Civil Engineering - Surveying - Discussion
Discussion Forum : Surveying - Section 6 (Q.No. 15)
15.
The sag of 50 m tape weighing 4 kg under 5 kg tension is roughly
Discussion:
24 comments Page 1 of 3.
Madhukar said:
10 years ago
What is the formula was used for calculation?
Sandeep said:
9 years ago
Please give the formula, to solve the question.
Adi said:
9 years ago
Anyone solve it.
Answer is not coming with formula w2l2/24p2.
Answer is not coming with formula w2l2/24p2.
RANJAN PATRA said:
9 years ago
It is LW2/24 * P2.
Raval chaitanya said:
8 years ago
@Adi
Formula = l3w2/24p2.
Formula = l3w2/24p2.
Hasan said:
8 years ago
Please, someone explain it clearly, I am not getting the exact answer.
Rpy said:
8 years ago
The actual formula is LW^2/24p^2.
Where l is the length of tape, W=wl=weightt of tape in kg and p is pull. But it comes 1.33.
Where l is the length of tape, W=wl=weightt of tape in kg and p is pull. But it comes 1.33.
Harish said:
8 years ago
(LW^2/24*10^4)*P^2.
((50*4^2)/(24*10^4))*5^2 = 0.0833.
Ans is right.
((50*4^2)/(24*10^4))*5^2 = 0.0833.
Ans is right.
Neelam said:
8 years ago
In formula why you multiply 24 with 10^4?
JAY PANCHAL said:
8 years ago
In formula why you multiply 24 with 10^4?
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