Answer:9.89 m/s
Explanation:
Given
Punctured hole is h=5 m below the top surface
Atmospheric Pressure [tex]P_{atm}=10^5 N/m^2[/tex]
Applying Bernoulli's theorem
Assuming Point 1 at top and 2 at Punctured hole
[tex]P_1+\rho gy_1+\frac{1}{2}\rho v_1^2=P_2+\rho gy_2+\frac{1}{2}\rho v_2^2[/tex]
Since Pressure at top and bottom is same therefore [tex]P_1=P_2[/tex]
velocity at top can be taken as zero as water is not flowing
[tex]\frac{1}{2}\rho v_2^2=\rho g(y_1-y_2) [/tex]
[tex]v_2=\sqrt{2g\times h}[/tex]
[tex]v_2=\sqrt{2\times 9.8\times 5}[/tex]
[tex]v_2=\sqrt{98}[/tex]
[tex]v_2=9.89 m/s[/tex]
