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Kapil Jha Class 9 Tuition trainer in Ahmedabad

Kapil Jha

IIT Foundation Expert

Chandlodia, Ahmedabad, India - 382481.

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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

I am an IIT Graduate, 15 Years of Experience. Good command over Physics and Maths.

Languages Spoken

Hindi

English Proficient

Education

IIT Roorkee 2000

Bachelor of Technology (B.Tech.)

Address

Chandlodia, Ahmedabad, India - 382481

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Teaches

Class 9 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 9 Tuition

15

Board

ICSE, CBSE

CBSE Subjects taught

Mathematics, Science

ICSE Subjects taught

Mathematics, Chemistry, Physics

Taught in School or College

No

Class 10 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 10 Tuition

15

Board

ICSE, CBSE

CBSE Subjects taught

Mathematics, Science

ICSE Subjects taught

Mathematics, Chemistry, 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

Engineering Entrance Exams

IIT JEE Coaching Classes

Courses

Reviews

No Reviews yet!

FAQs

1. Which school boards of Class 10 do you teach for?

ICSE, CBSE

2. Do you have any prior teaching experience?

No

3. Which classes do you teach?

I teach Class 10 Tuition, Class 9 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 15 years.

Answers by Kapil Jha (1)

Answered on 28/06/2017 Learn Exam Coaching/Engineering Entrance Coaching/IIT JEE Coaching +1 Exam Coaching/Engineering Entrance Coaching/IIT JEE Coaching/Physics

s=Ut+(1/2)at^2 Here U=0, t=0.2 Sec, and a=-12 m/s^2 {-ve sign is because its descending object} So Displacement S = (0*0.2) + 1/2 (-12*0.2*0.2) =-0.24 m
Answers 25 Comments
Dislike Bookmark

Teaches

Class 9 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 9 Tuition

15

Board

ICSE, CBSE

CBSE Subjects taught

Mathematics, Science

ICSE Subjects taught

Mathematics, Chemistry, Physics

Taught in School or College

No

Class 10 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 10 Tuition

15

Board

ICSE, CBSE

CBSE Subjects taught

Mathematics, Science

ICSE Subjects taught

Mathematics, Chemistry, 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

Engineering Entrance Exams

IIT JEE Coaching Classes

Courses

No Reviews yet!

Answers by Kapil Jha (1)

Answered on 28/06/2017 Learn Exam Coaching/Engineering Entrance Coaching/IIT JEE Coaching +1 Exam Coaching/Engineering Entrance Coaching/IIT JEE Coaching/Physics

s=Ut+(1/2)at^2 Here U=0, t=0.2 Sec, and a=-12 m/s^2 {-ve sign is because its descending object} So Displacement S = (0*0.2) + 1/2 (-12*0.2*0.2) =-0.24 m
Answers 25 Comments
Dislike Bookmark

Kapil Jha describes himself as IIT Foundation Expert. He conducts classes in Class 10 Tuition, Class 9 Tuition and Engineering Entrance Coaching. Kapil is located in Chandlodia, Ahmedabad. Kapil takes Regular Classes- at his Home and Online Classes- via online medium. He has 15 years of teaching experience . Kapil has completed Bachelor of Technology (B.Tech.) from IIT Roorkee in 2000. HeĀ is well versed in Hindi and English.

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