GATE ME 2021 Set 1

In English grammar, adverbs of degree like 'hardly' and 'barely' are typically placed after the auxiliary verb (in this case, 'could') and before the main verb ('walk'). Therefore, sentences (ii) "Raja could barely walk" and (iv) "Raja could hardly walk" are grammatically correct. Sentences (i) and (iii) incorrectly place the adverb before the auxiliary verb.

Given the mapping of symbols:

• $\Delta \rightarrow +$
• $\oplus \rightarrow -$
• $\otimes \rightarrow \div$
• $\nabla \rightarrow \times$

Substituting these into the expression:
$\Delta 2 \oplus 3 \Delta ((4 \otimes 2) \nabla 4)$
$= + 2 - 3 + ((4 \div 2) \times 4)$
$= 2 - 3 + (2 \times 4)$
$= 2 - 3 + 8$
$= 7$Hence, the correct option is (D).

The statement claims that increased consumption of leafy vegetables indicates a healthy lifestyle. This implies a direct relationship between the two.

• (A) is an overstatement; we only know consumption increased, not that it was zero before.
• (B) is a general truth but not an inference from the specific statement provided.
• (C) is a direct logical inference: if leafy vegetable consumption is an indicator of a healthy lifestyle, then the two are related.
• (D) introduces 'junk foods' and 'health hazards' which are not mentioned in the premise.

Thus, (C) is the correct inference.

Based on the passage:

• Oxpeckers feed on ticks $\rightarrow$ Benefit: food source.
• Oxpeckers warn rhinos about poachers $\rightarrow$ Benefit: safety from poachers.

Therefore, the benefits for oxpeckers and rhinos respectively are a food source and being saved from poachers. This matches option (C).

The given jigsaw piece has a specific profile with notches and protrusions. To form a rectangle, the missing piece must have a profile that is exactly complementary to the given one. By inspecting the options, Option A provides the shape that, when appropriately rotated or flipped, fits perfectly into the original piece to complete a rectangular boundary.

• Hens in $Q = \frac{5}{32} \times 416 = 5 \times 13 = 65$.
• Ducks in $Q = \frac{14}{32} \times 416 = 14 \times 13 = 182$.
• Goats in $Q = \frac{13}{32} \times 416 = 13 \times 13 = 169$.

3. New totals in Farm $P$ after transfer:

• Total Hens = $65 + 65 = 130$.
• Total Ducks = $91 + 182 = 273$.
• Total Goats = $169 + 169 = 338$.

4. New ratio in $P$: $130 : 273 : 338$.
Dividing by the common factor $13$:
$130/13 = 10$, $273/13 = 21$, $338/13 = 26$.
The ratio is $10:21:26$.

• Total executives in $C2 = 5\% \text{ of } 10,000 = 500$.
• Ratio (Management : Non-Management) = $1:4$.
• Management degree holders in $C2 = \frac{1}{1+4} \times 500 = \frac{1}{5} \times 500 = 100$.

3. Company $C5$:

• Total executives in $C5 = 20\% \text{ of } 10,000 = 2,000$.
• Ratio (Management : Non-Management) = $9:1$.
• Management degree holders in $C5 = \frac{9}{9+1} \times 2,000 = \frac{9}{10} \times 2,000 = 1,800$.

4. Total management degree holders in $C2$ and $C5$ = $100 + 1,800 = 1,900$.

Let the five positions in the row be 1, 2, 3, 4, and 5. The condition "$Q$ and $R$ are separated by one person" means the possible pairs of positions for $(Q, R)$ or $(R, Q)$ are $(1, 3), (2, 4), (3, 5), (3, 1), (4, 2),$ and $(5, 3)$.We also have the constraint that $S$ cannot be adjacent to $Q$.Let's analyze each case: Total arrangements = $2 \times 2 = 4$. Total arrangements = $1 \times 2 = 2$. Total arrangements = $1 \times 2 = 2$. Total arrangements = $1 \times 2 = 2$. Total arrangements = $1 \times 2 = 2$. Total arrangements = $2 \times 2 = 4$.Summing all cases: $4 + 2 + 2 + 2 + 2 + 4 = 16$.Therefore, there are 16 distinct seating arrangements possible.

To find the value of $y(\pi/6)$, we begin by analyzing the given first-order ordinary differential equation:$$(\sin x) \frac{dy}{dx} + y \cos x = 1$$

Step 1: Simplify the Differential Equation

We recognize that the left-hand side of the equation is the exact derivative of the product $y \sin x$ with respect to $x$ using the product rule:$$\frac{d}{dx} (y \sin x) = (\sin x) \frac{dy}{dx} + y \cos x$$Thus, the differential equation simplifies to:$$\frac{d}{dx} (y \sin x) = 1$$

Step 2: Integrate both sides

Integrating both sides with respect to $x$:$$\int \frac{d}{dx} (y \sin x) \, dx = \int 1 \, dx$$$$y \sin x = x + C$$where $C$ is the constant of integration.

