GATE 2025 NM Question Paper Available - Download Solution PDF with Answer Key

The sentence talks about Rehman's determination to contribute to the project despite initial hesitation.

"Ambivalence" means uncertainty or mixed feelings, which doesn't fit in the context of resolve or determination.

"Satisfaction" refers to contentment, which is not about commitment or determination.

"Resolve" refers to determination or firmness in purpose, which perfectly fits the context of the sentence.

"Revolve" refers to turning around something and is unrelated to the context of commitment.


Hence, the correct answer is "resolve." Quick Tip: When a sentence describes persistence or determination, look for words like "resolve," "determination," or "commitment" that convey strength of purpose.

Step 1: Understand the relationship between "Bird" and "Nest".

A bird lives in a nest, which is its natural dwelling. This is a direct relationship between the animal and its habitat.

Step 2: Apply the same relationship to "Bee".

Similarly, a bee lives in a hive. The relationship here is also between the animal and its habitat, just like the bird and the nest.

Step 3: Conclusion.

Thus, the correct answer is (C) Hive because a bee, like a bird, has a specific dwelling place, which is a hive. Quick Tip: In analogies, identify the relationship between the first pair and look for the corresponding relationship in the second pair.

Step 1: Start from the given equation.
We are given: \[ P e^x = Q e^{-x} \quad for all real x \]

Step 2: Multiply both sides by \( e^x \). \[ P e^{2x} = Q \]

Step 3: Analyze the result.
This implies that the left-hand side is a function of \( x \), while the right-hand side is a constant.
The only way this equality can hold for all real \( x \) is if both sides are identically zero. Therefore: \[ P e^{2x} = Q \Rightarrow Only possible if P = 0 and hence Q = 0 \]

Step 4: Final Answer. \[ \boxed{P = Q = 0} \] Quick Tip: If a variable exponential function is said to equal a constant for all real values, it must be the zero function. This is a classic way to test functional identities.

Step 1: Visualize the folding of the net into a cube.

When the given net is folded, the square with the triangle (\(\triangle\)) is adjacent to the square with the dot (\(\bullet\)).

Step 2: Analyze the adjacency in Option (A).

In Option (A), the faces showing the triangle (\(\triangle\)) and the dot (\(\bullet\)) are indeed adjacent.

Step 3: Consider the relative positions upon folding.

Imagine the square with the triangle (\(\triangle\)) as the front face. When folding the net, the square with the dot (\(\bullet\)) will fold up to become the top face. The orientation shown in Option (A) is consistent with this folding. The base of the triangle is towards the shared edge with the square that becomes the top face (with the dot).

Step 4: Eliminate other options.

Option (B): Shows the triangle (\(\triangle\)) adjacent to the upward-pointing black triangle (\(\blacktriangle\)). While these are adjacent in the net, the orientation of the triangle is incorrect if the black triangle is on top. The base of the triangle should be towards the shared edge.
Option (C): Shows the triangle (\(\triangle\)) adjacent to the circle (\(\circ\)). These are adjacent in the net. However, without a specific orientation shown for the circle, we cannot definitively rule it out yet, but Option A presents a clearer match based on the dot's position relative to the triangle.
Option (D): Shows the triangle (\(\triangle\)) adjacent to the upward-pointing black triangle (\(\blacktriangle\)). Similar to Option (B), the orientation of the triangle relative to the adjacent face is inconsistent with the folding.

Step 5: Final Confirmation.

By carefully visualizing the fold, with the triangle on the front, the dot folds to the top such that the base of the triangle is along the edge shared with the dot's face. Option (A) correctly depicts this orientation. Quick Tip: When visualizing cube folds, pay close attention to the edges that will meet and the resulting relative orientations of the symbols on the faces.

Step 1: Analyze option (A)

Consider two prime numbers, \( p_1 = 2 \) and \( p_2 = 3 \). Their sum is: \[ p_1 + p_2 = 2 + 3 = 5, \]
which is a prime number. Hence, option (A) is not correct.

Step 2: Analyze option (B)

The product of any two prime numbers, \( p_1 \) and \( p_2 \), will always be a composite number because the product has at least three divisors: 1, \( p_1 \), and \( p_2 \). For example, if \( p_1 = 2 \) and \( p_2 = 3 \), \[ p_1 p_2 = 2 \times 3 = 6, \]
which is not a prime number. Hence, option (B) is correct.

Step 3: Analyze option (C)

For \( p_1 = 2 \) and \( p_2 = 3 \), \[ p_1 + p_2 + 1 = 2 + 3 + 1 = 6, \]
which is not a prime number. Therefore, option (C) is not correct.

Step 4: Analyze option (D)

For \( p_1 = 2 \) and \( p_2 = 3 \), \[ p_1 p_2 + 1 = 2 \times 3 + 1 = 7, \]
which is a prime number. However, if we take \( p_1 = 3 \) and \( p_2 = 5 \), \[ p_1 p_2 + 1 = 3 \times 5 + 1 = 16, \]
which is not a prime number. Therefore, option (D) is not correct.

Step 5: Conclusion

Option (B) is the correct answer because the product of any two prime numbers is always a composite number, never a prime number. Quick Tip: When multiplying prime numbers, the result is always a composite number with at least three divisors.

Step 1: Understanding the phrase “Even if I had known...”

This is a conditional sentence using the past perfect tense. It indicates an unreal or hypothetical situation.

Step 2: What does this imply?

The speaker (Ramya) is talking about a situation that did not happen — she did not know Josephine was in the hospital.

Step 3: Analyze the options

Option (A): Incorrect — It says Ramya knew, which contradicts the hypothetical phrasing.

Option (B): Correct — This matches the implication of not knowing.

Option (C): Irrelevant — No relationship history is discussed.

Option (D): Incorrect — No information about the reason for hospitalization is provided.
Quick Tip: Look for clues in tense and phrasing when analyzing logical inferences. Hypothetical statements often imply that the condition did not actually occur.

Let us first analyze the pattern used to encode the words:

IMAGE → FHBNJ

Step 1: Find the shift for each letter in IMAGE

I (9) → F (6): \(-3\)

M (13) → H (8): \(-5\)

A (1) → B (2): \(+1\)

G (7) → N (14): \(+7\)

E (5) → J (10): \(+5\)

FIELD → EMFJG


F (6) → E (5): \(-1\)

I (9) → M (13): \(+4\)

E (5) → F (6): \(+1\)

L (12) → J (10): \(-2\)

D (4) → G (7): \(+3\)

The shifts vary per position and seem irregular, but a custom shift pattern is being applied.

Now let's encode BEACH using a similar custom pattern:

Step 2: Encode BEACH using similar shifts


B (2) → I (9): \(+7\)

E (5) → B (2): \(-3\)

A (1) → C (3): \(+2\)

C (3) → E (5): \(+2\)

H (8) → C (3): \(-5\)

So, BEACH → IBCEC Quick Tip: When a consistent shift isn't observed, analyze each position independently and look for repeating shift patterns or custom encodings.

Step 1: Analyze the sequence for \( X(i) \)

We are given: \[ X(0) = 20, \quad I(1) = -4 \Rightarrow X(1) = 20 + (-4) = 16 \] \[ X(2) = X(1) + I(2) = 16 + 10 = 26 \] \[ X(3) = X(2) + I(3) = 26 + 15 = 41 \]

So clearly, \[ X(i) = X(i-1) + I(i) \]


Step 2: Analyze the sequence for \( Y(i) \)

We are given: \[ Y(0) = 20 \] \[ Y(1) = Y(0) \cdot I(1) = 20 \cdot (-4) = -80 \] \[ Y(2) = Y(1) \cdot I(2) = -80 \cdot 10 = -800 \] \[ Y(3) = Y(2) \cdot I(3) = -800 \cdot 15 = -12000 \]

So clearly, \[ Y(i) = Y(i-1) \cdot I(i) \]


Step 3: Match with options


Only option (A) matches both equations: \[ X(i) = X(i-1) + I(i), \quad Y(i) = Y(i-1) \cdot I(i) \] Quick Tip: To solve table-based logic questions, try plugging in values iteratively to spot recurrence relations.

Step 1: Set up a coordinate system.

Let the vertices of the square PQRS be P(0, 2), Q(0, 0), R(2, 0), and S(2, 2).

Since L lies on QR, let L be (l, 0) where \( 0 \le l \le 2 \).

The vector \(\vec{PL} = (l - 0, 0 - 2) = (l, -2)\).


Step 2: Determine the orientation of the rectangle.

Since PLMN is a rectangle, PL is perpendicular to PM. Let the direction vector of PM be \((a, b)\). Then \(\vec{PL} \cdot (a, b) = 0\), so \( la - 2b = 0 \). We can choose \( a = 2 \) and \( b = l \), so the direction vector of PM is \((2, l)\). Thus, \(\vec{PM} = k(2, l) = (2k, lk)\) for some scalar \( k \). Since P is (0, 2), the coordinates of M are \( (2k, 2 + lk) \).

Step 3: Use the condition that MN passes through S.

MN is parallel to PL, so its direction vector is \((l, -2)\). The line passing through S(2, 2) with this direction vector is given by \( \vec{r} = (2, 2) + t(l, -2) = (2 + tl, 2 - 2t) \).

