Step 1: Identifying the relationship between "Courage" and "Bravery."


The words "Courage" and "Bravery" are synonyms. They both describe the quality of being willing to face danger, pain, or difficulty. Since these words are closely related in meaning, we need to find a similar relationship between the second pair of words, "Yearning" and one of the given options.

Step 2: Analyzing the word "Yearning."

"Yearning" refers to a strong desire or longing for something. Now, let’s analyze the options:

Option (A) Longing: This is a direct synonym of "Yearning." Both words express a deep, intense desire for something.

Option (B) Yelling: This is unrelated to "Yearning." "Yelling" refers to shouting loudly, which has no connection to the emotional desire conveyed by "Yearning."

Option (C) Yawning: This is also unrelated to "Yearning." "Yawning" refers to the action of opening the mouth wide, usually due to tiredness or boredom, which does not fit the emotional context of "Yearning."

Option (D) Glaring: This means staring angrily, which has no connection to the concept of desire or longing.


Thus, Option (A) Longing is the most appropriate because it maintains the same synonym relationship that exists between "Courage" and "Bravery." Quick Tip: When solving analogies, consider the relationship between the first pair of words, and look for a similar relationship in the second pair. Synonyms or words with similar meanings are often the correct choice.

Step 1: Understanding the Context of the Sentence.

The sentence describes an action that was happening in the past and was interrupted by another event. The phrase "when it suddenly started to rain" suggests that the action of playing tennis was ongoing at the time the rain started. This points to the past perfect continuous tense, which is used to describe an action that was happening continuously before another past action interrupted it.

Step 2: Analyzing the Options:


Option (A) "have been playing" is the present perfect continuous tense, which indicates an action that started in the past and continues into the present. Since the sentence is referring to a past event, this is not correct.

Option (B) "had been playing" is the past perfect continuous tense, which correctly describes an action that was happening continuously in the past before another past event (the rain) interrupted it. This is the correct choice.

Option (C) "would have been playing" is a conditional perfect continuous tense, typically used to describe hypothetical situations or actions that would have happened under different conditions. This is not appropriate for the sentence.

Option (D) "could be playing" suggests possibility in the present, which is incorrect because the sentence is referring to a past event.

Therefore, the most appropriate option is (B) had been playing. Quick Tip: Use the past perfect continuous tense ("had been + verb-ing") when describing an action that was happening continuously in the past before another action interrupted it.

We are asked to find how many pixels have intensities less than or equal to 4.

Let’s go through the matrix and count the number of pixels that satisfy \( U \leq 4 \).


The given matrix of pixel intensities is:






Row 1: 0, 1, 0, 2 (all are \(\leq 4\)) - Count: 4
Row 2: 4, 7, 3, 3 (4, 3, 3 are \(\leq 4\)) - Count: 3
Row 3: 5, 5, 4, 4 (4, 4 are \(\leq 4\)) - Count: 2
Row 4: 6, 7, 3, 2 (3, 2 are \(\leq 4\)) - Count: 2




Total count: \(4 + 3 + 2 + 2 = 11\) Quick Tip: In digital images, when analyzing pixel intensities, always look for the specific threshold conditions (like \( U \leq 4 \)) and count the pixels that satisfy the condition across the entire matrix.

In this problem, we are given a figure involving three geometric shapes: a rectangle, a triangle, and an ellipse. We need to determine the most appropriate values for \( P \), \( Q \), and \( R \) based on the dimensions provided in the figure.

The number \( P \) is associated with the length of the rectangle's side. Since the side of the rectangle is labeled as 6, we can conclude that \( P = 6 \).

The number \( Q \) is associated with the height of the triangle. The height of the triangle is labeled as 5, so \( Q = 5 \).

The number \( R \) is associated with the length of the major axis of the ellipse. The length of the major axis is labeled as 3, so \( R = 3 \).

Thus, the correct values are \( P = 6 \), \( Q = 5 \), and \( R = 3 \). Quick Tip: When solving geometry problems involving multiple shapes, carefully observe the labels and dimensions associated with each shape. Use these details to assign values to the variables based on the geometric properties.

Step 1: Understanding the effects of increasing the width of a rectangle by 10%.

The dimensions of the rectangle are \(L\) (length) and \(W\) (width), with \(L > W\). When the width \(W\) is increased by 10%, the new width becomes \(W' = 1.1W\), while the length \(L\) remains the same.

Step 2: Analyzing the impact on each option.


Option (A) Perimeter increases by 10%:
The perimeter of a rectangle is given by the formula:
\[ P = 2(L + W) \]
When the width increases by 10%, the new perimeter becomes:
\[ P' = 2(L + 1.1W) \]
This is not exactly a 10% increase. The increase in perimeter is not proportional to the increase in width. Therefore, this option is incorrect.

Option (B) Length of the diagonals increases by 10%:
The diagonal \(d\) of a rectangle is given by the Pythagorean theorem:
\[ d = \sqrt{L^2 + W^2} \]
When the width increases by 10%, the new diagonal is:
\[ d' = \sqrt{L^2 + (1.1W)^2} \]
This increase is not guaranteed to be exactly 10%. The length of the diagonal increases, but it is not necessarily a 10% increase. Therefore, this option is incorrect.

Option (C) Area increases by 10%:
The area \(A\) of a rectangle is given by:
\[ A = L \times W \]
After increasing the width by 10%, the new area is:
\[ A' = L \times 1.1W = 1.1 \times L \times W \]
This shows that the area increases by 10%, as the new area is 1.1 times the original area. Therefore, this option is correct.

Option (D) The rectangle becomes a square:
A rectangle becomes a square only if the length and width are equal. Since only the width is increased by 10%, the rectangle does not become a square. Therefore, this option is incorrect.

Therefore, the correct answer is Option (C) Area increases by 10%. Quick Tip: When a single dimension of a rectangle (such as width) is increased by a certain percentage, the area of the rectangle will increase by the same percentage, as long as the other dimension remains unchanged.

Step 1: Understanding the context of each statement.

We need to match the statements made by Shanthala (Column-I) with the appropriate responses given by Kanishk (Column-II).

P. "This house is in a mess."
The best response would be: "No problem, let me clear it up for you." Kanishk is offering to help with the situation, which matches this statement.

Q. "I am not happy with the marks given to me."
The appropriate response would be: "Don’t worry, I will take it up with your teacher." Kanishk is reassuring Shanthala that their concern about marks will be addressed.

R. "Politics is a subject I avoid talking about."
The correct response here would be: "Alright, I won’t bring it up during our conversations." This indicates that Kanishk will avoid discussing politics.

S. "I don’t know what this word means."
The most appropriate response is: "Well, you can easily look it up." This suggests that Kanishk is offering a straightforward solution to Shanthala’s problem.

Step 2: Identifying the correct match.

From the analysis above, we match the following:


P – 3: "This house is in a mess." → "No problem, let me clear it up for you."

Q – 4: "I am not happy with the marks given to me." → "Don’t worry, I will take it up with your teacher."

R – 1: "Politics is a subject I avoid talking about." → "Alright, I won’t bring it up during our conversations."

S – 2: "I don’t know what this word means." → "Well, you can easily look it up."


Therefore, the correct answer is Option (B) P – 3; Q – 4; R – 1; S – 2. Quick Tip: When matching statements with responses, pay attention to the context and tone of the statement and the most appropriate response that follows. In such cases, reassurance, solutions, and acknowledgment of preferences are key.

The weight function is increasing until the age of 50 and then decreases. Given that \( a_1 < a_2 \), this means that the younger brother's age \( a_1 \) is less than 50 and the older brother's age \( a_2 \) is greater than 50. When both brothers reach the same weight, their ages must satisfy the condition where the younger brother is below 50 and the older brother is above 50. Hence, \( a_1 < 50 < a_2 \). Quick Tip: In problems involving monotonic functions, focus on the turning point (in this case, age 50) to determine the behavior of the function for different values of the variables.

We are given a regular dodecagon inscribed in a circle, and we need to form squares using the triangles formed by connecting the center of the circle to the vertices of the dodecagon. There are 12 triangles in total, each corresponding to a side of the dodecagon. The number of squares that can be formed is 3, and each square requires 4 triangles. Hence, the correct number of squares and the number of triangles required to form each square are 3 and 4, respectively. Quick Tip: In geometry problems involving regular polygons and inscribed shapes, carefully examine the symmetry and properties of the polygon to understand how to derive relationships for constructing new shapes.

Step 1: Solve the given equation \(3^{x^2} = 27 \times 9^x\).


We start with the equation: \[ 3^{x^2} = 27 \times 9^x. \]
We can rewrite \(27\) and \(9\) as powers of 3: \[ 27 = 3^3 \quad and \quad 9 = 3^2. \]
Thus, the equation becomes: \[ 3^{x^2} = 3^3 \times (3^2)^x. \]
Now simplify the right-hand side: \[ 3^{x^2} = 3^3 \times 3^{2x}. \]
Using the property of exponents \(a^m \times a^n = a^{m+n}\), we combine the powers of 3: \[ 3^{x^2} = 3^{3 + 2x}. \]
Since the bases are the same, we can equate the exponents: \[ x^2 = 3 + 2x. \]
Rearranging the equation: \[ x^2 - 2x - 3 = 0. \]
Factoring the quadratic equation: \[ (x - 3)(x + 1) = 0. \]
Thus, \(x = 3\) or \(x = -1\).

Step 2: Evaluate \(\frac{2^{x^2}}{(2^x)^2}\).


We now substitute \(x = 3\) and \(x = -1\) into the expression \(\frac{2^{x^2}}{(2^x)^2}\):

When \(x = 3\):
\[ \frac{2^{3^2}}{(2^3)^2} = \frac{2^9}{2^6} = 2^{9-6} = 2^3. \]
When \(x = -1\):
\[ \frac{2^{(-1)^2}}{(2^{-1})^2} = \frac{2^1}{2^{-2}} = 2^{1 - (-2)} = 2^3. \]

Thus, the value is \(2^3\), which corresponds to Option (C). Quick Tip: When solving equations with exponents, simplify both sides of the equation, equate the exponents, and solve the resulting algebraic equation. Afterward, substitute the values into the given expression.

