A) 0.89 cm of Hg
B) 8.9 cm of Hg
C) 0.5 cm of Hg
D) 1 cm of Hg
Correct Answer: A
Solution :
In horizontal pipe \[{{p}_{1}}+\frac{1}{2}pv_{1}^{2}={{p}_{2}}+\frac{1}{2}pv_{2}^{2}\] Here, \[{{p}_{1}}={{p}_{m}}g{{h}_{1}}=13600\times 9.8\times {{10}^{-2}}\] \[{{p}_{2}}=13600\times 9.8\times h\] \[\rho =1000\,kg{{m}^{-3}}\] \[{{v}_{1}}=35\times {{10}^{-2}}m{{s}^{-1}}\] \[{{v}_{2}}=65\times {{10}^{-2}}m{{s}^{-1}}\] \[\therefore \]From Eq. (i) \[13600\times 9.8\times {{10}^{-2}}+\frac{1}{2}\times 1000\times {{(0.35)}^{2}}\] \[=13600\times 9.8\times h+\frac{1}{2}\times 1000\times {{(0.65)}^{2}}\] \[=0.86\,cm\,of\,Hg\]
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