UrbanPro
true

Abhinav Raj

Wagholi, Pune, India - 412207

Abhinav Raj Engineering Entrance trainer in Pune

Abhinav Raj

Maths Tutor

Wagholi, Pune, India - 412207.

Referral Discount: Get ₹ 500 off when you make a payment to start classes. Get started by Booking a Demo.

Details verified of Abhinav Raj

Identity

Education

Know how UrbanPro verifies Tutor details

Identity is verified based on matching the details uploaded by the Tutor with government databases.

Overview

I completed my B.Tech in Mechanical Engineering from NIT Silchar. Later I worked for Fiat India Automobile Pvt ltd. From 3rd year of my college I have been teaching Maths to 10+2 students for their board exams and various engineering entrance exam. My students have done exceptionally well in their exams. Once you get the concepts of maths you can easily solve problems. Maths enhances you imagination and analytical skills.

Languages Spoken

English

Hindi

Education

NIT Silchar 2017

Bachelor of Technology (B.Tech.)

Address

Wagholi, Pune, India - 412207

Verified Info

Phone Verified

Email Verified

Facebook Verified

Report this Profile

Is this listing inaccurate or duplicate? Any other problem?

Please tell us about the problem and we will fix it.

Please describe the problem that you see in this page.

Type the letters as shown below *

Please enter the letters as show below

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

2

Engineering Entrance Exams

IIT JEE Coaching Classes, BITSAT Coaching Classes, Delhi CEE Coaching Classes

IITJEE Coaching

IIT JEE Integrated Coaching, IIT JEE Foundation Course, IIT JEE Mains Coaching, IIT JEE Crash Course

Type of class

Regular Classes, Crash Course

IIT-JEE Subjects

Maths

Reviews

this is test message this is test message this is test message this is test message this is test message this is test message this is test message

No Reviews yet! Be the first one to Review

FAQs

1. Which classes do you teach?

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

2. Do you provide a demo class?

Yes, I provide a free demo class.

3. How many years of experience do you have?

I have been teaching for 2 years.

Answers by Abhinav Raj (2)

Answered on 09/01/2018 Tuition/Class XI-XII Tuition (PUC)

Sec(360)=1/cos(360). If we take a stick of length l and fix one of its end to the origin. It is free to rotate from its another end and will make circle at Cartesian coordinate plane. Rotate it by 90º it will lie on positive y axis. Similarly rotate it by 180° it will lie on negative x axis.... ...more

Sec(360)=1/cos(360). 

If we take a stick of length l and fix one of its end to the origin. It is free to rotate from its another end and will make circle at Cartesian coordinate plane. Rotate it by 90º it will lie on positive y axis. Similarly rotate it by 180° it will lie on negative x axis. If we rotate it by 360° we will find that it came in its original position i.e lies on positive x axis. Imagine a position where stick was just about complete. Take the projection of stick on x axis. We get a right angle triangle in third quadrant. Calling the angle between positive x axis and stick as θ. Now cosθ=(projection of stick on positive x axis) / (length of stick=l). 

As θ decreases. Projection of stick will increase and when stick completes its 360° rotation it will be equal to l. 

At that time 

Cosθ = l/l =1.

⇒secθ = 1/cosθ =1/1 =1 Ans. 

Answers 213 Comments
Dislike Bookmark

Answered on 09/01/2018 CBSE/Class 11/Science/Physics Tuition/Class XI-XII Tuition (PUC)

A particle is thrown upwards. It attains a height (h) after 5 seconds and again after 9s comes back.... ...more
A particle is thrown upwards. It attains a height (h) after 5 seconds and again after 9s comes back. What is the speed of the particle at a height h?

Total time taken by the particle to go up and come back = 5+9 =14s. Therefore it must take 7s to go up and 7s to come down. From first law, we know v=u+at At the top point velocity will be zero. Therefore v=0. 0=u-10*7 (taking a=g=-10m/s², negative sign indicates that acceleration is in downward... ...more

Total time taken by the particle to go up and come back = 5+9 =14s.

Therefore it must take 7s to go up and 7s to come down. 

From first law, we know

v=u+at

At the top point velocity will be zero. Therefore v=0.

0=u-10*7

(taking a=g=-10m/s²,  negative sign indicates that acceleration is in downward direction). Solving the equation we will get

u=70m/s.

It is given that after 5s its height was h. 

Again applying first law. 

v=u+at

v=70-10*5

v= 20m/s

Particle speed was 20m/s in upward direction at height h after 5s of throwing. 