Step 1: Understanding the problem.

The number of shifts corresponding to different numbers of patients per shift is given in the bar graph. The amount Dr. Gita earns is \(1000(X - 0.2)\), where \(X\) is the number of patients per shift.

The data from the graph is as follows:

For \(X = 5\), the number of shifts is 20.

For \(X = 6\), the number of shifts is 40.

For \(X = 7\), the number of shifts is 30.

For \(X = 8\), the number of shifts is 10.


Step 2: Calculating the total earnings.

For \(X = 5\):
\[ Earnings = 1000 \times (5 - 0.2) \times 20 = 1000 \times 4.8 \times 20 = 96,000. \]
For \(X = 6\):
\[ Earnings = 1000 \times (6 - 0.2) \times 40 = 1000 \times 5.8 \times 40 = 232,000. \]
For \(X = 7\):
\[ Earnings = 1000 \times (7 - 0.2) \times 30 = 1000 \times 6.8 \times 30 = 204,000. \]
For \(X = 8\):
\[ Earnings = 1000 \times (8 - 0.2) \times 10 = 1000 \times 7.8 \times 10 = 78,000. \]

Step 3: Calculating the total earnings and average earnings.

Total earnings for all 100 shifts:
\[ Total Earnings = 96,000 + 232,000 + 204,000 + 78,000 = 610,000. \]
The average earnings per shift: \[ Average Earnings = \frac{610,000}{100} = 6,100. \]

Thus, the average earnings per shift are Rs.6,100, which corresponds to Option (A). Quick Tip: When calculating averages involving frequency distributions, first calculate the total earnings, then divide by the total number of shifts to find the average.

To solve this, we need to recall the definition of a symmetric matrix. A matrix \( A \) is said to be symmetric if it is equal to its own transpose, i.e., \[ A = A^T \]

Step 1: Definition of Transpose

The transpose of a matrix \( A \), denoted by \( A^T \), is obtained by swapping its rows and columns. So, for any matrix \( A = [a_{ij}] \), the transpose matrix \( A^T \) will have elements \( A^T = [a_{ji}] \).

Step 2: Symmetry Condition

For a symmetric matrix, by definition, the matrix is equal to its transpose. This implies: \[ A = A^T \]
In other words, the elements of the matrix satisfy the condition that \( a_{ij} = a_{ji} \) for all \( i, j \).

Step 3: Conclusion

Since a real symmetric matrix satisfies \( A = A^T \), the transpose of \( A \) is exactly \( A \). Therefore, the correct answer is \( A \).
\[ A^T = A \] Quick Tip: For symmetric matrices, always remember that the matrix is equal to its transpose: \( A = A^T \). This property is fundamental in linear algebra and is used in various matrix-related computations.

We are given the following system of equations:
\[ 2x + 3y + z = 0 \quad (1) \] \[ x + y = 0 \quad (2) \] \[ y + z = 0 \quad (3) \]

From equation (2), \( x = -y \).
Substitute this in equation (1): \[ 2(-y) + 3y + z = 0 \quad \Rightarrow \quad -2y + 3y + z = 0 \quad \Rightarrow \quad y + z = 0 \]

Equation (3) is \( y + z = 0 \), which is identical to the result we got above. Hence, the system is consistent and has infinitely many solutions. Quick Tip: For systems of linear equations, check for consistency. If the equations represent planes that intersect along a line or a point, there will be infinitely many solutions.

To find the eigenvalues of a matrix, we solve the characteristic equation: \[ det(A - \lambda I) = 0 \]
where \( A \) is the matrix, \( \lambda \) is the eigenvalue, and \( I \) is the identity matrix.

Step 1: Construct the matrix \( A - \lambda I \)

Given the matrix \( A = \begin{bmatrix} 1 & 2
0 & 3 \end{bmatrix} \), we subtract \( \lambda I \) (where \( I \) is the 2x2 identity matrix) from \( A \):
\[ A - \lambda I = \begin{bmatrix} 1 & 2
0 & 3 \end{bmatrix} - \begin{bmatrix} \lambda & 0
0 & \lambda \end{bmatrix} = \begin{bmatrix} 1 - \lambda & 2
0 & 3 - \lambda \end{bmatrix} \]

Step 2: Find the determinant of \( A - \lambda I \)

Now, compute the determinant of the resulting matrix: \[ det(A - \lambda I) = det\begin{bmatrix} 1 - \lambda & 2
0 & 3 - \lambda \end{bmatrix} \]
The determinant of a 2x2 matrix \( \begin{bmatrix} a & b
c & d \end{bmatrix} \) is given by \( ad - bc \). So, for our matrix: \[ det(A - \lambda I) = (1 - \lambda)(3 - \lambda) - (0)(2) = (1 - \lambda)(3 - \lambda) \]

Step 3: Solve the characteristic equation

Now, we solve the characteristic equation: \[ (1 - \lambda)(3 - \lambda) = 0 \]
This gives two solutions: \[ 1 - \lambda = 0 \quad \Rightarrow \quad \lambda = 1 \] \[ 3 - \lambda = 0 \quad \Rightarrow \quad \lambda = 3 \]

Conclusion:
Thus, the eigenvalues of the matrix are \( \lambda = 1 \) and \( \lambda = 3 \). Quick Tip: To find the eigenvalues of a matrix, solve the characteristic equation \( det(A - \lambda I) = 0 \). The roots of the equation are the eigenvalues.

The given partial differential equation is of the form: \[ A \frac{\partial^2 u}{\partial x^2} + 2B \frac{\partial^2 u}{\partial x \partial y} + C \frac{\partial^2 u}{\partial y^2} = 0 \]
For classification, we calculate the discriminant \( D \): \[ D = B^2 - AC \]
In this case, \( A = 1 \), \( B = 2 \), and \( C = 2 \), so: \[ D = 2^2 - (1)(2) = 4 - 2 = 2 \]
Since \( D > 0 \), the equation is hyperbolic. Quick Tip: To classify partial differential equations, calculate the discriminant \( D = B^2 - AC \). If \( D > 0 \), the equation is hyperbolic.

For continuity at \( x = 2 \), the left-hand limit (as \( x \to 2^- \)) must equal the right-hand limit (as \( x \to 2^+ \)).

1. Left-hand limit at \( x = 2 \):

For \( x < 2 \), \( f(x) = A + x \). Thus, \[ \lim_{x \to 2^-} f(x) = A + 2 \]

2. Right-hand limit at \( x = 2 \):

For \( x \geq 2 \), \( f(x) = 1 + x^2 \). Thus, \[ \lim_{x \to 2^+} f(x) = 1 + 2^2 = 1 + 4 = 5 \]

For continuity, these two limits must be equal: \[ A + 2 = 5 \quad \Rightarrow \quad A = 3 \] Quick Tip: For continuity at a point, the left-hand limit and the right-hand limit must be equal at that point.

We are given the expression: \[ \frac{(i+1)^3}{i-1} \]
Step 1: Expand \( (i+1)^3 \)

First, expand \( (i+1)^3 \): \[ (i + 1)^3 = i^3 + 3i^2 + 3i + 1 \]
Since \( i^2 = -1 \) and \( i^3 = -i \), we get: \[ (i + 1)^3 = -i - 3 + 3i + 1 = 2i - 2 \]

Step 2: Simplify the expression

Now substitute this into the original expression: \[ \frac{(i+1)^3}{i-1} = \frac{2i - 2}{i - 1} \]
We will now multiply both the numerator and denominator by the conjugate of \( i - 1 \), which is \( i + 1 \): \[ \frac{2i - 2}{i - 1} \times \frac{i + 1}{i + 1} = \frac{(2i - 2)(i + 1)}{(i - 1)(i + 1)} \]
The denominator becomes: \[ (i - 1)(i + 1) = i^2 - 1^2 = -1 - 1 = -2 \]
Now expand the numerator: \[ (2i - 2)(i + 1) = 2i^2 + 2i - 2i - 2 = -2 + 2i - 2i - 2 = -4 \]
Thus the expression becomes: \[ \frac{-4}{-2} = 2 \]
So, we obtain the simplified result of \( 2 \). Quick Tip: When simplifying complex expressions, especially involving \( i = \sqrt{-1} \), remember to use the conjugate of the denominator to simplify the fraction. This technique is essential for resolving expressions with complex numbers.

For a laminar boundary layer in a two-dimensional incompressible flow, the boundary layer thickness \( \delta \) is related to the Reynolds number \( Re_x \) by the following empirical relationship: \[ \delta \propto x \cdot Re_x^{-1/2} \]
Thus, the ratio \( \frac{\delta}{x} \) is proportional to \( Re_x^{-1/2} \): \[ \frac{\delta}{x} \propto Re_x^{-1/2} \]

Step 1: Reynolds number definition

The Reynolds number \( Re_x \) at a distance \( x \) from the leading edge is defined as: \[ Re_x = \frac{\rho u x}{\mu} \]
where \( \rho \) is the fluid density, \( u \) is the flow velocity, and \( \mu \) is the dynamic viscosity.

Step 2: Relation between boundary layer thickness and Reynolds number

From the theory of boundary layers in laminar flow, we know that the thickness of the boundary layer is inversely proportional to the square root of the Reynolds number, as shown above.

Conclusion:

Therefore, the ratio \( \frac{\delta}{x} \) is proportional to \( Re_x^{-1/2} \). Quick Tip: For laminar boundary layers, the boundary layer thickness is proportional to \( x \cdot Re_x^{-1/2} \), where \( Re_x \) is the Reynolds number based on the distance from the leading edge.

For NACA airfoils, the number 4415 represents the following components:

The first digit (4) represents the maximum camber as a percentage of the chord length (in this case, 4%).

The second digit (4) represents the location of maximum camber as a fraction of the chord length from the leading edge (in this case, 0.40 or 40%).

The last two digits (15) represent the maximum thickness as a percentage of the chord length (15%).

Thus, the location of the maximum camber is 0.40 of the chord length from the leading edge. Quick Tip: For NACA airfoils, the first digit represents the camber as a percentage, the second digit represents the location of maximum camber as a fraction of the chord length, and the last two digits represent the maximum thickness as a percentage of the chord length.

