Hello

Choosing between these options depends on what you want for your future.
If your main goal is better placements and a flexible career, CSE at IIT Jodhpur or Ropar is the smarter choice.
Even newer IITs give strong packages in CSE because companies hire heavily from this branch. But if you truly love mechanical engineering and value the legacy of a top IIT, Mech at IIT Kanpur is great.
IIT Kanpur has higher prestige, but Mechanical will not give as many placement opportunities as CSE.
So overall, for most students, CSE at Jodhpur/Ropar is the more practical and career-friendly option.

Your question is not specific, but it seems like you are asking which IIT is better, so yes, I am providing you with some description.

IIT ROPAR

IIT JODHPUR

No, you cannot directly join the B.S. or B.Sc. in Applied AI and Data Science at IIT Jodhpur after completing a polytechnic diploma. This course needs either a 12th pass with Mathematics and at least 60% marks, or you must qualify JEE Main.

A polytechnic diploma is not equal to 12th standard, so it does not meet the required qualification for this program.

If you still want to study this course at IIT Jodhpur, you have two options. First, you can complete your 12th standard with Mathematics and good marks. Second, you can appear for the JEE Main exam after your diploma, and if you qualify, you may get admission without the 12th marks requirement.

Here is the official website where you can check more details about the course
https://futurense.iitj.ac.in

Hii There,

The upcoming engineering entrance exams include SRMJEEE, scheduled for April 20-22, 2024; MHT CET 2024, set to take place on April 22-24 and April 28-30, 2024; KCET, which will be conducted on April 18-19, 2024; and VITEEE, scheduled from April 19-30, 2024.

I hope this answers your question.

Thanks

Correct Answer: 43.75

Solution : The number of workers whose weekly earnings are below INR 7000
= 105 + 95 + 85 + 60 = 345
The number of workers whose weekly earnings are INR 7000 or above but less than INR 8500
= 90 + 80 + 70 = 240
Difference = 345 – 240 = 105
Required percentage = $\frac{105}{240}\times100$
= 43.75%
Hence, the correct answer is 43.75.

Hello,

To start your preparation, I would suggest familiarizing yourself with the exam pattern and syllabus of CAT. You can find this information on the official CAT website or through reliable study materials.

Next, it is  important to create a study plan and set a schedule for your preparation. Allocate dedicated time each day for studying different sections of the exam, such as Quantitative Ability, Verbal Ability, Data Interpretation, and Logical Reasoning.

And also consider joining a coaching institute it is better if offline. They can provide you with guidance, study materials, and mock tests to enhance your preparation.

Hope this helps you,

Thank you

Hi

This is a bit difficult to predict as your marks vs percentile vs rank in JEE main depends upon a number of factors like

-------------------------) number of students that appeared in the  jee main exam,

-------------------------) kind of marks obtained by them ,

-------------------------) toughness level of the jee main examination for different shift etc.

But yes ,

As per past year trends,

>>>>>>>> With    176 marks

--------------------) your percentile will be roughly around   98.98 - 99.06 considering paper of moderate level

--------------------)  considering the worst case if we go with 98.98 percentile then your

• crl rank would be roughly around 10435 - 11730
• and obc rank around 2.5k

To help you get the complete list of colleges that you can get, we have JEE main college predictor, the link for the same is given below:-

https://engineering.careers360.com/jee-main-college-predictor?icn=QnA&ici=qna_answer

To help you  further,

To give an idea of jee main 2022 qualifying cut off to be eligible for jee advanced 2022.

Minimum percentile in jee main 2021 to qualify it to  be eligible for jee advanced 2021 is given below :-

--------) general :-87.8992241 percentile

--------) EWS:- 66.2214845 percentile

--------) OBC-NCL :- 68.0234447 percentile

--------) SC :- 46.8825338 percentile

--------) ST:- 34.6728999 percentile

--------) PWD :- 0.0096375 percentile

You can check the same at our page the link is provided below

https://engineering.careers360.com/articles/jee-main-cutoff

Regards

ADITYA KUMAR