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UPSC Civil Service Exam Questions - Engineering

Q1:

The standard time meridian in India is 82°30'E. If the standard time at any instant is 20 hours 10 minutes, the local mean time for the place at a longitude of 20°E would be

A 4h PM

B 4h 10m PM

C lh 20m PM

D 0h 20m AM

ANS:A - 4h PM

82°30' - 20° = 62° 30'.

62*4 = 248 min
And for 30' = 0.5° *4 = 2.0 min
Total time we get = 248 + 2 = 250 min
250/60 = 4 hours and 10 min
Given standard time 20h 10 min
Therefore:- 20h 10 min - 4h 10 min = 16hours, Which is at 4 pm as per local mean time.