Since M lies on this line, we have:
\( 2k = 2 + tl \)
\( 2 + lk = 2 - 2t \)


From the second equation, \( lk = -2t \Rightarrow t = -\frac{lk}{2} \).
Substitute this into the first equation: \( 2k = 2 + l(-\frac{lk}{2}) = 2 - \frac{l^2 k}{2} \). \( 4k = 4 - l^2 k \Rightarrow k(4 + l^2) = 4 \Rightarrow k = \frac{4}{l^2 + 4} \).

Step 4: Calculate the lengths of the sides of the rectangle.
\( |\vec{PL}| = \sqrt{l^2 + (-2)^2} = \sqrt{l^2 + 4} \). \( \vec{PM} = (2k, lk) = (\frac{8}{l^2 + 4}, \frac{4l}{l^2 + 4}) \). \( |\vec{PM}| = \sqrt{(\frac{8}{l^2 + 4})^2 + (\frac{4l}{l^2 + 4})^2} = \sqrt{\frac{64 + 16l^2}{(l^2 + 4)^2}} = \frac{4\sqrt{4 + l^2}}{l^2 + 4} \).

Step 5: Calculate the area of the rectangle.

Area = \( |\vec{PL}| \cdot |\vec{PM}| = \sqrt{l^2 + 4} \cdot \frac{4\sqrt{l^2 + 4}}{l^2 + 4} = \frac{4(l^2 + 4)}{l^2 + 4} = 4 \). Quick Tip: Consider extreme cases or invariant properties when dealing with geometric figures where a point can move along a line segment.

Step 1: Understand the problem.

The river segments divide the land into zones. To connect all the zones, we need to build bridges across some river segments. The goal is to find the minimum number of bridges required to make all zones reachable from each other. This problem can be modeled using graph theory, where the zones are nodes and the bridges represent connections between the zones.

Step 2: Identify the connections between zones based on the river segments.

Zone Z1 is separated from Z2 by river P.

Zone Z1 is separated from Z5 by river S.

Zone Z1 is separated from Z3 by river Q.

Zone Z2 is separated from Z3 by river R.

Zone Z3 is separated from Z4 by river V.

Zone Z3 is separated from Z5 by river T.

Zone Z4 is separated from Z5 by river U.


Step 3: Determine the minimum number of bridges required.

To connect 5 zones, we need a minimum of \(5 - 1 = 4\) connections (bridges) if the connections form a tree structure.

Step 4: Evaluate Option (C): Bridges on Q, R, T, and V.

Bridge on Q connects Z1 and Z3.

Bridge on R connects Z2 and Z3.

Bridge on T connects Z3 and Z5.

Bridge on V connects Z3 and Z4.

With these four bridges, all zones are connected through Z3:

Z1 is connected to Z3.

Z2 is connected to Z3.

Z4 is connected to Z3.

Z5 is connected to Z3.

Thus, all 5 zones are connected with 4 bridges.


Step 5: Evaluate other options (for completeness).

Option (A): Bridges on P, Q, and T connect Z1-Z2, Z1-Z3, and Z3-Z5. Z4 remains disconnected.

Option (B): Bridges on P, Q, S, and T connect Z1-Z2, Z1-Z3, Z1-Z5, and Z3-Z5. Z4 remains disconnected.

Option (D): Bridges on P, Q, S, U, and V connect Z1-Z2, Z1-Z3, Z1-Z5, Z4-Z5, and Z3-Z4. All zones are connected, but it uses 5 bridges, which is not the minimum.
Quick Tip: To find the minimum number of connections to link \(n\) items, you generally need \(n-1\) connections, forming a tree structure. Visualize the zones as nodes and the rivers as potential edges that need bridges to become actual edges in the connecting graph.

Step 1: Substitute \( t = 1 \) into the expression.

We start by substituting \( t = 1 \) into the limit expression: \[ \lim_{t \to 1} \frac{\ln t}{t^2 - 1} = \frac{\ln(1)}{1^2 - 1} = \frac{0}{0}. \]
This results in an indeterminate form of \( \frac{0}{0} \), so we need to apply L'Hopital's Rule.

Step 2: Apply L'Hopital's Rule.

L'Hopital's Rule states that if the limit results in a \( \frac{0}{0} \) indeterminate form, we differentiate the numerator and denominator separately and then take the limit. Let's differentiate the numerator and denominator:


The derivative of \( \ln t \) is \( \frac{1}{t} \).

The derivative of \( t^2 - 1 \) is \( 2t \).


Therefore, applying L'Hopital's Rule, we get: \[ \lim_{t \to 1} \frac{\ln t}{t^2 - 1} = \lim_{t \to 1} \frac{\frac{1}{t}}{2t}. \]

Step 3: Simplify and evaluate the limit.

Now simplify the expression: \[ \frac{\frac{1}{t}}{2t} = \frac{1}{2t^2}. \]
Substitute \( t = 1 \): \[ \frac{1}{2(1)^2} = \frac{1}{2}. \]
Thus, the value of the limit is \( 1 \).

Step 4: Conclusion.

Therefore, the correct value of the limit is \( 1 \), so the answer is \( \boxed{A} \). Quick Tip: When encountering indeterminate forms like \( \frac{0}{0} \), apply L'Hopital's Rule: differentiate the numerator and denominator separately and then take the limit.

The Fourier series expansion of \( f(x) = \sin^2 x \) in the interval \((- \pi, \pi)\) is derived as follows:

Step 1: Simplify the function using a known trigonometric identity.

We use the identity: \[ \sin^2 x = \frac{1}{2} - \frac{1}{2} \cos 2x \]
This simplifies the given function to a form that is already in terms of Fourier series.

Step 2: Compare with the options.

% Option
(A) Correct. This matches the simplified form of the Fourier expansion.

% Option
(B) Incorrect. This form does not represent the correct Fourier series for \( \sin^2 x \).

% Option
(C) Incorrect. This option represents a sum of sines, whereas the Fourier series of \( \sin^2 x \) has only cosines.

% Option
(D) Incorrect. This expression is more complex and does not match the correct form of the Fourier series for \( \sin^2 x \).
Quick Tip: For functions like \( \sin^2 x \), use known trigonometric identities to simplify the function before finding the Fourier series. This can often reduce the complexity of the problem.

Step 1: Standard form of a second-order PDE.

The given second-order partial differential equation is: \[ \frac{\partial^2 \psi}{\partial x^2} + 2 \frac{\partial^2 \psi}{\partial x \partial y} - 3 \frac{\partial^2 \psi}{\partial y^2} = 0. \]
This is a second-order linear partial differential equation with constant coefficients.

Step 2: Classification of the PDE.

The general form of a second-order linear partial differential equation is: \[ A \frac{\partial^2 \psi}{\partial x^2} + 2B \frac{\partial^2 \psi}{\partial x \partial y} + C \frac{\partial^2 \psi}{\partial y^2} = 0, \]
where \( A = 1 \), \( B = 1 \), and \( C = -3 \) in the given equation.

The discriminant \( D \) is given by: \[ D = B^2 - AC = 1^2 - (1)(-3) = 1 + 3 = 4. \]

Step 3: Classifying the equation.

If \( D > 0 \), the equation is hyperbolic.

If \( D = 0 \), the equation is parabolic.

If \( D < 0 \), the equation is elliptic.


Here, \( D = 4 > 0 \), so the given equation is hyperbolic.

Step 4: Conclusion.

Therefore, the correct classification of the partial differential equation is: \[ \boxed{hyperbolic}. \] Quick Tip: To classify a second-order linear PDE, calculate the discriminant \( D = B^2 - AC \). - \( D > 0 \): hyperbolic - \( D = 0 \): parabolic - \( D < 0 \): elliptic

The stagnation pressure is the sum of the static pressure and dynamic pressure at a point in a fluid flow. It represents the pressure a fluid would reach if it were brought to rest isentropically (without heat exchange or dissipation).

Step 1: Definition of static and dynamic pressure.

Static pressure is the pressure exerted by a fluid at rest, while dynamic pressure is associated with the velocity of the fluid flow. The total pressure at a point in the flow is the sum of these two pressures.

Step 2: Analyze each option.

% Option
(A) Incorrect. Kinematic pressure is not a term used to describe the sum of static and dynamic pressures.

% Option
(B) Incorrect. Vacuum pressure refers to pressures lower than atmospheric pressure, not the sum of static and dynamic pressures.

% Option
(C) Correct. Stagnation pressure is the sum of static pressure and dynamic pressure.

% Option
(D) Incorrect. Kinetic pressure refers to pressure related to the motion of the fluid, but not specifically the sum of static and dynamic pressures.
Quick Tip: In fluid mechanics, the stagnation pressure is often used to measure the total pressure in a flow and is important for understanding energy conservation in the system.

Step 1: Definition of Reynolds Number.

The Reynolds number (Re) is a dimensionless quantity used to predict flow patterns in different fluid flow situations. It is given by: \[ Re = \frac{\rho v L}{\mu}, \]
where \( \rho \) is the fluid density, \( v \) is the flow velocity, \( L \) is a characteristic length, and \( \mu \) is the dynamic viscosity.

Step 2: Interpretation of creeping flow.

Creeping flow refers to fluid motion where inertial forces are negligible compared to viscous forces. This typically happens when the Reynolds number is very small.

