If $$\alpha$$ is the factor by which the women's world record average speedreduces with doubling of the running distance, the smallest value of $$\alpha$$ occurs for the pair of categories

Let us evaluate each of the options one by one.

In the first option, we need to check the speeds for 100 m and 200 m.

For 100 m, Average speed in women's category= $$\frac{\ Total\ dis\tan ce}{Total\ time}=\ \frac{\ 100}{10.5}=9.52\ metres\ per\ \sec.$$

For 200 m, Average speed in women's category= $$\frac{\ Total\ dis\tan ce}{Total\ time}=\ \frac{\ 200}{486.1}=9.39\ metres\ per\ \sec.$$

$$\alpha\ =1-\ \frac{\ 9.39}{9.52}=1-0.9863=0.0137$$

In the second option, we need to check the speeds for 1500 m and 3000 m.

For 1500 m, Average speed in women's category= $$\frac{\ Total\ dis\tan ce}{Total\ time}=\ \frac{\ 1500}{230.1}=6.52\ metres\ per\ \sec.$$

For 3000 m, Average speed in women's category= $$\frac{\ Total\ dis\tan ce}{Total\ time}=\ \frac{\ 3000}{586.1}=6.17\ metres\ per\ \sec.$$

$$\alpha\ =1-\ \frac{\ 6.17}{6.52}=1-0.9463=0.0536$$

In the third option, we need to check the speeds for 5000 m and 10000 m.

For 5000 m, Average speed in women's category= $$\frac{\ Total\ dis\tan ce}{Total\ time}=\ \frac{\ 5000}{851.2}=5.874\ metres\ per\ \sec.$$

For 10000 m, Average speed in women's category= $$\frac{\ Total\ dis\tan ce}{Total\ time}=\ \frac{\ 10000}{1757.5}=5.6912\ metres\ per\ \sec.$$

$$\alpha\ =1-\ \frac{\ 5.6912}{5.874}=1-0.9688=0.0311$$

In the fourth option, we need to check the speeds for 400 m and 800 m.

For 400 m, Average speed in women's category= $$\frac{\ Total\ dis\tan ce}{Total\ time}=\ \frac{\ 400}{47.6}=8.40336\ metres\ per\ \sec.$$

For 800 m, Average speed in women's category= $$\frac{\ Total\ dis\tan ce}{Total\ time}=\ \frac{\ 800}{113.3}=7.0609\ metres\ per\ \sec.$$

$$\alpha\ =1-\ \frac{\ 7.0609}{8.40336}=1-0.84024=0.16$$

Therefore, lowest $$\alpha\ $$ is in the first option