A cubical metal block of edge 12cm floats in mercury with one fifth of the height inside the mercury. Water is poured till the surface of the block is immersed in it. Height of the water column to be poured is [specific gravity of mercury=13.6]
A
6.4cm
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B
10.4cm
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C
8.4cm
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D
5.4cm
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Solution
The correct option is B10.4cm Let the height of the water column to be poured is x.
For initial case -
Applying equilibrium condition along vertical direction, Fbm−mg=0 [Vertical equilibrium] ⇒ρmgVd−ρbVg=0 ⎡⎢⎣ρm→density of mercury;Vd→Volume of displaced fluid;V→Volume of blockρb→density of block⎤⎥⎦
Here, Vd=2.4×12×12cm3
Substituting in above equation, ⇒ρmρw×2.4×(12)2−ρbρw×(12)3=0
[dividing by ρw,] and ρw=1g/cc,ρm=13.6g/cc ⇒ρbρw=13.6×2.4×1221×123 ⇒ρbρw=2.72...(1)
For final case, let x be the height of block immersed in water.
On applying ∑F=0 for equilibrium along vertical direction, Fbw+Fbm−mg=0 ⇒ρwgVw+ρmgVm−ρb(Vw+Vm)g=0....(2)
Total volume of block=(Vw+Vm)=total volume of liquid displaced
Dividing Eq.(2) by ρw, ⇒Vw+(ρmρw)Vm−(ρbρw)(Vw+Vm)=0 ⇒Vw+13.6Vm−2.72(Vw+Vm)=0
From Eq.(1) and substituting the values of volume, ⇒(x×122)+(13.6×(12−x)×122)−(2.72×123)=0 ⇒144x+23500.8−1958.4x−4700.16=0 ⇒1814.4x=18800.64 ⇒x≃10.4cm
Height of the water column to be poured is 10.4cm