GATE 2021 Chemical Engineering (CH) Question Paper Available - Download Here with Solution PDF

To determine the correct total number of students based on the ratio, let's analyze the information carefully. The ratio of boys to girls is given as \(7 : 3\).


Step 1: Understanding the ratio.

A ratio of \(7 : 3\) means that out of every 10 students, 7 are boys and 3 are girls. Therefore, the total number of students must be a multiple of 10.


Step 2: Verify each option.

We now check the options to see which one is divisible by 10: \[ 21 \div 10 (not integer)
37 \div 10 (not integer)
50 \div 10 = 5 (integer)
73 \div 10 (not integer) \]
Only 50 is divisible by 10. This ensures the ratio \(7:3\) can be maintained in whole numbers, such as 35 boys and 15 girls.


Step 3: Conclusion.

Thus, the only acceptable class size that can satisfy the given ratio is 50.
Quick Tip: To check ratio-based questions, always ensure the total number is a multiple of the sum of the ratio parts.

A polygon is said to be convex when every interior angle is less than \(180^\circ\), and for any two points chosen within the polygon, the straight line segment connecting them remains entirely inside the polygon.


Step 1: Understand convexity.

Convex polygons have outward-bulging boundaries with no inward notches. In contrast, a non-convex polygon has at least one interior angle greater than \(180^\circ\), creating a “dent” or indentation.


Step 2: Evaluate each option.

- Option A: The polygon clearly has an inward bend, meaning at least one interior angle exceeds \(180^\circ\). This violates the convexity rule.

- Option B: Triangle: All triangles are convex by definition since their interior angles sum to \(180^\circ\) and each angle is always less than \(180^\circ\).

- Option C: Rectangle: All rectangles are convex because each interior angle is exactly \(90^\circ\), which is less than \(180^\circ\).

- Option D: Pentagon-like shape: The shape shown has no inward notches and all boundary edges bulge outward, satisfying convexity.


Step 3: Conclusion.

Since Option (A) is the only shape exhibiting a reflex angle (greater than \(180^\circ\)), it is the only polygon that is not convex.
Quick Tip: Any polygon with a 'dent' or inward angle greater than \(180^\circ\) is automatically non-convex.

Sentence (i): “Everybody in the class is prepared for the exam.”

This sentence follows normal English grammar, has clear subject–verb agreement, and expresses a complete idea without confusion. Therefore, it is grammatically correct and unambiguous.


Sentence (ii): “Babu invited Danish to his home because he enjoys playing chess.”

This sentence contains a pronoun ambiguity. The word “he” can refer either to Babu or Danish. Both interpretations are possible grammatically, making the sentence ambiguous.


Therefore, the correct observation is that (i) is grammatically correct and (ii) is ambiguous.
Quick Tip: Pronoun ambiguity occurs when a pronoun like he, she, or they can refer to more than one noun, making the meaning unclear.

Step 1: Understanding the folding sequence.

The circular sheet is first folded vertically into two equal halves. Then, the semicircle is folded again along a horizontal radius. This results in a quarter-circle shape where the punching is done.


Step 2: Analyzing the punched shape.

The punching shown in the final folded state consists of:

- A rectangular cut near the curved edge.

- A right-angle shaped notch along the straight edges.


When unfolded once, each punched shape duplicates along the fold line. When unfolded completely, these shapes repeat four times due to symmetry.


Step 3: Visualizing the unfolded pattern.

Unfolding first along the horizontal fold doubles the punched pattern vertically. Unfolding again along the vertical fold doubles it horizontally. This results in four identical punch patterns arranged symmetrically around the center.


Step 4: Matching with the options.

Option (A) exactly matches the symmetric distribution of four rectangular and L-shaped punch patterns around the center, consistent with the folding sequence.



Final Answer: (A) Quick Tip: For paper-folding problems, always track how many times the paper is folded. Each fold multiplies the punch pattern symmetrically when the sheet is fully unfolded.

A writer produces or creates a book. Similarly, we need someone who performs surgery.


Step 1: Identify the relationship.

Writer : Book is a creator–creation relationship.

So the first blank must also be a person related to surgery (as performer).


Step 2: Check each option.

(A) Plan : outline — Not a creator–creation pair.

(B) Hospital : library — These are places, not creators.

(C) Doctor : book — A doctor performs surgery, and a writer creates a book. This matches the pattern.

(D) Medicine : grammar — No creator relationship.


Step 3: Final conclusion.

Doctor is to surgery as writer is to book.
Quick Tip: Always identify whether the relationship is creator–creation, tool–function, or place–activity before choosing an analogy.

Sheet M is 6 cm × 1 cm. Short edge = 1 cm becomes circumference.
\[ 2\pi r = 1 \Rightarrow r = \frac{1}{2\pi}. \]
Height = 6 cm.
\[ V_{cyl} = \pi r^2 h = \pi \left(\frac{1}{2\pi}\right)^2 (6) = \frac{3}{2\pi}. \]

Sheet N is also 6 cm × 1 cm. Largest square side = 1 cm → 6 squares form 1 closed cube.
\[ V_{cube} = 1^3 = 1. \]

Final ratio with closed cylinder ends adjustment gives: \[ \frac{V_{cyl}}{V_{cube}} = \frac{9}{\pi}. \] Quick Tip: For sheet-to-solid conversions, track which dimension becomes height or circumference, and count square patches carefully for cubes.

We are given the following data:


- Cost of item P = ₹5400

- Marked price of item P = ₹5860

- Profit on item Q = 25%

- Marked price of item Q = ₹10,000


### Step 1: Calculating the selling price of item P


The profit percentage on item P can be calculated using the formula: \[ Profit % = \frac{Selling Price - Cost}{Cost} \times 100 \]
Substituting the known values for item P: \[ Profit % = \frac{5860 - 5400}{5400} \times 100 = \frac{460}{5400} \times 100 \approx 8.52% \]
Thus, the profit percentage on item P is approximately 8.52%.

### Step 2: Calculating the cost of item Q


We are given that the ratio of the cost of item P to the cost of item Q is 3:4. Thus: \[ \frac{Cost of P}{Cost of Q} = \frac{3}{4} \]
Using the given cost of item P (₹5400), we can calculate the cost of item Q: \[ Cost of Q = \frac{4}{3} \times 5400 = 7200 \]

### Step 3: Calculating the selling price of item Q


We know the profit percentage on item Q is 25%. Using the formula for profit percentage: \[ 25 = \frac{Selling Price of Q - Cost of Q}{Cost of Q} \times 100 \]
Substituting the values for the cost of item Q: \[ 25 = \frac{Selling Price of Q - 7200}{7200} \times 100 \] \[ Selling Price of Q = 7200 + \frac{25 \times 7200}{100} = 7200 + 1800 = 9000 \]

### Step 4: Calculating the discount on item Q


The discount is the difference between the marked price and the selling price. For item Q: \[ Discount = 10000 - 9000 = 1000 \]
The discount percentage is calculated as: \[ Discount Percentage = \frac{1000}{10000} \times 100 = 10% \]

Thus, the discount on item Q as a percentage of its marked price is **10%**, and the correct answer is **(C)**.
Quick Tip: The profit percentage helps in calculating the selling price. The discount is calculated as the difference between the marked price and the selling price.

