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Question 1
ప్రశ్న 1
Using the convention u = ∂ψ/∂y and v = -∂ψ/∂x for a two-dimensional Cartesian flow, which statement about the stream function ψ is correct?
Explanation:
Under the stated convention, u = ∂ψ/∂y and v = -∂ψ/∂x. Along a curve ψ = constant, dψ = (∂ψ/∂x)dx + (∂ψ/∂y)dy = -v dx + u dy = 0. Rearrangement gives dy/dx = v/u wherever u is nonzero, which is the differential equation of a streamline. Thus constant-ψ curves coincide with streamlines for the two-dimensional incompressible-flow representation. Therefore Option A is correct.
వివరణ:
Question 2
ప్రశ్న 2
For the standard two-dimensional incompressible-flow stream function ψ and velocity potential φ, what are their SI dimensions?
Explanation:
Velocity components are obtained by differentiating either ψ or φ with respect to a length coordinate. Since velocity has dimensions LT^-1, both scalar functions must have dimensions L²T^-1 in a two-dimensional Cartesian formulation. The difference in stream-function values between two streamlines represents volumetric flow rate per unit depth, which also has units m²/s. Velocity potential has the same dimensions because its spatial gradient gives velocity. Therefore Option B is correct.
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Question 3
ప్రశ్న 3
Why does a sufficiently smooth stream function defined by u = ∂ψ/∂y and v = -∂ψ/∂x automatically satisfy the two-dimensional incompressible continuity equation?
Explanation:
For two-dimensional incompressible flow, continuity requires ∂u/∂x + ∂v/∂y = 0. Substituting u = ψ_y and v = -ψ_x gives ψ_yx - ψ_xy. For a sufficiently smooth function, the mixed partial derivatives ψ_yx and ψ_xy are equal. Their difference is therefore zero identically. Hence Option A correctly explains why the stream-function representation automatically satisfies continuity.
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Question 4
ప్రశ్న 4
Using the convention u = ∂φ/∂x and v = ∂φ/∂y, why is any sufficiently smooth velocity field derived from a scalar velocity potential φ irrotational in a two-dimensional region?
Explanation:
The z-component of vorticity in two-dimensional Cartesian flow is ∂v/∂x - ∂u/∂y. With u = φ_x and v = φ_y, this becomes φ_yx - φ_xy. For a sufficiently smooth scalar field, these mixed partial derivatives are equal. The vorticity therefore vanishes identically wherever the potential representation is valid. Thus Option B correctly connects velocity potential with irrotational flow.
వివరణ:
Question 5
ప్రశ్న 5
In a simply connected two-dimensional region with a continuously differentiable velocity field, which condition permits a single-valued velocity potential φ to be defined using u = ∂φ/∂x and v = ∂φ/∂y?
Explanation:
A velocity potential requires the velocity field to be the gradient of a scalar function. In a simply connected region, a continuously differentiable vector field is conservative when its curl is zero. For two-dimensional flow, the relevant curl condition is ∂v/∂x - ∂u/∂y = 0. The divergence-free condition instead relates to incompressibility and does not by itself guarantee a velocity potential. Therefore Option B is correct.
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Question 6
ప్రశ్న 6
Which statement about the existence of a conventional stream function ψ(x,y,t) is correct?
Explanation:
The conventional two-dimensional stream function is introduced so that the incompressible continuity equation is satisfied identically. The velocity components may still depend on time through ψ(x,y,t). Thus steadiness is not a necessary condition for the existence of this stream-function representation. Irrotationality becomes an additional condition when one also requires ψ to satisfy Laplace's equation or seeks a conjugate velocity potential. Therefore Option B is correct.
వివరణ:
Question 7
ప్రశ్న 7
If an incompressible flow has a velocity potential φ such that V = ∇φ, which governing equation must φ satisfy?
Explanation:
Incompressible flow with constant density requires the velocity field to have zero divergence. When V = ∇φ, the divergence becomes ∇·V = ∇·(∇φ) = ∇²φ. Setting the divergence equal to zero therefore gives Laplace's equation, ∇²φ = 0. The equation does not by itself require φ to be independent of time if an unsteady potential-flow formulation is considered. Hence Option A is correct.
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Question 8
ప్రశ్న 8
For a two-dimensional incompressible flow described by u = ∂ψ/∂y and v = -∂ψ/∂x, what additional condition makes the stream function ψ satisfy Laplace's equation?
Explanation:
The z-component of vorticity is ∂v/∂x - ∂u/∂y. Substituting the stream-function definitions gives -(∂²ψ/∂x² + ∂²ψ/∂y²). If the flow is irrotational, this vorticity is zero. Consequently ψ_xx + ψ_yy = 0, which is Laplace's equation in two dimensions. Therefore Option A is correct.
