GATE CE 2022 Set 1

The sentence is a common idiomatic expression. The first blank requires a verb indicating awareness or understanding, which is 'know'. The second blank refers to the word being spoken, which is 'no'. Thus, the complete sentence is: "You should know when to say no."

• Line 1 passes through $(0,0)$ and $(1,3)$. Its slope $m_1 = \frac{3-0}{1-0} = 3$. Equation: $y = 3x$.
• Line 2 passes through $(0,0)$ and $(1,2)$. Its slope $m_2 = \frac{2-0}{1-0} = 2$. Equation: $y = 2x$.

2. The area enclosed between these lines in the interval $x \in [0, 1]$ is given by the integral of the difference of the functions:
$$\text{Area} = \int_{0}^{1} (y_1 - y_2) dx = \int_{0}^{1} (3x - 2x) dx = \int_{0}^{1} x dx$$ $$\text{Area} = \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1^2}{2} - \frac{0^2}{2} = 0.5$$Alternatively, the area is a triangle with vertices $(0,0)$, $(1,3)$, and $(1,2)$.
Base length (along $x=1$) $= 3 - 2 = 1$.
Height (perpendicular distance from origin to $x=1$) $= 1$.
Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 1 = 0.5$.

Given ratios:
$\frac{p}{q} = \frac{1}{2}$
$\frac{q}{r} = \frac{4}{3}$
$\frac{r}{s} = \frac{4}{5}$Also, $u$ is $50\%$ more than $s$:
$u = s + 0.5s = 1.5s = \frac{3}{2}s \Rightarrow \frac{s}{u} = \frac{2}{3}$To find the ratio $p : u$, multiply all the intermediate ratios:$$\frac{p}{u} = \frac{p}{q} \times \frac{q}{r} \times \frac{r}{s} \times \frac{s}{u}$$$$\frac{p}{u} = \frac{1}{2} \times \frac{4}{3} \times \frac{4}{5} \times \frac{2}{3}$$$$\frac{p}{u} = \frac{1 \times 4 \times 4 \times 2}{2 \times 3 \times 5 \times 3} = \frac{32}{90} = \frac{16}{45}$$So, the ratio $p : u = 16 : 45$.

From (2) and (3), Q is the father of S.
From (1), P is the paternal aunt of S (father's sister).
From (4), T is the husband of S's paternal aunt, which makes T the uncle of S.

Let $r$ be the radius of the circle.Since $PQ$ is a tangent to the circle at $Q$, the radius $RQ$ is perpendicular to $PQ$ (\\angle PQR = 90^\\circ).
In right-angled triangle $\\triangle PQR$, by Pythagoras theorem:
$PR^2 = PQ^2 + RQ^2$Since $RQ = r$ (radius):
$PR^2 = PQ^2 + r^2$Substituting the areas:
Area(PXWR) = Area(SPQT) + Area(RUVZ)Rearranging gives:
Area(SPQT) = Area(PXWR) $-$ Area(RUVZ)

The passage states that healthy eating is critical for healthy aging and that it is "never too early" to start. It provides an example of babies starting in their first year to show long-term benefits. This directly supports the inference that healthy eating can be started at any age and that starting earlier is better for long-term health outcomes.

Total investment of P = $5000 \times 6 = 30000$.
Total investment of Q = $x \times 8 = 8x$.
The profit is shared in proportion to the total investment.
Profit share of Q = $\frac{\text{Investment of Q}}{\text{Total Investment}} = \frac{8x}{30000 + 8x}$.
Given that Q receives $\frac{4}{9}$ of the total profit:$$\frac{8x}{30000 + 8x} = \frac{4}{9}$$$$72x = 4(30000 + 8x)$$$$72x = 120000 + 32x$$$$40x = 120000$$$$x = 3000$$Therefore, the value of $x$ is ₹ 3000.

From the frequency chart, we can extract the following data:Comparing the values: $7 > 6 > 5.84$, which means mode > median > mean. This corresponds to option (B).

