can check the table below to know detail course-wise selection criteria at Alliance School of Computer Science and Engineering:

CoursesSelection Criteria
BTechJEE (Main)/JEE (Advanced)/Karnataka CET/COMED-K/AUEET  (Alliance Undergraduate Engineering Entrance Test)/ or any other State-Level engineering entrance examination + PI
MTechNationalised Management Test Scores and Alliance University M.Tech Computer Science and Engineering Entrance Test +

Academic Performance, Oral Extempore Presentation and Personal Interview Performance

MCACUET score/Alliance University Master of Computer Application Entrance Test + Oral Extempore Presentation
PhDAlliance Research Aptitude Test (ARAT) +

Personal Interview + Evaluation of the research synopsis on a proposed dissertation work + Other parameters such as scholastic and extracurricular record, and work experience

New Question

11 months ago

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P
Payal Gupta

Contributor-Level 10

14.14 A and B are the inputs of the given circuit. The output of the first NOR gate is A ? +B? . It can be observed from the following figure that the inputs of the second NOR gate become the output of the first one.

Hence, the output of the combination is given as:

Y = A + B ? + A + B ? ?  =  A ? . B ? ?  + A ? . B ? ?    = = A ? . B ? ?    = A ?  + B ?  = A + B ?

The truth table for this operation is given as:

This is the truth table of an or gate. Hence, this circuit functions as an or gate.

A

B

Y ( = A + B)

0

0

0

0

1

1

1

0

1

1

1

1

This is the truth table of an OR gate. Hence, this circuit functions as an OR gate.

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11 months ago

0 Follower 6 Views

P
Payal Gupta

Contributor-Level 10

14.13 The output of the left NAND gate will be A . B ? , as shown in the following figure:

Hence, the output of the combination of two NAND gates is given as:

Y = ( A . B ? ).( A . B ? ) = A . B ? + A . B ?  = AB

Hence the circuit functions as an AND gate.

A ? is the output of the upper left of the NAND gate and B ?  is the output of the lower half of the NAND gate, as shown in the following figure.

Hence, the output of the combination of the NAND gates will be given as:

Y = A ? . B ?  = A ? + B ?  = A + B

Hence, this circuit functions as an OR gate.

New Question

11 months ago

0 Follower 1 View

I
Ishita Uniyal

Contributor-Level 9

The tables given below provide an overview of some popular Mass Communication and Media courses' syllabi for candidates' reference, and given below is the semester-wise curriculum of the Mass Communication and Media course:

MCM Syllabus Semester I

Writing for Media

Introduction to Communication

Socio-Economic & Political Scenario

Introduction to Journalism (Reporting, Writing & Editing)

MCM Syllabus Semester II

History of Print & Broadcasting in India

Media Laws & Ethics

Print Journalism-I

Still Photography

MCM Syllabus Semester III

Development & Communication

Radio Journalism & Production

Basics of Camera, Lights & Sound

Print Journalism-II

MCM Syllabus Semester IV

Public Relations

Introduction to Advertising

New Media

Public Relations Lab

MCM Syllabus Semester V

Advertising Practices

Media Research

Event Management: Principles & Methods

Environment Communication

MCM Syllabus Semester VI

Media Organisation & Management

Contemporary Issues

Global Media Scenario

Value Education

Note: The information is sourced from external sites and may vary.

New Question

11 months ago

0 Follower 5 Views

N
Nishtha Shukla

Guide-Level 15

Mohammad Ali Jauhar University offers admission to the BCA course based on merit of the aspirants in the last qualifying exam. Candidates must complete Class 12 with a minimum of 45% aggregate in any stream from any recognised board for admission to the course. Selected candidates have to report at the campus for document verification and fee payment to confirm their seat.

New Question

11 months ago

0 Follower 20 Views

P
Payal Gupta

Contributor-Level 10

14.12 A acts as the two inputs of the NAND gate and Y is the output, as shown in the following figure.

Hence, the output can be written as:

Y = A + A ? = A ?  + A ? = A ?  ……………(i)

The truth table for equation (i) can be drawn as:

A               

Y = ( A ? )

0

1

1

0

This circuit functions as a NOT gate. The symbol for this logic circuit is as shown below:

New Question

11 months ago

0 Follower 2 Views

S
Shikha Arora

Contributor-Level 10

ASCSE offers BTech, MCA, and M.Tech courses at the UG and PG levels. Apart from the UG and PG courses, the institute also offers PhD courses. The selection criteria are entrance-based. The institute admits students based on various national-level entrance exam scores, such as JEE Main/JEE Advance/KCET/CUET, etc., for various courses such as BTech, MCA, among others. It also considers the candidates' academic performance for final selection. The college also uses other parameters, such as scholastic and extracurricular records, for admission.

New Question

11 months ago

0 Follower 3 Views

S
saurya snehal

Contributor-Level 10

Candidates wishing to pursue the MA course at CRSU can choose from the following specialisations: 

  • M.A. in History
  • M.A. in Hindi
  • M.A. in Psychology
  • M.A. in Fine Arts
  • M.A. in English
  • M.A. in Economics
  • M.A. in Political Science

Candidates can choose from the above mentioned and a plethora of other specialisations for the course. 

 

 

New Question

11 months ago

0 Follower 12 Views

P
Payal Gupta

Contributor-Level 10

14.11 A and B are the inputs and Y is the output of the given circuit. The left half of the given figure acts as the NOR gate, while the right half acts as the NOT gate

Hence the output of the NOR gate is A + B ?

This will be input for the NOT gate. Its output will be A + B ? = A + B

So Y = A + B

Hence, this circuit functions as an OR gate.

A and B are the inputs and Y is the output of the given circuit. It can be observed from the following figure that the inputs of the right half NOR gate are the outputs of the two NOT gates.

Hence, the output of the given circuit can be written as:

Y = A + B ? = A + B ? = A + BHence this circuit functions as an AND gate.

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