JSSC Recruitment 2024: Registrations begin for 4,919 constable posts at jssc.nic.in; selection process
Alivia Mukherjee | January 15, 2024 | 01:50 PM IST | 1 min read
JSSC Recruitment 2024: The selection process includes a physical efficiency test, a medical test and a written exam.
NEW DELHI: The Jharkhand Staff Selection Commission (JSSC) has commenced the registration process for the Constable Competitive Examination from today, January 15. Candidates can submit their JSSC recruitment 2024 application form by February 14. Candidates can apply for JSSC recruitment 2024 on the official website at jssc.nic.in. The JSSC recruitment drive is being conducted to fill 4,919 vacancies.
The JSSC recruitment 2024 registration fee can be paid until February 16. Candidates can upload the signature and photo by February 18. Candidates will also get the opportunity to edit their application form via the correction window which will be activated from February 20 to 22.
JSSC Recruitment 2024: age limit
The minimum age of the candidates applying for the JSSC recruitment 2024 should be 18 years. The upper age for the unreserved and economically weaker section is 25 years while for the extremely backward class and backward class, the upper age is 27 years. The maximum age limit for female candidates of the unreserved, economically weaker section, extremely backward classes the maximum age is 28 years. The upper age for the Scheduled Tribe (ST) and Scheduled Caste (SC) male and female candidates is 30 years.
Also read JKSSB stock assistant answer key 2024 out at jkssb.nic.in; raise objection by January 18
JSSC Recruitment 2024: Selection process
The JSSC recruitment selection process includes a physical efficiency test, a medical test and a written exam. In the physical efficiency test, a running race will be conducted. Men will have to cover 10 km in 60 minutes and women will have to cover 5 km in 30 minutes.
The written test will consist of two papers. The duration of each paper will be of 2 hours. Candidates will receive 3 marks for each correct answer and there will be a negative marking of 1 marks for each incorrect answer.
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