Question

A two dimensional flow has velocities in \( x \) and \( y \) directions given by \( u = 2xyt \) and \( v = -y^2 t \), where \( t \) denotes time. The equation for the streamline passing through \( x = 1, y = 1 \) is

• A. \( x^2 y = 1 \)
• B. \( x y^2 = 1 \)
• C. \( x^2 y^2 = 1 \)
• D. \( x/y^2 = 1 \)

Hint:

To find the equation of a streamline, use the differential relation \( \frac{dy}{dx} = \frac{v}{u} \), and integrate with respect to the coordinates.

Answer 1

The Correct Option is B

For the equation of streamline, we use the relationship between velocity components \( u \) and \( v \) and the differential equation for streamlines: \[ \frac{dy}{dx} = \frac{v}{u}. \] Given that \( u = 2xyt \) and \( v = -y^2 t \), we can substitute these into the equation: \[ \frac{dy}{dx} = \frac{-y^2 t}{2xyt} = \frac{-y}{2x}. \] This simplifies to: \[ \frac{dy}{dx} = -\frac{y}{2x}. \] Now, integrate both sides: \[ \int \frac{dy}{y} = \int -\frac{1}{2x} dx. \] On integrating, we get: \[ \ln y = -\frac{1}{2} \ln x + C. \] Exponentiating both sides, we obtain: \[ y = Cx^{-1/2}. \] Now, substitute \( x = 1 \) and \( y = 1 \) into the equation to find \( C \): \[ 1 = C \times 1^{-1/2} \Rightarrow C = 1. \] Thus, the equation for the streamline is: \[ y = \frac{1}{\sqrt{x}} \Rightarrow xy^2 = 1. \] Thus, the correct answer is Option (B). 
Final Answer: (B) \( xy^2 = 1 \)