Answer:
0.497
0.38806
Explanation:
According to the question normalizing data is done by
[tex]x=\frac{84}{169}\\\Rightarrow x=0.497[/tex]
The normalized value at 3.8 cm is 0.497
We have the relation
[tex]V_1r_1^2=V_2r_2^2\\\Rightarrow V_2=\frac{V_1r_1^2}{r_2^2}\\\Rightarrow V_2=\frac{1\times 3.8^2}{6.1^2}\\\Rightarrow V_2=0.38806[/tex]
Hence, the theoretically expected normalized value at 6.1 cm is 0.38806