A) 10.2 m of freshwater of \[\rho =998\,\,\text{kg/}{{\text{m}}^{\text{3}}}\]
B) 10.2 m of salt water \[\rho =1025\,\,\text{kg/}{{\text{m}}^{\text{3}}}\]
C) 25.5 m of kerosene of \[\rho =800\,\,\text{kg/}{{\text{m}}^{\text{3}}}\]
D) 6.4 m of carbon tetrachloride of \[\rho =1590\,\,\text{kg/}{{\text{m}}^{\text{3}}}\]
Correct Answer: B
Solution :
At sea level, h = 760 mm of Hg A. \[\frac{760\times {{10}^{-3}}\times 13.6\times 9.81\times {{10}^{3}}}{9.81\times 998}=10.35\,\,m\] B. \[\frac{760\times {{10}^{-3}}\times 13.6\times {{10}^{3}}}{1025}=10.08\,\,m=10.1\,m\] C. \[760\times {{10}^{-3}}\times 13.6/0.8=12.92\,m\] D. \[760\times {{10}^{-3}}\times 13.6/1.590=6.5\,m\]
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