వివరణ:
Question 9
ప్రశ్న 9
In a two-dimensional incompressible and irrotational region where both φ and ψ exist with u = φ_x = ψ_y and v = φ_y = -ψ_x, how do lines φ = constant and ψ = constant intersect at ordinary points where velocity is nonzero?
Explanation:
The gradient ∇φ = (u,v) is normal to a line of constant velocity potential. From the stream-function convention, ∇ψ = (ψ_x,ψ_y) = (-v,u). Their dot product is u(-v) + v(u) = 0. Thus the gradients are perpendicular, and the corresponding level curves are also orthogonal wherever the gradients are nonzero. Therefore Option B is correct.
వివరణ:
Question 10
ప్రశ్న 10
Using u = φ_x = ψ_y and v = φ_y = -ψ_x, which pair of relations is correct?
Explanation:
The chosen velocity-potential convention gives u = φ_x and v = φ_y. The chosen stream-function convention gives u = ψ_y and v = -ψ_x. Equating the two expressions for u gives φ_x = ψ_y. Equating the two expressions for v gives φ_y = -ψ_x. Therefore Option B states the correct conjugate-function relations for this sign convention.
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Question 11
ప్రశ్న 11
A uniform two-dimensional flow of speed U is directed along the positive x-axis. Under the conventions u = ψ_y, v = -ψ_x and u = φ_x, v = φ_y, which pair represents this flow, apart from additive constants?
Explanation:
For uniform flow along positive x, the velocity components are u = U and v = 0. If ψ = Uy, then ψ_y = U and -ψ_x = 0, giving the correct velocity components. If φ = Ux, then φ_x = U and φ_y = 0, again giving the same flow. Additive constants would not affect any velocity because spatial differentiation removes them. Hence Option B is correct.
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Question 12
ప్రశ్న 12
In a two-dimensional incompressible flow, two streamlines have stream-function values ψ1 = 2 m²/s and ψ2 = 7 m²/s. What is the magnitude of volumetric flow rate per unit depth between them?
Explanation:
For a two-dimensional incompressible flow, the difference in stream-function values between two streamlines equals the volumetric flow rate per unit depth crossing a curve joining them. The magnitude is therefore |ψ2 - ψ1|. Substituting the given values gives |7 - 2| = 5 m²/s. The sign would indicate orientation according to the chosen convention, while the question asks for magnitude. Therefore Option B is correct.
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Question 13
ప్రశ్న 13
A stream function is ψ = kxy, where k = 4.0 s^-1. Using u = ψ_y and v = -ψ_x, what are the velocity components and speed at (x,y) = (1 m, 2 m)?
Explanation:
The x-component is u = ∂ψ/∂y = kx. With k = 4.0 s^-1 and x = 1 m, this gives u = 4 m/s. The y-component is v = -∂ψ/∂x = -ky, so at y = 2 m, v = -8 m/s. The speed is sqrt(u² + v²) = sqrt(16 + 64) = sqrt(80) = 8.94 m/s approximately. Hence Option A is correct.
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Question 14
ప్రశ్న 14
For the stream function ψ = k(x² - y²), where k is a nonzero constant with units s^-1, which velocity potential generates the same velocity field under u = φ_x and v = φ_y?
Explanation:
From ψ = k(x² - y²), the velocity components are u = ψ_y = -2ky and v = -ψ_x = -2kx. A compatible velocity potential must therefore satisfy φ_x = -2ky. Integrating with respect to x gives φ = -2kxy + f(y). Differentiating this expression with respect to y and matching v = -2kx gives f'(y) = 0, so f is a constant. Therefore Option B is correct.
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Question 15
ప్రశ్న 15
A velocity potential is φ = k(x² - y²), where k = 1.0 s^-1. What are the velocity components at (x,y) = (1 m, 2 m), and what is the speed?
Explanation:
The velocity components derived from the potential are u = φ_x and v = φ_y. Differentiating gives u = 2kx and v = -2ky. With k = 1.0 s^-1 at the specified point, u = 2 m/s and v = -4 m/s. The speed is sqrt(2² + (-4)²) = sqrt(20) = 4.47 m/s approximately. Therefore Option A is correct.
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Question 16
ప్రశ్న 16
The velocity potential is φ = kxy, where k = 2.0 s^-1. Under the convention u = φ_x, v = φ_y and u = ψ_y, v = -ψ_x, which stream function represents the same flow?