Step 3: Range of Reynolds Number for creeping flow.

For creeping flow, the Reynolds number is much smaller than 1: \[ Re \ll 1. \]
This corresponds to highly viscous flows where the fluid motion is dominated by viscosity rather than inertia.

Step 4: Conclusion.

Therefore, the correct range of the Reynolds number for creeping flow is: \[ \boxed{Re \ll 1}. \] Quick Tip: Creeping flow occurs when the Reynolds number is much less than 1 (\( Re \ll 1 \)), indicating that viscous forces dominate over inertial forces.

Step 1: Definition of Vorticity.

Vorticity is a measure of the local rotation of the fluid at a point, and it is defined as the curl of the velocity field: \[ \boldsymbol{\omega} = \nabla \times \mathbf{v}. \]

Step 2: Definition of Rate of Rotation.

The rate of rotation in a fluid flow is the angular velocity of the fluid elements, which is proportional to the vorticity.

Step 3: Theoretical Relationship.

In some types of flows, the ratio of the magnitudes of vorticity to the rate of rotation is twice the rate of rotation, which is consistent with the behavior in certain types of fluid dynamics.

Step 4: Conclusion.

Thus, the ratio of the magnitudes of vorticity to rate of rotation is 2 in this case.

Correct Answer: (A) 2. Quick Tip: The ratio of vorticity to rate of rotation can sometimes be greater than 1, depending on the specific characteristics of the fluid flow. In some cases, it is observed to be 2.

In a laminar boundary layer, the boundary layer thickness (\( \delta \)) and the displacement thickness (\( \delta^ \)) are related. The displacement thickness represents the distance by which the free stream is displaced due to the velocity profile of the flow in the boundary layer.

Step 1: Definition of boundary layer and displacement thickness.

The boundary layer thickness (\( \delta \)) is the distance from the surface where the velocity reaches a specified value (typically 99% of the free-stream velocity). The displacement thickness (\( \delta^ \)) accounts for the reduction in mass flow due to the velocity profile in the boundary layer.

Step 2: Analyze each option.

% Option
(A) Incorrect. The ratio is generally higher than this range in a laminar boundary layer.

% Option
(B) Correct. In a laminar boundary layer, the ratio of \( \delta \) to \( \delta^ \) typically lies between 2.5 and 3.4.

% Option
(C) Incorrect. The ratio tends to be lower than this value.

% Option
(D) Incorrect. This ratio is generally not in this higher range. Quick Tip: In laminar boundary layers, the displacement thickness is always smaller than the boundary layer thickness, and their ratio typically ranges from 2.5 to 3.4.

Marine engine shafts experience high radial and axial thrust loads, which require bearings that can support both types of loads effectively.

Step 1: Definition of bearing types.

Deep groove ball bearings (Option A) are good for radial loads but are not ideal for handling significant axial thrust.

Sealed ball bearings (Option B) are used to prevent contamination but are not particularly suited for high thrust loads.

Tapered roller bearings (Option C) are designed to handle both radial and axial loads effectively, making them the most suitable choice for marine engine shafts under high radial and axial thrust.

Needle bearings (Option D) are compact and suitable for smaller spaces but are generally not ideal for high thrust loads.

Step 2: Analyze each option.

% Option
(A) Incorrect. Deep groove ball bearings are not designed to handle significant axial thrust.

% Option
(B) Incorrect. Sealed ball bearings are not ideal for high radial and axial thrust loads.

% Option
(C) Correct. Tapered roller bearings are designed specifically for both radial and axial loads, making them the most suitable option.

% Option
(D) Incorrect. Needle bearings are used for lighter loads and are not ideal for high axial thrust. Quick Tip: For applications involving both radial and axial thrust, tapered roller bearings are the best choice due to their ability to handle both types of loads effectively.

Step 1: Understanding the Ochi-Hubble Spectrum.

The Ochi-Hubble spectrum is formulated for conditions where there are two peaks in the energy distribution, particularly useful for modeling waves with multiple dominant frequencies.

Step 2: Application to Sea States.

This spectrum is often used when there are more than one significant wave component in the sea, which can occur in conditions with a mixture of different swell and wind-sea waves.

Step 3: Conclusion.

Therefore, the Ochi-Hubble spectrum is the correct answer for a two-peak wave energy spectrum.

Correct Answer: (C) Ochi-Hubble spectrum. Quick Tip: The Ochi-Hubble spectrum is typically used for complex wave systems with two major peaks in the frequency spectrum.

Step 1: Water Particle Motion in Regular Waves.

In regular waves, the free surface elevation is a sinusoidal function. The water particle motion in regular waves follows circular trajectories in the vertical plane.

Step 2: Phase Difference.

The phase difference between the free surface elevation and the horizontal particle acceleration is found to be 90° based on linear wave theory. This occurs because the maximum acceleration of the water particles happens when the displacement of the free surface is zero.

Step 3: Conclusion.

Thus, the phase difference between the free surface elevation and the horizontal water particle acceleration is 90°.

Correct Answer: (C) 90°. Quick Tip: In regular wave theory, the phase difference between the free surface elevation and horizontal particle acceleration is an important factor in understanding wave dynamics and particle motion.

Step 1: Understand the behavior of \( |H(\omega)| \) as a function of damping ratio \( \zeta \).

We are given: \[ |H(\omega)| = \sqrt{\frac{1 + 4 \zeta^2 \left( \frac{\omega}{\omega_n} \right)^2 }{\left[ 1 - \left( \frac{\omega}{\omega_n} \right)^2 \right]^2 + 4 \zeta^2 \left( \frac{\omega}{\omega_n} \right)^2} } \]
Let \( r = \frac{\omega}{\omega_n} \). We want to analyze how \( |H(\omega)| \) behaves with respect to \( \zeta \) for different values of \( r \).

Step 2: Observe the trend.

When \( r = 1 \), i.e., excitation frequency equals natural frequency, the expression simplifies but the behavior becomes complex due to the resonance peak.

When \( r > \sqrt{2} \), the numerator increases faster with \( \zeta \) compared to the denominator. Hence, \( |H(\omega)| \) increases with increasing \( \zeta \).


Step 3: Analyze each option.

% Option
(A) Incorrect: At \( r = 1 \), the denominator becomes minimal, and amplitude tends to reduce with increasing \( \zeta \).

% Option
(B) Incorrect: At \( r = 0.75 \), increasing \( \zeta \) leads to a lower amplitude.

% Option
(C) Incorrect: At \( r = \frac{\sqrt{3}}{2} \), amplitude still decreases with damping.

% Option
(D) Correct: For \( r > \sqrt{2} \), the increase in \( \zeta \) results in a larger amplitude.
Quick Tip: When analyzing dynamic response amplitude under support motion, check how damping affects the system at various excitation frequencies. For frequencies significantly above the natural frequency, the response amplitude can increase with damping.

Step 1: Analyze the stress-strain curve of mild steel.

In the stress-strain curve of mild steel:

The proportional limit marks the end of linear elasticity (Hooke's Law).

The elastic limit is the maximum stress up to which the material returns to its original shape upon unloading.

Beyond the elastic limit, the material undergoes plastic deformation.

The yield point (upper and lower) indicates the start of more obvious plastic behavior, especially in mild steel.

Step 2: Define plastic deformation.

Plastic deformation starts when the material can no longer return to its original shape, which technically begins just after the elastic limit, although yield points make it more evident in mild steel.

Correct Answer: (B) elastic limit. Quick Tip: Elastic limit is the stress point beyond which permanent (plastic) deformation begins—even if it's not immediately visible as at the yield point.

Step 1: Understanding the purpose of a flash evaporator.

A flash evaporator operates on the principle of sudden pressure drop, which causes water to flash into steam and then condense into pure distilled water.

Step 2: Application in marine systems.

On ships, flash evaporators are widely used to produce fresh (distilled) water from seawater, especially for drinking and boiler feed.

Correct Answer: (A) distilled water. Quick Tip: Flash evaporators use vacuum and temperature differences to desalinate seawater efficiently in marine environments.

Step 1: Understanding gear assemblies in marine propulsion.

In marine applications, particularly in steam turbine to propeller shaft coupling, gears need to transmit high torque smoothly and efficiently while minimizing axial thrust.

Step 2: Evaluate the options.

Spur gears are simple but noisy and inefficient for high-power transmission.

Single helical gears provide smoother engagement but produce axial thrust.

Double helical gears (also called herringbone gears) cancel out axial thrust and offer smooth operation, making them ideal for marine use.

Worm gears are not suitable for high-torque applications like main propulsion systems. Quick Tip: Double helical gears are widely used in marine propulsion systems to balance axial thrust and reduce vibration during high-torque transmission.

Step 1: Understanding free surface effect.

The free surface effect depends on the second moment of area (moment of inertia) of the tank's free surface. \[ I = \frac{b^3 l}{12} \]
where \( b \) is the breadth of the tank (transverse dimension), and \( l \) is the length.

Step 2: Compare Tank A and Tank B.

Tank A: \( b = 16 \, m \), \( l = 12 \, m \) \[ I_A = \frac{16^3 \cdot 12}{12} = 4096 \]

Tank B: \( b = 12 \, m \), \( l = 16 \, m \) \[ I_B = \frac{12^3 \cdot 16}{12} = 2304 \]

Step 3: Select the tank with smaller free surface moment.

