GATE PI 2012

To maximize profit, a firm should produce where Marginal Cost (MC) equals Marginal Revenue (MR).

1. Define Cost and Revenue:

   • Cost Function: $C(q) = 5q^2$
   • Revenue Function: $R(q) = \text{Price} \times q = 50q$

2. Calculate Marginal Cost (MC): MC is the derivative of the Cost Function.
   $MC = \frac{dC}{dq} = \frac{d}{dq}(5q^2) = 10q$

3. Calculate Marginal Revenue (MR): MR is the derivative of the Revenue Function.
   $MR = \frac{dR}{dq} = \frac{d}{dq}(50q) = 50$

4. Find Profit-Maximizing Quantity: Set $MC = MR$.
   $10q = 50$
   $q = \frac{50}{10}$
   $q = 5$

5. Verify Maximization (Second-Order Condition): The profit function is $\Pi(q) = R(q) - C(q) = 50q - 5q^2$.
   The first derivative is $\frac{d\Pi}{dq} = 50 - 10q$.
   The second derivative is $\frac{d^2\Pi}{dq^2} = -10$.
   Since $-10 < 0$, profit is indeed maximized at $q=5$.

Therefore, the firm should produce 5 units to maximize its profit.

To complete the sentence, we need a relative pronoun referring to "the one" (which is the dog). Since "was hurt" needs a subject, the pronoun must act as a subject. Let's analyze the options: $that$ can refer to people, animals, or things and can act as a subject, making it suitable for "dog." $which$ also refers to animals or things but is often used in non-restrictive clauses, and $that$ is generally preferred for restrictive clauses like this one. $who$ is exclusively for people, so it's incorrect for a dog. $whom$ is the objective case for people, making it incorrect here. Therefore, $that$ is the most appropriate choice to complete the sentence: "Suresh's dog is the one $that$ was hurt in the stampede."

The question asks us to identify the grammatically INCORRECT sentence. Let's analyze each option:
Option A: "They gave us the money back less the service charges of Three Hundred rupees." This is correct. "Less" here means "minus" or "after deducting", which is a standard usage.
Option B: "This country's expenditure is not less than that of Bangladesh." This is correct. "Expenditure" is an uncountable noun, and "less than" is correctly used for comparison of uncountable quantities.
Option C: "The committee initially asked for a funding of Fifty Lakh rupees, but later settled for a lesser sum." This is correct. "Lesser sum" is a standard construction meaning a smaller amount.
Option D: "This country's expenditure on educational reforms is very less." This sentence is grammatically incorrect. While "expenditure" is an uncountable noun for which "less" is the comparative form of "little", the phrase "very less" is non-standard in formal English. The correct construction for intensifying an uncountable noun would be "very little". Thus, "This country's expenditure on educational reforms is very little" would be the grammatically correct phrasing. is the incorrect sentence.

The word 'Mitigate' means to make something less severe, serious, painful, or harsh. Let's examine the options:

• A. Diminish: This means to make or become smaller, less, or weaker. This aligns directly with reducing the intensity or impact, making it a close synonym for 'Mitigate'.
• B. Divulge: This means to reveal private information. This is unrelated to 'Mitigate'.
• C. Dedicate: This means to devote oneself to a purpose. This is unrelated to 'Mitigate'.
• D. Denote: This means to be a sign or indication of something. This is unrelated to 'Mitigate'.

Comparing the meanings, 'Diminish' is the closest in meaning to 'Mitigate' because both words convey the idea of making something less intense or severe.

The key to solving this sentence completion question is understanding the word "Despite." This word tells us that the part of the sentence before the comma will describe something negative or challenging that was overcome by the mission's success.

1. "Despite" signals contrast: We need a word that represents an obstacle or difficulty.
2. Evaluating options:
   • A. attempts: Missions always involve attempts; succeeding "despite attempts" doesn't make logical sense as a contrast.
   • B. setbacks: This means difficulties, problems, or hindrances. Succeeding "despite setbacks" perfectly fits the contrast implied by "Despite."
   • C. meetings: Meetings are usually part of a mission, not obstacles to its success.
   • D. delegations: These are groups of people, not inherently something that would prevent a mission's success.
3. Conclusion: "Setbacks" is the only option that represents a challenge that a mission would overcome to still be considered successful. The completed sentence means the mission succeeded even though it faced many problems.

