Correct Answer - Option 1 : 0.0045
Concepts:
The correction due to sag of tape is given as:
\({C_{sag}} = \;\frac{{{W^2}L}}{{24\;{P^2}}}\)
Where,
L is the total unsupported length of tape
P is applied pull on tape
W is the total weight of tape.
Note:
The sag correction is always negative as noted down value or measured value is always greater than the actual value.
Calculation:
Given:
P = 100 N, W = 6 N,
No of spans = 3, Each Span length = 10 m
Therefore, L = 3 × 10 = 30 m.
Now, Sag correction is given as:
\({C_{sag}} = \;\frac{{{W^2}L}}{{24\;{P^2}}}\)
Or
\({C_{sag}} = \frac{{{6^2}\; \times \;30}}{{24\; \times\; {{100}^2}}}\)
∴ Csag = 0.0045 m