Answers 1 Comments
Dislike Bookmark
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

2

Engineering Entrance Exams

IIT JEE Coaching Classes, BITSAT Coaching Classes, Delhi CEE Coaching Classes

IITJEE Coaching

IIT JEE Integrated Coaching, IIT JEE Foundation Course, IIT JEE Mains Coaching, IIT JEE Crash Course

Type of class

Regular Classes, Crash Course

IIT-JEE Subjects

Maths

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

2

Board

CBSE, State, ISC/ICSE

ISC/ICSE Subjects taught

Mathematics

CBSE Subjects taught

Mathematics

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics

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

2

Board

CBSE, State, ISC/ICSE

ISC/ICSE Subjects taught

Mathematics

CBSE Subjects taught

Mathematics

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics

BTech Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in BTech Tuition

1

BTech Mechanical subjects

Dynamics of Machinery, Strength of Materials, Engineering Drawing & Graphics, Automobile Engineering, Fluid Mechanics, Heat & Mass Transfer, Manufacturing Technology, Analysis and Design of Machine Components, Kinematics of Machinery, Machine Design, Internal Combustion Engines and Emissions, Mechanics of Machines, Material Science and Metallurgy

BTech Branch

BTech Mechanical Engineering

Type of class

Regular Classes, Crash Course

Class strength catered to

Group Classes, One on one/ Private Tutions

Taught in School or College

Yes

this is test message this is test message this is test message this is test message this is test message this is test message this is test message

No Reviews yet! Be the first one to Review

Answers by Abhinav Raj (2)

Answered on 09/01/2018 Tuition/Class XI-XII Tuition (PUC)

Sec(360)=1/cos(360). If we take a stick of length l and fix one of its end to the origin. It is free to rotate from its another end and will make circle at Cartesian coordinate plane. Rotate it by 90º it will lie on positive y axis. Similarly rotate it by 180° it will lie on negative x axis.... ...more

Sec(360)=1/cos(360). 

If we take a stick of length l and fix one of its end to the origin. It is free to rotate from its another end and will make circle at Cartesian coordinate plane. Rotate it by 90º it will lie on positive y axis. Similarly rotate it by 180° it will lie on negative x axis. If we rotate it by 360° we will find that it came in its original position i.e lies on positive x axis. Imagine a position where stick was just about complete. Take the projection of stick on x axis. We get a right angle triangle in third quadrant. Calling the angle between positive x axis and stick as θ. Now cosθ=(projection of stick on positive x axis) / (length of stick=l). 

As θ decreases. Projection of stick will increase and when stick completes its 360° rotation it will be equal to l. 

At that time 

Cosθ = l/l =1.

⇒secθ = 1/cosθ =1/1 =1 Ans. 

Answers 213 Comments
Dislike Bookmark

Answered on 09/01/2018 CBSE/Class 11/Science/Physics Tuition/Class XI-XII Tuition (PUC)

A particle is thrown upwards. It attains a height (h) after 5 seconds and again after 9s comes back.... ...more
A particle is thrown upwards. It attains a height (h) after 5 seconds and again after 9s comes back. What is the speed of the particle at a height h?

Total time taken by the particle to go up and come back = 5+9 =14s. Therefore it must take 7s to go up and 7s to come down. From first law, we know v=u+at At the top point velocity will be zero. Therefore v=0. 0=u-10*7 (taking a=g=-10m/s², negative sign indicates that acceleration is in downward... ...more

Total time taken by the particle to go up and come back = 5+9 =14s.

Therefore it must take 7s to go up and 7s to come down. 

From first law, we know

v=u+at

At the top point velocity will be zero. Therefore v=0.

0=u-10*7

(taking a=g=-10m/s²,  negative sign indicates that acceleration is in downward direction). Solving the equation we will get

u=70m/s.

It is given that after 5s its height was h. 

Again applying first law. 

v=u+at

v=70-10*5

v= 20m/s

Particle speed was 20m/s in upward direction at height h after 5s of throwing. 

Answers 1 Comments
Dislike Bookmark

Abhinav Raj describes himself as Maths Tutor. He conducts classes in BTech Tuition, Class 11 Tuition and Class 12 Tuition. Abhinav is located in Wagholi, Pune. Abhinav takes at students Home, Regular Classes- at his Home and Online Classes- via online medium. He has 2 years of teaching experience . Abhinav has completed Bachelor of Technology (B.Tech.) from NIT Silchar in 2017. HeĀ is well versed in English and Hindi.

X
X

Post your Learning Need

Let us shortlist and give the best tutors and institutes.

or

Send Enquiry to Abhinav Raj

Let Abhinav Raj know you are interested in their class

Reply to 's review

Enter your reply*

1500/1500

Please enter your reply

Your reply should contain a minimum of 10 characters

Your reply has been successfully submitted.

UrbanPro.com is India's largest network of most trusted tutors and institutes. Over 55 lakh students rely on UrbanPro.com, to fulfill their learning requirements across 1,000+ categories. Using UrbanPro.com, parents, and students can compare multiple Tutors and Institutes and choose the one that best suits their requirements. More than 7.5 lakh verified Tutors and Institutes are helping millions of students every day and growing their tutoring business on UrbanPro.com. Whether you are looking for a tutor to learn mathematics, a German language trainer to brush up your German language skills or an institute to upgrade your IT skills, we have got the best selection of Tutors and Training Institutes for you. Read more