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(D) According to the equation of continuity, $A_{1} v_{1} = A_{2} v_{2}$.Given $A_{1} = 10 \,cm^{2}$, $v_{1} = 1 \,ms^{-1}$, $A_{2} = 5 \,cm^{2}$.Subs

Vedclass · Accepted Answer

$500$

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(D) According to the equation of continuity, $A_{1} v_{1} = A_{2} v_{2}$.Given $A_{1} = 10 \,cm^{2}$, $v_{1} = 1 \,ms^{-1}$, $A_{2} = 5 \,cm^{2}$.Substituting the values: $10 	imes 1 = 5 	imes v_{2} \implies v_{2} = 2 \,ms^{-1}$.Using Bernoulli's equation for a horizontal pipe: $P_{1} + \frac{1}{2} ho v_{1}^{2} = P_{2} + \frac{1}{2} ho v_{2}^{2}$.Here, $ho = 1000 \,kg/m^{3}$ (density of water).$2000 + \frac{1}{2} 	imes 1000 	imes (1)^{2} = P_{2} + \frac{1}{2} 	imes 1000 	imes (2)^{2}$.$2000 + 500 = P_{2} + 2000$.$2500 = P_{2} + 2000 \implies P_{2} = 500 \,Pa$.

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