Correct Answer - b
The frequency producted in a string of length of L mass per unit length m , and tension T is
`n = (1)/(2l) sqrt((T)/(m))`
Given `l_(1) = 50 cm , n_(1) = 800 Hz` M
and `n_(2) = 1000 Hz`
`:. (n_(1))/(n_(2)) = (l_(2))/(l_(1))`
`rArr l_(2) = (n_(1)l_(1))/(n_(2)) = (800 xx 50)/(100) = 40 cm`