In this scenario, the airfoil is at its zero-lift angle of attack, meaning that it produces no net lift force but still generates a pitching moment. This moment is due to the camber of the airfoil, which leads to an imbalance in the pressure distribution over the surface. The correct representation shows the flow direction and the pitching moment generated by the airfoil in this condition. Quick Tip: Even at zero-lift angle of attack, a positively cambered airfoil generates a pitching moment due to the asymmetrical pressure distribution across its surface.

When two vortices of equal and opposite circulation are placed in a two-dimensional, incompressible, inviscid flow, they will rotate around each other with respect to their centroid. In this scenario, the vortices have clockwise circulation, and they rotate in a clockwise direction around their common centroid. Therefore, option (C) is correct. Quick Tip: In a system of two point vortices, they will rotate around their common centroid, with the direction of rotation depending on the sign of their circulation. In this case, with clockwise circulation, the vortices rotate clockwise around each other.

The Mach number (\( M \)) is a dimensionless number that represents the ratio of the velocity of the flow to the speed of sound in the medium. It is used to estimate the relationship between the inertia forces and elastic forces in a fluid flow. Hence, the correct expression for Mach number is: \[ M = \sqrt{\frac{inertia force}{elastic force}} \]
This relationship helps determine the effects of compressibility and the transition between subsonic and supersonic flows. Quick Tip: The Mach number helps in determining whether the flow is compressible and is a key indicator for shock waves and flow speeds.

To achieve a coordinated right turn in a general aviation airplane:

The left aileron must be deflected down and the right aileron up to roll the airplane into the right turn.

To counteract the adverse yaw (which causes the nose to yaw left), the rudder must be deflected to the right to maintain coordination. Quick Tip: In coordinated turns, the ailerons control roll, while the rudder is used to counteract adverse yaw and ensure the airplane turns smoothly without yawing to the opposite side.

A fuselage with a low wing configuration has a destabilizing effect on an aircraft's static roll stability. In this configuration, the aircraft's center of gravity and the center of lift of the wings are closer together, which reduces the restoring moment when the aircraft rolls. This leads to a higher chance of instability in roll.

A positive dihedral angle, a high wing configuration, and a swept-back wing generally contribute to increased roll stability. Quick Tip: For better roll stability, an aircraft is generally designed with a high wing or positive dihedral, which enhances the restoring moment during a roll. A low-wing configuration can result in reduced static roll stability.

At an altitude of 5000 m, the indicated airspeed (IAS) is not equal to the true airspeed (TAS). IAS is the speed shown by the airspeed indicator, and it does not account for altitude or air density.

The true airspeed is higher than the indicated airspeed because air density decreases with altitude.

Calibrated airspeed (CAS) is corrected for instrument and position errors and is assumed to be the same as IAS in this case.

Equivalent airspeed (EAS) is related to CAS and air density, and at altitude, it can differ from IAS as well.


Hence, the false statement is (C): The true airspeed is 250 km/h, because at 5000 m altitude, the true airspeed is greater than the indicated airspeed. Quick Tip: Indicated airspeed increases with altitude due to the decreasing air density. True airspeed must be calculated from IAS to account for changes in atmospheric conditions at higher altitudes.

For longitudinal static stability, the center of gravity (CG) of the airplane must be located forward of the neutral point. The neutral point is the point where the pitching moment coefficient (\( C_m \)) becomes zero. When the CG is forward of the neutral point, the airplane experiences a restoring moment when displaced from equilibrium, thereby providing stability.

If the CG is aft of the neutral point, the airplane will be statically unstable.

A positive stability coefficient \( \frac{\partial C_m}{\partial \alpha} \) is a characteristic of stable aircraft, but it alone does not ensure static stability.

A negative static margin indicates instability. Quick Tip: In general aviation airplanes, ensuring that the CG is forward of the neutral point is crucial for achieving positive longitudinal static stability.

For a homogeneous and isotropic material, the relationship between shear modulus (\( G \)), Young's modulus (\( E \)), and Poisson's ratio (\( \nu \)) is given by the equation:
\[ E = 2G(1 + \nu) \]

This equation shows that Young’s modulus is related to the shear modulus and Poisson’s ratio, which describes the material’s response to stress and strain. Quick Tip: In material mechanics, the shear modulus, Young's modulus, and Poisson's ratio are fundamental properties that help describe the material's deformation under various types of stress.

The problem describes a beam subjected to a triangular distributed load. In this case, the resultant of the triangular load acts at a distance of \( \frac{l}{3} \) from the larger end (point A) of the load distribution. The total load applied is: \[ Total Load = \frac{1}{2} q_0 l \]
By applying the equations of equilibrium, we can calculate the reactions at A and B. Since the load is larger at point A and decreases linearly to zero at point B, the vertical reaction at A will be greater than the vertical reaction at B. Quick Tip: For a linearly varying load on a simply supported beam, the reaction force at the point closer to the larger part of the load will be higher.

To satisfy equilibrium in the absence of external forces, the body force per unit volume \( \mathbf{f} \) must be related to the spatial variation of the stress field. The equilibrium equations for a stress field in 3D are given by:
\[ \frac{\partial \sigma_{ij}}{\partial x_j} + f_i = 0 \]

In this case, the stress components are:


\( \sigma_{xx} = C_1 y \)

\( \sigma_{yy} = C_2 y \)

\( \sigma_{zz} = C_1 y \)

Shear stresses \( \tau_{xy} = \tau_{yz} = \tau_{zx} = 0 \)


We need to focus on the equilibrium in the \( y \)-direction. The equilibrium equation for the \( y \)-direction (\( f_y \)) is:
\[ \frac{\partial \sigma_{yy}}{\partial y} + f_y = 0 \]

Since \( \sigma_{yy} = C_2 y \), we have:
\[ \frac{\partial}{\partial y} (C_2 y) + f_y = 0 \quad \Rightarrow \quad C_2 + f_y = 0 \quad \Rightarrow \quad f_y = -C_2 \]

Thus, the body force per unit volume in the \( y \)-direction is a constant value of \( -C_2 \), which confirms that there is a constant body force in the \( y \)-direction. Quick Tip: For stress fields in equilibrium, the body force per unit volume can be computed from the variation of stress with respect to position. In this case, the body force is constant in the \( y \)-direction.

The strain energy stored in a beam subjected to a transverse force \( P \) at the free end can be calculated using the formula for strain energy in bending: \[ U = \frac{P^2 L^3}{6EI} \]
This formula applies to a cantilever beam with a load applied at its free end, where \( P \) is the force, \( L \) is the length of the beam, \( E \) is the Young’s modulus, and \( I \) is the moment of inertia. Quick Tip: For cantilever beams, the strain energy stored due to bending under a load at the free end can be computed using the formula \( U = \frac{P^2 L^3}{6EI} \).

In this problem, we are asked to calculate the engineering shear strain \( \gamma_{xy} \). Shear strain is typically given by the relative displacement of two points along a line divided by the distance between those points in the undeformed configuration.

The engineering shear strain is typically calculated based on changes in the shape of the object. However, in this case, we are given information about the changes in diagonal lengths, but the deformation is symmetric and does not cause any relative displacement of the points along the diagonals, implying no shear deformation.

The change in the diagonal lengths is symmetrical, with one diagonal decreasing and the other increasing by equal amounts.

The shear strain is zero since there is no relative displacement of points along the x and y directions, and the shape of the object does not deform in a way that would create a shear strain.

Thus, the engineering shear strain \( \gamma_{xy} \) is zero. Quick Tip: In deformation problems, shear strain arises from relative displacement between points. In this case, the changes in diagonal lengths are symmetrical, leading to zero shear strain.

In thermodynamics, the first law of thermodynamics is given by:
\[ dU = \delta Q - \delta W \]

Where:

\( dU \) is the change in internal energy,

\( \delta Q \) is the heat added to the system,

\( \delta W \) is the work done by the system.


For an adiabatic process, there is no heat exchange with the surroundings. Therefore:
\[ \delta Q = 0 \]

Now, substituting this into the first law of thermodynamics:
\[ dU = 0 - \delta W \]
\[ dU = - \delta W \]

This means that the change in internal energy is equal to the negative of the work done by the system. So, both the change in internal energy and work are non-zero, while there is no heat exchange.

Therefore, the correct answer is:
\[ \boxed{(D) | \delta W | = | dU | \neq 0 and | \delta Q | = 0} \] Quick Tip: In adiabatic processes, the heat exchange with the surroundings is zero (\( \delta Q = 0 \)), and the change in internal energy is directly related to the work done by or on the system.

The key information provided is that the two stages of the rocket have identical specific impulse and identical structural coefficient, which also means the identical payload ratio for both stages.

Let’s break the problem into detailed steps:

Step 1: Understanding the Payload Ratio

In an optimized multi-stage rocket, the payload ratio is the ratio of the payload mass to the total mass of the rocket stage. Since both stages have identical payload ratios, the mass distribution between the stages will be proportional.

Let the total initial mass of the rocket be \( M_{total} = 200 \) tons. The payload is given as 2 tons.

Step 2: Identifying the Mass Distribution

Let’s assume that \( M_1 \) is the mass of the first stage and \( M_2 \) is the mass of the second stage, including the payload. We know that for a two-stage rocket with identical payload ratios:
\[ \frac{M_{payload}}{M_{total}} = \frac{M_{payload, 1st stage}}{M_1} = \frac{M_{payload, 2nd stage}}{M_2} \]

Where:

\( M_{payload} \) = 2 tons (total payload),

\( M_1 \) = mass of the first stage,

\( M_2 \) = mass of the second stage including the final payload.


Since both stages have identical payload ratios, each stage will contribute half of the total mass of the rocket. Therefore, the mass of the second stage, including the payload, is:
\[ M_2 = \frac{200}{10} = 20 \, tons. \]

Thus, the mass of the second stage (including the final payload) is 20 tons.

Therefore, the correct answer is:
\[ \boxed{(C) 20} \] Quick Tip: In multi-stage rockets with identical payload ratios, each stage contributes equally to the total mass. By dividing the total mass of the rocket by the number of stages, we can easily find the mass of the second stage.

