Correct Answer - A
Equivalent of `K_(2)Cr_(2)O_(7)` = equivalent of `N_(2)H_(4)`
also equivalent of `KMnO_(4)` = equivalent of `N_(2)H_(4)`
So, equivalent of `K_(2)Cr_(2)O_(7)` = equivalent of `KMnO_(4)`
`0.1xx6xxV_(1)=0.3xx5xxV_(2) " " :. " so" V_(2)=2//5V_(1)`