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(B) $1$. Apply the equation of continuity: $A_{1} v_{1} = A_{2} v_{2}$.Since $A = \pi (d/2)^{2}$,we have $d_{1}^{2} v_{1} = d_{2}^{2} v_{2}$.Substitut

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$304500$

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(B) $1$. Apply the equation of continuity: $A_{1} v_{1} = A_{2} v_{2}$.Since $A = \pi (d/2)^{2}$,we have $d_{1}^{2} v_{1} = d_{2}^{2} v_{2}$.Substituting the values: $5^{2} 	imes 4 = 2^{2} 	imes v_{2} \implies 25 	imes 4 = 4 	imes v_{2} \implies v_{2} = 25 \, m/s$.$2$. Apply Bernoulli's equation (assuming horizontal flow): $P_{1} + \frac{1}{2} ho v_{1}^{2} = P_{2} + \frac{1}{2} ho v_{2}^{2}$.$3$. Rearrange to find the pressure difference $\Delta P = P_{1} - P_{2} = \frac{1}{2} ho (v_{2}^{2} - v_{1}^{2})$.$4$. Substitute the values: $\Delta P = \frac{1}{2} 	imes 1000 	imes (25^{2} - 4^{2})$.$\Delta P = 500 	imes (625 - 16) = 500 	imes 609 = 304500 \, Pa$.

Determine the pressure difference in a tube of non-uniform cross-sectional area as shown in the figure. $\Delta P = ?$ (in $Pa$)
$d_{1} = 5 \, cm, V_{1} = 4 \, m/s, d_{2} = 2 \, cm, V_{2} = ?$
Assume the fluid is water with density $\rho = 1000 \, kg/m^{3}$.

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Determine the pressure difference in a tube of non-uniform cross-sectional area as shown in the figure. $\Delta P = ?$ (in $Pa$)$d_{1} = 5 \, cm, V_{1} = 4 \, m/s, d_{2} = 2 \, cm, V_{2} = ?$Assume the fluid is water with density $\rho = 1000 \, kg/m^{3}$.