The movement described is identical to a knight's move in chess (an 'L' shape: 2 squares in one cardinal direction and 1 square perpendicular). Let's denote the grid positions by (Row, Column).
Start: P2 = (1, 2)
Target: P5 = (1, 5)
Shaded (blocked) squares: P1(1,1), P4(1,4), Q3(2,3), R1(3,1), R5(3,5), S2(4,2), S4(4,4), T3(5,3).We need to find the shortest path from (1, 2) to (1, 5) avoiding blocked squares.Step 1: From P2(1, 2), possible moves are Q4(2, 4) and R3(3, 3). Both are unshaded.
Let's try P2(1, 2) $\rightarrow$ Q4(2, 4).Step 2: From Q4(2, 4), possible moves are S3(4, 3), S5(4, 5), P2(back), P6(out). S3 and S5 are unshaded.
Let's try Q4(2, 4) $\rightarrow$ S3(4, 3).Step 3: From S3(4, 3), possible moves are T5(5, 5), T1(5, 1), Q1(2, 1), Q5(2, 5), R5(blocked), R1(blocked), P2(back), P4(blocked).
Let's try S3(4, 3) $\rightarrow$ T5(5, 5).Step 4: From T5(5, 5), possible moves are R4(3, 4) and S3(back). R4 is unshaded.
Let's try T5(5, 5) $\rightarrow$ R4(3, 4).Step 5: From R4(3, 4), possible moves are P3(1, 3), P5(1, 5), T3(blocked), T5(back), Q2(2, 2), Q6(out), S2(blocked), S6(out).
Target P5(1, 5) is reached in this step.Path: P2(1,2) $\rightarrow$ Q4(2,4) $\rightarrow$ S3(4,3) $\rightarrow$ T5(5,5) $\rightarrow$ R4(3,4) $\rightarrow$ P5(1,5).
Total steps = 5.Checking other paths will show that 5 is the minimum number of steps required to reach the target while avoiding shaded squares.

In the given 3D view, the cube has a black top face and a front face with a vertical line. The line is perpendicular to the edge shared with the black top face.

Given $z = \sin(y + it) + \cos(y - it)$.
Differentiating partially with respect to $y$:$$\frac{\partial z}{\partial y} = \cos(y + it) - \sin(y - it)$$$$\frac{\partial^2 z}{\partial y^2} = -\sin(y + it) - \cos(y - it) = -z$$Differentiating partially with respect to $t$:$$\frac{\partial z}{\partial t} = i \cos(y + it) - (-i) \sin(y - it) = i [\cos(y + it) + \sin(y - it)]$$$$\frac{\partial^2 z}{\partial t^2} = i^2 [-\sin(y + it)] + i(-i) [-\cos(y - it)] = \sin(y + it) + \cos(y - it) = z$$Adding the two second-order partial derivatives:$$\frac{\partial^2 z}{\partial t^2} + \frac{\partial^2 z}{\partial y^2} = z + (-z) = 0$$Thus, option (A) is correct.

The order of a differential equation is the highest derivative present, which is $\frac{d^3 y}{dx^3}$, so the order is 3.
The degree is the power of the highest order derivative when the equation is free from radicals and fractional powers in its derivatives.
Rearranging the equation:$$\frac{d^3 y}{dx^3} + x^2 y = -x \left( \frac{dy}{dx} \right)^{3/2}$$Squaring both sides to remove the fractional power:$$\left( \frac{d^3 y}{dx^3} + x^2 y \right)^2 = x^2 \left( \frac{dy}{dx} \right)^3$$The power of the highest order derivative $\frac{d^3 y}{dx^3}$ is now 2. Therefore, the degree is 2.

For a thin cylindrical pressure vessel, the principal stresses at a point on the surface are:The maximum shear stress ($τ_{max}$) can be calculated in two ways depending on the context:

• In-plane maximum shear stress: $\tau_{max, \text{in-plane}} = \frac{\sigma_1 - \sigma_2}{2} = \frac{30.0 - 15.0}{2} = 7.5$ MPa
• Absolute maximum shear stress: $\tau_{max, \text{abs}} = \max\left(\frac{\sigma_1 - \sigma_2}{2}, \frac{\sigma_2 - \sigma_3}{2}, \frac{\sigma_1 - \sigma_3}{2}\right) = \max(7.5, 7.5, 15.0) = 15.0$ MPa

In engineering practice, "maximum shear stress" usually refers to the absolute maximum value unless "in-plane" is specified. However, the official GATE 2022 answer key accepted both options (A) and (B) as correct.