Explanation:
From φ = kxy with k = 2.0 s^-1, the velocity components are u = ky = 2y and v = kx = 2x. Since u = ψ_y, integration with respect to y gives ψ = y² + f(x). The second relation requires -ψ_x = 2x, so ψ_x = -2x. Therefore f'(x) = -2x and f(x) = -x² plus a constant. Hence ψ = y² - x² + C and Option B is correct.
వివరణ:
Question 17
ప్రశ్న 17
A stream function is ψ = k(xy² - x³/3), where k = 1.0 m^-1 s^-1. Which velocity potential corresponds to the same two-dimensional flow under the conventions u = ψ_y, v = -ψ_x and u = φ_x, v = φ_y?
Explanation:
The stream function gives u = ψ_y = 2kxy. It also gives v = -ψ_x = k(x² - y²). A compatible potential must satisfy φ_x = 2kxy, which integrates to φ = kx²y + f(y). Matching φ_y = kx² + f'(y) to v = k(x² - y²) gives f'(y) = -ky² and therefore f(y) = -ky³/3. Thus Option A is correct.
వివరణ:
Question 18
ప్రశ్న 18
A two-dimensional velocity field is u = ax - by and v = bx + ay, where a = 2 s^-1 and b = 3 s^-1. Which statement is correct in a simply connected region?
Explanation:
The divergence is ∂u/∂x + ∂v/∂y = a + a = 4 s^-1, so the field is not divergence-free. The z-component of vorticity is ∂v/∂x - ∂u/∂y = b - (-b) = 6 s^-1, so the field is also rotational. A standard two-dimensional incompressible stream function requires the divergence-free condition. A velocity potential in a simply connected region requires zero curl. Therefore neither standard representation applies and Option D is correct.
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Question 19
ప్రశ్న 19
For the velocity field u = ax and v = -ay, where a = 2 s^-1, which pair of scalar functions represents the field under the conventions used in this test?
Explanation:
The field has zero divergence because ∂u/∂x + ∂v/∂y = a - a = 0. It also has zero vorticity because ∂v/∂x - ∂u/∂y = 0. Integrating u = φ_x gives φ = (a/2)x² + f(y), and matching v = φ_y gives φ = (a/2)(x² - y²) plus a constant. Likewise, ψ = axy gives ψ_y = ax and -ψ_x = -ay. Therefore Option A supplies both compatible scalar functions.
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Question 20
ప్రశ్న 20
A stream function is ψ = k(x² + y²), where k = 1.0 s^-1. Under u = ψ_y and v = -ψ_x, which statement about the resulting two-dimensional flow is correct?
Explanation:
The velocity components are u = 2ky and v = -2kx. Their divergence is ∂u/∂x + ∂v/∂y = 0, as expected from a valid stream-function representation. The vorticity is ∂v/∂x - ∂u/∂y = -2k - 2k = -4k, which is nonzero because k = 1.0 s^-1. Thus the flow is rotational even though it is incompressible. A scalar velocity potential cannot represent this rotational field throughout an ordinary simply connected region, so Option B is correct.
వివరణ:
Question 21
ప్రశ్న 21
A velocity potential is φ = k(x² + y²), where k = 1.0 s^-1. Under u = φ_x and v = φ_y, which statement is correct?
Explanation:
The velocity components are u = 2kx and v = 2ky. Because they are derivatives of a smooth potential, the curl is zero and the field is irrotational. Its divergence is ∂u/∂x + ∂v/∂y = 2k + 2k = 4k, which is nonzero because k = 1.0 s^-1. Equivalently, ∇²φ = 4k, so this particular potential is not harmonic and does not satisfy constant-density incompressible continuity. Therefore Option A is correct.
వివరణ:
Question 22
ప్రశ్న 22
At an ordinary point in a two-dimensional potential flow, which geometrical statement about an equipotential line φ = constant and the local velocity vector is correct?
Explanation:
The velocity vector in the adopted convention is V = ∇φ. A gradient is normal to the corresponding level curve φ = constant. Therefore the velocity is perpendicular to an equipotential line at any regular point where the gradient is nonzero. A streamline is tangent to the velocity vector at that point. This is also the geometric reason equipotential lines and streamlines are orthogonal when both families exist, so Option B is correct.
వివరణ:
Question 23
ప్రశ్న 23
Why is a curve ψ = constant a streamline under the convention u = ψ_y and v = -ψ_x?
Explanation:
Along a level curve of the stream function, dψ = ψ_x dx + ψ_y dy = 0. Using ψ_x = -v and ψ_y = u gives -v dx + u dy = 0. For u nonzero, this rearranges to dy/dx = v/u. That is exactly the differential equation requiring the curve tangent to align with the local velocity direction. Therefore Option A is correct.
