UPSC ESIC PA Online Form 2024 (323 Posts)

UPSC ESIC PA Recruitment 2024

Post Name – Personal Assistant

(Union Public Service Commission)



Start Date – 07-03-2024

Last Date – 27-03-2024

Fee Payment Last Date – 27-03-2024

Admit Card – Notify later

Exam Date – Notify Later


• General/OBC /EWS – Rs. 100/-

• SC/ST/PWD – Rs. 0/-

Fee can be paid only online through Net Banking, Credit or Debit cards


All India


(As on 01-08-2024)

Minimum Age – 18 Years

Maximum Age – 30 Years

Age Relaxation: As Per Rules

Total Posts :323 Posts

Post Description – UPSC Union Public Service Commission is inviting online applications for Personal Assistant for 323 posts. Interested people meet all the eligibility criteria, apply online

Details Of Recruitment Of UPSC ESIC Personal Assistant Online Form 2024

Posts Name –Personal Assistant

Number Of Posts – 323 Posts

Category Wise Posts –

General – 132 Posts

OBC – 87 Posts

SC – 48 Posts

ST – 24 Posts

EWS – 32 Posts

Pay Scale  Level 7

Education Qualification – Graduate Degree From Recognized University And Stenography And Typing.

How to Apply Online UPSC ESIC PA Online Form 2024– Candidates can apply through link provided below or they can also apply through official site of the Union Public Service Commission before 27/March/2024.

Required Documents For Online Application –

Educational Mark sheet And Certificates

Aadhar Card / Pan Card

Passport Size Photo


Selection Mode 

Written Exam

Skill Test

Merit List

Online Apply

Link Active On 07-03-2024

Download Notification

Link Active On 07-03-2024

Download Short Notice

Click Here

Join Telegram

Click Here

Official website

Click Here

Leave a Comment

Your email address will not be published. Required fields are marked *