Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 5 (Q.No. 15)
15.
A grinding wheel rotates 3000 rpm. When power supply is cut off, the wheel stops completely in 10 seconds. What is the number of revolutions made by the wheel before coming to rest ?
Discussion:
6 comments Page 1 of 1.
Haadi said:
10 years ago
How to solve it?
Vicky said:
8 years ago
=3000/60 = 50 * 10 = 500.
Mrinmoy Nath said:
8 years ago
The Answer Should be 500.
Phani said:
7 years ago
3000/12 = 250,
1minute = 12rpm.
1minute = 12rpm.
(1)
Omer said:
7 years ago
Average speed during deceleration is (3000 + 0)/2 = 1500 RPM.
Answer is 1500 rev/min * (1 min/60 sec) * 10 sec = 250 revolutions.
Answer is 1500 rev/min * (1 min/60 sec) * 10 sec = 250 revolutions.
Rakesh said:
6 years ago
No. of revolution =
Pic value is (3000/60)x10 = 500.
At the end of rotation it's 0 (zero).
So average is (500+0)/2 = 250.
Pic value is (3000/60)x10 = 500.
At the end of rotation it's 0 (zero).
So average is (500+0)/2 = 250.
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