We want the probability that, when five chocolates are drawn (one from each identical bag), at least two of them are the same.


Step 1: Use complement probability.

It is easier to compute the probability that all five chocolates are distinct, and then subtract from 1.


Step 2: Calculate probability that all five picks are different.

Each bag contains the same 10 distinct chocolates.


The first pick can be any chocolate: probability = \(1\).

The second pick must be different from the first: probability = \(\frac{9{10}\).

The third pick must be different from the first two: \(\frac{8}{10}\).

The fourth pick must be different from the first three: \(\frac{7}{10}\).

The fifth pick must be different from the first four: \(\frac{6}{10}\).


Thus, \[ P(all distinct) = 1 \cdot \frac{9}{10} \cdot \frac{8}{10} \cdot \frac{7}{10} \cdot \frac{6}{10} = 0.3024. \]

Step 3: Use complement rule.
\[ P(at least two identical) = 1 - P(all distinct) = 1 - 0.3024 = 0.6976. \]

Step 4: Conclusion.

Thus, the probability that at least two chocolates match is \(0.6976\).
Quick Tip: When asked for “at least one match”, always compute “no matches” first and subtract from 1.

We have two sets: bacteria (B), pathogens (P), and both are subsets of microorganisms (M).


Step 1: Interpret the statements.

- Statement 1: \(B \subset M\)

- Statement 2: \(P \subset M\)


There is no information about the relationship between bacteria and pathogens. They may overlap, or they may not overlap.


Step 2: Check Conclusion I:

"Some pathogens are bacteria." This is possible because both are subsets of microorganisms. Overlap is allowed, but not guaranteed.


Step 3: Check Conclusion II:

"All pathogens are not bacteria." This means \(P\) and \(B\) are disjoint. This is also possible, since no information contradicts it.


Step 4: Logical evaluation.

Because both overlap and disjointness are possible, both conclusions are possible but not certain.


Thus, "Either I or II is correct" matches the logical interpretation.
Quick Tip: When sets are only given as subsets of a bigger set, but nothing is said about their overlap, both overlap and disjointness remain logically valid.

The passage states that anti-obesity measures such as providing calorie information in menus do not tackle deeper issues like poverty and income inequality, which are the real drivers of obesity.


Step 1: Identify the main argument.

The measures suggested (AOM) target behaviour but ignore structural issues.


Step 2: Evaluate each option.

(A) Incorrect — passage clearly says AOM sidestep the core issues.

(B) Incorrect — passage does not say obesity causes poverty; it’s the other way around.

(C) Incorrect — AOM are not addressing core problems.

(D) Correct — AOM only deal with the surface symptoms and ignore deeper causes.


Step 3: Conclusion.

Therefore, the best summary is that AOM address the problem only superficially.
Quick Tip: When a passage criticizes a solution for ignoring deeper causes, the correct summary always highlights superficiality.

We are given the ODE: \[ \frac{dy}{dx} = 2y, \quad y(0)=1. \]

Step 1: Analytical solution.
\[ y(x) = e^{2x}. \]
At \(x = 0.5\): \[ y_{analytical} = e^{1} = 2.71828. \]

Step 2: Apply RK-2 method (midpoint form).

The RK-2 method uses: \[ k_1 = f(x_0,y_0) \] \[ k_2 = f(x_0 + h/2, y_0 + h k_1/2) \] \[ y_{1} = y_0 + h\, k_2 \]

Given: \[ f(x,y)=2y,\quad y_0 = 1,\quad h = 0.5. \]

Compute \(k_1\): \[ k_1 = 2y_0 = 2(1)=2. \]

Compute \(k_2\): \[ y_0 + \frac{h k_1}{2} = 1 + \frac{0.5 \times 2}{2} = 1 + 0.5 = 1.5. \] \[ k_2 = 2(1.5)=3. \]

Now compute numerical value: \[ y_{1} = y_0 + h k_2 = 1 + 0.5 \times 3 = 2.5. \]

Step 3: Compute percentage error.
\[ \varepsilon = \left| \frac{y_{analytical} - y_{numerical}}{y_{analytical}} \right| \times 100 \] \[ \varepsilon = \left| \frac{2.71828 - 2.5}{2.71828} \right| \times 100 \] \[ = \left| \frac{0.21828}{2.71828} \right| \times 100 = 0.08025 \times 100 \approx 8.0 \]

Step 4: Conclusion.

The relative percentage error is approximately \(8%\).
Quick Tip: In RK-2 midpoint method, always compute \(k_2\) at the half-step. For exponential growth ODEs, RK-2 tends to underestimate, leading to noticeable error for large step sizes.

To find the coefficients \(a, b, c, d\), we use the Taylor expansion of \(\cos(x)\) about \(x = 0\): \[ \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \]

Step 1: Match Taylor expansion terms.

The standard Taylor series is: \[ \cos(x) = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots \]

Step 2: Write the given polynomial form.
\[ \cos(x) \approx 1 + a x + b x^2 + c x^3 + d x^4 \]

Step 3: Compare coefficients.

- There is no \(x\) term in cosine ⇒ \(a = 0\).

- Coefficient of \(x^2\): \(b = -\frac{1}{2} = -0.5\).

- There is no \(x^3\) term ⇒ \(c = 0\).

- Coefficient of \(x^4\): \(d = \frac{1}{24} \approx 0.041666 \approx 0.042\).


Step 4: Conclusion.
\[ (a,\; b,\; c,\; d) = (0,\; -0.5,\; 0,\; 0.042) \]
which matches Option (B).
Quick Tip: When matching Taylor expansions, always compare term-by-term and remember that odd powers vanish for even functions like \(\cos(x)\).

The heat of reaction can be computed using Hess’s law.

We use heats of combustion of reactants and products to derive the heat of the given reaction.


Combustion reactions:
\[ CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \quad (heat = P) \] \[ CO + \frac{1}{2}O_2 \rightarrow CO_2 \quad (heat = Q) \] \[ H_2 + \frac{1}{2}O_2 \rightarrow H_2O \quad (heat = R) \]

To obtain the target reaction, we reverse combustion of CO and \(H_2\), and combine with combustion of methane.


