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IFTM Moradabad News & Updates 2025: Notifications, Notice, Result

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Updated on - Oct 31, 2023 07:51 PM IST

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VIT Vellore Category 1 Expected Cutoff Rank 2025: Vellore Institute of Technology (VIT), Vellore, is generally considered a renowned Tier 2 college in India offering B.Tech courses in various branches. The institute has been ranked 10th under the Universities category by NIRF 2024. Based on last year’s figures, VIT Vellore Category 1 Expected Cutoff Rank 2025 for B.Tech CSE is likely to range from 500 to 550. The branch is considered highly popular, and Category 1 has the highest cutoff compared to subsequent categories. While the Category 1 fee at VIT Vellore is the lowest, i.e., Rs 2,00,000/-. Find the anticipated Category 1 cutoff ranks for other branches below, and know your probable selection chances. 

KCET 2025 Result Likely After CBSE, ISC results
April 30, 2025 11:47 AM

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KCET 2025 Result: Candidates who have appeared for the KCET 2025 exams on April 16 and 17, 2025, are eagerly waiting for the KCET 2025 Results to be announced. The KCET final answer key 2025 has been released by the Karnataka Examinations Authority through its official website, and the results are expected soon. However, as per the format, KCET ranks are calculated as 50% KCET marks and 50% 12th marks, so the results are most likely to be released after the class 12th board results are announced. The KCET result will likely be released only after CBSE and ISC results. 

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April 30, 2025 11:47 AM

4 Grace Marks Awarded in KCET Final Answer Key 2025; 21 total changes

KCET Final Answer Key 2025: The Karnataka Examination Authority had successfully conducted KCET 2025 exams on April 16 and 17, the provisional answer key for which was released on April 18, 2025. The window to file objections for applicants was opened until April 22, and after carefully analysing the objections, KEA released KCET Final Answer Key 2025. As per the final answer key, a total of 21 changes have been noted in Physics, Chemistry, and Biology. Grace marks for 4 questions from Physics are awaited. It should be noted that there are no changes to the Mathematics answer key, and the provisional answer key will be regarded as the final answer key. Based on the KCET final answer key 2025, the KCET results will be declared. 

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MS Ramaiah KCET Expected Cutoff Rank 2025: Aspirants aiming for admission to MS Ramaiah Institute of Technology (Code: E006) through the KCET 2025 examination can refer to the following expected cutoff ranks across the 18 academic programs offered by the institute. For the Computer Science and Engineering (CSE) program, the expected cutoff range for General Merit (GM) candidates is between 1200 and 1300, and for general category, the cutoff is likely to fall between 2300 and 2500. For candidates belonging to reserved categories, the cutoff for SC and ST are anticipated to be in the range of 10400 to 10600,and 13600 and 13800. Additionally, GM candidates securing ranks below 3000 are expected to have better chances of gaining admission to high-demand specializations such as Artificial Intelligence, Machine Learning, Data Science, and Cybersecurity. 

BMSCE Bangalore KCET Expected Cutoff Rank 2025
April 30, 2025 11:46 AM

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BMSCE Bangalore KCET Expected Cutoff Rank 2025: Candidates aspiring to secure admission into BMSCE Bangalore (Code: E003) through the KCET 2025 exam can refer to the following estimated cutoff ranks for different academic programs. For the Electronics and Communication Engineering (ECE) program, the expected cutoff ranks are between 1850 to 1950 for the general merit, i.e., the candidates that qualify for the exam based on their overall merit without any reservation or relaxation. However, for the 1G candidates, it might be expected to fall between 8000 to 8300. The scheduled caste categories will have a lower cutoff for all the academic programs compared to the general category. It is important to note that these estimates are based on past trends, and the actual cutoff ranks could vary once the official results are released.

Related Questions

S

Samiksha Rautela, Content Team

-Answered on Apr-30-2025

Dear Student,

The NEET application password can be generated through a security question, OTP verification, E-mail address. Here are the steps to retrieve the password of NEET application:

Step 1: Visit the official registration portal and click on 'Forgot your password'

Step 2: Select the preferred option- through a security question, OTP verification or E-mail address

Step 3: If you select a security question, you will have to answer a question before submitting the same.

Step 4: If you select an OTP verification, an OTP will be sent either to the E-mail/ Mobile No. of the registered candidate

Step 5: If you select E-mail address, a reset link will be sent to the registered email id.

It is normal to forget not only the password but also the application number. Here is how to retrieve the NEET 2025 Login Application Number and Password.

