A) \[\sqrt{200}m/s\]
B) \[\sqrt{400}m/s\]
C) \[\sqrt{500}m/s\]
D) \[\sqrt{800}m/s\]
Correct Answer: B
Solution :
Applying Bernoulli's theorem \[{{p}_{1}}+0+\rho ghH={{p}_{2}}+\frac{1}{2}\rho {{v}^{2}}+\rho gH\] \[{{p}_{1}}-{{p}_{2}}=\frac{1}{2}\rho {{v}^{2}}\] \[3\times {{10}^{5}}-1\times {{10}^{5}}=\frac{1}{2}\rho {{v}^{2}}\] \[2\times {{10}^{5}}=\frac{1}{2}\rho {{v}^{2}}\] \[2\times {{10}^{5}}=\frac{1}{2}\times {{10}^{3}}\times {{v}^{2}}\] \[{{v}^{2}}=400\] \[v=\sqrt{400}\]You need to login to perform this action.
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