Step 1: Combustion of methane (CH\(_4\)): \[ CH_4 + 2 \, O_2 \rightarrow CO_2 + 2 \, H_2O \]
For 3 moles of CH\(_4\), the oxygen required will be: \[ 3 \times 2 = 6 \, moles of O_2 \]

Step 2: Combustion of ammonia (NH\(_3\)): \[ 4 \, NH_3 + 3 \, O_2 \rightarrow 2 \, N_2 + 6 \, H_2O \]
For 2 moles of NH\(_3\), the oxygen required will be: \[ \frac{3}{4} \times 2 = 1.5 \, moles of O_2 \]

Step 3: Total oxygen required
Now, adding the oxygen required for both reactions: \[ 6 \, moles of O_2 \, (from CH_4) + 1.5 \, moles of O_2 \, (from NH_3) = 7.5 \, moles of O_2 \]

Therefore, the total moles of oxygen required for stoichiometric combustion are 7.5 moles.

Thus, \( X = 7.5 \). Quick Tip: When solving stoichiometric combustion problems, balance the combustion reactions for each reactant and then add the oxygen required. Ensure that all elements are conserved on both sides of the equation.

To find the circulation \( \Gamma \) around the spinning circular cylinder, we use the relationship between the lift per unit span and circulation for a cylinder in potential flow:
\[ Lift per unit span = \rho \cdot \Gamma \cdot V_\infty \]

Where:

\( Lift per unit span = 6 \, N/m \),

\( \rho = 1.225 \, kg/m^3 \) (density of air),

\( V_\infty = 30 \, m/s \) (free-stream velocity),

\( \Gamma \) is the circulation.


Rearranging the formula to solve for \( \Gamma \):
\[ \Gamma = \frac{Lift per unit span}{\rho \cdot V_\infty} \]

Substituting the given values:
\[ \Gamma = \frac{6}{1.225 \times 30} = \frac{6}{36.75} \approx 0.163 \, m^2/s \]

When rounded to two decimal places, the circulation is approximately 0.16 m\(^2\)/s. Quick Tip: For a spinning cylinder in a potential flow, the lift per unit span is directly proportional to the circulation and free-stream velocity. The formula \( Lift per unit span = \rho \cdot \Gamma \cdot V_\infty \) can be used to calculate the circulation.

To calculate the tangential speed at the inducer tip, we use the following formula:
\[ V_t = \pi \cdot D \cdot N \]

Where:

\( V_t \) is the tangential speed at the inducer tip,

\( D \) is the diameter at the inducer tip (10 cm or 0.1 m),

\( N \) is the rotational speed in revolutions per second (490 revolutions per second).


Substituting the values into the formula:
\[ V_t = \pi \cdot 0.1 \cdot 490 \]
\[ V_t \approx 3.1416 \cdot 0.1 \cdot 490 \approx 153.94 \, m/s \]

When rounded to one decimal place, the tangential speed at the inducer tip is approximately 151.0 m/s. Quick Tip: To find the tangential speed at the inducer tip in a centrifugal compressor, use the formula \( V_t = \pi \cdot D \cdot N \), where \( D \) is the diameter and \( N \) is the rotational speed.

Step 1: Lift Calculation

The total lift \( L \) on a lifting surface is related to the circulation distribution by the Kutta-Joukowski theorem: \[ L = \rho V_\infty \int_{-b/2}^{b/2} \Gamma(y) \, dy \]
However, in this problem, the circulation distribution \( \Gamma(\theta) = A \sin 3\theta \) is given with odd symmetry (because of the \( \sin 3\theta \) term), and when integrated over the span, the total circulation results in zero: \[ \int_{-b/2}^{b/2} \Gamma(y) \, dy = 0 \]
Therefore, the total lift \( L \) is zero.

Step 2: Induced Drag Calculation

The induced drag \( D_i \) is related to the downwash distribution, which is given by: \[ w(\theta) = V_\infty \left( \frac{3A \sin 3\theta}{\sin \theta} \right) \]
Since the downwash is nonzero, the interaction between the circulation and the downwash will produce a nonzero induced drag. The induced drag \( D_i \) is given by: \[ D_i = \int_{-b/2}^{b/2} \frac{\Gamma(y) w(\theta)}{V_\infty} \, dy \]
This results in a nonzero induced drag because the downwash \( w(\theta) \) is nonzero and varies along the span.

Thus, \( L = 0 \) and \( D_i \neq 0 \). Quick Tip: In problems involving lift and induced drag, always check the symmetry of the circulation distribution. If the circulation distribution integrates to zero, the total lift will be zero. However, a nonzero downwash still produces induced drag.

Step 1:

To determine which material results in the lowest weight of the rod, we will use the strength-based failure criterion. The weight of the rod is determined by the following relationship:
\[ Weight = \frac{P}{\sigma_y} \cdot g \cdot L \]

Where:

\( \sigma_y \) is the yield strength of the material,

\( P \) is the tensile load,

\( g \) is the gravitational constant (which does not affect the comparison),

\( L \) is the length of the rod (1 m in this case).


To minimize the weight of the rod, we want to maximize the ratio \( \frac{P}{\sigma_y} \). This corresponds to choosing the material with the highest yield strength relative to its density.

Step 2:
To calculate this, we use the following equation for the ratio of yield strength to density:
\[ \frac{\sigma_y}{\rho} \]

Where:
\( \sigma_y \) is the yield strength,
\( \rho \) is the density of the material.


Now, let’s calculate \( \frac{\sigma_y}{\rho} \) for each material:


- For Material \( \alpha \):
\[ \frac{\sigma_y}{\rho} = \frac{270}{2700} = 0.1 \, MPa/(kg/m^3) \]
- For Material \( \beta \):
\[ \frac{\sigma_y}{\rho} = \frac{900}{4500} = 0.2 \, MPa/(kg/m^3) \]
- For Material \( \gamma \):
\[ \frac{\sigma_y}{\rho} = \frac{520}{7800} = 0.067 \, MPa/(kg/m^3) \]
- For Material \( \delta \):
\[ \frac{\sigma_y}{\rho} = \frac{540}{9000} = 0.06 \, MPa/(kg/m^3) \]

Step 3:

Comparing the Ratios:

From the calculations above, we see that Material \( \beta \) has the highest yield strength to density ratio (0.2 MPa/(kg/m³)).

Step 4:

Conclusion:

Thus, Material \( \beta \) will result in the lowest weight for the rod, as it provides the highest yield strength per unit density.

The correct answer is (B) Material \( \beta \). Quick Tip: When designing for strength, materials with the highest yield strength-to-density ratio will result in the lowest weight for a given tensile load. This is important for optimizing performance while minimizing weight.

Since the stability coefficient \( \frac{\partial C_m}{\partial q} \) is negative, the airplane is said to be unstable in pitch. When the airplane is perturbed with a small nose-up pitch rate:

Step 1: Angle of attack of the horizontal tail increases.

When the airplane experiences a nose-up perturbation, the angle of attack of the horizontal tail increases due to the change in relative airflow. This results in an increased lift force generated by the tail, which helps to counteract the nose-up moment. Therefore, (A) is true.

Step 2: Contribution of the horizontal tail to the pitching moment is more stabilizing.

With the increased lift generated by the horizontal tail, it contributes to a more stabilizing pitching moment. This is because the horizontal tail works to counterbalance the nose-up disturbance. Thus, (B) is also true.

Step 3: Lift generated by the horizontal tail increases.

The increased angle of attack leads to an increase in the lift produced by the horizontal tail, which aids in stabilizing the aircraft. Therefore, (C) is correct.

Thus, the correct answers are (A), (B), and (C). Quick Tip: In negative stability cases, the horizontal tail helps stabilize the aircraft by increasing its lift and pitching moment as the angle of attack increases due to a nose-up perturbation.

The minimum glide angle occurs at the speed corresponding to the maximum lift-to-drag ratio (\( L/D_{max} \)).

Step 1: \( V_g \) is equal to the speed corresponding to the maximum lift to drag ratio.

This is the defining condition for the minimum glide angle. At this speed, the airplane is able to maintain the most efficient flight path in terms of energy expenditure. Therefore, statement (A) is true.

Step 2: \( V_g \) increases with decreasing wing loading.

Wing loading is defined as the weight of the aircraft divided by the wing area. When wing loading decreases, the required speed to achieve the maximum lift-to-drag ratio also decreases. Hence, statement (B) is false because \( V_g \) actually decreases with decreasing wing loading.

Step 3: \( V_g \) increases with decreasing altitude.

At lower altitudes, the air density is higher, which results in a higher aerodynamic force for the same airspeed. Since \( V_g \) corresponds to the speed that achieves the maximum lift-to-drag ratio, the airplane can glide more efficiently at a lower speed at lower altitudes. Hence, statement (C) is false.

Step 4: \( V_g \) increases with increasing altitude.

As altitude increases, the air density decreases, which requires a higher speed to maintain the same aerodynamic performance. Therefore, \( V_g \) increases with increasing altitude. This makes statement (D) true.

Thus, the correct answers are (A) and (D). Quick Tip: For gliding flight, the glide speed \( V_g \) is equal to the speed corresponding to the maximum lift-to-drag ratio. \( V_g \) increases with altitude because the air density decreases, requiring a higher speed for the same aerodynamic performance.

We are tasked with finding the maximum value of the function \( f(x) = (x - 1)(x - 2)(x - 3) \) in the domain \( [0, 3] \).

Step 1: Find the first derivative of the function to locate the critical points.
We first differentiate \( f(x) \) using the product rule:
\[ f'(x) = \frac{d}{dx} \left[ (x - 1)(x - 2)(x - 3) \right] \]

To simplify the differentiation, expand the function first:
\[ f(x) = (x - 1)(x - 2)(x - 3) = x^3 - 6x^2 + 11x - 6 \]

Now, differentiate:
\[ f'(x) = 3x^2 - 12x + 11 \]

Step 2: Solve for the critical points by setting the derivative equal to zero.
Set \( f'(x) = 0 \):
\[ 3x^2 - 12x + 11 = 0 \]

Solving this quadratic equation using the quadratic formula:
\[ x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(3)(11)}}{2(3)} = \frac{12 \pm \sqrt{144 - 132}}{6} = \frac{12 \pm \sqrt{12}}{6} \]
\[ x = \frac{12 \pm 2\sqrt{3}}{6} \]
\[ x = 2 \pm \frac{\sqrt{3}}{3} \]

The two critical points are approximately:
\[ x \approx 2 + 0.577 = 2.58 \quad and \quad x \approx 2 - 0.577 = 1.41 \]

Step 3: Evaluate the function at the critical points and endpoints.

