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UPSC Civil Service Exam Questions - Engineering

Q1:

A train starts from rest on a curved track of radius 800 m. Its speed increases uniformly and after 3 minutes it is 72 km/hr. Its normal acceleration after 2 minutes is

A

B

C

D

ANS:D -

V = 72km/hr = 72 * 5/18 =2 0m/sec. u = 0m/sec.
t = 3minutes = 3*60 = 180sec ,t = 2 * 60minutes=120sec r=800m.
V=u+at . 20=0+a*180. a=20/180=1/9m/sec^2.
Again V=u+at=0+1/9*120=40/3m/sec.
Normal centripetal acceleration (a)=V^2/r =(40/3)^2/800=1600/9/800=1600/9*800=4/18 m/sec^2.