Chaining — sag correction for a suspended tape: A tape of length l and weight W (kg per m) is supported at its ends under a pull P (kg). What is the sag correction to be applied to the measured length?

Introduction / Context:
When a steel tape is supported only at its ends, it sags under its own weight, making the measured horizontal distance longer than the true straight-line distance. A sag correction must be subtracted to obtain the correct length. This question asks for the standard formula in consistent units.

Given Data / Assumptions:

• Tape length = l (m).
• Tape weight per unit length = W (kg/m) representing weight units consistent with pull P.
• Pull (tension) applied at the ends = P (kg).
• Small-sag approximation and symmetric suspension are assumed.

Concept / Approach:
Under uniform weight, the tape assumes a catenary; for small sag, the correction is derived from catenary approximations and proportional to W^2 and l^3, and inversely proportional to P^2. The correction is always positive in value and is subtracted from the measured length (because sag makes the measured length too long).

Step-by-Step Solution:

Verification / Alternative check:
Surveying texts list this formula; note that increasing tension P or shortening span l lowers the correction rapidly (P^2 and l^3 dependence).

Why Other Options Are Wrong:

• (a) and (d): Wrong dimensional dependence; do not match catenary-based form.
• (c): Inverted dependence on W and P; physically incorrect.

Common Pitfalls:
Forgetting that the sag correction is subtractive; mixing force/weight units—keep W and P consistent (both treated as kg-force units here).

Final Answer:
c_sag = (W^2 * l^3) / (24 * P^2)