(a) (i) Considering vertical equilibrium of cylinderWeight of cylinder = upthrust due to upper liquid + upthrust due to lower liquid∴(5A​)(L)D⋅g=(5A​)(43L​)(d)g+(5A​)(4L​)(2d)(g)∴D=(43​)d+(41​)(2d)⇒D=45​d​(ii) Considering vertical equilibrium of two liquids and the cylinder.(p−p0​)A= weight of two liquids + weight of cylinder∴P=P0​+A weight of two liquids + weight of cylinder ​...(i) Now, weight of cylinder=(5A​)(L)(D)(g)=(5A​Lg)(45​d)=4ALdg​Weight of upper liquid =(2H​Adg) andWeight of lower liquid =2H​A(2d)g=HAgd∴ Total weight of two liquids =23​HAdg∴ From Eq. (i) pressure at the bottom of the container will bep=p0​+A(23​)HAdg+4ALdg​​or p=p0​+4dg(6H+L)​(b) (i) Applying Bernoulli's theorem,p0​+dg(2H​)+2dg(2H​−h)=p0​+21​(2d)v2​Here, v is velocity of efflux at 2 .Solving this equation, we getv=(3H−4h)2g​​(ii) Time taken to reach the liquid to the bottom will bet=2h/g​∴ Horizontal distance x travelled by the liquid isx=vt=(3H−4h)2g​)​(g2h​​)x=h(3H−4h)​​(iii) For x to be maximum dhdx​=02h(3H−4h)​1​(3H−8h)=0h=83H​​Therefore, x will be maximum at h=83H​The maximum value of x will bexm​=(83H​)[3H−4(83H​)]​xm​=43​H​