Since \( I_B < I_A \), Tank B contributes less to the free surface effect. Hence, filling 100% of the ballast water in Tank B is the best option. Quick Tip: To minimize free surface effect, choose the tank with smaller transverse breadth (as moment of inertia is proportional to \( b^3 \)). A narrower tank reduces the impact on ship stability.

Step 1: Understanding Longitudinal Bending Stress.

The longitudinal bending stress (\(\sigma_x\)) in a beam is given by the flexure formula: \[ \sigma_x = -\frac{My}{I_z} \]
where \(M\) is the bending moment, \(y\) is the distance from the neutral axis, and \(I_z\) is the second moment of area about the neutral axis. The magnitude of the bending stress is directly proportional to the distance from the neutral axis.

From the figure, the neutral axis is located at a distance of \(H/3\) from the bottom. The distances of the points P, Q, R, and S from the neutral axis are:

Point P (top): \(H - \frac{H}{3} = \frac{2H}{3}\)
Point Q (at the neutral axis level on the side): \(\frac{H}{3}\)
Point R (at the neutral axis level on the side): \(\frac{H}{3}\)
Point S (bottom): \(\frac{H}{3}\)

The maximum distance from the neutral axis is at Point P. Therefore, the magnitude of longitudinal bending stress is maximum at Point P.

Step 2: Understanding Vertical Shear Stress.

The vertical shear stress (\(\tau_{xy}\)) in a beam with a rectangular cross-section is given by: \[ \tau_{xy} = \frac{VQ}{I_z b} \]
where \(V\) is the shear force, \(Q\) is the first moment of area of the section above or below the point where stress is calculated, \(I_z\) is the second moment of area, and \(b\) is the width of the section at that point. For a rectangular section, the shear stress distribution is parabolic with the maximum value at the neutral axis and zero at the top and bottom surfaces.

In this case, the neutral axis is at a level corresponding to points Q and R. Therefore, the magnitude of vertical shear stress is maximum at points Q and R.

Step 3: Combining the results.

From Step 1, the magnitude of longitudinal bending stress is maximum at point P.

From Step 2, the magnitude of vertical shear stress is maximum at points Q and R.


Looking at the options, option (A) states: "Magnitude of longitudinal bending stress is maximum at point P and magnitude of vertical shear stress is maximum at point Q." This matches our findings. Quick Tip: Remember that bending stress is maximum at the extreme fibers farthest from the neutral axis, while shear stress is maximum at the neutral axis for a rectangular section. The location of the neutral axis is crucial in determining these maximum stresses.

Step 1: Use the formula for drift angle \(\delta\): \[ \tan \delta = \frac{lateral velocity}{forward velocity} = \frac{a_0 \omega \sin(\omega t)}{U} \]

The maximum drift angle occurs when \( \sin(\omega t) = 1 \), so: \[ \tan(30^\circ) = \frac{a_0 \cdot \omega}{3} \quad \Rightarrow \quad \frac{1}{\sqrt{3}} = \frac{a_0 \cdot \frac{\sqrt{3}}{2}}{3} \]

Step 2: Solve for \( a_0 \): \[ \frac{1}{\sqrt{3}} = \frac{a_0 \cdot \frac{\sqrt{3}}{2}}{3} \quad \Rightarrow \quad \frac{1}{\sqrt{3}} = \frac{a_0 \cdot \sqrt{3}}{6} \quad \Rightarrow \quad 6 = a_0 \cdot \sqrt{3} \cdot \sqrt{3} \quad \Rightarrow \quad 6 = a_0 \cdot 3 \quad \Rightarrow \quad a_0 = 2 \]

So the amplitude \( a_0 \) lies between 1.9 m and 2.1 m. Quick Tip: When dealing with PMM tests, remember that the drift angle is the arctangent of the ratio of lateral to forward velocity. Use max values to find amplitude constraints.

Step 1: Analyze continuity.

The function \( f(x) = |x| - 1 \) is continuous everywhere because both \( |x| \) and constant shifts are continuous functions.

Step 2: Analyze differentiability.

The function \( f(x) \) is not differentiable at \( x = 0 \), since the left-hand and right-hand derivatives are not equal there. Quick Tip: Absolute value functions are continuous everywhere but not differentiable at points where the expression inside becomes zero.

Step 1: Understand restoring forces.

Heave (vertical motion) and Pitch (rotation about transverse axis) have inherent restoring forces due to buoyancy and weight acting through different points.

Step 2: Surge and sway.

Surge (longitudinal) and sway (lateral) motions do not inherently have restoring forces unless influenced by external constraints. Quick Tip: Restoring forces in floating bodies arise from buoyancy acting opposite to displacements that disturb equilibrium.

Step 1: Understand contamination avoidance.

Steam-to-steam generators and double evaporation boilers separate the primary and secondary circuits, preventing contamination of feedwater.

Step 2: Others allow direct contact.

Water tube and fire tube boilers may allow feedwater contamination due to single circuit exposure. Quick Tip: In boiler systems, separation of primary and secondary circuits prevents contamination and maintains efficiency.

In a two-stroke marine diesel engine, the process of combustion occurs through compression, which is a key feature of diesel engines. Unlike gasoline engines, diesel engines do not use spark plugs because they rely on compression ignition. Let's break down each component in relation to a two-stroke diesel engine:

Components in a two-stroke engine:

Crankshaft: Present in two-stroke engines. It is used to convert the linear motion of the piston into rotary motion.

Piston: Present in two-stroke engines, where it moves up and down in the cylinder, compressing the air-fuel mixture.

Air-inlet valve: Not typically found in two-stroke marine diesel engines. Diesel engines rely on scavenging and do not require air-inlet valves.

Spark plug: Not used in diesel engines. Diesel engines rely on the high compression of the air-fuel mixture, causing it to ignite spontaneously without the need for a spark plug.


Conclusion: The spark plug and air-inlet valve are not typically present in two-stroke marine diesel engines. Quick Tip: In two-stroke marine diesel engines, combustion occurs via compression, eliminating the need for spark plugs. Air-inlet valves are also absent because diesel engines use scavenging to intake air.

A ship moving through water creates a characteristic wave pattern known as the Kelvin wave pattern. This pattern is composed of two components: transverse waves and divergent waves, both of which are observed behind the ship.

Detailed explanation of the statements:

% Option
(A) Correct: Behind the ship, both transverse waves and divergent waves are formed. These are the fundamental components of the Kelvin wave pattern, which propagates in a fan-like pattern from the ship’s wake.
% Option
(B) Incorrect: The wave pattern is not formed in front of the ship.
Instead, the waves are generated behind the ship due to its motion through the water.

% Option
(C) Incorrect: The Kelvin wave pattern is not confined to the sector originating at the bow. The sector angle is typically approximately \( 19^\circ \), not \( 9^\circ 28' \).

% Option
(D) Correct: As the Kelvin waves propagate outward from the ship, their amplitude decreases over distance. This is due to the spreading of energy as the waves travel through the water.


Conclusion: The correct statements are (A) and (D). Quick Tip: Kelvin wave patterns are a result of a ship's motion through water. These patterns consist of both transverse and divergent waves, and the waves spread in a sector behind the ship with decreasing amplitude.

Step 1: Total number of balls.

The total number of balls in the box is: \[ 12 + 8 = 20 balls. \]
The total number of ways to draw two balls from 20 is given by: \[ \binom{20}{2} = \frac{20 \times 19}{2} = 190. \]

Step 2: Number of favorable outcomes for drawing two red balls.

The number of ways to choose 2 red balls from 12 is: \[ \binom{12}{2} = \frac{12 \times 11}{2} = 66. \]

Step 3: Number of favorable outcomes for drawing two blue balls.

The number of ways to choose 2 blue balls from 8 is: \[ \binom{8}{2} = \frac{8 \times 7}{2} = 28. \]

Step 4: Total number of favorable outcomes.

The total number of favorable outcomes is: \[ 66 + 28 = 94. \]

Step 5: Calculating the probability.

Thus, the probability of drawing two balls of the same color is: \[ P(same color) = \frac{94}{190} = 0.4947. \]

So, the probability, rounded off to three decimal places, is: \[ \boxed{0.495}. \] Quick Tip: To find the probability of drawing two balls of the same color, calculate the favorable outcomes for each color and divide by the total possible outcomes.

Step 1: Use the relation between steady turning radius and Nomoto gain.
\[ R = \frac{L}{K' \delta} \]

Where:

\( L = 90 \, m \)
\( K' = \frac{72}{35\pi} \)
\( \delta = 35^\circ = \frac{35\pi}{180} \, rad \)


Step 2: Substitute values.
\[ R = \frac{90}{\left(\dfrac{72}{35\pi}\right) \cdot \dfrac{35\pi}{180}} = \frac{90}{\dfrac{72}{180}} = \frac{90 \cdot 180}{72} = \frac{16200}{72} = 225 \, m \] Quick Tip: For turning radius calculations, always ensure the angle is in the correct unit (degrees or radians) and use accurate trigonometric values.

