Content Curator | Updated on - Jan 28, 2026

M.Des Industrial Design at IIT Roorkee is a 2 years, Full Time Degree program offered On Campus. It offers a limited number of 20 seats.

The total fees for the M.Des Industrial Design program at IIT Roorkee is ₹ 65,200.
Students need to pay ₹ 34,100 in the first year.

IIRF Design ranking 2025, ranks IIT Roorkee at 8 out of 30 in India.

Find detailed IIT Roorkee M.Des Industrial Design fees, eligibility, admission dates, cutoff and ranking below.

Course Highlights	Details
duration	2 Years (Full Time)
course level	PG (Degree)
mode of study	On Campus
total fees	₹ 2.18 Lakhs
eligibility	Graduation+CEED

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IIT Roorkee M.Des Industrial Design Fees

M.Des Industrial Design Fees at IIT Roorkee covers a range of expenses, including tuition fee, course materials, and access to campus facilities.

The total fees for M.Des Industrial Design is ₹ 65,200.
This includes a total tuition fee of ₹ 20,000, Caution Fees of ₹ 1,000, Other Fees of ₹ 44,200.
Candidates need to pay ₹ 34,100 in the first year.
For SC/ST category students, the total fees is ₹ 46,200.
IIT Roorkee Hostel Fee is ₹ 29,250.

Earlier, M.Des Industrial Design fees at IIT Roorkee was ₹ 71,200 (in 2024) for the entire course duration. This shows a fee decrease of ₹ 6,000 over the year.

Compare IIT Roorkee M.Des Industrial Design Fees with Similar Colleges

Students have an option to pursue M.Des Industrial Design from similar colleges such as IIT Delhi, IIT Bombay, IIT Roorkee. Similar to IIT Roorkee, M.Des Industrial Design at these group colleges is offered Full Time for a duration of 2 Years.

Comparing the course fees at these colleges, the total course fee at IIT Delhi is the highest with ₹ 1.6 Lakhs. This is followed by IIT Bombay with ₹ 1.24 Lakhs and IIT Roorkee with ₹ 65,200. Among these, IIT Roorkee has the lowest fee for M.Des Industrial Design with ₹ 65,200.