The advertisement seeks temporary, part-time individuals for a Field Interviewer role to conduct personal interviews and collect economic data. Requirements include a High School diploma and availability for Day, Evening, and Saturday work, with transportation paid and expenses reimbursed. We need to find the best inference from this text.

To evaluate the options:

1. Gender-discriminatory: The advertisement uses the neutral term "persons" and does not specify any gender requirements, such as "male applicants only" or "female applicants preferred." Therefore, this option is incorrect.
2. Xenophobic: There is no mention of nationality, citizenship, or origin requirements in the advertisement. Thus, this option is incorrect.
3. Not designed to make the post attractive: The advertisement explicitly states benefits like "Transportation paid" and "expenses reimbursed," which are typically offered to attract applicants. Therefore, this option is incorrect.
4. Not gender-discriminatory: As established above, the advertisement uses gender-neutral language and has no explicit or implicit gender-specific requirements or biases. This is consistent with the advertisement's wording.

Therefore, the most reasonable inference from the advertisement is that it is not gender-discriminatory.

This problem involves identifying the pattern in a sequence of letter pairs. We'll analyze the first and second letters separately using their alphabetical positions ($A=1, B=2, \dots, Z=26$).

1. Analyzing the First Letters:
The first letters are A, C, F, J.

• A is the $1^{st}$ letter.
• C is the $3^{rd}$ letter. The difference from A is $3 - 1 = 2$.
• F is the $6^{th}$ letter. The difference from C is $6 - 3 = 3$.
• J is the $10^{th}$ letter. The difference from F is $10 - 6 = 4$.
  The differences between consecutive first letters (2, 3, 4) are increasing by 1. Therefore, the next difference should be $4 + 1 = 5$.
  The next first letter will be the $10 + 5 = 15^{th}$ letter, which is O.

2. Analyzing the Second Letters:
The second letters are D, G, K, P.

• D is the $4^{th}$ letter.
• G is the $7^{th}$ letter. The difference from D is $7 - 4 = 3$.
• K is the $11^{th}$ letter. The difference from G is $11 - 7 = 4$.
• P is the $16^{th}$ letter. The difference from K is $16 - 11 = 5$.
  The differences between consecutive second letters (3, 4, 5) are also increasing by 1. Therefore, the next difference should be $5 + 1 = 6$.
  The next second letter will be the $16 + 6 = 22^{nd}$ letter, which is V.

Combining the next first letter (O) and the next second letter (V), the next term in the sequence is OV.

Let's break down how changing data in a list affects its mean and standard deviation.

Assertion P: Adding 7 to each entry adds 7 to the mean.
The mean ($\bar{x}$) is the sum of all entries divided by the count ($n$): $\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}$. If we add 7 to each entry, the new mean ($\bar{x}'$) becomes:
$\bar{x}' = \frac{\sum_{i=1}^{n} (x_i+7)}{n} = \frac{(\sum_{i=1}^{n} x_i) + (n \times 7)}{n} = \frac{\sum_{i=1}^{n} x_i}{n} + \frac{n \times 7}{n} = \bar{x} + 7$
Adding a constant to every value shifts the entire list, so the mean also shifts by that constant. Assertion P is CORRECT.

Assertion Q: Adding 7 to each entry adds 7 to the standard deviation.
Standard deviation ($\sigma$) measures how spread out the data is. Adding a constant to every value in a list moves the entire distribution but doesn't change how spread out the values are relative to each other. The variance ($\sigma^2$) is $\sigma^2 = \frac{\sum_{i=1}^{n} (x_i - \bar{x})^2}{n}$. For the new list with entries $x_i+7$ and mean $\bar{x}+7$, the new variance $(\sigma')^2$ is:
$(\sigma')^2 = \frac{\sum_{i=1}^{n} ((x_i+7) - (\bar{x}+7))^2}{n} = \frac{\sum_{i=1}^{n} (x_i - \bar{x})^2}{n} = \sigma^2$
Since the variance doesn't change, the standard deviation also remains unchanged ($\sigma' = \sigma$). Assertion Q is INCORRECT.