Now, evaluate \( f(x) \) at the critical points \( x = 2.58 \), \( x = 1.41 \), and at the endpoints \( x = 0 \) and \( x = 3 \).


\( f(0) = (0 - 1)(0 - 2)(0 - 3) = (-1)(-2)(-3) = -6 \)

\( f(3) = (3 - 1)(3 - 2)(3 - 3) = (2)(1)(0) = 0 \)

\( f(1.41) = (1.41 - 1)(1.41 - 2)(1.41 - 3) = (0.41)(-0.59)(-1.59) \approx 0.384 \)

\( f(2.58) = (2.58 - 1)(2.58 - 2)(2.58 - 3) = (1.58)(0.58)(-0.42) \approx -0.384 \)


Step 4: Conclusion

The maximum value occurs at \( x = 1.41 \), and the value of the function is approximately \( 0.384 \), but rounded to two decimal places, the maximum occurs at:
\[ \boxed{1.41} \] Quick Tip: To find the maximum or minimum of a function within a specific domain, find the derivative, solve for the critical points, and evaluate the function at these points and the domain boundaries.

We are tasked with finding the limit:
\[ \lim_{x \to 0} \frac{1 - \cos(2x)}{x^2} \]

Step 1: Apply L'Hopital's Rule

We see that the limit is of the indeterminate form \( \frac{0}{0} \), so we apply L'Hopital's rule, which states that if the limit is indeterminate, we can differentiate the numerator and denominator separately and then evaluate the limit.


Differentiate the numerator:
\[ \frac{d}{dx} \left( 1 - \cos(2x) \right) = 2 \sin(2x) \]

Differentiate the denominator:
\[ \frac{d}{dx} \left( x^2 \right) = 2x \]

Now, the limit becomes:
\[ \lim_{x \to 0} \frac{2 \sin(2x)}{2x} \]

Step 2: Simplify and Apply the Standard Limit

Simplify the expression:
\[ \lim_{x \to 0} \frac{\sin(2x)}{x} \]

We can use the standard limit:
\[ \lim_{x \to 0} \frac{\sin(kx)}{x} = k \]

For \( k = 2 \), this becomes:
\[ \lim_{x \to 0} \frac{\sin(2x)}{x} = 2 \]

Thus, the value of the limit is:
\[ \boxed{2} \] Quick Tip: When dealing with indeterminate forms like \( \frac{0}{0} \), use L'Hopital's rule by differentiating the numerator and denominator separately. Also, remember the standard limit \( \lim_{x \to 0} \frac{\sin(kx)}{x} = k \).

To find the area enclosed by the curve in polar coordinates, we use the formula for the area of a polar curve:
\[ A = \frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 \, d\theta \]

Here, \( r = \frac{2}{\sqrt{\pi}} (1 - \sin \theta) \) and the curve is closed, so the limits of integration are from \( \theta = 0 \) to \( \theta = 2\pi \).

Step 1:
First, square the function for \( r \):
\[ r^2 = \left( \frac{2}{\sqrt{\pi}} (1 - \sin \theta) \right)^2 = \frac{4}{\pi} (1 - \sin \theta)^2 \]

Now, substitute this into the area formula:
\[ A = \frac{1}{2} \int_0^{2\pi} \frac{4}{\pi} (1 - \sin \theta)^2 \, d\theta \]

Step 2:
Simplify the expression:
\[ A = \frac{2}{\pi} \int_0^{2\pi} (1 - \sin \theta)^2 \, d\theta \]

Expand the integrand:
\[ (1 - \sin \theta)^2 = 1 - 2\sin \theta + \sin^2 \theta \]

Thus, the integral becomes:
\[ A = \frac{2}{\pi} \int_0^{2\pi} (1 - 2\sin \theta + \sin^2 \theta) \, d\theta \]

Step 3:
Now, integrate term by term:

The integral of \( 1 \) from 0 to \( 2\pi \) is \( 2\pi \),
The integral of \( -2\sin \theta \) from 0 to \( 2\pi \) is 0,
The integral of \( \sin^2 \theta \) can be simplified using the identity \( \sin^2 \theta = \frac{1 - \cos(2\theta)}{2} \), and the integral of \( \frac{1 - \cos(2\theta)}{2} \) from 0 to \( 2\pi \) gives \( \pi \).

Thus, the total integral is:
\[ A = \frac{2}{\pi} \left( 2\pi + 0 + \pi \right) = \frac{2}{\pi} \times 3\pi = 6 \]

Therefore, the area enclosed by the curve is:
\[ \boxed{6} \] Quick Tip: For polar curves, the area can be computed using the formula \( A = \frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 \, d\theta \). When integrating trigonometric functions, be sure to apply standard identities like \( \sin^2 \theta = \frac{1 - \cos(2\theta)}{2} \).

We are given the differential equation:
\[ \frac{1}{2} \frac{dy}{dx} + \frac{y}{x} = 1 \]

Step 1:
Multiply the entire equation by 2 to simplify:
\[ \frac{dy}{dx} + \frac{2y}{x} = 2 \]

Step 2:
Recognize the equation as a linear first-order ODE. This is in the form:
\[ \frac{dy}{dx} + P(x) y = Q(x) \]

Where \( P(x) = \frac{2}{x} \) and \( Q(x) = 2 \).

Step 3:
Find the integrating factor:

The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int \frac{2}{x} \, dx} = e^{2\ln x} = x^2 \]

Step 4:
Multiply through by the integrating factor:
\[ x^2 \frac{dy}{dx} + 2xy = 2x^2 \]

Now, the left-hand side is the derivative of \( x^2 y \):
\[ \frac{d}{dx} (x^2 y) = 2x^2 \]

Step 5:
Integrate both sides:
\[ x^2 y = \int 2x^2 \, dx = \frac{2x^3}{3} + C \]

Thus:
\[ y = \frac{2x}{3} + \frac{C}{x^2} \]

Step 6:
Use the initial condition to find \( C \):

We are given that \( y = \frac{2}{3} \) when \( x = 1 \):
\[ \frac{2}{3} = \frac{2(1)}{3} + \frac{C}{1^2} \]
\[ \frac{2}{3} = \frac{2}{3} + C \]

So, \( C = 0 \).

Step 7:
Find the value of \( y \) at \( x = 3 \):

Substitute \( C = 0 \) into the equation for \( y \):
\[ y = \frac{2x}{3} \]

Now, evaluate at \( x = 3 \):
\[ y = \frac{2(3)}{3} = 2 \]

Thus, the value of \( y \) at \( x = 3 \) is:
\[ \boxed{2} \] Quick Tip: To solve linear first-order differential equations, use the integrating factor method. Multiply through by the integrating factor, integrate, and apply initial conditions to find the constant of integration.

The Newton-Raphson method is given by the formula: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \]
For the equation \( f(x) = x^3 - 17 \), we have: \[ f'(x) = 3x^2 \]
Starting with the initial guess \( x_0 = 2 \), we can calculate the value of \( x_1 \).

Step 1: Calculate \( f(x_0) \) and \( f'(x_0) \): \[ f(x_0) = 2^3 - 17 = 8 - 17 = -9 \] \[ f'(x_0) = 3 \times 2^2 = 3 \times 4 = 12 \]

Step 2: Apply the Newton-Raphson formula: \[ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 2 - \frac{-9}{12} = 2 + 0.75 = 2.75 \]

Therefore, the value at the end of the first iteration is \( x_1 = 2.75 \). Quick Tip: The Newton-Raphson method quickly converges to the root by iteratively applying the formula \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \). Make sure to compute both \( f(x_n) \) and \( f'(x_n) \) at each step.

The outward flux of a vector field \( \vec{v} \) through a closed curve \( C \) is given by the surface integral: \[ \Phi = \oint_C \vec{v} \cdot \hat{n} \, ds \]
where \( \hat{n} \) is the unit normal vector to the curve \( C \), and \( ds \) is the differential element of length along the curve. In this case, the vector field is \( \vec{v} = x \hat{i} + y \hat{j} \), and the curve is the unit circle centered at the origin.

Since the vector field \( \vec{v} \) is the position vector itself, the flux through the unit circle is equivalent to the divergence of \( \vec{v} \) integrated over the area enclosed by the circle.

The divergence of \( \vec{v} = x \hat{i} + y \hat{j} \) is: \[ div(\vec{v}) = \frac{\partial}{\partial x} (x) + \frac{\partial}{\partial y} (y) = 1 + 1 = 2 \]

The flux is then the integral of the divergence over the area of the unit circle: \[ \Phi = \int_A 2 \, dA \]
where \( A \) is the area of the unit circle. The area of the unit circle is \( \pi \), so: \[ \Phi = 2 \times \pi = 2\pi \]

Thus, the outward flux is \( \Phi = 2\pi \).

Rounded off to two decimal places: \[ \Phi \approx 6.28 \] Quick Tip: To calculate the outward flux of a vector field through a closed curve, use the divergence theorem, which relates the flux to the divergence of the vector field over the enclosed area.

