ISRO Scientist or Engineer Mechanical 2007

$$\triangledown \times V = 0$$

Increases

Decreases

Remains constant

Mayincrease or decrease depending on temperature

$$\left(\frac{p_1}{p_2}\right)^{\frac{\gamma - 1}{\gamma}}$$

$$\left(\frac{p_2}{p_1}\right)^{\frac{\gamma - 1}{\gamma}}$$

$$\left(\frac{v_1}{v_2}\right)^{\frac{\gamma - 1}{\gamma}}$$

$$\left(\frac{v_2}{v_1}\right)^{\frac{\gamma - 1}{\gamma}}$$

Local pressure becomes very high

Local pressure falls below the vapour pressure

Thoma cavitation parameter exceeds a certain limit

Local temperature becomes low

$$\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z} = 0$$

$$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} = \frac{\partial w}{\partial z} = 0$$

$$\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z} = 1$$

$$\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z} = u.v.w$$

400 kg/cm$$^2$$

4000 kg/cm$$^2$$

$$40 \times 10^5 kg/cm^2$$

$$40 \times 10^6 kg/cm^2$$

Mass of liquid displaced

Viscosity of the liquid

Pressure of the liquid displaced

Depth of immersion

0.8

1.25

0.2

0.64

$$\frac{\rho AV^2}{g}$$

$$\frac{\rho AV}{g}\left(V - \frac{\pi DN}{60}\right)$$

$$\frac{\rho A}{g} \times \frac{\pi DN}{60}$$

$$\rho A\left(V - \frac{\pi DN}{60}\right)$$

$$\frac{\rho}{10}$$ meters

$$\frac{10(1 + \rho)}{\rho}$$ meters

$$\frac{10}{1 - \rho}$$ meters

$$\frac{10}{\rho}$$ meters