It's a Question of your Career!

HI Manali Uttam More,

Please find the JEE Mains 2026 Official Question Paper - All Shifts link here in a free PDF format

https://engineering.careers360.com/download/ebooks/jee-main-2026-january-session-official-question-paper

Hi Harigovind, can you please mention which college you are asking details for?

HI Varshitha,

Please find the JEE Mains 2026 Official Question Paper - All Shifts link here in a free PDF format

https://engineering.careers360.com/download/ebooks/jee-main-2026-january-session-official-question-paper

Parul University fee for the BBA programme varies as per the specialisations. However, the BBA fee ranges from Rs 2,40,000 to Rs 4,34,000. Candidates interested in taking BBA admission at Parul University need to pay the fee and confirm their admission.

Hi Jayesh,

Here is the 21 January 2026 Jee paper analysis
https://engineering.careers360.com/download/ebooks/jee-main-2026-memory-based-questions-and-analysis-of-21st-january-shift-1

Hi Shivaks,

Please find the JEE Mains 2026 Official Question Paper - All Shifts link here in a free PDF format

https://engineering.careers360.com/download/ebooks/jee-main-2026-january-session-official-question-paper

HI Avvari Anuradha,

Please refer to these links here

https://engineering.careers360.com/download/sample-papers/ap-eapcet-sample-paper

https://engineering.careers360.com/download/ebooks/ap-eapcet-syllabus

https://engineering.careers360.com/download/ebooks/ap-eapcet-chapter-wise-weightage

https://engineering.careers360.com/download/ebooks/ap-eamcet-previus-year-question-paper

Completing Mathematics through NIOS On-Demand examinations is a valid way to meet the Class 12 eligibility for the B.Sc. B.Ed. (ITEP) at IIT Bhubaneswar, as long as you maintain the required aggregate of 60% (55% for OBC).

Validity of NIOS On-Demand (ODE) for IITs

The National Institute of Open Schooling

Hi Deepthi rajesh,

Molar mass of $\mathrm{H}_2 \mathrm{O}$

$$
\begin{aligned}
& =2 \times 1+16=18 \mathrm{~g} / \mathrm{mol} \\
& \text { Mass }=\text { moles } \text { × } \text { molar mass } \\
& =1.2 \times 18
\end{aligned}
$$

Hi Vandana Singh,

Please provide the college or exam name for which you want to apply for. So that I can help you out with the application process.