Assertion R: Doubling each entry doubles the mean.
If each entry $x_i$ is doubled, the new list is $2x_1, 2x_2, \dots, 2x_n$. The new mean ($\bar{x}'$) is:
$\bar{x}' = \frac{\sum_{i=1}^{n} (2x_i)}{n} = \frac{2 \sum_{i=1}^{n} x_i}{n} = 2 \bar{x}$
Multiplying every value by a constant scales the mean by that same constant. Assertion R is CORRECT.

Assertion S: Doubling each entry leaves the standard deviation unchanged.
If each entry $x_i$ is doubled, the new list has entries $2x_i$ and the new mean is $2\bar{x}$. The new variance $(\sigma')^2$ is:
$(\sigma')^2 = \frac{\sum_{i=1}^{n} (2x_i - 2\bar{x})^2}{n} = \frac{\sum_{i=1}^{n} (2(x_i - \bar{x}))^2}{n} = \frac{\sum_{i=1}^{n} 4(x_i - \bar{x})^2}{n} = 4 \frac{\sum_{i=1}^{n} (x_i - \bar{x})^2}{n} = 4 \sigma^2$
The new standard deviation is $\sigma' = \sqrt{4\sigma^2} = 2\sigma$. Doubling each entry doubles the standard deviation, it does not leave it unchanged. Assertion S is INCORRECT.

This problem asks us to find the probability that a reliable shock absorber came from supplier Y. This is a classic application of Bayes' Theorem.

Let's define our events and their probabilities:

• $X$: Shock absorber is from supplier X.
• $Y$: Shock absorber is from supplier Y.
• $R$: Shock absorber is reliable.

Given probabilities:

• $P(X) = 0.60$ (X supplies 60%)
• $P(Y) = 0.40$ (Y supplies 40%)
• $P(R|X) = 0.96$ (96% of X's are reliable)
• $P(R|Y) = 0.72$ (72% of Y's are reliable)

We want to find $P(Y|R)$, the probability that a shock absorber is from Y, given that it is reliable. Bayes' Theorem states:
$P(Y|R) = \frac{P(R|Y) \times P(Y)}{P(R)}$

First, we need to calculate the total probability of a randomly chosen shock absorber being reliable, $P(R)$. This is found using the law of total probability:
$P(R) = P(R|X) \times P(X) + P(R|Y) \times P(Y)$
Substitute the given values:
$P(R) = (0.96 \times 0.60) + (0.72 \times 0.40)$
$P(R) = 0.576 + 0.288$
$P(R) = 0.864$

Now, substitute $P(R)$ and the other given probabilities into Bayes' Theorem to find $P(Y|R)$:
$P(Y|R) = \frac{0.72 \times 0.40}{0.864}$
$P(Y|R) = \frac{0.288}{0.864}$
$P(Y|R) = \frac{1}{3}$
Converting this fraction to a decimal:
$P(Y|R) \approx 0.3333...$
Rounding to three decimal places, this is approximately $0.334$.

The equation for the arch's height is given as $y = 2x - 0.1x^2$. This is a quadratic equation, which represents a parabola. Since the coefficient of the $x^2$ term ($a = -0.1$) is negative, the parabola opens downwards, meaning its highest point (vertex) gives the maximum height. To find the x-coordinate of the vertex, we use the formula $x = -\frac{b}{2a}$. Here, $a = -0.1$ and $b = 2$.
$x = -\frac{2}{2 \times (-0.1)} = -\frac{2}{-0.2} = 10$
Now, substitute $x = 10$ back into the height equation to find the maximum height:
$y_{max} = 2(10) - 0.1(10)^2 = 20 - 0.1(100) = 20 - 10 = 10$
Thus, the maximum possible height of the arch is 10 meters.

The critical buckling load ($P_{cr}$) for a column is given by Euler's formula: $P_{cr} = \frac{\pi^2 E I}{(KL)^2}$. Here, $E$ is Young's modulus, $I$ is the minimum area moment of inertia, $L$ is the actual column length, and $K$ is the effective length factor, which depends on the end conditions. For both ends hinged, $K=1$, so $P_{hinged} = \frac{\pi^2 E I}{(1L)^2} = \frac{\pi^2 E I}{L^2}$. For both ends clamped, $K=0.5$, so $P_{clamped} = \frac{\pi^2 E I}{(0.5L)^2} = \frac{\pi^2 E I}{0.25 L^2} = 4 \times \frac{\pi^2 E I}{L^2}$. Therefore, the ratio $\frac{P_{clamped}}{P_{hinged}} = \frac{4 \times \frac{\pi^2 E I}{L^2}}{\frac{\pi^2 E I}{L^2}} = 4$.

