Here , `(dm)/dt =1 "quintal"//sec = 100kg//s, u=-5km//s = -5xx 10^(3) m//s `
As `F = - u ((dm)/(dt)) :. F= - (-5xx10^(3)) (100) =5xx 10^(5) N`
Now `upsilon =?, (m)/(m_(0)) = (1)/(100)` or `m_(0)/(m)=100`
From `upsilon = u log_(e) ((m_(0))/(m ))`
`upsilon = 5 xx 10^(3) log_(e) 100 = (5xx10^(3)) xx 2.303 log_(10) 100 = 5xx 10^(3) xx 2.303 xx 2`
` =2.303 xx 10^(4) m//s `.