1. **State the problem:** The sag correction for a 100 m tape is 30 cm when supported only at the ends. We need to find the sag correction for the same tape when it is supported at the ends and at the 50 m mark (middle). 2. **Understand sag correction:** Sag correction is due to the tape forming a catenary curve when supported only at the ends. When the tape is supported also at the midpoint, it divides the tape into two segments, each behaving like a shorter tape segment sagging between supports. 3. **Formula for sag correction:** The sag correction $S$ for a tape length $L$ supported only at the ends is proportional to $L^2$. If the sag correction for length $L=100$ m is $S=0.30$ m, then $S \propto L^2$. 4. **New supports:** When supported at the ends and at the midpoint (at 50 m), the tape essentially behaves like two tapes of length 50 m each with sag supported at ends only. 5. Calculate sag correction for one half length $L'=50$ m using proportionality: $$ S' = S \times \left(\frac{L'}{L}\right)^2 = 0.30 \times \left(\frac{50}{100}\right)^2 = 0.30 \times \left(0.5\right)^2 = 0.30 \times 0.25 = 0.075 \text{ m} $$ 6. Since the tape is supported at the 50 m mark, the total sag correction is the sum of sag corrections of two halves: $$ S_{total} = 2 \times S' = 2 \times 0.075 = 0.15 \text{ m} = 15 \text{ cm} $$ **Final answer:** The sag correction when supported at the ends and midpoint is 15 cm.