Hello mam,

IIM admission criteria specifies the key parameters which the candidates must meet such as the minimum qualifying marks, admission process, how to get admission in premier colleges of IIMs Every year around two lakh candidates appear for Common Admission Test (CAT) conducted by Indian Institute of Management (IIM).

The eligibility criteria are -

The candidates must have done their bachelor’s degree with at least 50% marks (for general category) and 45 percent marks (for SC, ST and PWD category) or equivalent CGPA for recognized university. Those candidates who are pursuing their final year bachelor’s degree may also apply for CAT.

Indian Institutes of Management (IIMs) will release their admission criteria for batch 2020-22. All the 20 IIMs will shortlist candidates for final admission round on the basis of CAT 2019 score and on other parameters such as past academic qualification, diversity, work experience etc. After shortlisting candidates all the IIMs conduct written ability test (WAT) and personal interview (PI) round.

Many of the IIMs give maximum weightage to CAT score and personal interview for admission. In the final selection factors like past academic performance, academic cut offs, gender diversity, work experience etc. plays plays major role in admission process.

Candidates who qualify the first stage of admission criteria (CAT) are called for writing ability test (WAT) and personal interview (PI) rounds. Admission criteria in IIMs is independent from each other. Have a look at the admission criteria of different IIMs for the batch 2020-22.

For detailed information you can refer to the link given below -

https://www.google.com/url?sa=t&source=web&rct=j&url=https://bschool.careers360.com/articles/iim-admission-criteria/amp&ved=2ahUKEwjSzJ_D6P_mAhURyDgGHdfXD9kQFjAKegQIAxAF&usg=AOvVaw2PS5YUHetdxNgSi3ef4_9N&ampcf=1

Good luck,

Hope it helps :)

Hello,

Given that,

Length of cylinder = 300mm

diameter of cylinder = 150mm

Shear Stress = 12 MPa = 12 N / mm^2

We know that,

Stress = Force / Area

So, 12 = Force / ( 3.14 x 150^2 )

So, Force = 848230 N

So, Force = 848.23 kN

So, The safe compressive load is 848.23 kN

Best Wishes.