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Right option is (B) False

The explanation: Vorticity of the fluid is GIVEN by the FORMULA ∇ × V which is twice as much as the angular VELOCITY of the fluid.

∇ × V = 2ω

Where, ω is the angular velocity of the fluid.

Right choice is (a) True

For EXPLANATION: For the fluid to FLOW, it is essential for the velocity potential ϕ to satisfy the Laplace equation. If the CONDITION is not MET, the fluid does not flow and has ZERO velocity.

Thus, the condition for the fluid to flow is:

\(\frac {∂^2 ϕ}{∂x} + \frac {∂^2 ϕ}{∂y} + \frac {∂^2 ϕ}{∂z}\) = 0

Or

∇^2ϕ = 0

The correct ANSWER is (b) Irrotational

For explanation: For irrotational flow, curl of VELOCITY vector yields ZERO. In case the curl of any vector is zero i.e.∇ × V = 0, where V is a vector, it is also expressed in the FORM of ∇ζ where ζ is a scalar function. In case of irrotational flow, velocity potential ϕ is the scalar function. Hence if the flow has a velocity potential, it automatically implies that it is irrotational.

Correct option is (b) ∇ × V

The best I can explain: Vorticity REPRESENTS the CIRCULATION motion of the fluid as it flows. It is given by the curl of the VELOCITY vector (∇ × V). The curl operation determines the circulatory NATURE. For an irrotational flow vorticity is not equal to zero.

Correct choice is (c) \(\FRAC {∂v}{∂x} = \frac {∂u}{∂z}\)

For explanation: The CARTESIAN FORM of irrotational flow is given by:

∇ × V = \(\begin{vmatrix}

i & j & k\\

\frac {\partial }{\partial x} & \frac {\partial }{\partial y} & \frac {\partial }{\partial z}\\

u & v & w\\

\end{vmatrix} \)

On expanding this we get,

i(\(\frac {∂w}{∂y} – \frac {∂v}{∂z}\)) – j(\(\frac {∂w}{∂x} – \frac {∂u}{∂z}\)) + k(\(\frac {∂v}{∂x} – \frac {∂u}{∂y}\)) = 0

For irrotational flow SINCE vorticity = 0, ∇ × V = 0

\(\frac {∂w}{∂y} = \frac {∂v}{∂z}\) and \(\frac {∂w}{∂x} = \frac {∂u}{∂z}\) and \(\frac {∂v}{∂x} = \frac {∂u}{∂y}\)

The correct option is (b) False

Explanation: The velocity potential is GIVEN by ϕ = 2x + 3Y – 4y^2 + 8x^2

The velocity components u and v are calculated as FOLLOWS:

u = –\(\frac {∂ϕ}{∂x} = -\frac {∂}{∂x}\)(2x + 3y – 4y^2 + 8x^2) = -2 – 16x

v = –\(\frac {∂ϕ}{∂y} = -\frac {∂}{∂y}\)(2x + 3y – 4y^2 + 8x^2) = -3 + 8y

The continuity equation is given by:

\(\frac {∂u}{∂x} + \frac {∂v}{∂y}\) = 0

Substituting the values of u and v,

\(\frac {∂}{∂x}\)(-2 – 16x) + \(\frac {∂}{∂y}\)(-3 + 8y) = -16 + 8 = -8

Since this is not equal to ZERO, HENCE continuity equation is not satisfied.

The correct answer is (c) 36 m/s

The BEST EXPLANATION: The velocity potential is given as ϕ = 2x^2 – 4xy^2 + \( \frac {8x^2}{y}\)

Velocity component in x – direction is given by u = –\( \frac {∂ϕ}{∂x}\)

u = –\( \frac {∂}{∂x} \big ( \)2x^2 – 4xy^2 + \( \frac {8x^2}{y} \big ) \) = 4x – 4y^2 + \( \frac {16x}{y}\)

For calculating the velocity component at point (2,1), we substitute these POINTS in the above equation

u = 4(2) – 4(1)^2 + \( \frac {16(2)}{(1)}\) = 8 – 4 + 32 = 36 m/s

Correct answer is (a) Flow BEHIND curved shock wave

For explanation I would say: The flow within the boundary LAYER and behind curved shock wave is a rotational flow as the curl of velocity vector is not equal to zero. But, flow over a WEDGE, cone, in a two – dimensional NOZZLE flow and over a slender body is irrotational since the curl of velocity vector is zero.

Correct option is (b) RESULTS in ZERO drag

The explanation: In REAL life scenario, it is impossible to have only irrotational region over the body. The flow COMPRISES of both rotational and irrotational region. According to d’Alember’s paradox, having irrotational and inviscid flow THROUGHOUT results in zero drag which is impossible in real life hence it’s a paradox.

The correct option is (b) False

To elaborate: The special form of Euler’s EQUATION DP = – ρVdV is applicable for irrotational FLOW which is derived by adding the x, y, z components of the Euler’s equation for irrotational CONDITION. This would be applicable for ROTATIONAL flow only along a streamline.

The correct option is (b) ∇ϕ

To ELABORATE: The irrotational flow is given by the curl of velocity vector. If we TAKE a GRADIENT of the scalar function, we get zero as a result.

∇ × (∇ϕ) = 0

Thus the velocity potential is described as the scalar function ∇ϕ. It satisfies the Laplace EQUATIONS as well.

Right choice is (a) Laplace EQUATION

The best explanation: The VELOCITY component is the NEGATIVE derivative of the velocity potential in that direction. According to this,

u = –\(\frac {∂ϕ}{∂x}\), v = –\(\frac {∂ϕ}{∂y}\), w = –\(\frac {∂ϕ}{∂z}\)

The continuity equation for THREE – dimensional flow is given by:

\(\frac {∂u}{∂x} + \frac {∂v}{∂y} + \frac {∂w}{∂z}\) = 0

Substituting the velocity components in the continuity equation, we get

\(\frac {∂}{∂x} \BIG ( – \frac {∂ϕ}{∂x} \big ) + \frac {∂}{∂y} \big ( – \frac {∂ϕ}{∂x} \big ) + \frac {∂}{∂z} \big ( – \frac {∂ϕ}{∂x} \big )\) = 0

\(\frac {∂^2 ϕ}{∂x} + \frac {∂^2 ϕ}{∂y} + \frac {∂^2 ϕ}{∂z}\) = 0

The above final equation is known as Laplace equation, thus velocity potential satisfies the Laplace equation.

Right option is (a) Creates vorticity downstream

Explanation: When the FLUID flows through a CURVED shock wave, it undergoes a strong entropy change due to the PRESENCE of shock wave. This CAUSES the flow to rotate leading to the FORMATION of vorticity in the downstream region of the shock wave.

Right answer is (c) Zero

The explanation: When the FLUID flows towards a solid surface, there is a boundary layer FORMATION with zero velocity at its surface and as we proceed normally, it increases to FINALLY BECOME EQUAL to the free stream velocity. At this particular point there is no vorticity i.e. the flow is irrotational. For an irrotational flow the angular velocity is zero.