Union Public Service Commission(UPSC) has issued recruitment notification for the post of Livestock, Medical Officer and Other. UPSC as about 13 vacant positions to fulfill in New Delhi. Online applications are invited from eligible candidates having Bachelor's Degree, Medical Qualification. However, eligible can apply online for UPSC Recruitment 2019 on or before the last date.
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Vacancy Details and Eligibility Criteria - UPSC Notification 2019:
|Name of the post||No.of Vacancies||Educational Qualification|
|Medical Officer / Research Officer||01||Degree in Siddha / MBBS|
|Assistant Director, Central Poultry||01||Bachelor's Degree in Veterinary Science and Animal Husbandry|
|Assistant Legal Adviser||05||Degree in law|
|Deputy Fire Adviser||01||Bachelor's Degree in Fire Eng.|
Applicants age should be.
- Minimum age of 33 years.
- Maximum age of 40 years.
Pay Matrix level - 10,11, 7(Rs.32000 - 75000) pert month.
How to apply for UPSC Recruitment 2019:
Eligible can apply online at UPSC official website.
- Log on to upsc.gov.in.
- Go to "Current Openings" & select the post to apply.
- Register Online, if you are a new user.
- Complete the registration process and click on Submit.
- Log in with Registration number and password.
- Fill the details in the application form and upload the photo, signature.
- Pay the application fee and submit the online application.
- For General / OBC candidates: Rs.25.
- For SC / ST / PH / Women candidtaes: Nil.
|Online application starts from||23rd September 2019|
|Last date for online application||12th September 2019|
To download the official; notification. Click Here
For Online application form. Go Here