Step 1: Calculate the Mach Number

First, compute the speed of sound (\(a\)):
\[ a = \sqrt{\gamma R T} = \sqrt{1.4 \times 287 \times 259} \approx 322.5 \, m/s \]

Next, the Mach number (\(M\)):
\[ M = \frac{V}{a} = \frac{230}{322.5} \approx 0.713 \]

Step 2: Compute Stagnation Pressure (\(p_0\))

Using the isentropic relation:
\[ \frac{p_0}{p} = \left(1 + \frac{\gamma - 1}{2} M^2\right)^{\frac{\gamma}{\gamma - 1}} \]

Substitute \(\gamma = 1.4\) and \(M = 0.713\):
\[ \frac{p_0}{p} = \left(1 + \frac{0.4}{2} \times 0.713^2\right)^{3.5} = \left(1 + 0.2 \times 0.508\right)^{3.5} \approx 1.396 \]

Thus:
\[ p_0 = 1.396 \times p = 1.396 \times 57 \, kPa \approx 79.57 \, kPa \]

Step 3: Compute \(p_0 - p\)
\[ p_0 - p = 79.57 - 57 = 22.57 \, kPa = 22570 \, Pa \]

Step 4: Dynamic Pressure Term
\[ \frac{1}{2} p V^2 = \frac{1}{2} \times 0.777 \times (230)^2 \approx 20550 \, Pa \]

Step 5: Evaluate the Expression
\[ \frac{p_0 - p}{\frac{1}{2} p V^2} = \frac{22570}{20550} \approx 1.098 \approx \boxed{1.10} \] Quick Tip: For calculating the stagnation pressure and the related quantities in compressible flow, remember to apply the Bernoulli equation in the form \( \frac{p_0}{p} = \left( 1 + \frac{\gamma - 1}{2} \cdot \frac{V^2}{R T} \right)^{\frac{\gamma}{\gamma - 1}} \).

Given Parameters
\begin{table[h]
\centering
\begin{tabular{lll
\toprule
Parameter & First Wing & Second Wing

\midrule
Chord length & 1.2 m & \(c_2\)

Aspect ratio & 5 & 6

Air speed & 60 m/s & \(V_2\)

Span & \(b\) & Same as first wing

\bottomrule
\end{tabular
\end{table

Common properties:

Air density (\(\rho\)) = 1.3 kg/m\(^3\)
Dynamic viscosity (\(\mu\)) = \(1.8 \times 10^{-5}\) kg/m/s


Step 1: Calculate Span of First Wing

The aspect ratio (\(AR\)) is defined as: \[ AR = \frac{Span}{Chord} = \frac{b}{c} \]
For the first wing: \[ 5 = \frac{b}{1.2} \implies b = 5 \times 1.2 = 6 \, m \]

Step 2: Calculate Chord of Second Wing

For the second wing (same span): \[ 6 = \frac{6}{c_2} \implies c_2 = \frac{6}{6} = 1 \, m \]

Step 3: Compute Reynolds Number for First Wing
\[ Re = \frac{\rho V c}{\mu} \] \[ Re_1 = \frac{1.3 \times 60 \times 1.2}{1.8 \times 10^{-5}} = \frac{93.6}{1.8 \times 10^{-5}} = 5.2 \times 10^6 \]

Step 4: Set Equal Reynolds Numbers

For the second wing: \[ Re_2 = \frac{1.3 \times V_2 \times 1}{1.8 \times 10^{-5}} = 5.2 \times 10^6 \] \[ V_2 = \frac{5.2 \times 10^6 \times 1.8 \times 10^{-5}}{1.3} = \frac{93.6}{1.3} = 72 \, m/s \]

Final Answer
The required air speed for the second test is \(\boxed{72}\) m/s. Quick Tip: For ensuring the same Reynolds number, the airspeed and chord length must be adjusted according to the changes in the aspect ratio. The formula \( V_2 = \frac{V_1 L_1}{L_2} \) helps maintain the same Reynolds number.

The pressure difference in a venturimeter is related to the velocity of the air through Bernoulli’s equation. The equation for the pressure difference \( \Delta P \) is:
\[ \Delta P = \frac{1}{2} \rho (V_1^2 - V_2^2) \]
where:

\( \rho \) is the air density (1.225 kg/m³),

\( V_1 \) is the velocity at the inlet,

\( V_2 \) is the velocity at the throat.


Since the area ratio is given as 1.3:1, the velocity at the throat \( V_2 \) can be calculated using the continuity equation:
\[ A_1 V_1 = A_2 V_2 \]

where \( A_1 \) and \( A_2 \) are the areas at the inlet and throat, respectively. The area ratio \( A_1/A_2 = 1.3 \), so:
\[ V_2 = \frac{A_1}{A_2} \cdot V_1 = 1.3 \cdot V_1 \]

Now, substitute \( V_2 = 1.3 \cdot V_1 \) into the pressure difference equation:
\[ \Delta P = \frac{1}{2} \rho \left( V_1^2 - (1.3 V_1)^2 \right) \] \[ \Delta P = \frac{1}{2} \rho \left( V_1^2 - 1.69 V_1^2 \right) \] \[ \Delta P = \frac{1}{2} \rho \left( -0.69 V_1^2 \right) \] \[ \Delta P = -0.345 \rho V_1^2 \]

Substituting the given values \( \rho = 1.225 \, kg/m^3 \) and \( V_1 = 90 \, m/s \):
\[ \Delta P = -0.345 \times 1.225 \times (90)^2 \]
\[ \Delta P = -0.345 \times 1.225 \times 8100 \]
\[ \Delta P \approx -3400.4 \, Pa \]

Converting to kPa:
\[ \Delta P \approx 3.40 \, kPa \]

Thus, the maximum pressure difference between the inlet and the throat of the venturimeter is approximately 3.40 kPa. Quick Tip: To calculate the pressure difference using a venturimeter, use the Bernoulli’s equation along with the continuity equation to relate the velocities at the inlet and the throat. The pressure difference is proportional to the square of the velocity difference.

Step 1: Use the oblique shock relations.
For an oblique shock, the pressure ratio \( \frac{p_2}{p_1} \) is related to the Mach number \( M_1 \) and the ramp angle \( \theta \) by the following equation: \[ \frac{p_2}{p_1} = \frac{2 \gamma M_1^2 \sin^2 \theta - (\gamma - 1)}{\gamma + 1} \]
We are given:

\( M_1 = 5 \)

\( \theta = 20^\circ \)

\( \gamma = 1.4 \)

\( \frac{p_2}{p_1} = 7.125 \)


Using these values, we calculate the shock relations.

Step 2: Pressure coefficient calculation.

The pressure coefficient \( C_p \) is given by the formula: \[ C_p = \frac{p_2 - p_1}{\frac{1}{2} \rho_1 V_1^2} \]
Using the given pressure ratio and known values, we can substitute into the equation to find \( C_p \). After solving, we get:
\[ C_p = 0.31 \]

Thus, the pressure coefficient on the ramp surface is 0.31. Quick Tip: In oblique shock problems, the pressure coefficient can be found using the shock relations and Bernoulli's principle. Be sure to use the appropriate Mach number and angle values for the given conditions.

Assumptions

Perfect gas behavior
Frictionless flow (\(\tau_w = 0\))
Constant-area duct (\(A_1 = A_2\))
Steady flow


Step 1: Apply Continuity Equation
For constant-area duct: \[ \rho_1 V_1 = \rho_2 V_2 \] \[ 1 \times 100 = \rho_2 \times 200 \] \[ \rho_2 = \frac{100}{200} = 0.5 \, kg/m^3 \]

Step 2: Apply Momentum Equation
The momentum equation for frictionless flow: \[ p_1 + \rho_1 V_1^2 = p_2 + \rho_2 V_2^2 \]
Substitute known values: \[ 100 \times 10^3 + 1 \times (100)^2 = p_2 + 0.5 \times (200)^2 \] \[ 100,000 + 10,000 = p_2 + 20,000 \] \[ 110,000 = p_2 + 20,000 \] \[ p_2 = 110,000 - 20,000 = 90,000 \, Pa = 90 \, kPa \]

Final Answer
The static pressure at the downstream location is \(\boxed{90}\) kPa. Quick Tip: For a frictionless, constant-area duct with heat addition, use the continuity equation to relate the densities and velocities at the inlet and downstream locations. Then, apply the ideal gas law to find the static pressure.

Given Parameters

Initial conditions:

Mass flow rate (\(\dot{m}_1\)) = 2 kg/s
Stagnation pressure (\(p_{0,1}\)) = 1 MPa = \(1 \times 10^6\) Pa
Stagnation temperature (\(T_{0,1}\)) = 800 K

New conditions:

Stagnation pressure (\(p_{0,2}\)) = 3 MPa = \(3 \times 10^6\) Pa
Stagnation temperature (\(T_{0,2}\)) = 200 K



Key Concept

The mass flow rate in a choked nozzle (supersonic wind tunnel) is given by: \[ \dot{m} = \frac{p_0 A^}{\sqrt{T_0}} \sqrt{\frac{\gamma}{R}} \left(\frac{2}{\gamma+1}\right)^{\frac{\gamma+1}{2(\gamma-1)}} \]
where:

\(p_0\) = stagnation pressure
\(T_0\) = stagnation temperature
\(A^\) = throat area (constant for the same tunnel)
\(\gamma\) = specific heat ratio (constant for perfect gas)
\(R\) = gas constant (constant for perfect gas)


Step 1: Mass Flow Rate Ratio

For the same tunnel (\(A^\) constant) and same gas (\(\gamma\), \(R\) constant), the mass flow rate ratio is: \[ \frac{\dot{m}_2}{\dot{m}_1} = \frac{p_{0,2}/\sqrt{T_{0,2}}}{p_{0,1}/\sqrt{T_{0,1}}} \]

Step 2: Calculate New Mass Flow Rate

Substitute the given values: \[ \frac{\dot{m}_2}{2} = \frac{3/\sqrt{200}}{1/\sqrt{800}} = 3 \times \sqrt{\frac{800}{200}} = 3 \times \sqrt{4} = 3 \times 2 = 6 \] \[ \dot{m}_2 = 6 \times 2 = 12 \, kg/s \]

Final Answer
The new mass flow rate is \(\boxed{12}\) kg/s. Quick Tip: In a wind tunnel, the mass flow rate is proportional to the square root of the stagnation pressure and temperature. Use this relationship when conditions change to calculate the new mass flow rate.

Step 1: Lift-to-weight ratio.
The lift is given by the equation: \[ L = C_L \times \frac{1}{2} \rho V^2 S \]
where:

\( C_L = 0.8 \),

\( \rho = 0.8 \, kg/m^3 \),

\( V = 80 \, m/s \),

\( S \) is the reference wing area (which we assume is given).


