Let's solve the problem step by step: **Given:** - Velocity of flow: \( V = 2x\hat{i} + 3y\hat{j} \) - Find the equation of the stream line passing through point (4,3). **For a streamline:** \[ \frac{dx}{u} = \frac{dy}{v} \] where \(u = 2x\), \(v = 3y\). So, \[ \frac{dx}{2x} = \frac{dy}{3y} \] Cross-multiplied: \[ \frac{dx}{x} = \frac{2}{3} \frac{dy}{y} \] Integrate both sides: \[ \int \frac{dx}{x} = \frac{2}{3} \int \frac{dy}{y} \] \[ \ln x = \frac{2}{3} \ln y + C \] Exponentiate both sides: \[ x = e^C y^{2/3} \] Let \(e^C = k\), so: \[ x = k y^{2/3} \] **Find \(k\) using the point (4,3):** \[ 4 = k \cdot 3^{2/3} \] \[ k = \frac{4}{3^{2/3}} \] Plug \(k\) back in: \[ x = \frac{4}{3^{2/3}} y^{2/3} \] Or, \[ x \cdot 3^{2/3} = 4 y^{2/3} \] Or, \[ \left(\frac{x}{4}\right) = \left(\frac{y}{3}\right)^{2/3} \] Or, \[ x^{3/2} = 4^{3/2} \left(\frac{y}{3}\right) \] But let's match the given options. **Looking at the options:** Option 2: \(x^{3/2} = 0.72 y\) Let's check: \(4^{3/2} = 8\) So at \((4,3)\): \(x^{3/2} = 8\), \(y = 3\), so \(x^{3/2} = 0.72 \times 3 = 2.16\), which does not match. Let's check the other options by plugging \(x = 4, y = 3\): Option 1: \(0.72 x = y^{3/2}\) \(0.72 \times 4 = 2.88\), \(y^{3/2} = 3^{3/2} \approx 5.196\) Option 3: \(0.72 x^{1/2} = y^{1/2}\) \(x^{1/2} = 2, y^{1/2} \approx 1.732\), so \(0.72 \times 2 = 1.44\), not matching. Option 4: None of these. **Conclusion:** The correct answer is **None of these**. The correct equation should be: \[ x = \frac{4}{3^{2/3}} y^{2/3} \] which does **not** directly match any of the given options. --- **Final Answer:** **None of these**