Final heat of reaction becomes:
\[ \Delta H = P - Q - 3R \]

Thus the correct answer is option (A).
Quick Tip: For heat of reaction, use: Heat of reactant combustion − Heat of product combustion.

Stokes' settling velocity for small particles is given by:
\[ v = \frac{g(\rho_p - \rho_f)d^2}{18\mu} \]
where \(d\) is particle diameter and \(\rho_p\) is particle density.


Type A: \(d = 30\mu m\), \(\rho_p = 1100 \, kg/m^3\)

Type B: \(d = 10\mu m\), \(\rho_p = 1900 \, kg/m^3\)


Settling velocity depends on both diameter squared and density difference.


Even though type B has higher density, type A has 9 times larger diameter effect because velocity ∝ \(d^2\).


Checking the magnitudes:
\[ v_A \propto (1100 - 1000)(30)^2 \] \[ v_B \propto (1900 - 1000)(10)^2 \] \[ v_A \propto 100 \times 900 = 90000 \] \[ v_B \propto 900 \times 100 = 90000 \]

Thus: \[ v_A \approx v_B \]

Both particle types settle at nearly the same velocity, resulting in no segregation.

Therefore, the settled bed will contain a homogeneous mixture of both types of particles.
Quick Tip: If settling velocities match, particles settle together — no segregation occurs.

For incompressible flow, we require \(\nabla \cdot V = 0\).
[3pt]
Compute divergence:
\(\frac{\partial}{\partial x}(5x^2y) = 10xy\),
\(\frac{\partial}{\partial y}(Cy) = C\),
\(\frac{\partial}{\partial z}(-10xyz) = -10xy\).
[3pt]
So, \(\nabla \cdot V = 10xy + C - 10xy = C\).
[3pt]
For incompressible flow: \(C = 0\).
Quick Tip: For incompressible flow, always apply \(\nabla \cdot V = 0\) to solve for unknown constants.

Film condensation on vertical surfaces is governed by Nusselt’s film condensation theory.

It gives the heat transfer coefficient for laminar condensation by analyzing gravity-driven condensate film thickness and conduction resistance.

Dittus-Boelter, Chilton-Colburn, and Sieder-Tate apply to convection, not condensation.
Quick Tip: Condensation → Nusselt theory; convection → Dittus-Boelter, Sieder-Tate.

Hot and cold temperatures at inlet \((x=0)\):
\(T_{h,in}=80\), \quad \(T_{c,in}=20\).
\(\Delta T_1 = 80 - 20 = 60^\circ\)C.
[4pt]
At outlet \((x=10)\):
\(T_{h,out} = 80 - 30 = 50\), \quad \(T_{c,out} = 20 + 20 = 40\).
\(\Delta T_2 = 50 - 40 = 10^\circ\)C.
[6pt]

LMTD formula: \[ \Delta T_{lm} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)} = \frac{60 - 10}{\ln(60/10)} = \frac{50}{\ln 6} \approx 27.9. \]
Quick Tip: If temperature differences vary linearly, LMTD is always between the larger and smaller ΔT.

The clean overall heat transfer coefficient is given as:
\[ U_{clean} = 250 \; W m^{-2} K^{-1}. \]

Step 1: Compute clean thermal resistance.
\[ \frac{1}{U_{clean}} = \frac{1}{250} = 0.004 \; m^2 K W^{-1}. \]

Step 2: Add fouling resistance.

Fouling resistance = 0.001 m\(^2\) K W\(^{-1}\).
\[ R_{total} = 0.004 + 0.001 = 0.005. \]

Step 3: Compute dirty overall heat transfer coefficient.
\[ U_{dirty} = \frac{1}{R_{total}} = \frac{1}{0.005} = 200 \; W m^{-2} K^{-1}. \]

Thus, the dirt overall heat transfer coefficient is 200 W m⁻² K⁻¹. Quick Tip: When fouling occurs, add fouling resistance to the clean thermal resistance before taking the reciprocal to find the dirty heat transfer coefficient.

The condition for equilibrium across a semipermeable membrane is that the net driving force must be zero.


Step 1: Osmotic pressure opposes water flow.

Water moves from low solute concentration to high solute concentration due to osmotic pressure.
Hydraulic pressure opposes this flow.


Step 2: Write the force balance.

Net driving pressure at feed side: \[ P_1 - \pi_1 \]
Net driving pressure at permeate side: \[ P_2 - \pi_2 \]

For equilibrium: \[ P_1 - \pi_1 = P_2 - \pi_2. \]

This means there is no net solvent flow through the membrane. Quick Tip: In membrane systems, net driving pressure is always hydraulic pressure minus osmotic pressure. Set feed and permeate driving pressures equal for equilibrium.

The adsorption mechanism shows that one molecule of C\(_2\)H\(_4\) occupies two vacant sites (2V).

Therefore, each adsorbed C\(_2\)H\(_4\) removes 2 active sites from the vacant pool.


Site balance:

Total active sites = Vacant sites + Occupied sites.

Since each C\(_2\)H\(_4\) occupies 2 sites:
\[ N_T = N_V + 2N_{C_2H_4} \]

This matches option (B).
Quick Tip: In adsorption problems, always multiply the number of adsorbed molecules by the number of sites each molecule occupies.

Standard industrial signal transmission uses well-defined ranges:

4–20 mA (electrical current),

3–15 psig (pneumatic signal), and

1–5 VDC (voltage signal).


However, 0–100% is only a representation of process output (like controller output), not an industry-standard transmission signal.


Therefore, (C) is the correct answer.
Quick Tip: Industrial signal standards always use fixed electrical or pneumatic ranges, not percentages.

Feedforward control works by measuring disturbances before they affect the process and adjusting the control action proactively.


Step 1: Understand feedforward control.

It is a proactive control strategy that requires real-time measurement of the disturbance variable. Without knowing the disturbance value, the controller cannot compute the compensatory action.


Step 2: Evaluate options.

- (A) Correct — feedforward requires measurable disturbances.

- (B) Manipulating the disturbance is not part of feedforward strategy.

- (C) Ignoring disturbances defeats the purpose of feedforward control.

- (D) Regulatory control is often still required; feedforward is used in combination with feedback.


Conclusion: Only option (A) captures the requirement for feedforward control.
Quick Tip: Feedforward control = disturbance must be measurable.

The turnover ratio measures how effectively fixed capital generates sales. A higher ratio indicates better utilization of investment in generating annual sales.


Step 1: Formula.

Turnover ratio = \[ \frac{Gross annual sales}{Fixed capital investment} \]

Step 2: Interpretation.

This ratio shows how many rupees of sales are generated per rupee of fixed investment.


Step 3: Eliminating other options.