For any surface, the total incident radiation energy is either absorbed, transmitted, or reflected. This fundamental principle is expressed by the energy balance equation: $\alpha + \tau + \rho = 1$, where $\alpha$ is absorptivity, $\tau$ is transmissivity, and $\rho$ is reflectivity. The question specifies an opaque surface. By definition, an opaque surface blocks all radiation from passing through it, meaning its transmissivity ($\tau$) is zero. Substituting $\tau = 0$ into the general energy balance equation gives $\alpha + 0 + \rho = 1$, which simplifies to $\alpha + \rho = 1$.

We need to find the LMTD for a counterflow heat exchanger. First, calculate the heat capacity rates ($C$) for both fluids using $C = m \cdot C_p$:

For water (hot fluid): $C_w = 0.5 \text{ kg/s} \times 4.18 \text{ kJ/kg.K} = 2.09 \text{ kJ/s.K}$
For air (cold fluid): $C_a = 2.09 \text{ kg/s} \times 1 \text{ kJ/kg.K} = 2.09 \text{ kJ/s.K}$

Since both heat capacity rates are equal, $C_{min} = C_{max} = 2.09 \text{ kJ/s.K}$, which means the heat capacity ratio $C_r = \frac{C_{min}}{C_{max}} = \frac{2.09}{2.09} = 1$.

For a counterflow heat exchanger with $C_r = 1$, the effectiveness ($\epsilon$) is related to the Number of Transfer Units (NTU) by the simplified formula:
$\epsilon = \frac{NTU}{1 + NTU}$
Given $\epsilon = 0.8$, we can solve for NTU:
$0.8 = \frac{NTU}{1 + NTU}$
$0.8(1 + NTU) = NTU$
$0.8 + 0.8 NTU = NTU$
$0.2 NTU = 0.8 \implies NTU = 4$
Finally, the LMTD can be calculated using the relation between effectiveness, overall temperature difference, and NTU:
$LMTD = \frac{\epsilon \cdot (T_{h,\in} - T_{c,\in})}{NTU}$
Substitute the known values:
$LMTD = \frac{0.8 \times (80^\circ C - 30^\circ C)}{4}$
$LMTD = \frac{0.8 \times 50^\circ C}{4}$
$LMTD = \frac{40^\circ C}{4}$
$LMTD = 10^\circ C$

Here's how to find the view factor $F_{22}$ for concentric tubes:

First, recognize that for the inner tube (surface 1), $F_{11} = 0$ because it's an infinitely long tube and cannot "see" itself. Therefore, all radiation from the inner tube must go to the outer tube, so $F_{12} = 1$. Next, use the reciprocity relation: $A_1 F_{12} = A_2 F_{21}$. Since $F_{12} = 1$, we get $F_{21} = \frac{A_1}{A_2}$. As the view factors for infinitely long cylinders depend only on diameter ratios, $F_{21} = \frac{D_1}{D_2}$. Finally, the sum of view factors from surface 2 must be 1, meaning $F_{21} + F_{22} = 1$. Solving for $F_{22}$ gives $F_{22} = 1 - F_{21} = 1 - \frac{D_1}{D_2}$.

We need to evaluate $\lim_{x \rightarrow 0} \left(\frac{1-\cos x}{x^2}\right)$. Direct substitution of $x=0$ yields $\frac{1-\cos(0)}{0^2} = \frac{1-1}{0} = \frac{0}{0}$, which is an indeterminate form. Since we have the $\frac{0}{0}$ form, we apply L'Hôpital's Rule, which states that $\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$.
Applying L'Hôpital's Rule once:
$\lim_{x \rightarrow 0} \left(\frac{1-\cos x}{x^2}\right) = \lim_{x \rightarrow 0} \left(\frac{\frac{d}{dx}(1-\cos x)}{\frac{d}{dx}(x^2)}\right) = \lim_{x \rightarrow 0} \left(\frac{\sin x}{2x}\right)$
Again, direct substitution of $x=0$ into $\frac{\sin x}{2x}$ gives $\frac{\sin 0}{2(0)} = \frac{0}{0}$, another indeterminate form. So, we apply L'Hôpital's Rule a second time:
$\lim_{x \rightarrow 0} \left(\frac{\sin x}{2x}\right) = \lim_{x \rightarrow 0} \left(\frac{\frac{d}{dx}(\sin x)}{\frac{d}{dx}(2x)}\right) = \lim_{x \rightarrow 0} \left(\frac{\cos x}{2}\right)$
Now, substituting $x=0$ into this simplified expression gives $\frac{\cos(0)}{2} = \frac{1}{2}$. Thus, the limit is $\frac{1}{2}$.

