Answer:
Here, \[=45\cdot 8\times
{{10}^{-11}}p{{a}^{-1}}.\]\[=1\cdot 013\times {{10}^{5}}pa.\]
\[p=80\cdot 0\times 1\cdot 013\times
{{10}^{5}}pa;\]
For narrow tube
\[\frac{1}{B}=45\cdot 8\times
{{10}^{-11}}P{{a}^{-1}}\] or \[p=1\cdot 03\times {{10}^{3}}kg{{m}^{-3}}\]
For wider tube,
\[\text{V=}\frac{\text{M}}{\text{P}}\text{
and V }\!\!'\!\!\text{ =}\frac{\text{M}}{\text{P }\!\!'\!\!\text{ }}\] or \[\therefore
\]
Let \[\vartriangle V=V-V'=M\left( \frac{1}{P}-\frac{1}{P'}
\right)\] be the heights to which water rises in narrow tube and wider tube
respectively
Then, \[\therefore \] and \[\frac{\vartriangle
V}{V}=M\left( \frac{1}{P}-\frac{1}{P'} \right)\]
\[\times
\frac{P}{M}=1-\frac{P}{P'}\]Difference in levels of water in two limbs of tube
is, \[\frac{\vartriangle V}{V}=1-\frac{1\cdot 03\times {{10}^{3}}}{P'}\]
\[B=\frac{pV}{\vartriangle V}\]\[\frac{\vartriangle
V}{V}=\frac{P}{B}\]\[\frac{\vartriangle V}{V}=\left( 80\cdot 0\times 1\cdot
013\times {{10}^{5}} \right)\times 45\cdot 8\times {{10}^{-11}}\]
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