1. **State the problem:** We are given a 100 m tape with a sag correction of 30 cm when supported at the ends only. 2. The sag correction when a tape is supported only at the ends is calculated for a tape length $L=100$ m as $s=0.30$ m. 3. When the tape is supported at the ends and at the midpoint (50 m), the sag effect is divided because the tape behaves like two shorter tape segments each of length $L/2 = 50$ m. 4. Sag correction $s$ is proportional to the square of the tape length supported between points, so for half the length (50 m), sag correction $s_1$ is: $$s_1 = s \times \left(\frac{L/2}{L}\right)^2 = 0.30 \times \left(\frac{50}{100}\right)^2 = 0.30 \times \left(0.5\right)^2 = 0.30 \times 0.25 = 0.075 \text{ m}$$ 5. Since there are two equal segments each with sag correction 0.075 m, total sag correction when supported at ends and midpoint is: $$s_{total} = 2 \times s_1 = 2 \times 0.075 = 0.15 \text{ m}$$ 6. **Final answer:** The sag correction reduces to 0.15 m or 15 cm when supported at the ends and at the 50 m mark.