A cubical metal block of edge 12 cm floats in mercury with one-fifth of the height in 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. The specific gravity of mercury = 13.6.
Answers
x = 12 cm
Length of the edge of the block
ρHg= 13.6 gm/cc
Given that,
initially 1/5 of block is inside mercury.
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
Volume of the Metal cube = 5³
= 125 cm³
Weight of the mercury displaced by the metal cube = Volume of the metal immersed × density of mercury × g.
∴ Weight of the mercury displaced by the metal cube = 125/5 × 13.6 × 980 dyne.
By the law of Flotation,
Weight of metal cube = weight of mercury displaced
= 12³/5 × 13.6 × 980 dyne.
Now, let the height of the metal cube in water be x cm.
∴ Volume of the metal cube immersed in water = 12² × x
Therefore, upthrust offered by the water to metal cube = Volume of metal cube immersed × density of water × g.
∴ Upthrust by water = 144x × 1 × 980 dyne.
Upthrust by Mercury = 12²(12 - x) × 13.6 × 980
Now, We know that upthrust always act in upward direction.
∴ Upthrust by water + Upthrust by mercury = Weight of metal cube
144x × 1 × 980 + 144(12 - x) × 13.6 × 980 = 12³/5 × 13.6 × 980
144x + 144(12 - x) × 13.6 = 144 × 12 × 13.6/5
144x + (1728 - 144x)13.6 = 1728 × 13.6/5
144x - 144x × 13.6 = 1728 × 13.6/5 - 1728 × 13.6
144x(1 - 13.6) = 1728 × 13.6[1/5 - 1]
x(1 - 13.6) = 12 × 13.6[-4/5]
x(-12.6) = 12 × 13.6(-4/5)
12.6x = 12 × 13.6 × 4/5
x = 10.36 cm.
Therefore, height of the water column to be poured = 10.36 cm.
Hope it helps.