# A cubical metal block of edge 12 cm floats in mercury with one fifth of the height inside the mercury. Water is poured till the surface of the block is just immersed in it. Find the height of the water column to be poured. Specific gravity of mercury = 13.6.

Navjyot Kalra
8 years ago
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
Raunaq Mehta
39 Points
5 years ago
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