A) \[1.4\times {{10}^{5}}Pa\]
B) \[12\times {{10}^{5}}Pa\]
C) \[1.9\times {{10}^{5}}Pa\]
D) \[1.9Pa\]
Correct Answer: C
Solution :
[c] Since pressure is transmitted undiminished throughout the fluid (Pascal's law) \[{{F}_{1}}=\frac{{{A}_{1}}}{{{A}_{2}}}{{F}_{2}}=\frac{\pi (5\times 5)}{\pi (15\times 15)}(1350\times 9.81)\] \[\approx \,\,1.5\times {{10}^{3}}N\] The air pressure that will produce this force is \[P=\frac{{{F}_{1}}}{{{A}_{1}}}=\frac{1.5\times {{10}^{3}}}{\pi {{(5\times {{10}^{-2}}m)}^{2}}}=1.9\times {{10}^{5}}Pa\]
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