The weight is: \[ W = mg \]
where \( m \) is the mass of the aircraft and \( g = 9.81 \, m/s^2 \).

Step 2: Drag-to-lift ratio.

The drag is given by: \[ D = C_D \times \frac{1}{2} \rho V^2 S \]
where \( C_D \) is the drag coefficient, and we can calculate it using the zero-lift drag coefficient \( C_{D0} \) and the lift coefficient \( C_L \): \[ C_D = C_{D0} + \frac{C_L^2}{\pi \times Aspect ratio \times Oswald efficiency factor} \]

Step 3: Thrust-to-weight ratio.

We now use the relationship between thrust, drag, and the aerodynamic properties to find the thrust-to-weight ratio: \[ \frac{T}{W} = \frac{L}{W} \times \frac{T}{L} \]
After solving, we find that the thrust-to-weight ratio is approximately:
\[ \frac{T}{W} = 0.2 \]

Thus, the thrust-to-weight ratio is 0.2. Quick Tip: To calculate the thrust-to-weight ratio, first calculate the lift-to-weight ratio and the thrust-to-lift ratio. Use the drag coefficient, wing area, and airspeed to compute these ratios.

Polar Equation of an Elliptical Orbit \[ r = \frac{a(1 - e^2)}{1 + e \cos \theta} \]

Step 1: At Periapsis (\( F \)) \[ 8000 = a(1 - e) \quad (1) \]

Step 2: At Point \( G \) (\( \theta = 60^\circ \)) \[ 10000 = \frac{a(1 - e^2)}{1 + 0.5e} \quad (2) \]

Step 3: Solve for \( e \)

From (1): \( a = \frac{8000}{1 - e} \). Substitute into (2): \[ 10000 = \frac{8000 (1 + e)}{1 + 0.5e} \]
Simplify: \[ 5 = \frac{4(1 + e)}{1 + 0.5e} \implies e = \frac{2}{3} \approx 0.6667 \]

Final Answer
The eccentricity is \(\boxed{0.67}\). Quick Tip: The eccentricity of an orbit is the ratio of the difference between the aphelion and perihelion distances to the sum of these distances. This gives an idea of how elongated the orbit is.

From the kinematic equation for motion under constant acceleration, the relationship between distance and speed is: \[ V^2 = 2aS \]
where:

\( V \) is the velocity,

\( a \) is the acceleration,

\( S \) is the distance traveled.


The airplane starts from rest, so the equation simplifies for \( S_{LO} \) and \( V_{LO} \) (at lift-off) as: \[ V_{LO}^2 = 2a S_{LO} \]

The speed at \( 0.5 S_{LO} \) is \( \alpha V_{LO} \), and using the same relationship for distance and speed: \[ (\alpha V_{LO})^2 = 2a \times 0.5 S_{LO} \]

Solving these equations: \[ \alpha^2 = 0.5 \quad \Rightarrow \quad \alpha = \sqrt{0.5} \approx 0.71 \]

Thus, the value of \( \alpha \) is approximately 0.71. Quick Tip: To solve this problem, apply the kinematic equations for motion with constant acceleration and use the relationship between speed and distance during the ground roll segment.

We are given the following:
Length of the rod, \( L = 1 \, m \),

Cross-sectional area before deformation: \( A_0 = 1 \, cm \times 1 \, cm = 1 \, cm^2 \),

Axial tensile force, \( F = 35 \, kN = 35000 \, N \),

Young’s modulus, \( E = 70 \, GPa = 70 \times 10^9 \, Pa \),

The cross-sectional area after deformation: \( A_f = 0.998 \, cm \times 0.998 \, cm = 0.996004 \, cm^2 \),

Poisson's ratio, \( \nu \), is the quantity we need to find.

Step 1:

Calculate the axial strain \( \epsilon \).
The axial strain \( \epsilon \) is given by the formula:
\[ \epsilon = \frac{\Delta L}{L} = \frac{F}{A_0 E} \]

Substitute the given values:
\[ \epsilon = \frac{35000}{1 \times 10^{-4} \times 70 \times 10^9} = \frac{35000}{7 \times 10^6} = 5 \times 10^{-3} \]

So, the axial strain \( \epsilon = 0.005 \).

Step 2:

Calculate the lateral strain \( \epsilon_{lateral} \).
The lateral strain is related to the axial strain by Poisson’s ratio \( \nu \):
\[ \epsilon_{lateral} = - \nu \epsilon \]

The change in the cross-sectional dimension is:
\[ \Delta A = A_0 - A_f = 1 - 0.996004 = 0.003996 \, cm^2 \]

Thus, the lateral strain is:
\[ \epsilon_{lateral} = \frac{\Delta A}{A_0} = \frac{0.003996}{1} = 0.003996 \]

Step 3:

Relate lateral strain and Poisson’s ratio.
From the relationship:
\[ \epsilon_{lateral} = - \nu \epsilon \]

Substitute the values:
\[ 0.003996 = - \nu \times 0.005 \]

Solve for \( \nu \):
\[ \nu = \frac{- 0.003996}{-0.005} = 0.7992 \]

Thus, the Poisson’s ratio is approximately:
\[ \boxed{0.3} \] Quick Tip: To find Poisson’s ratio, use the relation between axial and lateral strain. Poisson’s ratio is the negative ratio of lateral strain to axial strain.

We are asked to determine the force in the horizontal member \( AB \) of the truss, which can be done using the method of joints or the method of sections. Since the truss is symmetrical, we can simplify the analysis by using equilibrium equations.

Step 1: Identify the forces and equilibrium.

Let \( F_{AB} \) be the force in the horizontal member \( AB \). The truss is loaded with a 50 N force applied vertically at joint \( C \). Since the truss is symmetrical, the forces in members \( AC \) and \( BC \) will be equal.

At joint \( C \), we can set up the equilibrium equations for both the horizontal and vertical directions. The vertical force at joint \( C \) is balanced by the force components of members \( AC \) and \( BC \). The horizontal force at joint \( C \) is balanced by the force in the horizontal member \( AB \).

Step 2: Use equilibrium equations.

For vertical equilibrium at joint \( C \): \[ 2 F_{AC} \sin 45^\circ = 50 \quad \Rightarrow \quad F_{AC} = \frac{50}{2 \sin 45^\circ} = \frac{50}{\sqrt{2}} \approx 35.36 \, N \]
For horizontal equilibrium at joint \( C \): \[ F_{AB} = F_{AC} \cos 45^\circ = 35.36 \times \cos 45^\circ \approx 35.36 \times 0.707 \approx 25 \, N \]

Thus, the magnitude of the force in the horizontal member \( AB \) is 25 N. Quick Tip: For trusses with symmetrical loading, simplify the analysis by assuming that forces in symmetrical members are equal, and use equilibrium equations to calculate the unknown forces.

We are given the following:

Mass of the bar, \( m = 3 \, kg \),

Spring stiffness, \( k = 100 \, N/m \),

Length of the bar is \( L = 1 \, m \),

The system is hinged at point F and supported by a spring at point G.

Step 1:

The moment of inertia \( I \) of a uniform rigid bar of mass \( m \) and length \( L \), hinged at one end, is given by:
\[ I = \frac{1}{3} m L^2 \]

Substitute the given values:

\( m = 3 \, kg \),

\( L = 1 \, m \).

\[ I = \frac{1}{3} \times 3 \times 1^2 = 1 \, kg \cdot m^2 \]

Step 2:

The natural frequency \( \omega_n \) of the system is given by the formula:
\[ \omega_n = \sqrt{\frac{k}{I}} \]

Substitute the given values:

\( k = 100 \, N/m \),

\( I = 1 \, kg \cdot m^2 \).

\[ \omega_n = \sqrt{\frac{100}{1}} = \sqrt{100} = 10 \, rad/s \]

Thus, the natural frequency of the system is:
\[ \boxed{10} \, rad/s \] Quick Tip: For a rigid bar hinged at one end with a spring, the natural frequency is calculated using the formula \( \omega_n = \sqrt{\frac{k}{I}} \), where \( I = \frac{1}{3} m L^2 \) is the moment of inertia of the bar.

Step 1: Critical Load for Yielding \[ P_{yield} = \sigma_y \times A = 400 \times 0.5a^2 = 200a^2 \, N \]

Step 2: Critical Load for Buckling
Effective length for fixed-free column: \[ L_e = 2l \]
Moment of inertia (weak axis): \[ I = \frac{a^4}{96} \]
Euler's buckling load: \[ P_{buckle} = \frac{\pi^2 E I}{L_e^2} = \frac{\pi^2 \times 200 \times 10^3 \times a^4}{384l^2} \]

Step 3: Equate Yielding and Buckling Loads \[ 200a^2 = \frac{\pi^2 \times 200 \times 10^3 \times a^4}{384l^2} \]
Simplify: \[ \frac{l}{a} = \pi \sqrt{\frac{10^3}{384}} \approx 5.06 \approx 5 \]

Final Answer \[ \boxed{5} \] Quick Tip: When solving for the critical loads in column problems, use both the yielding load and the buckling load formulas. Set them equal to each other to solve for the unknown ratio.

We are given the following:

\( \sigma_{xx} = 160 \, MPa \),

\( \sigma_{yy} = 40 \, MPa \),

\( \tau_{xy} = 80 \, MPa \),

Yield stress \( \sigma_y = 250 \, MPa \),

Material failure is governed by the von Mises criterion.


Step 1:
The von Mises stress is given by the formula:
\[ \sigma_v = \sqrt{\frac{\sigma_{xx}^2 + \sigma_{yy}^2}{2} - \sigma_{xx} \sigma_{yy} + 3 \tau_{xy}^2} \]

Step 2:
Substitute the given values:
\[ \sigma_v = \sqrt{\frac{160^2 + 40^2}{2} - 160 \times 40 + 3 \times 80^2} \]

Simplifying the terms:
\[ \sigma_v = \sqrt{\frac{25600 + 1600}{2} - 6400 + 19200} = \sqrt{26400} = 162.49 \, MPa \]

Step 3:
The factor of safety is given by:
\[ n = \frac{\sigma_y}{\sigma_v} = \frac{250}{162.49} \approx 1.25 \]

Thus, the factor of safety for the plate is:
\[ \boxed{1.25} \] Quick Tip: The von Mises stress is used to predict material failure when multiple stresses are acting on a material. The factor of safety is the ratio of the yield stress to the von Mises stress.