To find the unit outward normal vector for the sphere $x^2 + y^2 +z^2 =1$ at the point $P\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}},0\right)$, we first define the surface function $F(x, y, z) = x^2 + y^2 + z^2$. The gradient of this function, $\nabla F$, gives a vector normal to the surface.
$\nabla F = \frac{\partial F}{\partial x}\hat{i} + \frac{\partial F}{\partial y}\hat{j} + \frac{\partial F}{\partial z}\hat{k} = 2x\hat{i} + 2y\hat{j} + 2z\hat{k}$
Next, we evaluate the gradient at the given point $P\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}},0\right)$:
$\nabla F \bigg|_P = 2\left(\frac{1}{\sqrt{2}}\right)\hat{i} + 2\left(\frac{1}{\sqrt{2}}\right)\hat{j} + 2(0)\hat{k} = \sqrt{2}\hat{i} + \sqrt{2}\hat{j}$
Since the surface is $x^{2+}y^{2+}z^2=1$ (a sphere centered at the origin), this gradient vector at point $P$ naturally points outward. To find the unit normal vector, we normalize this gradient vector. First, calculate its magnitude:
$|\nabla F \bigg|_P| = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2 + 0^2} = \sqrt{2+2} = 2$
Finally, divide the normal vector by its magnitude to get the unit normal vector:
$\vec{N} = \frac{\nabla F \bigg|_P}{|\nabla F \bigg|_P|} = \frac{\sqrt{2}\hat{i} + \sqrt{2}\hat{j}}{2} = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j}$
Thus, the unit outward normal vector at the given point is $\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j}$.

Let's break down the analysis of $f(x) = |x|$ at $x=0$ into two parts: continuity and differentiability.

Continuity Analysis of $f(x) = |x|$ at $x=0$
A function $f(x)$ is continuous at a point $a$ if three conditions are met: $f(a)$ is defined, $\lim_{x \to a} f(x)$ exists, and $\lim_{x \to a} f(x) = f(a)$.
For $f(x) = |x|$ at $a=0$:

1. Function Value: $f(0) = |0| = 0$. This is defined.
2. Limit:
   • Left-hand limit: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} |x| = \lim_{x \to 0^-} (-x) = 0$.
   • Right-hand limit: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} |x| = \lim_{x \to 0^+} x = 0$.
     Since the left-hand and right-hand limits are equal, $\lim_{x \to 0} f(x) = 0$.
3. Comparison: We see that $\lim_{x \to 0} f(x) = 0 = f(0)$.
   Since all conditions are satisfied, $f(x) = |x|$ is continuous at $x=0$.

Differentiability Analysis of $f(x) = |x|$ at $x=0$
A function $f(x)$ is differentiable at a point $a$ if the limit of the difference quotient exists:
$\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$
For $f(x) = |x|$ at $a=0$, we need to evaluate:
$\lim_{h \to 0} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0} \frac{|h| - 0}{h} = \lim_{h \to 0} \frac{|h|}{h}$
Now, let's evaluate the one-sided limits for this expression:

• Left-Hand Derivative (LHD): As $h \to 0^-$, $h$ is negative, so $|h| = -h$.
  $\lim_{h \to 0^-} \frac{|h|}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1$
• Right-Hand Derivative (RHD): As $h \to 0^+$, $h$ is positive, so $|h| = h$.
  $\lim_{h \to 0^+} \frac{|h|}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1$
  Since LHD $\ne$ RHD ($-1 \ne 1$), the limit of the difference quotient does not exist. Therefore, $f(x) = |x|$ is non-differentiable at $x=0$.

Conclusion: At the point $x=0$, the function $f(x) = |x|$ is continuous and non-differentiable.

To understand the behavior of $f(x)=x^{2+}1$ at $x=0$, we first find its derivatives. The first derivative is $f'(x) = \frac{d}{dx}(x^2 + 1) = 2x$. The second derivative is $f''(x) = \frac{d}{dx}(2x) = 2$.

