Correct Answer - D
`A rArr (u^(2))/(4)-u^(2)=-2gh_(1)`
`B rArr (u^(2))/(9)-u^(2)=-2gh_(2)`
`C rArr (u^(2))/(16)-u^(2)=-2gh_(3)`
`therefore AB=(u^(2))/(2g)((8)/(9)-(3)/(4))=(u^(2))/(2g).(5)/(36)`
`BC=(u^(2))/(2g)((15)/(16)-(8)/(9))=(u^(2))/(2g).(7)/(144)`
`therefore (AB)/(BC)=(5)/(36)xx(144)/(7)=(20)/(7)`