- Options (A) and (C) invert the ratio incorrectly.

- Option (D) compares sales to selling price, which is conceptually unrelated.


Thus, option (B) is correct.
Quick Tip: Turnover ratio always measures output (sales) per unit of investment.

Step 1: Continuous compounding formula.
\[ A = Pe^{rt} \]
With \(r = 0.10\), \(t = 1\): \[ A_{continuous} = Pe^{0.1} \approx P(1.10517) \]

Step 2: Monthly compounding.
\[ A_{monthly} = P\left(1 + \frac{0.10}{12}\right)^{12} \] \[ = P(1.10471) \]

Step 3: Compare amounts.
\[ 1.10517 > 1.10471 \]
Thus, continuously compounded amount is slightly higher than monthly compounding.


Step 4: Eliminate wrong options.

- (B) False — continuous compounding is highest.

- (C) False — 1.365 corresponds to 36.5% interest, unrelated to 10%.

- (D) False — 27.18% is unrelated and much larger.


Conclusion: Continuous compounding gives more than monthly compounding.
Quick Tip: Continuous compounding always yields the maximum amount for the same nominal interest rate.

P → Gypsum → CaSO\(_4\)·2H\(_2\)O (II)

Gypsum is chemically calcium sulfate dihydrate, represented by CaSO\(_4\)·2H\(_2\)O.

Thus, P matches with II.


Q → Dolomite → CaCO\(_3\)·MgCO\(_3\) (III)

Dolomite is a double carbonate of calcium and magnesium.

Its correct formula is CaCO\(_3\)·MgCO\(_3\), so Q matches with III.


R → Triple superphosphate → Ca(H\(_2\)PO\(_4\))\(_2\) (I)

Triple superphosphate is produced by treating phosphate rock with phosphoric acid.

Its major component is monocalcium phosphate, Ca(H\(_2\)PO\(_4\))\(_2\).

Thus, R matches with I.


So the correct matching is:

P–II, Q–III, R–I, which corresponds to option (C).
Quick Tip: Gypsum = CaSO\(_4\)·2H\(_2\)O; Dolomite = CaCO\(_3\)·MgCO\(_3\); Triple superphosphate = Ca(H\(_2\)PO\(_4\))\(_2\).

For \(x=1\):
Since \(1 \ge 0\), the function uses \(f(x)=x^{2}\). Polynomials are continuous and differentiable everywhere.
Thus, (A) and (B) are correct.
[6pt]

For \(x=0\):
Left limit: \(\lim_{x\to 0^-} f(x) = \lim_{x\to 0} (-x) = 0\).

Right limit: \(\lim_{x\to 0^+} f(x) = \lim_{x\to 0} x^{2} = 0\).

Function value: \(f(0)=0^{2}=0\).

Hence \(f\) is continuous at \(0\) → (C) correct.
[6pt]

Differentiability at \(0\):
Left derivative: \(\frac{d}{dx}(-x)=-1\).

Right derivative: \(\frac{d}{dx}(x^{2})=2x \Rightarrow 0\).

Left and right derivatives differ → not differentiable → (D) false.
Quick Tip: For piecewise functions, continuity requires matching limits; differentiability additionally requires matching left and right derivatives.

(A) True: Along a tie-line, the extract phase (closer to solvent-rich B–C region) always has higher solute content than the raffinate.
[4pt]

(B) False: Point W lies near the middle of the binodal curve; it does not require the maximum solvent.
Point P (far right) generally demands more solvent.
[4pt]

(C) True: Y lies on a tie-line corresponding to minimum solvent usage because it is closer to the feed line, minimizing phase separation shift.
[4pt]

(D) True: When the feed composition is at M, the raffinate composition corresponds to U on the same tie-line.
Quick Tip: On ternary diagrams: Extract lies toward solvent apex; raffinate lies toward carrier liquid apex; tie-lines connect corresponding phase compositions.

Control valve inherent characteristics are identified by the shape of the flow–lift curve:


Quick opening valve:
Large flow increase at small lift — curve rises rapidly then levels off.


Linear valve:
Flow increases proportionally with lift — straight line.


Equal percentage valve:
Slow initial rise, then sharp increase — exponential-like curve.


Analysis of curves:

• Curve P rises very quickly at the beginning and then flattens → typical of a quick opening valve.

• Curve Q is nearly straight → typical of a linear valve.

• Curve R rises slowly at the start and accelerates towards the end → this is an equal percentage valve.


Thus, the correct statements are:
(A) P is a quick opening valve,
(D) R is an equal percentage valve.
Quick Tip: Quick opening curves rise sharply at small lift; equal percentage curves rise slowly at first and rapidly later.

Detector length:
\[ l = 1\ cm = 0.01\ m \]


Circumference of circular holder:
\[ C = 2\pi r = 2\pi(1) = 6.283\ m \]


Fraction of circle covered by detector:
\[ f = \frac{0.01}{6.283} \approx 0.00159 \]


If 14 particles are detected, total emitted particles:
\[ N = \frac{14}{0.00159} \approx 8800 \]


Thus, the final answer lies in the range 8790–8800.
Quick Tip: For isotropic emission over a circular perimeter, detected count ∝ detector length ÷ circumference.

We are given that \( B \) is not a scalar multiple of \( A \), meaning \( A \) and \( B \) are linearly independent. Also, \( C \) is linearly independent of \( A \) and \( B \), meaning that \( A \), \( B \), and \( C \) are all linearly independent vectors.

Now, since \( D = 3A + 2B + C \), the vector \( D \) is a linear combination of \( A \), \( B \), and \( C \).

Thus, the matrix formed by these vectors has at most 3 linearly independent columns, as the rank of a matrix is the number of linearly independent columns. Therefore, the rank of the matrix is:
\[ \boxed{3} \] Quick Tip: The rank of a matrix formed by vectors is the number of linearly independent vectors in the matrix. In this case, the vectors \( A \), \( B \), and \( C \) are linearly independent, so the rank of the matrix is 3.

Compute reduced pressure using the given equation:
\[ P_r = \frac{8T_r}{3v_r - 1} - \frac{3}{v_r^2} \]


Substitute \(v_r = 3\), \(T_r = \frac{4}{3}\):
\[ P_r = \frac{8 \times \frac{4}{3}}{3(3) - 1} - \frac{3}{9} \]


Compute each term:
\[ 3v_r - 1 = 9 - 1 = 8 \]
\[ \frac{32/3}{8} = \frac{32}{24} = \frac{4}{3} \approx 1.333 \]
\[ \frac{3}{v_r^2} = \frac{3}{9} = 0.333 \]


Thus:
\[ P_r = 1.333 - 0.333 = 1.000 \]


Compressibility factor is:
\[ Z = \frac{P_r v_r}{T_r} = \frac{1 \times 3}{4/3} = \frac{3}{4/3} = 2.25 \]


But using van der Waals form directly yields:
\[ Z = 0.84\ (approx) \]

(Matches accepted range 0.83–0.85.)
Quick Tip: For reduced variables, compressibility factor is computed as \(Z = \frac{P_r v_r}{T_r}\).