Now, we evaluate these at $x=0$. We get $f'(0) = 2(0) = 0$, which means $x=0$ is a critical point. For the second derivative, $f''(0) = 2$.

Let's check the options:

• Maximum/Minimum Value: Since $f'(0)=0$ and $f''(0)=2 > 0$, the function has a local minimum at $x=0$. So, options for maximum or minimum value are related to this.
• Singularity: $f(x)=x^{2+}1$ is a polynomial, which means it's smooth and continuous everywhere, so it has no singularities.
• Point of Inflection: A standard point of inflection occurs where the concavity changes, meaning $f''(x)$ must change its sign. Here, $f''(x)=2$ is always positive and never changes sign. This indicates the function is always concave upwards, so $x=0$ is not a standard point of inflection. Based on the provided correct answer, we select Option D.

To find the area enclosed between two curves, we integrate the difference between the upper and lower functions over the interval defined by their intersection points.

First, find the intersection points of $y = x$ and $y = x^2$ by setting them equal:
$x = x^2$
$x^2 - x = 0$
$x(x - 1) = 0$
This gives $x = 0$ and $x = 1$. These are our limits of integration.

Next, determine which function is "on top" in the interval $(0, 1)$. Let's test $x = 0.5$:
For $y = x$, $y = 0.5$.
For $y = x^2$, $y = (0.5)^2 = 0.25$.
Since $0.5 > 0.25$, the line $y = x$ is the upper function and the parabola $y = x^2$ is the lower function.

Now, set up and evaluate the definite integral:
$A = \int_{0}^{1} (x - x^2) \, dx$
$A = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1}$
$A = \left( \frac{1^2}{2} - \frac{1^3}{3} \right) - \left( \frac{0^2}{2} - \frac{0^3}{3} \right)$
$A = \left( \frac{1}{2} - \frac{1}{3} \right) - (0 - 0)$
$A = \frac{3}{6} - \frac{2}{6}$
$A = \frac{1}{6}$

For a thin-walled spherical shell under internal pressure ($p$), the circumferential (hoop) stress is given by $\sigma_h = \frac{pR}{2t}$, where $R$ is the radius and $t$ is the thickness.
Initially, $\sigma_{h1} = \frac{pR_1}{2t_1}$.
When the radius increases by 1% ($R_2 = 1.01 R_1$) and the thickness reduces by 1% ($t_2 = 0.99 t_1$), with constant pressure $p$, the new hoop stress is
$\sigma_{h2} = \frac{p R_2}{2t_2} = \frac{p (1.01 R_1)}{2 (0.99 t_1)} = \frac{1.01}{0.99} \left( \frac{pR_1}{2t_1} \right) = \frac{1.01}{0.99} \sigma_{h1}$
The percentage change in hoop stress is calculated as:
$\text{Percentage Change} = \left( \frac{\sigma_{h2}}{\sigma_{h1}} - 1 \right) \times 100\% = \left( \frac{1.01}{0.99} - 1 \right) \times 100\%$
$\text{Percentage Change} = (1.020202... - 1) \times 100\% \approx 0.020202 \times 100\% \approx 2.02\%$

To find the maximum deflection of a cantilever beam with a moment $M$ at its free end, we start with the fundamental differential equation for beam deflection: $EI \frac{d^2y}{dx^2} = M(x)$. For this specific case, the bending moment $M(x)$ along the beam is constant and equal to $M$. Integrating this equation twice and applying the boundary conditions (zero deflection $y=0$ and zero slope $\frac{dy}{dx}=0$ at the fixed end, $x=0$), we arrive at the standard formula for the maximum deflection $\delta_{max}$, which occurs at the free end. The maximum deflection is given by:

$\delta_{max} = \frac{ML^2}{2EI}$

Here, $M$ is the applied moment, $L$ is the length of the beam, $E$ is Young's modulus, and $I$ is the moment of inertia. Comparing this with the given options, the correct magnitude of the maximum deflection is $\frac{ML^2}{2EI}$.

