Manoj J. -  Tutor in Mumbai

Manoj J.   (On Spanedea since May 12, 2013)

IIT Bombay - Physics Tutor


B-Tech + M-Tech from IIT Bombay in Aerospace Engineering, 2014

Tutoring Experience: 5 Years 4 Months

Total Hours Taught on Spanedea: 162 Hours

Tutor Expertise:



SAT II - Subject Tests

 & 1 others

Contact: Not Shared

About Manoj Jaiswal

I am a Final year Aerospace Engineering student at IIT Bombay. I have a teaching experience of more than 3 years.
Have secured AIR 621 in JEE and AIR 231 in AIEEE. Have worked on an International project by MIT and EU, and in addition have authored a Paper which is going to be published in the International 19th APDR Congress. I love teaching Physics and Maths.


Physics is a concept based subject and so I want to impart concepts from the scratch rather than just making students to mug up the solutions. So that they would be able to solve any seen/unseen problem at a glance.

Educational Qualification

B-Tech + M-Tech from IIT Bombay in Aerospace Engineering, 2014

Rate (INR/hr):  650

Rate could vary upwards depending on exact student requirement ( e.g. if requirement is very specialised, or if requirement demands a lot of background preparation)

Manoj J.'s answer to student's question

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Hello Rashmi . As far as Maths is concerned, if you are preparing for IIT JEE, then you don't need to worry about your 11th/ 12th school syllabus because the later is subset of IIT JEE itself. If you are concerned about Physics/ Chemistry, then almost 80% syllabus is common in both and remaining 20% (like Semiconductors, Biomolecule...

The stopping distance would be P^2/(2xgXM^3). You can derive it in following manner : Total loss of Kinetic Energy would be equal to the work done by friction force. i.e 1/2 x M x V^2 = M(coeff. of friction) x M(mass) x g(gravity) x S (distance travelled). Now P(momentum) = MxV so you can put V = P/M in above equations and get value of 'S...

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