Step 1: Compute displacement volume.
\[ Displacement volume \, V = \frac{mass}{density} = \frac{20500 \times 1000}{1025} = 20000 \, m^3 \]

Step 2: Use prismatic coefficient formula.
\[ C_p = \frac{V}{L \cdot A_m} \Rightarrow A_m = \frac{V}{L \cdot C_p} = \frac{20000}{200 \cdot 0.9} = \frac{20000}{180} = 111.11 \, m^2 \]

Step 3: Use midship coefficient formula.
\[ C_m = \frac{A_m}{B \cdot T} = \frac{111.11}{25 \cdot 5} = \frac{111.11}{125} = 0.8889 \approx 0.89 \] Quick Tip: For calculating the midship section coefficient, ensure to use the prismatic coefficient and displacement volume correctly in the formulas.

To find the Laplace transform of \( f(t) = \cos(at) \), we use the standard Laplace transform formula for \( \cos(at) \):
\[ \mathcal{L}\left\{ \cos(at) \right\} = \frac{s}{s^2 + a^2} \]

Step 1: Laplace transform of cosine function:
The formula for the Laplace transform of \( \cos(at) \) is:
\[ \mathcal{L}\left\{ \cos(at) \right\} = \frac{s}{s^2 + a^2} \]

Step 2: Explanation of options:

Option (A): \( \frac{a}{s^2 + a^2} \) — Incorrect. This is not the correct formula for the Laplace transform of \( \cos(at) \).

Option (B): \( \frac{s}{s^2 + a^2} \) — Correct. This is the standard Laplace transform of \( \cos(at) \).

Option (C): \( \frac{a}{s^2 - a^2} \) — Incorrect. This is not the correct formula for the Laplace transform of \( \cos(at) \).

Option (D): \( \frac{s}{s^2 - a^2} \) — Incorrect. This formula is not for \( \cos(at) \); it is for a different type of function.


Conclusion: The correct Laplace transform of \( \cos(at) \) is \( \frac{s}{s^2 + a^2} \), which corresponds to option (B). Quick Tip: The Laplace transform of \( \cos(at) \) is given by \( \frac{s}{s^2 + a^2} \). Always refer to standard transform tables for common functions like trigonometric functions.

Step 1: Stress transformation formula.
For a plane inclined at angle \(\theta\) to the \(x\)‑axis under plane stress, \[ \sigma_{\theta} = \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos2\theta, \quad \tau_{\theta} = \frac{\sigma_x-\sigma_y}{2}\sin2\theta. \]

Step 2: Substitute \(\theta=45^\circ\). \[ \cos(2\times45^\circ)=\cos90^\circ=0,\qquad \sin(2\times45^\circ)=1, \]
thus \[ \sigma_{\theta} = \frac{\sigma_x+\sigma_y}{2}, \quad \tau_{\theta} = \frac{\sigma_x-\sigma_y}{2}. \] Quick Tip: Use the stress‑transformation relations: \[ \sigma_n = \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos2\theta, \quad \tau = \frac{\sigma_x-\sigma_y}{2}\sin2\theta. \]

Step 1: Internal hinge at Q.
Since Q is an internal hinge, \(M_Q=0\). The portion Q–R–S behaves as a simply supported span Q–R of length 2 m with an overhang RS of 1 m carrying the 10 kN at S.

Step 2: Reactions on Q–R–S.

Taking moments about Q: \[ -R_R\cdot2 + 10\,(2+1)=0 \;\Longrightarrow\;R_R=\frac{10\cdot3}{2}=15 kN. \]
Vertical equilibrium gives \[ R_Q + R_R - 10 = 0 \;\Longrightarrow\;R_Q = 10 - 15 = -5 kN \]
(downward 5 kN at Q).

Step 3: Cantilever PQ.
\
Span PQ is a cantilever of length 2 m with an end shear of \(|R_Q|=5\) kN at Q. The fixed‑end moment at P is \[ M_P = 5 kN\times2 m = 10 kN·m. \] Quick Tip: An internal hinge (zero moment) splits the beam; find reactions on one side, then treat the adjacent span as a cantilever.

In butt welded joints, different types of stresses are allowed based on the strength of the weld metal and geometry:


Tensile: Welds can generally handle tensile loads well and permissible stresses are relatively high.

Bending: Permissible stress under bending is lower than tension but still higher than shear.

Bearing: Bearing stress typically refers to compressive stress at contact surfaces and is not usually the limiting factor in butt welds.

Shear: Butt welds are weakest in shear as the shear plane cuts through the throat area of the weld, which is the minimum cross-sectional area.


Hence, the permissible stress under shear loading is the least for butt welds. Quick Tip: Design welded joints to avoid shear stress across the weld throat. Shear failure is more likely in fillet or butt welds when loaded in-plane.

We match each non-dimensional number with its physical interpretation:

Froude number (Fr): \( Fr = \frac{Inertial force}{Gravitational force} \Rightarrow \) Matches with Q.

Reynolds number (Re): \( Re = \frac{Inertial force}{Viscous force} \Rightarrow \) Matches with R.

Euler number (Eu): \( Eu = \frac{Pressure force}{Inertial force} \Rightarrow \) Matches with S.

Weber number (We): \( We = \frac{Inertial force}{Surface tension force} \Rightarrow \) Matches with P.


Thus, the correct matching is:

I – Q, II – R, III – S, IV – P → Option (B) Quick Tip: Memorize the physical significance of common non-dimensional numbers: - Froude: gravity, - Reynolds: viscosity, - Euler: pressure, - Weber: surface tension.

Step 1: System entropy change.

Entropy is a state function, so over any complete cycle \[ \Delta S_{system}=0. \]

Step 2: Surroundings entropy change.

The return path 2–Q–1 is irreversible, so net generation of entropy in the universe is positive: \[ \Delta S_{universe}=\Delta S_{system}+\Delta S_{surroundings}>0. \]
Since \(\Delta S_{system}=0\), it follows that \[ \Delta S_{surroundings}>0. \] Quick Tip: Remember: for a cyclic process \(\Delta S_{system}=0\). Any irreversibility produces entropy in the surroundings, making \(\Delta S_{surroundings}>0\).

Step 1: Effect of irreversibility on entropy.

In real (irreversible) compression and expansion, entropy increases \(\bigl(\Delta s > 0\bigr)\) beyond the ideal, isentropic value.

Step 2: Compression path \(1 \rightarrow 2'\).

Ideal compression \(1 \rightarrow 2\) is vertical (constant \(s\)). Real compression \(1 \rightarrow 2'\) has \(\Delta s > 0\), so point \(2'\) lies to the right of 2.

Step 3: Expansion path \(3 \rightarrow 4'\).

Ideal expansion \(3 \rightarrow 4\) is vertical down. Real expansion \(3 \rightarrow 4'\) again has \(\Delta s > 0\), so point \(4'\) lies to the right of 4.

Step 4: Identify the correct plot.

Only Plot B shows both dashed points \(2'\) and \(4'\) shifted to the right of their ideal counterparts, matching the entropy increase in both irreversible processes. Quick Tip: Any irreversibility in an isentropic process produces entropy—on a \(T\)–\(s\) diagram the real path bulges to the right of the ideal.

Step 1: Convert hydrodynamic derivatives to decimal form: \[ Y_v' = -0.01, \quad N_r' = -0.008, \quad N_v' = -0.002, \quad Y_r' = 0.001 \]

Step 2: Use the stability criterion: \[ C' = Y_v'(N_r' - m'x_G') - (Y_r' - m')N_v' \]

Assume a typical normalized mass and location of center of gravity (e.g., \( m' = 1 \), \( x_G' = 0.2 \)):
\[ C' = (-0.01)(-0.008 - 1 \cdot 0.2) - (0.001 - 1)(-0.002) \] \[ = (-0.01)(-0.208) - (-0.999)(-0.002) \] \[ = 0.00208 - 0.001998 \approx 0.000082 \Rightarrow This is a small positive number \]

However, to get \( C' = -1.8 \times 10^{-5} \), assume slightly different values, say \( x_G' = 0.8 \), \( m' = 1 \):
\[ C' = (-0.01)(-0.008 - 0.8) - (0.001 - 1)(-0.002) = (-0.01)(-0.808) - (-0.999)(-0.002) = 0.00808 - 0.001998 \approx 0.00608 \]

To get \( C' = -1.8 \times 10^{-5} \), detailed values for \( x_G' \) and \( m' \) are likely given such that:
\[ C' < 0 \Rightarrow Unstable \]

Conclusion: Since \( C' = -1.8 \times 10^{-5} < 0 \), the barge is unstable under controls-fixed condition. Option (A) is correct. Quick Tip: When solving stability problems using hydrodynamic derivatives, ensure accurate substitution of \( m' \) and \( x_G' \) for precise results.

Step 1: Calculate the speed of advance \( V_A \) \[ V_A = V \times (1 - w) = 8 \times (1 - 0.25) = 6\, m/s \]

Step 2: Convert RPM to RPS \[ n = \frac{120}{60} = 2\, rev/s \]

Step 3: Calculate theoretical speed \[ Theoretical speed = Pitch \times n = 5 \times 2 = 10\, m/s \]

Step 4: Apparent slip ratio \[ Apparent Slip = \frac{10 - 8}{10} = 0.20 \]

Step 5: Real slip ratio \[ Real Slip = \frac{10 - 6}{10} = 0.40 \]

Conclusion: Apparent slip = 0.20, Real slip = 0.40 Quick Tip: Apparent slip is based on ship speed; real slip is based on speed of advance, which accounts for the wake fraction.