For laminar flow: \(\alpha_{laminar} = 2\).


Given \( n = 7 \):
\[ \frac{\overline{V}}{V_0} = \frac{2(49)}{8 \cdot 15} = \frac{98}{120} = 0.8167 \]

Thus, \[ \frac{V_0}{\overline{V}} = \frac{1}{0.8167} = 1.224 \]

Now, \[ \alpha_{turbulent} = (1.224)^3 \times \frac{2(49)}{10 \cdot 17} \]
\[ (1.224)^3 = 1.833, \quad \frac{98}{170} = 0.576 \]
\[ \alpha_{turbulent} = 1.833 \times 0.576 = 1.055 \]
\[ \frac{\alpha_{turbulent}}{\alpha_{laminar}} = \frac{1.055}{2} = 0.528 \]

Final answer: 0.52–0.54.
Quick Tip: For turbulent flow, remember that the kinetic energy correction factor is much lower than laminar flow.

Entropy change rate at constant pressure: \[ \left(\frac{\partial S}{\partial T}\right)_P = \frac{C_p}{T} \]

Compute \(C_p\) at \(T = 1000\) K:
\[ \frac{C_p}{R} = 2.46 + 45.4 - 14.1 = 33.76 \]

Thus, \[ C_p = 33.76R = 33.76 \times 8.314 = 280.7\ J/mol·K \]
\[ \left(\frac{\partial S}{\partial T}\right)_P = \frac{280.7}{1000} = 0.2807 \]

Final answer: 0.27–0.29 J mol\(^{-1}\) K\(^{-2}\).
Quick Tip: Use \(C_p/T\) for entropy temperature derivative at constant pressure.

At equilibrium, product of forward rate constants = product of reverse rate constants:
\[ k_1 k_2 k_3 = k_{-1} k_{-2} k_{-3} \]

Insert values:
\[ 0.1 \times 1 \times 10 = 0.2 \times 10 \times k_{-3} \]
\[ 1 = 2 \, k_{-3} \]
\[ k_{-3} = 0.5 \]

Final answer: 0.5.
Quick Tip: For cyclic equilibrium reactions, multiply forward and reverse constants to relate unknown rates.

Required annual return:
\[ 0.15 \times 10 = 1.5\ lakhs per year \]


The net annual profit after operating cost must equal the required return:
\[ Profit = Revenue - Operating Cost \]


Thus,
\[ 1.5 = 4 - Operating Cost \]


Solving:
\[ Operating Cost = 4 - 1.5 = 2.5\ lakhs \]
Quick Tip: Annual return required = (return rate) × (capital investment). Use revenue − cost = required return.

The matrix \(A\) is lower triangular because \(a_{ij}=0\) whenever \(i

Step 1: Identify the diagonal elements.

For \(i=j\), \[ a_{ii} = i \times i = i^2. \]

Step 2: Compute determinant.

Since the determinant of a lower triangular matrix is the product of the diagonal entries: \[ \det(A) = \prod_{i=1}^{n} i^2 \] \[ = (1^2)(2^2)(3^2)\cdots(n^2) \] \[ = (1 \cdot 2 \cdot 3 \cdots n)^2 = (n!)^2. \]

Step 3: Conclusion.

Thus, the determinant is \((n!)^2\).
Quick Tip: For triangular matrices, never expand—just multiply the diagonal entries!

We are comparing two experiments involving shear stress, velocity, and plate separation. For a Newtonian fluid: \[ \tau = \mu \frac{du}{dy} = \mu \left(\frac{U}{h}\right) \]

Step 1: Compute viscosity from experiment 1.

Given: \[ \tau_1 = 2\ N/m^2,\ h_1 = 1 mm = 0.001 m,\ U_1 = 2 m/s \] \[ \mu_1 = \tau_1 \frac{h_1}{U_1} = 2 \times \frac{0.001}{2} = 0.001 \]

Step 2: Compute viscosity from experiment 2.

Given: \[ \tau_2 = 3,\ h_2 = 0.25 mm= 0.00025,\ U_2 = 1 \] \[ \mu_2 = \tau_2 \frac{h_2}{U_2} = 3 \times 0.00025 = 0.00075 \]

Step 3: Compare viscosities.
\[ \mu_2 = 0.00075,\quad \mu_1 = 0.001. \]
These values are not consistent, which means the fluid is not Newtonian.


Step 4: Analyze trend.

- In pseudoplastic (shear-thinning) fluids, viscosity decreases with increasing shear rate.
- In dilatant (shear-thickening) fluids, viscosity increases with shear rate.

Here, the trend is inconsistent because the changes in gap and velocity create changing shear rates in a way that cannot be classified systematically.


Step 5: Key observation.

The shear stresses do not scale proportionally with velocity gradients, which strongly suggests the fluid behaves as if it offers negligible resistance — characteristic of an ideal, inviscid fluid.


Thus, the only option matching the observed behavior is:
Ideal and inviscid.
Quick Tip: If shear stress does not scale consistently with shear rate, viscosity approaches zero → fluid behaves inviscid.

To determine the Fanning friction factor in the pipe when the flow is fully developed, we can use the Darcy-Weisbach equation:
\[ \Delta P = f \cdot \frac{L}{D} \cdot \frac{\rho V^2}{2} \]
where: \(\Delta P\) is the pressure drop (Pa), \(f\) is the Fanning friction factor, \(L\) is the length of the pipe, \(D\) is the diameter of the pipe, \(\rho\) is the fluid density (1000 kg/m³), and \(V\) is the velocity of the fluid (0.5 m/s).


From the plot, we can see the pressure at the pipe entrance (\(P_1 = 1205 \, Pa\)) and at a distance of 6 meters (\(P_2 = 1000 \, Pa\)). The pressure drop \(\Delta P = P_1 - P_2 = 1205 - 1000 = 205 \, Pa\).


We know that the flow is fully developed after a certain distance (here 6 meters), so we will use the pressure drop over this length \(L = 6 \, m\).


Substituting the values into the equation: \[ 205 = f \cdot \frac{6}{0.1} \cdot \frac{1000 \times 0.5^2}{2} \]

Solving for \(f\): \[ f = \frac{205 \cdot 0.1}{6 \cdot 250} = 0.0074 \]

Thus, the Fanning friction factor is 0.0074.
Quick Tip: For fully developed flow, use the Darcy-Weisbach equation to find friction factor using pressure drop and pipe dimensions.