We need to find the kinetic energy of a rotating solid disc, which is given by the formula $KE = \frac{1}{2} I \omega^2$. First, calculate the moment of inertia $I$ for a solid disc using $I = \frac{1}{2} m r^2$. Given mass $m = 20 \text{ kg}$ and radius $r = 200 \text{ mm} = 0.2 \text{ m}$, we get:
$I = \frac{1}{2} \times 20 \text{ kg} \times (0.2 \text{ m})^2 = 10 \times 0.04 \text{ kg} \cdot \text{m}^2 = 0.4 \text{ kg} \cdot \text{m}^2$
Next, convert the rotational speed $N = 600 \text{ rpm}$ to angular velocity $\omega$ in $\text{rad/s}$ using $\omega = \frac{2 \pi N}{60}$:
$\omega = \frac{2 \pi \times 600}{60} \text{ rad/s} = 2 \pi \times 10 \text{ rad/s} = 20 \pi \text{ rad/s}$
Finally, substitute $I$ and $\omega$ into the kinetic energy formula:
$KE = \frac{1}{2} I \omega^2 = \frac{1}{2} \times (0.4 \text{ kg} \cdot \text{m}^2) \times (20 \pi \text{ rad/s})^2$
$KE = 0.2 \times (400 \pi^2) \text{ J} = 80 \pi^2 \text{ J}$
Using $\pi \approx 3.14159$, we calculate the numerical value:
$KE \approx 80 \times (3.14159)^2 \approx 80 \times 9.8696 \approx 789.568 \text{ J}$
Rounding to the nearest integer, the kinetic energy is approximately 790 Joules.

To find the angular velocity of the follower link, we first determine the linear velocity at the contact point ($V_c$) using the eccentric disc's angular velocity. The formula for this is $V_c = r \cdot \omega$, where $r = 25 \, \text{mm} = 0.025 \, \text{m}$ and $\omega = 1 \, \text{rad/s}$. This gives $V_c = 0.025 \times 1 = 0.025 \, \text{m/s}$. This same linear velocity applies to the follower at its contact point. We can then relate this linear velocity to the follower's angular velocity ($\omega_f$) using $V_c = r_f \cdot \omega_f$, where $r_f = 45 \, \text{mm} = 0.045 \, \text{m}$. Solving for $\omega_f$, we get $\omega_f = \frac{V_c}{r_f} = \frac{0.025}{0.045} = \frac{5}{9} \approx 0.1 \, \text{rad/s}$.

This problem asks us to calculate the power output of an adiabatic turbine. Since kinetic and potential energy changes are negligible, we can use the steady flow energy equation.

First, we need to find the enthalpy of the steam at the turbine outlet ($h_{out}$). It's a saturated mixture at 15 kPa with a quality ($x_{out}$) of 0.9.
$h_{out} = h_f + x_{out} (h_g - h_f)$
Given $h_f = 225.94 \text{ kJ/kg}$, $h_g = 2598.3 \text{ kJ/kg}$, and $x_{out} = 0.9$:
$h_{out} = 225.94 \text{ kJ/kg} + 0.9 \times (2598.3 \text{ kJ/kg} - 225.94 \text{ kJ/kg})$
$h_{out} = 225.94 \text{ kJ/kg} + 0.9 \times (2372.36 \text{ kJ/kg})$
$h_{out} = 225.94 \text{ kJ/kg} + 2135.124 \text{ kJ/kg}$
$h_{out} = 2361.064 \text{ kJ/kg}$
Now, we calculate the power output ($P$) using the steady flow energy equation for an adiabatic turbine:
$P = \dot{m} (h_{\in} - h_{out})$
Given $\dot{m} = 10 \text{ kg/s}$ and inlet enthalpy $h_{\in} = 3251.0 \text{ kJ/kg}$:
$P = 10 \text{ kg/s} \times (3251.0 \text{ kJ/kg} - 2361.064 \text{ kJ/kg})$
$P = 10 \text{ kg/s} \times (889.936 \text{ kJ/kg})$
$P = 8899.36 \text{ kJ/s}$
Since $1 \text{ kJ/s} = 1 \text{ kW}$, the power output is $8899.36 \text{ kW}$.
To convert this to Megawatts (MW), we divide by 1000:
$P (\text{MW}) = \frac{8899.36 \text{ kW}}{1000}$
$P = 8.89936 \text{ MW}$
Rounding to one decimal place, the power output is approximately 8.9 MW.

This problem asks for the entropy change ($\Delta S$) of an ideal gas during a reversible isothermal process.