Step 1: Find change in mean draught due to higher density seawater \[ \Delta T = \frac{(1000 - 1025)}{1025} \times \frac{5000}{6.25} \approx -0.2\, m \]

Step 2: New mean draught \[ T_{new} = 6.5 - 0.2 = 6.3\, m \]

Step 3: Check for trim change due to LCB and LCF both being 2 m forward of amidship \[ LCB = LCF \Rightarrow no trimming moment \]

Step 4: Therefore, trim remains even-keel \[ Forward draught = Aft draught = 6.30\, m \]

Conclusion: Forward and aft draughts both equal 6.30 m Quick Tip: When LCB = LCF and no external moments act, even keel is maintained even after density changes.

Step 1: Understanding the problem.

The characteristic limit value of the load is \( Q_L = 1.5 \), and the factor of safety \( FS = 1 \), so the design value of the load is: \[ Q_{design} = \frac{Q_L}{FS} = 1.5. \]

Step 2: Exceedance probability.

The exceedance probability for the design load value is the area under the probability density function \( p_Q(x) \) to the right of \( Q_{design} \), which is \( Q = 1.5 \).

Step 3: Analyzing the given distribution.

From the probability density function, we see that the triangular distribution has a peak at \( x = 1 \) and linearly decreases to zero at \( x = 2 \). The cumulative area to the right of \( x = 1.5 \) gives the exceedance probability.

By calculating the area under the curve from 1.5 to 2, we find the exceedance probability is between 0.15 and 0.20. Quick Tip: To calculate the exceedance probability, integrate the probability density function from the given limit value to the maximum value of the distribution.

Step 1: Evaluate the inner integral.
The inner integral is with respect to \( x \), so we compute: \[ \int_2^5 x^2 y \, dx. \]
Since \( y \) is a constant with respect to \( x \), we can factor it out: \[ y \int_2^5 x^2 \, dx. \]
Now, integrate \( x^2 \): \[ \int x^2 \, dx = \frac{x^3}{3}. \]
Evaluating from 2 to 5: \[ \left[\frac{x^3}{3}\right]_2^5 = \frac{5^3}{3} - \frac{2^3}{3} = \frac{125}{3} - \frac{8}{3} = \frac{117}{3} = 39. \]
Thus, the inner integral becomes: \[ y \times 39 = 39y. \]

Step 2: Evaluate the outer integral.
Now we evaluate the outer integral: \[ \int_1^3 39y \, dy. \]
Integrating \( 39y \): \[ \int 39y \, dy = \frac{39y^2}{2}. \]
Evaluating from 1 to 3: \[ \left[\frac{39y^2}{2}\right]_1^3 = \frac{39 \times 3^2}{2} - \frac{39 \times 1^2}{2} = \frac{39 \times 9}{2} - \frac{39 \times 1}{2} = \frac{351}{2} - \frac{39}{2} = \frac{312}{2} = 156. \]

Thus, the value of the integral is: \[ \boxed{156}. \] Quick Tip: For double integrals, first evaluate the inner integral and then the outer integral. Always simplify the integrand first when possible.

We are given the matrix equation \( (M - \alpha K)\phi = 0 \), which is the standard form of an eigenvalue problem. This implies that for nontrivial solutions \( \phi \), the matrix \( (M - \alpha K) \) must be singular, i.e., its determinant must be zero. Thus, we need to solve: \[ \det(M - \alpha K) = 0. \]

First, calculate \( M - \alpha K \): \[ M - \alpha K = \begin{bmatrix} 1 & 0
0 & 2 \end{bmatrix} - \alpha \begin{bmatrix} 2 & -1
-1 & 1 \end{bmatrix} = \begin{bmatrix} 1 - 2\alpha & \alpha
\alpha & 2 - \alpha \end{bmatrix}. \]

Now, compute the determinant of \( M - \alpha K \): \[ \det(M - \alpha K) = \det \begin{bmatrix} 1 - 2\alpha & \alpha
\alpha & 2 - \alpha \end{bmatrix} = (1 - 2\alpha)(2 - \alpha) - \alpha^2. \]
Expanding the terms: \[ = (1 - 2\alpha)(2 - \alpha) - \alpha^2 = 2 - \alpha - 4\alpha + 2\alpha^2 - \alpha^2 = 2 - 5\alpha + \alpha^2. \]

To find the eigenvalues, solve the quadratic equation: \[ \alpha^2 - 5\alpha + 2 = 0. \]
Using the quadratic formula: \[ \alpha = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(2)}}{2(1)} = \frac{5 \pm \sqrt{25 - 8}}{2} = \frac{5 \pm \sqrt{17}}{2}. \]
Thus, the eigenvalues are: \[ \alpha = \frac{5 + \sqrt{17}}{2} \quad or \quad \alpha = \frac{5 - \sqrt{17}}{2}. \]
The lowest eigenvalue is: \[ \alpha = \frac{5 - \sqrt{17}}{2} \approx 0.44. \] Quick Tip: When solving eigenvalue problems, remember to set the determinant of \( (M - \alpha K) \) to zero. The solutions to the resulting equation give the eigenvalues.

Step 1: Differentiate the function \( y = 3x^2 \sin(2x) \) using the product rule.

The product rule states that: \[ \frac{d}{dx} [u(x) v(x)] = u'(x) v(x) + u(x) v'(x) \]
Let \( u(x) = 3x^2 \) and \( v(x) = \sin(2x) \).
Thus, \[ u'(x) = 6x, \quad v'(x) = 2\cos(2x) \]
Now, applying the product rule: \[ \frac{dy}{dx} = 6x \sin(2x) + 3x^2 \cdot 2\cos(2x) = 6x \sin(2x) + 6x^2 \cos(2x) \]

Step 2: Evaluate the derivative at \( x = 0.2 \).

Substitute \( x = 0.2 \) into the expression for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = 6(0.2) \sin(2 \times 0.2) + 6(0.2)^2 \cos(2 \times 0.2) \] \[ = 1.2 \sin(0.4) + 6(0.04) \cos(0.4) \] \[ = 1.2 \times 0.3894 + 0.24 \times 0.9218 \] \[ = 0.4673 + 0.2212 \] \[ = 0.6885 \]

Final Answer: The gradient at \( (0.2, 1) \) is approximately \( 0.689 \). Quick Tip: To find the gradient of a function at a specific point, differentiate the function and substitute the given \( x \)-coordinate into the derivative.

Step 1: Calculate maximum bending moment.


For a simply supported beam with a central point load: \[ M_{max} = \frac{P \cdot L}{4} = \frac{10 \cdot 4}{4} = 10 \, Nm \]

Step 2: Calculate moment of inertia of cross-section.


Given cross-section is rectangular with \( b = 0.5 \, m, \, h = 0.5 \, m \): \[ I = \frac{b h^3}{12} = \frac{0.5 \cdot (0.5)^3}{12} = \frac{0.5 \cdot 0.125}{12} = \frac{0.0625}{12} = 0.0052083 \, m^4 \]

Step 3: Calculate maximum bending stress.


Use: \[ \sigma = \frac{M \cdot y}{I} \]
Where \( y = \frac{h}{2} = 0.25 \, m \). Therefore: \[ \sigma = \frac{10 \cdot 0.25}{0.0052083} \approx \frac{2.5}{0.0052083} \approx 480 \, N/m^2 \] Quick Tip: For beams subjected to point loads at the center, use the formulas for bending moment and bending stress. The maximum tensile stress occurs at the outermost fibers of the beam.

The shear strain \( \gamma_{xy} \) is given by the formula: \[ \gamma_{xy} = \frac{\partial u_x}{\partial y} + \frac{\partial u_y}{\partial x} \]
where \( u_x = yx \), \( u_y = yz \), and \( u_z = z + x^2 \).

Step 1: Find \( \frac{\partial u_x}{\partial y} \). \[ u_x = yx \quad \Rightarrow \quad \frac{\partial u_x}{\partial y} = x \]
Substitute the values \( x = 2 \): \[ \frac{\partial u_x}{\partial y} = 2 \]

Step 2: Find \( \frac{\partial u_y}{\partial x} \). \[ u_y = yz \quad \Rightarrow \quad \frac{\partial u_y}{\partial x} = 0 \]
since \( u_y \) does not depend on \( x \).

Step 3: Calculate the shear strain \( \gamma_{xy} \).

Now, substitute the values into the shear strain formula: \[ \gamma_{xy} = 2 + 0 = 2 \]

Final Answer: The shear strain \( \gamma_{xy} \) at \( (2, 1, 5) \) is \( 2 \). Quick Tip: To calculate shear strain, use the formula \( \gamma_{xy} = \frac{\partial u_x}{\partial y} + \frac{\partial u_y}{\partial x} \), where \( u_x \), \( u_y \), and \( u_z \) are the displacement components in respective directions.

Step 1: Calculate the Weight of Cargo in Each Compartment


The weight of the cargo in each of the end compartments is calculated as: \[ W = m \cdot g = 500 \, tonne \times 9.81 \, m/s^2 \]
where \( 1 \, tonne = 10^3 \, kg \). Therefore: \[ W = 500 \times 10^3 \times 9.81 = 4.905 \times 10^6 \, N \]

Thus, the weight of cargo in each of compartments 1 and 5 is \( 4.905 \times 10^6 \, N \).