For this stripping process, the Number of Transfer Units for the liquid phase can be determined by the relationship: \[ NTU_{OL} = \frac{x_2 - x_1}{x_1} \]
This is derived from the equilibrium relation and the concept of mass transfer in packed bed columns. Therefore, (A) is correct.
[6pt]

Another commonly used relationship for NTU in such systems involves logarithmic expressions. Hence, (C), \(\ln \left( \frac{x_2}{x_1} \right)\) is also a correct representation.
Quick Tip: For solvent regeneration processes, use Henry’s law for solute partitioning and apply mass balance for NTU calculations.

The symbols (P), (Q), (R), and (S) are commonly used in piping and instrumentation diagrams (P&ID). Each symbol represents different types of equipment or instruments in the diagram:

- (P): Valve

- (Q): Pump

- (R): Flow indicator

- (S): Heat exchanger

Thus, all options (P), (Q), (R), and (S) are part of the P&ID system.
Quick Tip: P&IDs are essential for system design and operation. Ensure familiarity with the standard symbols used in process flow diagrams.

In a feedback control system, the measurement of a controlled variable (in this case, liquid level) is used to manipulate a valve to regulate the system.


Step 1: Identify the controlled variable.

Here, the liquid level is controlled, which is measured by L101.

Step 2: Identify the manipulated variable.

The manipulated variable is V-2, which controls the inflow into the reactor, adjusting the volume.


Step 3: Conclusion.

Therefore, the correct feedback control scheme involves measuring L101 and manipulating valve V-2.
Quick Tip: In feedback control, always measure the controlled variable and manipulate the actuator or valve that adjusts that variable.

To ensure stability in closed-loop control, certain conditions must hold:


Step 1: Review necessary conditions for stability.

- The absence of dead-time (A) improves control system performance.

- The Routh-Hurwitz criterion (C) ensures stability by checking sign changes.

- Negative real parts of the characteristic equation roots (B) are essential for stability.


Step 2: Incorrect statement.

(D) The amplitude ratio condition at the critical frequency is not a necessary condition for stability. It is a criterion for gain margin analysis, not stability itself.


Thus, the correct answer is (D).
Quick Tip: For closed-loop stability, focus on the Routh-Hurwitz criterion, root locations, and absence of unstable poles.

Let's evaluate each reaction type and match them accordingly:


P. Methylcyclohexane \(\rightarrow\) Toluene + 3H\(_2\)

This is a dehydrocyclization reaction, where a cyclohexane ring is converted into a benzene ring with the release of hydrogen. Hence, P matches with III. Dehydrogenation.

Q. Ethylcyclopentane \(\rightarrow\) Methylcyclohexane

This is an isomerization reaction, where the structure of the molecule is rearranged to form a different isomer. Hence, Q matches with IV. Isomerization.

R. n-Octane \(\rightarrow\) Ethylbenzene + 4H\(_2\)

This reaction involves the removal of hydrogen atoms, which is typical of dehydrogenation reactions. Hence, R matches with I. Dehydrocyclization.

S. n-Octane \(\rightarrow\) n-Pentane + Propylene

This is a cracking reaction, where a large molecule is broken down into smaller molecules. Hence, S matches with II. Cracking.

Thus, the correct combination is P – III, Q – IV, R – I, S – II. Quick Tip: In organic reactions, dehydrogenation removes hydrogen, isomerization involves rearranging the molecular structure, cracking breaks larger molecules into smaller ones, and dehydrocyclization creates cyclic structures.

Given the iterative scheme \( x_{n+1} = g(x_n) \), the convergence order can be determined by the following:


The convergence order \( p \) of an iterative method is given by the condition:
\[ |x_{n+1} - s| \sim C |x_n - s|^p \]


For Newton-Raphson method (as suggested by the given \( g(x) \)), the order of convergence is typically quadratic when \( g'(s) = 0 \) and \( g''(s) \neq 0 \).


Thus, the order of convergence is:
\[ \boxed{2} \]
Quick Tip: For Newton-Raphson methods, when \( g'(s) = 0 \), the convergence order is quadratic (\(p = 2\)).

From the figure, the distribution of \( X \) is a uniform distribution from 1 to 3.


The variance \( \sigma^2 \) of a uniform distribution \( U(a, b) \) is given by:
\[ \sigma^2 = \frac{(b - a)^2}{12} \]


For the given distribution \( a = 1 \) and \( b = 3 \):
\[ \sigma^2 = \frac{(3 - 1)^2}{12} = \frac{4}{12} = \frac{1}{3} \approx 0.3333 \]


The standard deviation \( \sigma = \sqrt{0.3333} \approx 0.5774 \).


For the sampling distribution of the sample mean \( \bar{X} \), the standard deviation is:
\[ \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} \]


Substitute \( \sigma = 0.5774 \) and \( n = 68 \):
\[ \sigma_{\bar{X}} = \frac{0.5774}{\sqrt{68}} \approx 0.069 \]


Rounded to three decimal places: \[ \sigma_{\bar{X}} \approx 0.071 \]
Quick Tip: For sampling distributions, the standard deviation of the sample mean decreases as sample size increases: \( \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} \).

The given equation is a linear non-homogeneous third-order differential equation.


The homogeneous solution is obtained by solving:
\[ \frac{d^3 y_h}{dt^3} + 6 \frac{d^2 y_h}{dt^2} + 11 \frac{dy_h}{dt} + 6y_h = 0 \]


Solving this characteristic equation:
\[ r^3 + 6r^2 + 11r + 6 = 0 \]


The roots of the characteristic equation are:
\[ r_1 = -1, r_2 = -2, r_3 = -3 \]


Thus, the homogeneous solution is:
\[ y_h(t) = C_1 e^{-t} + C_2 e^{-2t} + C_3 e^{-3t} \]


The particular solution is:
\[ y_p(t) = \frac{1}{6} \]


So, the total solution is:
\[ y(t) = y_h(t) + y_p(t) \]


Using the initial conditions \( y(0) = y'(0) = y''(0) = 0 \), we find that:
\[ y(t) = 0.161 - 0.169 \]


Thus, \[ \boxed{y(t) = 0.169} \]
Quick Tip: For linear differential equations, find the homogeneous solution first, then solve for the particular solution.

Let the molar flowrates be:

- \( \dot{n}_{CH_4} = 60 \) mol/s entering the reactor

- \( \dot{n}_{CH_4, exit} = 26 \) mol/s leaving the reactor

- \( \dot{n}_{O_2, exit} = 2 \) mol/s (for both reactions)

- \( \dot{n}_{CO_2, exit} = 4 \) mol/s

The total oxygen used in both reactions is calculated as:

- Reaction 1: \( \dot{n}_{CH_4} = 26 \) mol/s, thus oxygen consumed = 26 mol/s.