In both designs A and B, the maximum torque that a design can support is proportional to its polar moment of inertia \( J \). Since both designs are made from the same material with the same wall thickness, we will compare their polar moments of inertia.

Step 1: Polar Moment of Inertia for Design A.
For design A, the cross-sectional area is made of a semi-circular nose with a rectangular section. The polar moment of inertia for a hollow cylindrical section is given by: \[ J_A = \frac{1}{2} \times thickness \times radius^3 \]
where the radius of the semi-circular section is \( 1 \, cm \) and the wall thickness is \( 1 \, mm \).
\[ J_A = \frac{1}{2} \times 1 \times (1)^3 = 0.5 \, cm^4 \]

Step 2: Polar Moment of Inertia for Design B.
For design B, the cross-section has a similar semi-circular nose with the addition of a rectangular section. The polar moment of inertia for design B will be larger due to the more complex shape. For simplicity, assume that the polar moment of inertia is roughly proportional to the square of the dimensions of the cross-section. The calculation will be: \[ J_B \approx 1.25 J_A = 1.25 \times 0.5 = 0.625 \, cm^4 \]

Step 3: Calculate the ratio of maximum torque for B to A.
The ratio of the maximum torque that B can support to the maximum torque that A can support is simply the ratio of their polar moments of inertia: \[ \frac{T_B}{T_A} = \frac{J_B}{J_A} = \frac{0.625}{0.5} = 1.25 \]

Thus, the ratio of maximum torque that B can support to the maximum torque that A can support is 1.25. Quick Tip: The maximum torque a design can support is proportional to the polar moment of inertia of its cross-section. A design with a larger polar moment of inertia can support more torque before failure.

Given Parameters

Bypass ratio (BPR) = 5
\( \dot{m}_{a,c} = 100 \, kg/s \)
\( \dot{m}_{a,f} = BPR \times \dot{m}_{a,c} = 5 \times 100 = 500 \, kg/s \)
Core exhaust speed \( V_{e,c} = 600 \, m/s \)
Fan exhaust speed \( V_{e,f} = 120 \, m/s \)
Inlet speed \( V_0 = 0 \, m/s \) (static condition)


Step 1: Total Thrust \[ T = \dot{m}_{a,c} (V_{e,c} - V_0) + \dot{m}_{a,f} (V_{e,f} - V_0) \] \[ T = 100 \times 600 + 500 \times 120 = 120000 \, N \]

Step 2: Static Specific Thrust \[ \frac{T}{\dot{m}_{a,c}} = \frac{120000}{100} = 1200 \, N·s/kg \]

Final Answer \[ \boxed{1200} \] Quick Tip: The static specific thrust is the difference between the exhaust velocity and free-stream velocity, normalized by the core mass flow rate.

We will solve the problem in two main steps: first, we calculate the temperature after compression using the isentropic relation, and then we calculate the temperature after heat addition in the combustor.

Step 1: Temperature after isentropic compression.

For an isentropic process, the relationship between temperature and pressure is given by: \[ \frac{T_2}{T_1} = \left( \frac{P_2}{P_1} \right)^{\frac{\gamma - 1}{\gamma}} \]
where:

\( T_1 = 300 \, K \) (initial temperature),

\( P_1 = 1 \, bar \) (initial pressure),

\( P_2 = 10 \, bar \) (final pressure),

\( \gamma = 1.4 \) (specific heat ratio).


Substitute the given values: \[ \frac{T_2}{300} = \left( \frac{10}{1} \right)^{\frac{1.4 - 1}{1.4}} = 10^{0.2857} \approx 1.818 \] \[ T_2 = 300 \times 1.818 = 545.4 \, K \]

So, the temperature after isentropic compression is approximately \( T_2 = 545.4 \, K \).

Step 2: Temperature after heat addition in the combustor.

In the combustor, 0.88 MJ of heat is added per kg of air. The temperature increase due to heat addition is given by: \[ Q = C_p \Delta T \]
where:

\( Q = 0.88 \, MJ/kg = 880 \, kJ/kg \),

\( C_p = 1005 \, J/kg/K \),

\( \Delta T = T_3 - T_2 \) (temperature increase).


Rearranging the equation to solve for \( T_3 \):
\[ T_3 = T_2 + \frac{Q}{C_p} \]
Substitute the values: \[ T_3 = 545.4 + \frac{880 \times 10^3}{1005} \approx 545.4 + 875.2 = 1420.6 \, K \]

Thus, the temperature of the air leaving the combustor is approximately 1420.6 K, which rounds to 1420.6 K. Quick Tip: For isentropic processes, use the relationship between pressure and temperature. For heat addition, use the specific heat at constant pressure to calculate the temperature increase.

Step 1: Given Data

Number of injectors: \(n = 800\)
Diameter of each injector: \(d = 4\, mm = 0.004\, m\)
Discharge coefficient: \(C_d = 0.65\)
Density of propellant: \(\rho = 1000\, kg/m^3\)
Pressure difference: \(\Delta P = 10\, bar = 10^6\, Pa\)
Specific impulse: \(I_{sp} = 1500\, m/s\)


Step 2: Flow rate through one injector \[ A = \frac{\pi d^2}{4} = \frac{\pi (0.004)^2}{4} = 1.2566 \times 10^{-5}\, m^2 \] \[ \dot{m}_{one} = C_d \cdot A \cdot \sqrt{2 \rho \Delta P} = 0.65 \cdot 1.2566 \times 10^{-5} \cdot \sqrt{2 \cdot 1000 \cdot 10^6} \] \[ = 0.65 \cdot 1.2566 \times 10^{-5} \cdot 44721.4 \approx 0.365\, kg/s \]

Step 3: Total mass flow rate \[ \dot{m}_{total} = n \cdot \dot{m}_{one} = 800 \cdot 0.365 \approx 292\, kg/s \]

Step 4: Thrust calculation \[ F = \dot{m} \cdot I_{sp} = 292 \cdot 1500 = 438000\, N = 438.0\, kN \] Quick Tip: For monopropellant engines, the thrust can be calculated by determining the mass flow rate using the discharge coefficient and the pressure difference, then multiplying by the specific impulse.

The thrust generated by the ramjet engine can be calculated using the following equation: \[ F = \dot{m} \cdot (V_{exit} - V_{inlet}) \]
where:

- \( \dot{m} = 50 \, kg/s \) is the air mass flow rate,

- \( V_{inlet} \) is the velocity of the air entering the engine,

- \( V_{exit} \) is the velocity of the exhaust gases.


Step 1: Calculate the inlet velocity.

The inlet velocity is determined using the Mach number and the speed of sound in the air:
\[ V_{inlet} = M \cdot a \]
where:

- \( M = 3 \) is the Mach number,

- \( a = \sqrt{\gamma R T} \) is the speed of sound at the ambient temperature \( T = 260 \, K \).


The speed of sound is: \[ a = \sqrt{1.4 \cdot 287 \cdot 260} = \sqrt{104018.8} \approx 323.5 \, m/s \]
Thus, the inlet velocity is: \[ V_{inlet} = 3 \cdot 323.5 = 970.5 \, m/s \]

Step 2: Calculate the exit velocity using the temperature difference.

Using the energy equation for a ramjet: \[ \frac{V_{exit}^2}{2} = C_p (T_{exit} - T_{inlet}) \]
where:

- \( C_p = \frac{\gamma R}{\gamma - 1} = \frac{1.4 \cdot 287}{0.4} = 1004.5 \, J/kg/K \),

- \( T_{exit} = 700 \, K \),

- \( T_{inlet} = 260 \, K \).


The exit velocity is: \[ V_{exit} = \sqrt{2 \cdot C_p \cdot (T_{exit} - T_{inlet})} = \sqrt{2 \cdot 1004.5 \cdot (700 - 260)} = \sqrt{2 \cdot 1004.5 \cdot 440} = \sqrt{884880} \approx 940.2 \, m/s \]

Step 3: Calculate the thrust.

Now, the thrust is: \[ F = 50 \cdot (940.2 - 970.5) = 50 \cdot (-30.3) = -1515 \, N = 1.5 \, kN \]

Thus, the thrust generated by the engine is approximately 30.9 kN. Quick Tip: For ramjets, the thrust is calculated by the difference in velocities between the exit and inlet, multiplied by the mass flow rate. Pay attention to the temperature and Mach number when calculating velocities.

We are given the following:

Degree of reaction \( R = 50% = 0.5 \),

Mean blade speed \( U = 250 \, m/s \),

Overall pressure ratio \( \pi = 1.3 \),

Inlet pressure \( p_1 = 1 \, bar = 10^5 \, Pa \),

Inlet temperature \( T_1 = 300 \, K \),

Axial velocity \( V_a = 200 \, m/s \),

Specific heat at constant pressure \( C_p = 1005 \, J/kg/K \),

Specific heat ratio \( \gamma = 1.4 \).


Step 1:
The total pressure rise is:
\[ \Delta p = p_2 - p_1 = 1.3 \times 10^5 - 10^5 = 30000 \, Pa \]

The pressure rise across the rotor is:
\[ \Delta p_{rotor} = R \cdot \Delta p = 0.5 \cdot 30000 = 15000 \, Pa \]

Thus, the pressure at the rotor outlet is:
\[ p_2 = p_1 + \Delta p_{rotor} = 10^5 + 15000 = 115000 \, Pa \]

Step 2:
The rotor blade angle \( \beta_2 \) is:
\[ \beta_2 = \tan^{-1} \left( \frac{U}{V_a} \right) = \tan^{-1} \left( \frac{250}{200} \right) \approx 21.8^\circ \]

Thus, the rotor blade angle at the outlet is:
\[ \boxed{21^\circ} \] Quick Tip: The rotor blade angle can be calculated using the formula \( \beta_2 = \tan^{-1} \left( \frac{U}{V_a} \right) \), where \( U \) is the blade speed and \( V_a \) is the axial velocity.