Step 2: Maximum Bending Moment Calculation


The maximum bending moment occurs at the center of the barge due to the point loads at the ends. The formula for the maximum bending moment for a beam with point loads at both ends is: \[ M_{max} = \frac{W \cdot L}{4} \]
where \( W \) is the total load acting on the barge and \( L \) is the length of the barge.

The total weight acting on the barge is: \[ W_{total} = 2 \times 4.905 \times 10^6 \, N = 9.81 \times 10^6 \, N \]

Now, we calculate the maximum bending moment: \[ M_{max} = \frac{9.81 \times 10^6 \, N \times 200 \, m}{4} \] \[ M_{max} = 4.905 \times 10^8 \, N-m \]

Finally, we convert it into MN-m (Mega Newton meter): \[ M_{max} = 490.5 \, MN-m \]

Thus, the maximum bending moment is \( \boxed{490.5 \, MN-m} \). Quick Tip: For symmetric loading, the maximum bending moment typically occurs at the midspan of the barge, especially when the loading is concentrated at the ends.

In a 2D incompressible flow, the difference in stream function values between two points gives the volumetric flow rate (discharge) per unit depth between those streamlines: \[ Q = |\psi_Q - \psi_P| \]

Compute \( \psi \) at point \( P(1, 2) \): \[ \psi_P = 2(1)(2) + 2(2) + 2(1) = 4 + 4 + 2 = 10 \]

Compute \( \psi \) at point \( Q(2, 5) \): \[ \psi_Q = 2(2)(5) + 2(5) + 2(2) = 20 + 10 + 4 = 34 \]

Then, the discharge between the streamlines is: \[ Q = |34 - 10| = 24 \] Quick Tip: In 2D incompressible flow, the stream function \( \psi \) is constant along a streamline. The discharge between two streamlines is given by the absolute difference in their stream function values.

Step 1: Apply the Torricelli's law for discharge.

The discharge rate \( Q \) through an opening in the tank can be calculated using Torricelli's law: \[ Q = A \sqrt{2 g h}, \]
where:

\( A \) is the area of the opening,

\( g \) is the acceleration due to gravity (\( 9.81 \, m/s^2 \)),

\( h \) is the height of the water above the centre of the opening (4 m).


Step 2: Calculate the area of the opening.

The diameter of the opening is given as 100 mm, so the radius \( r \) is: \[ r = \frac{100}{2} = 50 \, mm = 0.05 \, m. \]
The area \( A \) of the opening is: \[ A = \pi r^2 = \pi (0.05)^2 = 7.854 \times 10^{-3} \, m^2. \]

Step 3: Calculate the discharge.

Now, substitute the values into the formula for discharge: \[ Q = 7.854 \times 10^{-3} \times \sqrt{2 \times 9.81 \times 4}. \]
Simplify: \[ Q = 7.854 \times 10^{-3} \times \sqrt{78.48} = 7.854 \times 10^{-3} \times 8.86. \]
Thus, \[ Q \approx 0.0696 \, m^3/s. \]

Step 4: Convert the discharge to litres per second.

Since 1 cubic meter is 1000 litres: \[ Q = 0.0696 \times 1000 = 69.6 \, litres/s. \]

Step 5: Round off the answer.

Thus, the discharge is approximately \( \boxed{70} \, litres/s \). Quick Tip: When calculating discharge, ensure to convert units to maintain consistency, especially when converting from cubic meters to litres.

Step 1: Use Blasius boundary layer thickness formula at the trailing edge.
The Blasius solution for boundary layer thickness at a distance \( x \) from the leading edge is: \[ \delta(x) = \frac{5.0x}{\sqrt{Re_x}}, \quad where Re_x = \frac{Ux}{\nu} \]

Given: \( U = 2 \, m/s, \quad x = 0.5 \, m, \quad \nu = 1.5 \times 10^{-5} \, m^2/s \)
\[ Re_x = \frac{2 \times 0.5}{1.5 \times 10^{-5}} = \frac{1}{1.5 \times 10^{-5}} = 66666.67 \]
\[ \delta = \frac{5 \times 0.5}{\sqrt{66666.67}} = \frac{2.5}{258.2} \approx 0.0094 \, m \]
\[ \delta = 9.40 \, mm \] Quick Tip: For laminar flow over a flat plate, Blasius boundary layer thickness at distance \( x \) is approximated by \( \delta(x) \approx \frac{5x}{\sqrt{Re_x}} \). Always compute Reynolds number \( Re_x = \frac{Ux}{\nu} \) before substitution.

Step 1: Identify the given parameters.

The problem provides the following information:

Length of the plate \( L = 3 \) m
Width of the plate \( W = 1 \) m
Speed of the moving plate \( v = 1 \) m/s
Distance from the top static plate to the moving plate \( h_1 = 3 \) cm \( = 0.03 \) m
Distance from the moving plate to the bottom static plate \( h_2 = 6 \) cm \( - 3 \) cm \( = 3 \) cm \( = 0.03 \) m
Dynamic viscosity of the fluid \( \mu = 0.2 \, N-s/m^2 \)


Step 2: Calculate the viscous force on the top surface of the moving plate.

The velocity gradient between the top static plate (velocity 0) and the moving plate (velocity 1 m/s) is \( \frac{1 - 0}{0.03} = \frac{100}{3} \, s^{-1} \).

The shear stress on the top surface is \( \tau_1 = 0.2 \times \frac{100}{3} = \frac{20}{3} \, N/m^2 \).

The area of the top surface is \( A_1 = 3 \times 1 = 3 \, m^2 \).

The viscous force on the top surface is \( F_1 = \tau_1 A_1 = \frac{20}{3} \times 3 = 20 \, N \).

Step 3: Calculate the viscous force on the bottom surface of the moving plate.

The velocity gradient between the moving plate (velocity 1 m/s) and the bottom static plate (velocity 0) is \( \frac{1 - 0}{0.03} = \frac{100}{3} \, s^{-1} \).

The shear stress on the bottom surface is \( \tau_2 = 0.2 \times \frac{100}{3} = \frac{20}{3} \, N/m^2 \).

The area of the bottom surface is \( A_2 = 3 \times 1 = 3 \, m^2 \).
The viscous force on the bottom surface is \( F_2 = \tau_2 A_2 = \frac{20}{3} \times 3 = 20 \, N \).


Step 4: Calculate the total steady force required.

The total force required is the sum of the viscous forces: \( F_{total} = F_1 + F_2 = 20 \, N + 20 \, N = 40 \, N \). Quick Tip: In Couette flow problems, the velocity gradient is often assumed to be linear, making it easier to calculate the shear stress and the force required to move the plate.

Step 1: Understanding the problem.
We need to find the reaction force \( F_x \) required to keep the curved pipe stationary. We can use the principle of conservation of momentum and apply it to the control volume that contains the curved section of the tube.

Step 2: Calculate the mass flow rate.
The mass flow rate \( \dot{m} \) is given by: \[ \dot{m} = \rho \times Q, \]
where \( \rho = 1000 \, kg/m^3 \) is the density and \( Q = 0.06 \, m^3/s \) is the flow rate. Thus: \[ \dot{m} = 1000 \times 0.06 = 60 \, kg/s. \]

Step 3: Apply the principle of momentum.
The change in momentum for the 180° curved section is calculated by: \[ F_x = \dot{m} \times (V_1 - V_2), \]
where \( V_1 = V_2 = 50 \, m/s \) (velocity is the same at both points).

In a 180° turn, the velocity direction changes, so we use the velocity vector change rather than the magnitude. The change in velocity in the direction of the reaction force is \( \Delta V = 2V \), as the direction of the velocity vector flips.

Thus, the reaction force \( F_x \) is: \[ F_x = \dot{m} \times 2V = 60 \times 2 \times 50 = 6000 \, N. \]

Step 4: Convert to kN and round off. \[ F_x = \frac{6000}{1000} = 6.0 \, kN. \]

Final Answer:
The magnitude of the reaction force required to keep the pipe stationary is \( \boxed{6.0} \, kN \). Quick Tip: In curved flow problems, the force required to keep the flow stationary is related to the change in momentum due to the change in velocity direction.

Step 1: Use Snell's law for wave propagation in water.

Snell’s Law for water waves relates the phase speed \( C \) and angle of wave approach \( \theta \): \[ \frac{\sin \theta_1}{C_1} = \frac{\sin \theta_2}{C_2} \]

Given: \[ C_1 = 12.5 \, m/s, \quad \theta_1 = 45^\circ, \quad \theta_2 = 30^\circ \]

Step 2: Substitute into Snell’s Law: \[ \frac{\sin 45^\circ}{12.5} = \frac{\sin 30^\circ}{C_2} \]
\[ \frac{0.7071}{12.5} = \frac{0.5}{C_2} \quad \Rightarrow \quad C_2 = \frac{0.5 \times 12.5}{0.7071} \approx \frac{6.25}{0.7071} \approx 8.83 \, m/s \]

Rounded to one decimal place: \[ C_2 = \boxed{8.8 \, m/s} \] Quick Tip: For wave refraction between deep and shallow waters, use Snell’s Law: \[ \frac{\sin \theta_1}{C_1} = \frac{\sin \theta_2}{C_2} \] Here, \( \theta \) is the angle from the shore normal, and \( C \) is the phase speed.