వివరణ:
Question 24
ప్రశ్న 24
For φ = k(x²y - y³/3), where k = 1.0 m^-1 s^-1, which statement is correct for the two-dimensional field defined by u = φ_x and v = φ_y?
Explanation:
Differentiating the potential gives u = 2kxy and v = k(x² - y²). The divergence is ∂u/∂x + ∂v/∂y = 2ky - 2ky = 0. The vorticity is ∂v/∂x - ∂u/∂y = 2kx - 2kx = 0. Thus the field is both divergence-free and irrotational. Therefore Option A is correct.
వివరణ:
Question 25
ప్రశ్న 25
Consider the following statements for a sufficiently smooth two-dimensional flow using u = ψ_y, v = -ψ_x and u = φ_x, v = φ_y: I. A stream-function representation automatically satisfies incompressible continuity. II. A velocity-potential representation automatically gives zero vorticity. III. If an incompressible potential flow exists, φ satisfies Laplace's equation. IV. If the same flow is both incompressible and irrotational, constant-φ and constant-ψ lines are orthogonal at regular points. Which option is correct?
Explanation:
Statement I is correct because the stream-function derivatives make the two-dimensional divergence cancel through equality of mixed partial derivatives. Statement II is correct because the curl of the gradient of a sufficiently smooth scalar potential is zero. Statement III is correct because incompressibility requires ∇·V = 0 and V = ∇φ therefore gives ∇²φ = 0. Statement IV is correct because ∇φ = (u,v) and ∇ψ = (-v,u) have zero dot product. Consequently all four statements are valid and Option D is correct.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: A. Curves ψ = constant are streamlines, and the velocity components are obtained from the stated cross-derivatives సమాధానం: A.
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Question 2 ప్రశ్న 2Answer: B. Both have dimensions L²T^-1 సమాధానం: B.
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Question 3 ప్రశ్న 3Answer: A. Because ∂u/∂x + ∂v/∂y reduces to equal mixed partial derivatives with opposite signs సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: B. Because ∂v/∂x - ∂u/∂y reduces to equal mixed partial derivatives with opposite signs సమాధానం: B.
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Question 5 ప్రశ్న 5Answer: B. ∂v/∂x - ∂u/∂y = 0 సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: B. It may be defined for a two-dimensional incompressible flow even when the velocity field is unsteady సమాధానం: B.
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Question 7 ప్రశ్న 7Answer: A. ∇²φ = 0 సమాధానం: A.
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Question 8 ప్రశ్న 8Answer: A. The flow must be irrotational సమాధానం: A.
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Question 9 ప్రశ్న 9Answer: B. They intersect orthogonally సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: B. φ_x = ψ_y and φ_y = -ψ_x సమాధానం: B.
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Question 11 ప్రశ్న 11Answer: B. ψ = Uy and φ = Ux సమాధానం: B.
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Question 12 ప్రశ్న 12Answer: B. 5 m²/s సమాధానం: B.
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Question 13 ప్రశ్న 13Answer: A. u = 4 m/s, v = -8 m/s, speed = 8.94 m/s సమాధానం: A.
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Question 14 ప్రశ్న 14Answer: B. φ = -2kxy + C సమాధానం: B.
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Question 15 ప్రశ్న 15Answer: A. u = 2 m/s, v = -4 m/s, speed = 4.47 m/s సమాధానం: A.
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Question 16 ప్రశ్న 16Answer: B. ψ = y² - x² + C సమాధానం: B.
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Question 17 ప్రశ్న 17Answer: A. φ = k(x²y - y³/3) + C సమాధానం: A.
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Question 18 ప్రశ్న 18Answer: D. It has both nonzero divergence and nonzero vorticity, so it satisfies neither the standard incompressible stream-function condition nor the velocity-potential condition సమాధానం: D.
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Question 19 ప్రశ్న 19Answer: A. φ = (a/2)(x² - y²) and ψ = axy సమాధానం: A.
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Question 20 ప్రశ్న 20Answer: B. It is incompressible but rotational, so a velocity potential does not exist over an ordinary simply connected region సమాధానం: B.
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Question 21 ప్రశ్న 21Answer: A. The field is irrotational but has divergence 4k, so it does not satisfy the constant-density incompressible continuity equation సమాధానం: A.
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Question 22 ప్రశ్న 22Answer: B. The velocity is normal to the equipotential line సమాధానం: B.
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Question 23 ప్రశ్న 23Answer: A. Because dψ = 0 gives -v dx + u dy = 0, hence dy/dx = v/u సమాధానం: A.
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Question 24 ప్రశ్న 24Answer: A. The field has divergence zero and vorticity zero సమాధానం: A.
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Question 25 ప్రశ్న 25Answer: D. I, II, III and IV సమాధానం: D.
