Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 16 (Q.No. 44)
44.
A 20 m chain was found to be 10 cm too long after chaining a distance of 2000 m. It was found to be 18 cm too long at the end of the day's work after chaining a total distance of 4000 m. What is the true distance if the chain was correct before the commencement of the day's work ?
Discussion:
4 comments Page 1 of 1.
Rahul verma said:
2 years ago
Thanks @Rahul Chopra.
Rahul guleria said:
8 years ago
T*T' = F' * F.
F' = 20(.10+.18/2) = 20.14,
Now T * 20 = 20.14 * 4000.
T = 20.14 * 4000/20 = 4029.
So the answer is B.
F' = 20(.10+.18/2) = 20.14,
Now T * 20 = 20.14 * 4000.
T = 20.14 * 4000/20 = 4029.
So the answer is B.
Rahul chopra said:
9 years ago
L = 20m
L' = 20 + {(0 + 0.10)/2} = 20.05 mean elongation.
Measured lngth = 2000m(M.L).
True length(T.L) = L'/L * M.L.
T.L = 20.05/20 * 2000.
= 2005m.
2nd part
2nd M.L = 4000 - 2000 = 2000m.
mean elongation for part 2nd = 20{(.10 + .18)/2} = 20.14.
T.L = 20.14/20 * 2000(2nd M.L).
= 2014m.
Therefore, the True distance = 2005 + 2014 = (part1 + part2) = 4019m.
L' = 20 + {(0 + 0.10)/2} = 20.05 mean elongation.
Measured lngth = 2000m(M.L).
True length(T.L) = L'/L * M.L.
T.L = 20.05/20 * 2000.
= 2005m.
2nd part
2nd M.L = 4000 - 2000 = 2000m.
mean elongation for part 2nd = 20{(.10 + .18)/2} = 20.14.
T.L = 20.14/20 * 2000(2nd M.L).
= 2014m.
Therefore, the True distance = 2005 + 2014 = (part1 + part2) = 4019m.
(1)
Parushrut said:
9 years ago
Please describe it.
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