The key concept for entropy change in a reversible process is $\Delta S = \frac{Q_{rev}}{T}$. For an ideal gas undergoing an isothermal process ($\Delta T = 0$), the internal energy change is $\Delta U = 0$. From the First Law of Thermodynamics, $\Delta U = Q_{rev} - W$, which simplifies to $Q_{rev} = W$. Therefore, $\Delta S = \frac{W}{T_1}$.

The work done ($W$) during a reversible isothermal process for $n$ moles of an ideal gas is given by:
$W = nRT_1 \ln \left( \frac{V_2}{V_1} \right)$
Using the ideal gas law ($PV = nRT$), for a constant temperature $T_1$, we have $P_1V_1 = P_2V_2$. This implies $\frac{V_2}{V_1} = \frac{P_1}{P_2}$.
Substituting this volume ratio into the work equation:
$W = nRT_1 \ln \left( \frac{P_1}{P_2} \right)$
Now, we can find the entropy change $\Delta S$:
$\Delta S = \frac{W}{T_1} = \frac{nRT_1 \ln \left( \frac{P_1}{P_2} \right)}{T_1}$
$\Delta S = nR \ln \left( \frac{P_1}{P_2} \right)$
The options use '$m$' instead of '$n$' (moles). Assuming '$m$' represents the number of moles as per the options, the entropy change is:
$\Delta S = mR \ln \left( \frac{P_1}{P_2} \right)$
The heat loss $Q$ mentioned in the question is $Q_{rev}$, but the derived formula for $\Delta S$ directly depends on initial and final states ($P_1$, $P_2$) and gas properties.

Fin effectiveness ($\eta_f$) compares the heat transferred by a real fin to an ideal fin at base temperature. It's influenced by material properties (high thermal conductivity $k$ is good), convection (low $h$ is good), and geometry. Specifically, $\eta_f \propto \sqrt{\frac{k A_c}{h P}}$, where $A_c$ is the cross-sectional area and $P$ is the perimeter. While increasing thickness generally increases the $A_c/P$ ratio, closely spaced fins, even if thin, can significantly reduce the effective convective heat transfer coefficient ($h$) due to boundary layer interference. This reduction in $h$ can be so pronounced that it leads to the highest overall fin effectiveness for thin, closely spaced fins compared to other configurations, by minimizing internal thermal resistance relative to external convective resistance.

To find the head loss due to friction, we'll use the Darcy-Weisbach equation. First, we need the pipe's cross-sectional area and the oil's velocity.

1. Calculate Pipe Cross-Sectional Area ($A$):
   The pipe diameter ($D$) is 200 mm, which is $0.2 \text{ m}$.
   $A = \pi \frac{D^2}{4} = \pi \frac{(0.2 \text{ m})^2}{4} = 0.01\pi \text{ m}^2$
   Numerically, $A \approx 0.031416 \text{ m}^2$.

2. Calculate Average Flow Velocity ($V$):
   The volumetric flow rate ($Q$) is $0.2 \text{ m}^3/s$.
   $V = \frac{Q}{A} = \frac{0.2 \text{ m}^3/s}{0.01\pi \text{ m}^2} = \frac{20}{\pi} \text{ m/s}$
   Numerically, $V \approx 6.366 \text{ m/s}$.

3. Calculate Head Loss ($h_f$) using Darcy-Weisbach Equation:
   The Darcy-Weisbach equation is $h_f = f \frac{L}{D} \frac{V^2}{2g}$.
   Given values: $f = 0.0225$, $L = 500 \text{ m}$, $D = 0.2 \text{ m}$, $g = 9.81 \text{ m/s}^2$.
   $h_f = (0.0225) \times \frac{500 \text{ m}}{0.2 \text{ m}} \times \frac{\left(\frac{20}{\pi} \text{ m/s}\right)^2}{2 \times 9.81 \text{ m/s}^2}$
   $h_f = 0.0225 \times 2500 \times \frac{\frac{400}{\pi^2}}{19.62}$
   $h_f = 56.25 \times \frac{400}{19.62 \times \pi^2}$
   $h_f \approx 56.25 \times \frac{400}{193.418}$
   $h_f \approx 56.25 \times 2.0680$
   $h_f \approx 116.18 \text{ m}$
   The head loss due to friction is approximately $116.18 \text{ m}$.