Correct Answer - D
`2P_(0)=(h_(2)+h_(0))rhog+p_(0)`
(since liquids at the same level have the same pressure) `P_(0)=h_(2)rhog+h_(0)rhog,h_(2)rhog=P_(0)-h_(0)rhog`
`h_(2)=(P_(0))/(rhog)+(h_(0)rhog)/(rhog)=(P_(0))/(rhog)-h_(0)`
KE to the water `=` pressure energy of the water at that layer `(1)/(2)mV^(2)=mxx(P)/(rho)`
`V^(2)=(2P)/(rho)=(2)/(rho)[P_(0)+rhog(h_(1)+h_(0))]`
`V=[(2)/(rho){P_(0)+rhog(h_(1)-h_(0))}]^(1//2)`
We know `2P_(0)+rhog(h_(1)-h_(0))=P_(0)+rhogX`
`impliesX=(P_(0))/(rhog)+(h_(1)-h_(0))=h_(2)+h_(1)`
i.e., X is `h_(1)` meter below the top or X is `h_(2)` above the top.