Sol. Given, x = 12 cm
Length of the edge of the block ρHg= 13.6 gm/cc
Given that, initially 1/5 of block is inside mercuty.
Let ρb→density of block in gm/cc.
∴(x)3× ρb× g = (x)2× (x/5) × ρHg× g
⇒123 × ρb= 122× 12/5 × 13.6
⇒ ρb = 13.6/5 gm/cc
After water poured, let x = height of water column.
Vb= VHg+ Vw= 123
Where VHgand Vware volume of block inside mercury and water respectively
∴ (Vb× ρb× g) = (VHg× ρHg× g) + (Vw× ρw× g)
⇒(VHg+ Vw)ρb= VHg× ρHg+ Vw× ρw.
⇒(VHg+ Vw) × 13.6/5 VHg× 13.6 + Vw× 1
⇒(12)3 × 13.6/5 = (12 – x) × (12)2× 13.6 + (x) × (12)^2× 1
⇒x = 10.4 cm
