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