An incompressible fluid flows steadily throug | vedclass

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(D) According to the equation of continuity for an incompressible fluid, the product of the cross-sectional area $(A)$ and the velocity $(v)$ remains

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$4v$

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(D) According to the equation of continuity for an incompressible fluid, the product of the cross-sectional area $(A)$ and the velocity $(v)$ remains constant throughout the flow: $A_A v_A = A_B v_B$.Given the radius at point $A$ is $r_A = 3R$ and at point $B$ is $r_B = 1.5R$.The area of a circular cross-section is given by $\pi r^2$.Therefore, $A_A = \pi (3R)^2 = 9\pi R^2$ and $A_B = \pi (1.5R)^2 = 2.25\pi R^2$.Substituting these values into the continuity equation: $(9\pi R^2) \cdot v = (2.25\pi R^2) \cdot v_B$.Solving for $v_B$: $v_B = \frac{9\pi R^2}{2.25\pi R^2} \cdot v = \frac{9}{2.25} \cdot v = 4v$.Thus, the velocity at point $B$ is $4v$.

An incompressible fluid flows steadily through a horizontal cylindrical pipe. The pipe has radius $3R$ at point $A$ and $1.5R$ at point $B$, further along the flow direction at the same level. If the velocity at point $A$ is $v$, then the velocity at point $B$ is:

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