Quick Guide of Smart Courseclose
The Etoosindia SD card learning is very easy and simple.
A few simple steps, you can learning to anytime, anywhere.
※ Etoosindia learning of SD card is only available for Samsung Tablet.
- Mount SD card: tInsert SD card on the Samsung tablet.
- Connect internet: Connect to the Internet using WiFi or 3G.
- Run application: Run the Shelf Reader Application.
- Create account: Click New user to create a new account.
- Login & Activation: After login, select the desired course.
(Activation is the first one you enter only once.)
RG SIR LECTURE SERVICES ARE CLOSED
14 Nov 2017
ETOOSINDIA is closing the video lecture services for Mr. Rajat Goyal (RG Sir) courses from i.e. 19-11-2017 (Sunday).
From now onwards, RG Sir video lectures who was the faculty of JEE Physics will not be available on our website. Purchased user can avail their course till the end of their validity period.
Disclaimer: We do not entertain or sell any youtube or any video lectures related RG sir JEE courses (Physics) !!
Complete Mathematics for JEE MAIN & ADVANCED in 3 Month by GB SIR
09 Nov 2017
Etoosindia brought you a new Crash Course of Mathematics for class 12th & 13th by GB Sir named as "Complete Mathematics for JEE MAIN & ADVANCED in 3 Month by GB SIR". It includes more than 1500 questions.
He is highly experienced faculty with over 13 years of experience in teaching. His success stories are well known by students and he tends to give you a good quality and very well prepared content for the preparation of JEE examination and explains from the basic level.
You can get command on Mathematics with GB sir in just 90 hours . He has tried to provide you best quality content with Previous Year JEE questions discussion.
To avail this course, go to: http://www.etoosindia.com/courses/jee/500409/Complete-Mathematics-Course-for-JEE-MAINS-ADVANCED-in-3-Month-by-GB-sir/detail.do
JEE ADVANCED 2017: IIT will give 11 bonus marks to students for enigmatic questions
06 Jun 2017
Indian Institute of Technology (IIT) Madras, the exam conducting authority for JEE Advanced 2017 going to award 11 bonus marks for 3 ambiguous questions to all the candidates who had taken JEE Advanced 2017 on May 21.
Two questions were from the mathematics section and one from physics. In two other questions - from physics and chemistry - marks will be given for any of the two answers given by the IITs. In the year 2016, students were awarded bonus marks for three questions, while in 2015, they received bonus marks for one question.
In mathematics one question, for which bonus marks were given, was incorrect; no answer was matching, said a teacher. "In the other question, we were not anticipating bonus marks, but IIT played it safe as there was ambiguity," added the teacher. In one question in chemistry, students will get marks for any one of the two right options.
Best of Luck for JEE Advanced 2017
20 May 2017
We know all you champs out there must have prepared thoroughly for your JEE Advanced 2017 by now and just as you came out with flying colours in your JEE Main you shall be able to crack this one too. Here’s wishing you All the Very Best for your JEE Advanced 2017 from all of us at Etoosindia.
- Chemistry Jitendra Hirwani new JEE Complete Physical Chemistry with Past Year's Problem Solving Techniques for JEE Main & Advanced by JH Sir OT
- Mathematics Gavesh Bhardwaj new Complete Mathematics for JEE Main and Advanced in 3 month by GB Sir OT DEMO COURSE
- Chemistry Jitendra Hirwani new Problem Solving Techniques Complete Past Years JEE Main & Advanced (Physical Chemistry) Exam Papers by JH Sir OT
- Physics Nitin Vijay JEE Complete Physics for Main & Advanced by NV Sir OT
question more than one correct
Two heaters desinged for the same voltage V have different power ratings when connectede individually across a source of voltage V they produce H amount of heat each in time t1 and t2 respectively when used together accross the same source they produce same amount of heat
A) if they are in series t= t1 + t2
B) if they are in series t = 2(t1 + t2)
C) if they are in parallel t = t1 t2 / t1 + t2
D) if they are in parallel t = t1 t2 /2( t1 + t2)JEE, Physics, Current Electricity
A conductor has a temp independent resistance R and a total heat capacity C, at the moment t=0 it is conected to a DC source voltage find the time dependence of the conductor's temperature T assuming the thermal power dissipated into surrounding space to vary as q=k ( T-T0 ), where k is a constant, T0 is the surrounding temperature ( equal to the conductor's temperature at the initial moment )JEE, Physics, Current Electricity
How is da=.. ? Sir said da is the area of the shaded outer region...JEE, Physics, Modern Physics
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- ★★★★★ Amit Verma Best Sir in India Vector , Calculus by AV Sir
- ★★★★★ Nipun Mittal bad Capacitor by NM Sir