A) \[2.8\times {{10}^{5}}\] Pa
B) \[2.6\times {{10}^{5}}\] Pa
C) \[2.4\times {{10}^{5}}\] Pa
D) \[2.1\times {{10}^{5}}\] Pa
Correct Answer: B
Solution :
From Bernaullis theorem \[{{p}_{1}}+\rho gh+\frac{1}{2}pv_{1}^{2}={{p}_{2}}+\frac{1}{2}pv_{2}^{2}\] \[15\times {{10}^{5}}+{{10}^{3}}\times 10\times 10+\frac{1}{2}\times {{10}^{3}}\times 25\] \[={{p}_{2}}+\frac{1}{2}\times {{10}^{3}}\times \frac{25}{4}\] \[15\times {{10}^{5}}+{{10}^{5}}+\frac{25}{2}\times {{10}^{3}}={{p}_{2}}+\frac{25}{8}\times {{10}^{3}}\] \[{{p}_{2}}=\frac{25\times 83\times {{10}^{3}}}{8}\] \[=259.375\times {{10}^{3}}=2.6\times {{10}^{5}}Pa\]You need to login to perform this action.
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