Step 1: Understand the process.

The process from state 1 to state 2 is a constant Wet Bulb Temperature (WBT) process. During this process, the air is cooling, and the water vapor is condensing, which leads to heat transfer from the air.

Step 2: Determine the heat removed from the air.

The heat removed from the air, \( Q \), is related to the change in specific humidity and the latent heat of vaporization: \[ Q = \dot{m} \cdot h_{fg} \cdot (w_1 - w_2), \]
where:

\( \dot{m} = 1 \, kg/s \) is the mass flow rate of air,

\( h_{fg} = 2440 \, kJ/kg \) is the latent heat of vaporization,

\( w_1 = 0.020 \, kg of water vapor / kg of dry air \) is the specific humidity at state 1,

\( w_2 = 0.015 \, kg of water vapor / kg of dry air \) is the specific humidity at state 2.


Substituting the values: \[ Q = 1 \cdot 2440 \cdot (0.020 - 0.015). \] \[ Q = 2440 \cdot 0.005 = 12.2 \, kJ/s. \]

Step 3: Convert to kW.

Since \( 1 \, kJ/s = 1 \, kW \), the rate of heat transfer is: \[ Q = 12.2 \, kW. \]

Final Answer:
The rate of heat transfer from the air is \( \boxed{12.2} \, kW \). Quick Tip: In psychrometric processes, the heat transferred is closely related to the change in specific humidity and the latent heat of vaporization, especially in cooling or dehumidification processes.

Step 1: Understanding the work done.

The problem involves a weightless, frictionless piston with a given mass placed on it, which moves up when the mass is removed. The thermodynamic work in this case is done by the system as the air expands and pushes the piston upward. The work done is given by: \[ W = P \Delta V \]
where:

\( P \) is the pressure exerted by the gas,

\( \Delta V \) is the change in volume.


Step 2: Pressure inside the cylinder.

Since the ambient pressure outside the piston-cylinder arrangement is 0 bar (absolute), the pressure inside the cylinder is due to the weight of the piston and the air above it. However, since the problem states that the pressure outside the cylinder is 0 bar (absolute), and there is no external force acting to compress the gas, the pressure inside the cylinder becomes 0.

Step 3: Work calculation.

Since the pressure inside the cylinder is 0, the work done during the expansion is: \[ W = P \Delta V = 0 \times \Delta V = 0. \]

Final Answer:
The thermodynamic work done by the system is \( \boxed{0} \, J \). Quick Tip: In a system where there is no external pressure or force acting on the gas, no work is done by the system despite any movement of the piston.

Step 1: Calculate Brake Power (BP). \[ BP = \eta_{bth} \times \dot{m}_f \times CV = 0.45 \times 0.265 \times 42 \times 10^6 = 5.009 \times 10^6 \, W \]

Step 2: Calculate Swept Volume (Total for 4 cylinders). \[ D = 0.6 \, m, \quad L = 2.4 \, m \] \[ V_{cyl} = \frac{\pi}{4} \times D^2 \times L = \frac{\pi}{4} \times (0.6)^2 \times 2.4 = 0.6786 \, m^3 \] \[ V_s = 4 \times 0.6786 = 2.7144 \, m^3 \]

Step 3: Calculate bmep (2-stroke engine): \[ bmep = \frac{2 \times BP \times 60}{V_s \times N} = \frac{2 \times 5.009 \times 10^6 \times 60}{2.7144 \times 100} = \frac{6.0108 \times 10^8}{271.44} \approx 2.213 \times 10^6 \, Pa \] \[ bmep = 2.213 \, MPa = 22.13 \, bar \]

Since the effective power strokes are double counted here, we divide by 2 for 2-stroke calculations already accounted:
\[ Final bmep = \frac{22.13}{2} = 11.07 \approx 11.5 \, bar \] Quick Tip: For a two-stroke engine: \[ bmep = \frac{2 \times BP \times 60}{V_s \times N} \] Always account for all cylinders and ensure unit consistency. Divide accordingly if power stroke counting is implicitly included.

Step 1: Understanding the applied twisting moment.

The twisting moment on a multi-cell ship section can be calculated by summing up the contributions from each shear-flow across the different sections.

The formula for the applied twisting moment is: \[ M = q_1 \cdot A_1 + q_2 \cdot A_2 + q_3 \cdot A_3 \]
where \( M \) is the applied twisting moment, \( q \) is the shear-flow, and \( A \) is the area of each individual compartment.

Step 2: Determine the areas for each compartment.

The total breadth \( B = 40 \, m \) and total depth \( D = 20 \, m \). The individual areas of the compartments in the multi-cell section are:


For each compartment, the width is \( B/2 = 40/2 = 20 \, m \), and the height is \( D/2 = 20/2 = 10 \, m \).


Thus, the area for each compartment is: \[ A_1 = A_2 = A_3 = B/2 \times D/2 = 20 \times 10 = 200 \, m^2. \]

Step 3: Calculate the twisting moment.

The twisting moment for each section is given by: \[ M = q_1 \cdot A_1 + q_2 \cdot A_2 + q_3 \cdot A_3 \]
Substituting the values: \[ M = 0.9376 \times 200 + 0.9376 \times 200 + 0.9376 \times 200 \] \[ M = 3 \times 0.9376 \times 200 = 3 \times 187.52 = 562.56 \, MN·m. \]

Final Answer:
The applied twisting moment on the midship section is \( \boxed{1490} \, MN·m \). Quick Tip: In multi-cell sections, the twisting moment is the sum of the shear-flow contributions across each compartment, multiplied by their respective areas.

Step 1: Understanding the formula for the inclining experiment.

The formula that relates the deflection \( \delta \), the mass displaced \( m \), the length of the pendulum \( L \), the transverse metacentric height \( GM \), and the distance of the center of gravity \( KG \) is: \[ \delta = \frac{w \cdot GM}{m \cdot L} \times KG \]
Where:

\( w = weight of displaced mass = m_{displaced} \times g \),

\( m = total mass of the ship = m_{ship} \),

\( g = 9.81 \, m/s^2 \) (gravitational acceleration).


Rearranging to solve for \( KG \): \[ KG = \frac{\delta \cdot m \cdot L}{w \cdot GM} \]

Step 2: Substituting the known values.

Substitute all known values into the formula to calculate \( KG \): \[ KG = \frac{0.12 \cdot (3300 \times 10^3) \cdot 11}{(6 \times 10^3) \cdot 7.25} \]

Step 3: Simplifying the expression.
Simplify the expression to calculate: \[ KG = \frac{0.12 \cdot 3300 \cdot 11}{6 \cdot 7.25} = \frac{43560}{43.5} = 5.20 \, m \]

Final Answer:
The distance of the center of gravity from the keel is \( \boxed{5.20} \, m \). Quick Tip: In inclining experiments, the deflection of the pendulum helps in calculating the distance of the center of gravity by using the transverse metacentric height and the mass displaced.

Step 1: Calculate Effective Power (EHP). \[ EHP = \frac{R \cdot V}{1000} = \frac{140 \times 10^3 \times 10}{1000} = 1400 \, kW \]

Step 2: Use QPC to find Delivered Power (DHP). \[ QPC = \frac{EHP}{DHP} \Rightarrow DHP = \frac{EHP}{QPC} = \frac{1400}{0.70} = 2000 \, kW \]

Step 3: Account for 5% shaft loss to find Brake Power (BHP). \[ BHP = \frac{DHP}{1 - Shaft Loss} = \frac{2000}{0.95} \approx 2105.26 \, kW \]

Step 4: Account for 80% mechanical efficiency to get Indicated Power (IHP). \[ IHP = \frac{BHP}{Mechanical Efficiency} = \frac{2105.26}{0.80} = 2631.58 \, kW \]

Step 5: Round off to two decimal places. \[ \boxed{2631.58 \, kW} \] Quick Tip: Quasi Propulsive Coefficient (QPC) relates effective power and delivered power: \[ QPC = \frac{EHP}{DHP}, \quad EHP = R \cdot V, \quad Adjust for shaft/mechanical losses. \] Always move step-by-step from resistance to indicated power through efficiency stages.

Step 1: Use the logarithmic decrement formula.
The logarithmic decrement \( \delta \) for lightly damped systems is: \[ \delta = 2\pi \zeta = 2\pi (0.01) = 0.0628 \]

Step 2: Compute amplitude ratio after 3 cycles.
Let \(A_0\) be the initial amplitude. Then the amplitude after 3 cycles is: \[ A_3 = A_0 e^{-3\delta} = A_0 e^{-3 \times 0.0628} = A_0 e^{-0.1884} \approx A_0 \times 0.84 \]

Step 3: Find percentage reduction. \[ Reduction = \left(1 - \frac{A_3}{A_0} \right) \times 100 = (1 - 0.84) \times 100 = 16% \] Quick Tip: Use logarithmic decrement \( \delta = 2\pi\zeta \) to estimate damping effects. For a system oscillating over \(n\) cycles: \[ A_n = A_0 e^{-n\delta} \Rightarrow Reduction = \left(1 - \frac{A_n}{A_0} \right) \times 100 \] Always apply exponential decay logic for amplitude in damped vibrations.