Thank You

R

Rupsa, Content Team

-Answered on Apr-14-2025

Dear Student,

JEECUP Group I new syllabus UP Board ke Class 11 and 12 course curriculum ke anusar frame kiya gaya hain jis mein Physics, Chemistry, and Mathematics subjects added hain. Aap entrance exam preparation ke liye UP Board Class 11th and 12th ke new syllabus follow kar sakte hain. Poora JEECUP 2025 syllabus detail mein humare page pe diya hua hain. JEECUP Group I syllabus ke through Diploma in Aircraft Maintenance Engineering, Diploma in Aircraft Maintenance Engineering (Avionics), and Diploma in Aircraft Maintenance Engineering (Helicopter and Powerplants) courses mein admission diya jayega. Iss saal JEECUP 2025 exam May 20 se May 28 tak conduct hoga. Hum umeed karte hain yeh jaankari aap ke liye useful rahi ho. Good luck!

If you have further queries regarding Diploma in Engineering course admission, you can write to hello@collegedekho.com or call our toll free number 18005729877, or simply fill out our Common Application Form on the website.

JEECUP kis tarah ka exam hota hai
-asheesh -Updated on Apr-14-20251 Answers
R

Rupsa, Content Team

-Answered on Apr-14-2025

Dear Student,

JEECUP exam ek state-level entrance exam hain jo Uttar Pradesh ki government and private engineering colleges mein Diploma/ Polytechnic course admission ke liye conduct kiya jata hain. Iss exam ke through aap Engineering, Technology, or Pharmacy streams mein Diploma kar sakte hain. JEECUP 2025 exam pattern ke anusaar yeh test online mode mein conduct hoga total 400 marks ke liye. As a student, aap ko 150 minutes (2 hours 30 minutes) mein total 100 MCQ type questions attempt karne hoge. JEECUP exam paper Hindi aur English dono langauge mein rahega. Har question ke sahi answers ke liye aap ko 4 marks milenge; Galat answer ke liye koi negative marking nahi hogi. Saare questions JEECUP syllabus ke anusar hi hoge, jo ki Class 10th/ 11th/ 12th level ka hain.Iss saal JEECUP 2025 exam May 20 se May 28 tak conduct hoga. Hum umeed karte hain yeh jaankari aapke liye useful rahi ho. Good luck!

If you have further queries regarding Diploma/ Polytechnic course admission, you can write to hello@collegedekho.com or call our toll free number 18005729877, or simply fill out our Common Application Form on the website.

J

Jayita Ekka, Content Team

-Answered on Feb-11-2025

Dear student,

You are asking if the focal length of a lens is 20 cm, what is its power in diopters?

Calculating Lens Power:

  • Power of a lens is defined as the reciprocal of its focal length when expressed in meters
  • The unit of power of a lens is diopter (D)

Formula:

  • Power (P) = 1 / focal length (in meters)

Given:

  • Focal length (f) = 20 cm = 0.2 meters

Calculation:

  • Power (P) = 1 / 0.2 = 5 diopters

Therefore, the power of the lens is +5 diopters

Why is it positive?

  • A positive power indicates a convex lens (converging lens)
  • Convex lenses are thicker at the center and thinner at the edges
  • They converge light rays to a point called the focus

convex lens

Additional Notes:

  • Concave Lens: If the lens were concave (diverging lens), the power would be negative
  • Units: Always ensure that the focal length is converted to meters before calculating the power in diopters

Therefore, if a lens has a focal length of 20 cm, its power is +5 diopters. This means it is a convex lens capable of converging light rays.

S

Samiksha Rautela, Content Team

-Answered on Oct-17-2024

Dear Student,

Candidates can start early, create a strategy and have a structured schedule to level up their preparation. To effectively prepare for NEET candidates must follow the below given tips and a study schedule. 

  • Understand the Exam Syllabus: Candidates must familiarise NEET syllabus and the marking scheme. 
  • Craft a Study Plan: Students must allocate specific time slots for each subject and study in a similar schedule. 
  • Read, Revise & Repeat: Candidates must schedule weekly revisions to check things that they have learnt and revise the last read chapters. 
  • Practice Mock Papers: Candidates must regularly attempt mock tests and previous year's papers to keep a check on their preparation.
  • Maintain a Healthy Lifestyle: Students must maintain a balanced diet and ensure adequate sleep to keep their minds sharp.

Further, candidates can follow the NEET 2025 Preparation Time Table and study for 7-8 hours everyday.

Thank You

Admission Updates for 2025

ITS Engineering College

Greater Noida (Uttar Pradesh)

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Adhunik Group of Institutions

Ghaziabad (Uttar Pradesh)

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Sharda University

Greater Noida (Uttar Pradesh)

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IILM University Greater Noida

Greater Noida (Uttar Pradesh)

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