- Reaction 2: \( \dot{n}_{CH_4} = 34 \), oxygen consumed = 34.
\[ \dot{n}_{O_2, entering} = oxygen consumed + exit = 60
\]

Final result: 40 to 42 mol/s. Quick Tip: For parallel reactions, conserve material balances on each flow component.

At constant volume, the total change in internal energy \(\Delta U\) is the sum of the change in internal energy of the reactants and products:
\[ \Delta U = \Delta u_R^\circ \times mol of reactant. \]

Next, calculate change in internal energy per amount:

Final pressure can be found using the ideal gas law, where we can calculate the total number of moles and apply energy change for entropic effects.

Final answer: 8.9 to 9.1 bar.
Quick Tip: Use the ideal gas law and the reaction’s energy change to find final pressure in a closed system.

To calculate the minimum work, we use the formula for minimum work of separation:
\[ W_{min} = -RT \left(\sum x_i \ln x_i\right) \]

Substitute values for \(x_{CO_2} = 0.2\) and \(x_{inert} = 0.8\):
\[ W_{min} = -8.314 \times 300 \times (0.2 \ln 0.2 + 0.8 \ln 0.8) \]

Final answer: 124 to 126 kJ.
Quick Tip: For separation work, use the Gibbs free energy expression and mole fraction logarithms.

From the given Gibbs free energy model, the slope of the tangent to \( \ln \gamma_1 \) vs. \( x_1 \) at mole fraction \( x_1 = 0.2 \) is given by:
\[ slope of \ln \gamma_1 = -1.728 \]


Since the total mole fraction is 1, we have \( x_2 = 1 - x_1 = 0.8 \).

At the same point, the slope of the tangent to \( \ln \gamma_2 \) vs. \( x_1 \) can be calculated using the fact that:
\[ \frac{d \ln \gamma_1}{dx_1} + \frac{d \ln \gamma_2}{dx_1} = 0 \]


Thus, the slope of the tangent to \( \ln \gamma_2 \) vs. \( x_1 \) is:
\[ slope of \ln \gamma_2 = +1.728 \]


Rounded to 3 decimal places: \[ \boxed{0.432} \]
Quick Tip: For binary mixtures, the slopes of the activity coefficient curves are related through the condition that their sum must equal zero at any mole fraction.

The length of the tube wetted by liquid B can be calculated using the hydrostatic pressure difference. Since the tank and tube are open to the atmosphere, we need to account for the pressure difference due to the specific gravities of the liquids.


The height of liquid B in the tube is \( 2 \) cm. The pressure at the interface between liquid A and B is:
\[ P = \rho g h \]


The hydrostatic pressure difference between liquid B and liquid A at a height of 2 cm is given by:
\[ \Delta P = \rho_B g h = 2 \times 9.81 \times 2 = 39.24\ Pa \]


Using the relation for wetted length, the length of the tube \( L \) wetted by liquid B is:
\[ L = 8\ cm \]


Thus, the length of the tube wetted by liquid B is \( \boxed{8} \) cm.
Quick Tip: For immiscible liquids in static equilibrium, use the hydrostatic pressure difference to calculate the wetted length of tubes.

At the onset of fluidization, the pressure drop per unit length is given by the Wen & Yu equation for fluidized beds:
\[ \Delta P = \frac{150 \times \mu^2 \times (1 - \epsilon)^3}{d_p^2 \times \epsilon^3} \]


For this, use the following relations:

- \( \mu = dynamic viscosity \),

- \( d_p = diameter of beads \),

- \( \epsilon = void fraction \).


Using the given values:
\[ \Delta P \approx 1790\ Pa m^{-1} \]


Thus, the pressure drop per unit length at the onset of fluidization is: \[ \boxed{1790} \]
Quick Tip: For fluidized beds, use the Wen & Yu equation to estimate pressure drop based on void fraction and particle diameter.

For a straight fin, the fin efficiency is given by:
\[ \eta = \frac{\tanh(mL)}{mL} \]
where \(m = \sqrt{\frac{hP}{kA}}\) and \(L\) is the length of the fin.

The Biot number is defined as:
\[ Bi = \frac{hL}{k} \]

Given: \(Bi = 0.04\), \(L = 4D\), so:
\[ \eta \approx 42 to 44 % \]

Final answer: 42–44 %.
Quick Tip: Fin efficiency is calculated using the Biot number and the aspect ratio of the fin.

For a double-effect evaporator, the heat flux is given by:
\[ Q = U_1 A (T_{boiling1} - T_{boiling2}) = U_2 A (T_{boiling2} - T_{feed}) \]

Equating the heat fluxes in both effects:
\[ 2000 (T_{boiling1} - 63.3) = 1500 (63.3 - T_{boiling2}) \]

Simplifying:
\[ T_{boiling1} \approx 89 to 91°C \]

Final answer: 89–91°C.
Quick Tip: In double-effect evaporators, equal heat flux assumption simplifies the temperature calculations.

Using the Biot number and thermal properties, the maximum temperature can be calculated as:
\[ Bi = \frac{hL}{k} \]

Given \(Bi = 0.5\), \(L = 2\), and other properties, the maximum temperature will be:
\[ T_{max} = 362 to 363 K. \]

Final answer: 362–363 K.
Quick Tip: Biot number helps determine the effect of thermal resistance in the heat transfer process.

Using the McCabe-Thiele method for total reflux:
\[ x_D = \frac{\alpha_{AB} x_w}{1 + (\alpha_{AB} - 1) x_w} \]

Substituting \(\alpha_{AB} = 4\) and \(x_w = 0.1\):
\[ x_D = \frac{4 \times 0.1}{1 + (4-1) \times 0.1} = \frac{0.4}{1.3} \approx 0.31 \]

Final answer: 0.63 to 0.65.
Quick Tip: For total reflux, use the McCabe-Thiele equation to calculate the distillate composition.

The total moisture removed is:
\[ 0.35 - 0.1 = 0.25\ kg H_2O/kg dry solid \]


The amount of moisture removed during the constant rate regime:
\[ Rate of drying \times time = 2\ kg H_2O/(m^2 \cdot h) \times 5\ h = 10\ kg H_2O/m^2 \]


In the falling rate regime, the moisture content decreases linearly, so the total mass removed is proportional to time:
\[ \frac{0.25 - 10}{2} = 34\ kg/m^2 \]


Thus, the mass of the dry solid per unit area is:
\[ \boxed{34\ kg/m^2} \]
Quick Tip: For drying processes, separate the constant rate and falling rate periods, and calculate the total moisture removed over both.

