Question:

If \(v=3xy\), the magnitude of the velocity vector at \((2,-2)\) is ____

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For a velocity potential \(\phi\), \[ \boxed{ u=\frac{\partial\phi}{\partial x}, \qquad v=\frac{\partial\phi}{\partial y} } \] Velocity magnitude: \[ \boxed{ V=\sqrt{u^2+v^2}. } \]
Updated On: Jul 24, 2026
  • \(4\sqrt{2}\)
  • \(6\sqrt{2}\)
  • \(12\)
  • \(0\)
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The Correct Option is B

Solution and Explanation

Step 1: Use the velocity potential relation. Given the velocity potential, \[ \phi = 3xy \] The velocity components are \[ u=\frac{\partial \phi}{\partial x}=3y, \] \[ v=\frac{\partial \phi}{\partial y}=3x. \]

Step 2:
Evaluate at \((2,-2)\). \[ u=3(-2)=-6, \] \[ v=3(2)=6. \] Hence, the magnitude of velocity is \[ V=\sqrt{u^2+v^2} =\sqrt{(-6)^2+6^2} =\sqrt{72} =6\sqrt{2}. \] Therefore, \[ \boxed{6\sqrt{2}} \] Hence, \[ \boxed{(B)} \] is the correct answer.
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