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(A) According to Bernoulli's principle,for a horizontal pipe,the sum of pressure energy and kinetic energy remains constant: $P_1 + \frac{1}{2} \rho v

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$10$

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(A) According to Bernoulli's principle,for a horizontal pipe,the sum of pressure energy and kinetic energy remains constant: $P_1 + \frac{1}{2} ho v_1^2 = P_2 + \frac{1}{2} ho v_2^2$.When the tap is closed,the velocity of water $v_1 = 0$. Let the pressure be $P_1 = 3.5 	imes 10^5 \, N/m^2$.When the tap is open,the pressure is $P_2 = 3 	imes 10^5 \, N/m^2$ and the velocity is $v_2 = v$.Substituting the values into the equation: $3.5 	imes 10^5 + 0 = 3 	imes 10^5 + \frac{1}{2} 	imes 10^3 	imes v^2$.$(3.5 - 3) 	imes 10^5 = \frac{1}{2} 	imes 10^3 	imes v^2$.$0.5 	imes 10^5 = 500 	imes v^2$.$50000 = 500 	imes v^2$.$v^2 = 100$.$v = 10 \, m/s$.

The pressure of water in a water pipe when the tap is closed and opened is $3.5 \times 10^5 \, N/m^2$ and $3 \times 10^5 \, N/m^2$ respectively. With the tap open,the velocity of water flowing is ........... $m/s$. (Take density of water $\rho = 10^3 \, kg/m^3$)

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