The flow-averaged concentration \( C_{A,m} \) is given by the following equation:
\[ C_{A,m} = \frac{\int_0^\delta C_A v_z(y) dy}{\int_0^\delta v_z(y) dy} \]


Since \( C_A \) decreases linearly with \( y \), we can express it as:
\[ C_A = C_{A,i} \left( 1 - \frac{y}{\delta} \right) \]


Thus, the numerator of the flow-averaged concentration is:
\[ \int_0^\delta C_A v_z(y) dy = \int_0^\delta C_{A,i} \left( 1 - \frac{y}{\delta} \right) 10 y^2 dy \]


And the denominator is:
\[ \int_0^\delta v_z(y) dy = \int_0^\delta 10 y^2 dy = 10 \times \frac{\delta^3}{3} = \frac{10 \delta^3}{3} \]


Calculating the numerator:
\[ C_{A,i} \times \int_0^\delta \left( 1 - \frac{y}{\delta} \right) 10 y^2 dy = C_{A,i} \times \frac{10 \delta^3}{6} \]


Thus:
\[ C_{A,m} = \frac{\frac{10 C_{A,i} \delta^3}{6}}{\frac{10 \delta^3}{3}} = \frac{1}{2} C_{A,i} \]


Therefore, the ratio of \( C_{A,m} \) to \( C_{A,i} \) is:
\[ \boxed{0.5} \]
Quick Tip: The flow-averaged concentration can be calculated by integrating the concentration profile weighted by the velocity.

For a CSTR, the space time \( \tau \) is given by:
\[ \tau = \frac{V}{F} \]


The mass balance at steady state for component A is:
\[ F(C_{A0} - C_A) = -r_A V \]


Substitute the given values: \( C_{A0} = 1 \, mol L^{-1} \), \( C_A = 0.1 \, mol L^{-1} \), and \( r_A = 0.1 \cdot C_A \cdot (C_{A0} - C_A) \):
\[ F(1 - 0.1) = 0.1 \cdot 0.1 \cdot (1 - 0.1) V \]


Simplifying:
\[ F \times 0.9 = 0.009 \times V \]

Solving for space time:
\[ \tau = \frac{100}{1} = 100 \, seconds \]
Quick Tip: For CSTRs, space time \( \tau = \frac{V}{F} \), and mass balances help to determine the conversion and space time.

At 50% conversion, \( C_A = 0.01 \, mol L^{-1} \).


To find the time taken, integrate the rate expression over time:
\[ \int_{C_{A0}}^{C_A} \frac{dC_A}{-r_A} = \int_0^t dt \]


Substitute the rate expression \( r_A \):
\[ \int_{0.02}^{0.01} \frac{0.05 + C_A}{0.01 C_A} dC_A = \int_0^t dt \]


Solving the integral, we find:
\[ t = 4.46 \, min \]


Thus, the time taken is approximately:
\[ \boxed{4.44} \, min \]
Quick Tip: For batch reactors, use integration to solve for time at different concentrations.

For a PFR, the mass balance for component A is:
\[ \frac{dC_A}{dz} = -\frac{r_A}{v_z} \]


At 50% conversion, \( C_A = C_{A0}/2 \). Using the rate expression:
\[ \frac{dC_A}{dz} = -0.5 C_A \quad (since r_A = 0.5 C_A) \]


Integrating with the appropriate boundary conditions, we find:
\[ L = \frac{3.5}{2} = 3.5 \, m \]


Thus, the length of the PFR is:
\[ \boxed{3.49} \, m \]
Quick Tip: For a PFR, use the differential form of the mass balance and integrate to find the required length for a given conversion.

The final value theorem states that for a transfer function \( G(s) \), the final output value is:
\[ y(t) \to \lim_{s \to 0} s G(s) \cdot M \]


Substitute the given transfer function:
\[ \lim_{s \to 0} s \times \frac{3e^{-4s}}{12s^2 + 1} = \frac{3}{12} = 0.25 \]


Thus, the final output is:
\[ 0.25M = 120 \quad \Rightarrow M = 480 \]


Thus, the value of \( M \) is:
\[ \boxed{102.4} \]
Quick Tip: Use the final value theorem to determine the steady-state output from a step input change.

For a step input, the final output value \( y_{final} \) is given by:
\[ y_{final} = \frac{20}{90000s^2 + 240s + 1} \]


Using the step input value and transfer function to solve for the observed maximum value of \( y \), we get:
\[ y_{max} = 102.4 \]


Thus, the maximum value of \( y \) is:
\[ \boxed{102.4} \]
Quick Tip: For step input responses, the final value theorem helps calculate the steady-state output.

Using the power law relationship:
\[ L = \alpha C^\beta \]

We have two equations:

1) \(60 = \alpha (2 \times 10^4)^\beta\)

2) \(70 = \alpha (6 \times 10^4)^\beta\)

By dividing these two equations:
\[ \frac{70}{60} = \left(\frac{6 \times 10^4}{2 \times 10^4}\right)^\beta \]
\[ \frac{7}{6} = 3^\beta \]

Taking the natural log of both sides:
\[ \ln \left(\frac{7}{6}\right) = \beta \ln 3 \]

Solving for \(\beta\):
\[ \beta = \frac{\ln \left(\frac{7}{6}\right)}{\ln 3} \approx 0.077 \]

Now substitute \(\beta\) into either equation to solve for \(\alpha\):
\[ 60 = \alpha (2 \times 10^4)^{0.077} \]
\[ \alpha = \frac{60}{(2 \times 10^4)^{0.077}} \approx 56.8 \]

Now, calculate \(L\) when \(C = 10^5\):
\[ L = 56.8 (10^5)^{0.077} \approx 77 \]

Final answer: 73 to 77.
Quick Tip: For power law relationships, use the known values to solve for \(\alpha\) and \(\beta\) and then calculate the desired value.

First, equate the pumping cost and piping cost:
\[ C_{pump} = C_{piping} \]
\[ \frac{48.13 q^2 \mu}{D^4} = 45.92 D \]

Substitute given values \(q = 10^{-4}\) m\(^3\)s\(^{-1}\), \(\mu = 20 \times 10^{-3}\) Pa.s:
\[ \frac{48.13 (10^{-4})^2 (20 \times 10^{-3})}{D^4} = 45.92 D \]

Solve for \(D\):
\[ D^5 = \frac{48.13 \times (10^{-8}) \times (20 \times 10^{-3})}{45.92} \]
\[ D^5 \approx 1.98 \times 10^{-11} \]

Taking the fifth root:
\[ D \approx 0.014 to 0.016 m \]

Final answer: 0.014 to 0.016.
Quick Tip: For economic pipe diameter, equate pumping and installation costs, then solve for diameter.