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Dilip Jha Class 11 Tuition trainer in Delhi/>

Dilip Jha

Physics Expert

Ganesh Nagar, Delhi, India - 110092.

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Referral Discount: Get ₹ 250 off when you make a payment to start classes. Get started by Booking a Demo.

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Overview

Excellent problem solving skills in physics, In-depth knowledge of physics.

Languages Spoken

Bengali

Hindi

English

Education

Meerut university

Master of Science (M.Sc.)

Address

Ganesh Nagar, Delhi, India - 110092

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Teaches

Class 11 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 11 Tuition

17

Board

CBSE

CBSE Subjects taught

Mathematics, Physics

Taught in School or College

No

Class 12 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 12 Tuition

17

Board

CBSE

CBSE Subjects taught

Mathematics, Physics

Taught in School or College

No

Engineering Entrance Coaching classes

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Engineering Entrance Coaching classes

17

Engineering Entrance Exams

IIT JEE Coaching Classes

Type of class

Regular Classes, Crash Course

Reviews

No Reviews yet!

FAQs

1. Which school boards of Class 12 do you teach for?

CBSE

2. Have you ever taught in any School or College?

No

3. Which classes do you teach?

I teach Class 11 Tuition, Class 12 Tuition and Engineering Entrance Coaching Classes.

4. Do you provide a demo class?

Yes, I provide a free demo class.

5. How many years of experience do you have?

I have been teaching for 17 years.

Answers by Dilip Jha (3)

Answered on 01/04/2016 Learn Exam Coaching/Engineering Entrance Coaching/IIT JEE Coaching

T=KN^3, T=TIME PERIOD, N=PRINCIPAL QUANTUM NUMBER, K=CONSTANT. Therefore the sought ratio is 1/8.
Answers 4 Comments
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Answered on 01/04/2016 Learn Exam Coaching/Engineering Entrance Coaching/IIT JEE Coaching

working in the frame of the heavy(so assumed relative to the electron) nucleus and there fore non accelerating-the centripetal force is provided by the inward electrostatic force of attraction between the electron and the nucleus, so equating these two we get one equation connecting speed of electron... ...more
working in the frame of the heavy(so assumed relative to the electron) nucleus and there fore non accelerating-the centripetal force is provided by the inward electrostatic force of attraction between the electron and the nucleus, so equating these two we get one equation connecting speed of electron v and radius of orbit r. we can get another equation between these two from Bohr's second postulate of quantization of angular momentum. this later equation will involve the principal quantum number n whose value for first and second orbit is 1 and 2 respectively. from these two equations we can solve for v and r.. then the time period would be T=2(pi)r/v. it will then turn out that T is directly proportional to n^3. THUS THE SOUGHT RATION WOULD BE 1^3/2^3=1/8.
Answers 4 Comments
Dislike Bookmark

Teaches

Class 11 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 11 Tuition

17

Board

CBSE

CBSE Subjects taught

Mathematics, Physics

Taught in School or College

No

Class 12 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 12 Tuition

17

Board

CBSE

CBSE Subjects taught

Mathematics, Physics

Taught in School or College

No

Engineering Entrance Coaching classes

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Engineering Entrance Coaching classes

17

Engineering Entrance Exams

IIT JEE Coaching Classes

Type of class

Regular Classes, Crash Course

No Reviews yet!

Answers by Dilip Jha (3)

Answered on 01/04/2016 Learn Exam Coaching/Engineering Entrance Coaching/IIT JEE Coaching

T=KN^3, T=TIME PERIOD, N=PRINCIPAL QUANTUM NUMBER, K=CONSTANT. Therefore the sought ratio is 1/8.
Answers 4 Comments
Dislike Bookmark

Answered on 01/04/2016 Learn Exam Coaching/Engineering Entrance Coaching/IIT JEE Coaching

working in the frame of the heavy(so assumed relative to the electron) nucleus and there fore non accelerating-the centripetal force is provided by the inward electrostatic force of attraction between the electron and the nucleus, so equating these two we get one equation connecting speed of electron... ...more
working in the frame of the heavy(so assumed relative to the electron) nucleus and there fore non accelerating-the centripetal force is provided by the inward electrostatic force of attraction between the electron and the nucleus, so equating these two we get one equation connecting speed of electron v and radius of orbit r. we can get another equation between these two from Bohr's second postulate of quantization of angular momentum. this later equation will involve the principal quantum number n whose value for first and second orbit is 1 and 2 respectively. from these two equations we can solve for v and r.. then the time period would be T=2(pi)r/v. it will then turn out that T is directly proportional to n^3. THUS THE SOUGHT RATION WOULD BE 1^3/2^3=1/8.
Answers 4 Comments
Dislike Bookmark

Dilip Jha describes himself as Physics Expert. He conducts classes in Class 11 Tuition, Class 12 Tuition and Engineering Entrance Coaching. Dilip is located in Ganesh Nagar, Delhi. Dilip takes Regular Classes- at his Home and Online Classes- via online medium. He has 17 years of teaching experience . Dilip has completed Master of Science (M.Sc.) from Meerut university. He is well versed in Bengali, Hindi and English.

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