In the elastic theory (Working Stress Method) of reinforced concrete, the modular ratio ($m$) is defined as the ratio of the modulus of elasticity of steel ($E_s$) to the modulus of elasticity of concrete ($E_c$).$$m = \frac{E_s}{E_c}$$Therefore, it is the ratio of Young's modulus of elasticity of the reinforcement material to that of concrete.

The pozzolanic reaction is a chemical reaction that occurs in portland cement concrete upon the addition of pozzolans. It involves the reaction of calcium hydroxide ($Ca(OH)_2$), which is a byproduct of cement hydration, with reactive silica (silicon dioxide, $SiO_2$) from the pozzolan in the presence of water ($H_2O$) to form additional calcium silicate hydrate (C-S-H) gel, which contributes to strength and durability.$$\text{Pozzolan (Reactive } SiO_2) + Ca(OH)_2 + H_2O \rightarrow \text{C-S-H}$$

For a thin-walled section consisting of two intersecting legs, the shear center is located at the point of intersection of the centerlines of the legs. In this cross-section, the vertical leg (web) lies along the line $x = 0$ and the horizontal leg (flange) lies along the line $y = 3a$. Their intersection point is $(0, 3a)$. Therefore, the coordinates of the shear center are $x = 0, y = 3a$.

Hydraulic conductivity ($k$) depends on the size of the pores, which is directly related to the particle size of the soil. Coarser soils have larger pores and thus higher hydraulic conductivity. The general order of decreasing hydraulic conductivity is Gravel > Sand > Silt > Clay.

• SW (Well-graded sand) and SP (Poorly-graded sand) are coarse-grained soils (Sands).
• ML (Low plasticity silt) is a fine-grained soil (Silt).
• CH (High plasticity clay) is a fine-grained soil (Clay).

Therefore, the decreasing order is Sand > Silt > Clay, which corresponds to SW, SP, ML, CH.

In Rankine's passive earth pressure state, the soil is compressed horizontally as a retaining structure moves towards the backfill. This causes the horizontal effective stress $\sigma_h'$ to increase. At the state of plastic failure, the horizontal stress becomes the major principal stress ($\sigma_1' = \sigma_h'$) and the vertical effective stress remains the minor principal stress ($\sigma_3' = \sigma_v'$). For failure to occur, $\sigma_1' > \sigma_3'$, which means $\sigma_h' > \sigma_v'$ or $\sigma_v' < \sigma_h'$.

The dimensionless numbers are defined based on the ratio of forces:Correct combination: (P)-(II), (Q)-(IV), (R)-(I), (S)-(III).

Therefore, $\psi$ decreases and $K$ increases.

Total hardness is given as 500 mg/L as $CaCO_3$.
To express this in milligram equivalent per liter (meq/L), we use the formula:$$\text{Hardness in meq/L} = \frac{\text{Hardness in mg/L as } CaCO_3}{\text{Equivalent weight of } CaCO_3}$$The molecular weight of $CaCO_3 = 40 + 12 + (3 \times 16) = 100$ g/mol.
The valency of $Ca^{2+}$ is 2, so the equivalent weight of $CaCO_3 = \frac{100}{2} = 50$ g/eq or 50 mg/meq.$$\text{Hardness in meq/L} = \frac{500}{50} = 10 \text{ meq/L}$$Therefore, the correct option is (A).

The scale of an aerial photograph is given by the formula:$$S = \frac{f}{H - h}$$where:
$f$ = focal length of the camera = 152 mm = 0.152 m
$H$ = flight height above mean sea level = 3.5 km = 3500 m
$h$ = average ground elevation above mean sea level = 460 m
Substituting the values:$$S = \frac{0.152}{3500 - 460} = \frac{0.152}{3040}$$$$S = \frac{1}{20000}$$So, the scale is 1 : 20000. The correct option is (A).

Step 1: Correct the measured slope length for the tape error.
Nominal length of tape ($L_{nom}$) = 50 m
Actual length of tape ($L'$) = $50 - 0.1 = 49.9$ m
Measured slope length ($l$) = 350 m
Corrected slope length ($L$) = $l \times \frac{L'}{L_{nom}} = 350 \times \frac{49.9}{50} = 349.3$ mStep 2: Calculate the horizontal length from the corrected slope length.
The slope is 1 in 5 (vertical : horizontal). Let the horizontal distance be $x$ and vertical be $y$. Then $y/x = 1/5$.
The slope distance $s = \sqrt{x^2 + y^2} = \sqrt{x^2 + (x/5)^2} = x\sqrt{1 + 1/25} = x\sqrt{26}/5$.
We have the corrected slope distance $L = 349.3$ m.
Horizontal length $H = L \times \frac{5}{\sqrt{26}} = 349.3 \times \frac{5}{5.0990} \approx 342.517$ m.
Rounding to two decimal places, we get 342.52 m. The correct option is (A).

If $\lambda$ is an eigenvalue of matrix $\mathbf{M}$, then $f(\lambda)$ is an eigenvalue of the matrix polynomial $f(\mathbf{M})$.
Given $f(\mathbf{M}) = \mathbf{M}^3 - 4\mathbf{M}^2 - 2\mathbf{M}$, we define the scalar function $f(x) = x^3 - 4x^2 - 2x$.
The eigenvalues of $\mathbf{M}$ are $\lambda_1 = 5$ and $\lambda_2 = -2$.$f(5) = 5^3 - 4(5^2) - 2(5) = 125 - 100 - 10 = 15$.$f(-2) = (-2)^3 - 4(-2)^2 - 2(-2) = -8 - 16 + 4 = -20$.
Thus, the eigenvalues of $\mathbf{Q}$ are 15 and -20.

Chemical Oxygen Demand (COD) measures the total oxygen required to chemically oxidize both biodegradable and non-biodegradable organic matter. Biochemical Oxygen Demand (BOD) measures only the oxygen required for biological oxidation of biodegradable organic matter by microorganisms.
In almost all cases, $\text{COD} \geq \text{BOD}_5$. For industrial wastewater like that from a wood pulping industry, there is a significant amount of non-biodegradable organic matter (such as lignin), which makes the COD significantly higher than the $\text{BOD}_5$. Therefore, $\text{COD} > \text{BOD}_5$ is correct, which also implies $\text{COD} \neq \text{BOD}_5$.

Desalination is the process of removing dissolved salts from water to produce fresh water.

The radius of curvature $R$ for a horizontal curve is given by the formula:$$e + f = \frac{v^2}{gR} \implies R = \frac{v^2}{g(e + f)}$$where:

• $e$ is the superelevation
• $f$ is the coefficient of lateral friction
• $v$ is the design speed
• $g$ is the acceleration due to gravity

To decrease the radius of curvature $R$:

The given iteration scheme is the Newton-Raphson method for finding the square root of $P$, derived from $f(x) = x^2 - P = 0$.The iteration formula is:$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x_n - \frac{x_n^2 - P}{2x_n} = \frac{2x_n^2 - x_n^2 + P}{2x_n} = \frac{1}{2} \left( x_n + \frac{P}{x_n} \right)$$For $P = 10$ and initial guess $x_1 = 1$:Rounding to three decimal places, the value of $x_5$ is 3.162.

We need to express $f(x) = \cos^4 x$ in terms of multiple angles using trigonometric identities:$$\cos^2 x = \frac{1 + \cos 2x}{2}$$$$\cos^4 x = (\cos^2 x)^2 = \left( \frac{1 + \cos 2x}{2} \right)^2 = \frac{1}{4} (1 + 2 \cos 2x + \cos^2 2x)$$Now, substitute $\cos^2 2x = \frac{1 + \cos 4x}{2}$:$$\cos^4 x = \frac{1}{4} \left( 1 + 2 \cos 2x + \frac{1 + \cos 4x}{2} \right)$$$$\cos^4 x = \frac{1}{4} + \frac{1}{2} \cos 2x + \frac{1}{8} + \frac{1}{8} \cos 4x$$$$\cos^4 x = \frac{3}{8} + \frac{1}{2} \cos 2x + \frac{1}{8} \cos 4x$$Comparing this with the Fourier cosine series $f(x) = \sum_{n=0}^{\infty} f_n \cos nx$:

• $f_0 = \frac{3}{8}$
• $f_2 = \frac{1}{2}$
• $f_4 = \frac{1}{8} = 0.125$
• All other $f_n = 0$ for $n \neq 0, 2, 4$.

Therefore, $f_5 = 0$.
$f_4 + f_5 = 0.125 + 0 = 0.125$.