Step 3: Apply the boundary condition

We are given the condition $y(\pi/2) = \pi/2$. Substituting $x = \pi/2$ and $y = \pi/2$ into the integrated equation:$$\left(\frac{\pi}{2}\right) \sin\left(\frac{\pi}{2}\right) = \frac{\pi}{2} + C$$Since $\sin(\pi/2) = 1$:$$\frac{\pi}{2} (1) = \frac{\pi}{2} + C \implies C = 0$$

Step 4: Find the particular solution

Substituting $C = 0$ back into the equation gives:$$y \sin x = x$$$$y(x) = \frac{x}{\sin x}$$

Step 5: Calculate $y(\pi/6)$

Now, we evaluate $y(x)$ at $x = \pi/6$:$$y\left(\frac{\pi}{6}\right) = \frac{\pi/6}{\sin(\pi/6)}$$Since $\sin(\pi/6) = 1/2$:$$y\left(\frac{\pi}{6}\right) = \frac{\pi/6}{1/2} = \frac{\pi}{3}$$

Conclusion

The value of $y(\pi/6)$ is $\frac{\pi}{3}$.Correct Option: C

To find the limit $$L = \lim_{x \to 0} \left( \frac{1 - \cos x}{x^2} \right)$$we first evaluate the limit directly by substituting $x = 0$:$$\frac{1 - \cos(0)}{0^2} = \frac{1 - 1}{0} = \frac{0}{0}$$Since this is an indeterminate form of type $\frac{0}{0}$, we can apply L'Hôpital's Rule, which states that if $\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{0}{0}$, then:$$\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$$Differentiating the numerator and the denominator with respect to $x$:

• Derivative of the numerator: $\frac{d}{dx}(1 - \cos x) = \sin x$
• Derivative of the denominator: $\frac{d}{dx}(x^2) = 2x$

Applying L'Hôpital's Rule:$$L = \lim_{x \to 0} \frac{\sin x}{2x}$$We can rewrite this as:$$L = \frac{1}{2} \lim_{x \to 0} \left( \frac{\sin x}{x} \right)$$Using the standard trigonometric limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$:$$L = \frac{1}{2} \times 1 = \frac{1}{2}$$

Alternative Method (Using Trigonometric Identity)

Using the half-angle identity $1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right)$:$$L = \lim_{x \to 0} \frac{2 \sin^2\left(\frac{x}{2}\right)}{x^2}$$Multiply and divide the denominator by $4$:$$L = \lim_{x \to 0} \frac{2 \sin^2\left(\frac{x}{2}\right)}{4 \left(\frac{x}{2}\right)^2} = \frac{2}{4} \lim_{x \to 0} \left[ \frac{\sin\left(\frac{x}{2}\right)}{\frac{x}{2}} \right]^2$$Since $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ (where $\theta = \frac{x}{2}$):$$L = \frac{1}{2} \times (1)^2 = \frac{1}{2}$$Thus, the correct option is C.

To find the Laplace transform of the Dirac-delta function $\delta(t - a)$ for $a > 0$, we use the definition of the Laplace transform.The Laplace transform of a function $f(t)$ is defined as:$$\mathcal{L}(f(t)) = \int_{0}^{\infty} f(t) e^{-st} dt$$Substituting $f(t) = \delta(t - a)$ into the definition:$$\mathcal{L}(\delta(t - a)) = \int_{0}^{\infty} \delta(t - a) e^{-st} dt$$We are given the shifting property of the Dirac-delta function:$$\int_a^b \phi(t)\delta(t - t_0)dt = \begin{cases} \phi(t_0) & a < t_0 < b \\ 0 & \text{otherwise} \end{cases}$$Comparing our integral with the property:

• $\phi(t) = e^{-st}$
• $t_0 = a$
• The limits of integration are from $0$ to $\infty$.

Since $a > 0$, the point $t_0 = a$ lies strictly within the interval of integration $(0, \infty)$. Applying the shifting property:$$\int_{0}^{\infty} e^{-st} \delta(t - a) dt = e^{-s(a)} = e^{-sa}$$Thus, the Laplace transform is:$$\mathcal{L}(\delta(t - a)) = e^{-sa}$$Correct Option: D

To find the interval of the time step $h$ for which the numerical scheme is stable, we analyze the given finite difference equation:\frac{y(t_{n+1}) - y(t_n)}{h} = -$\pi$ y(t_n)This is the Forward Euler (explicit) discretization of the ordinary differential equation. Let $y_n$ denote the numerical approximation of $y(t_n)$. Rearranging the equation to express $y_{n+1}$ in terms of $y_n$:y_{n+1} - y_n = -$\pi$ h y_ny_{n+1} = (1 - $\pi$ h) y_nFor the numerical scheme to be stable, the amplification factor $\lambda = 1 - \pi h$ must satisfy the stability criterion:|$\lambda$| < 1Substituting the expression for $\lambda$:|1 - $\pi$ h| < 1This inequality can be written as:-1 < 1 - $\pi$ h < 1Subtracting $1$ from all parts of the inequality:-2 < -$\pi$ h < 0Dividing by $-\pi$ (which reverses the inequality signs):0 < h < \frac{2}{$\pi$}Thus, the numerical scheme is stable for all values of $h$ in the interval $0 < h < \frac{2}{\pi}$.Correct Option: A

To find the limiting distribution of the sum $Y = \sum_{i=1}^{n} X_i$ as $n \to \infty$, we apply the Central Limit Theorem (CLT).

Step-by-Step Derivation

We are given that $X_1, X_2, \dots, X_n$ are independent and identically distributed (i.i.d.) random variables.
Each $X_i$ follows a binomial distribution, $X_i \sim \text{Binomial}(m, p)$, with a finite mean $\mu$ and a finite, non-zero variance $\sigma^2$:
$\mu$ = E[X_i] = mp $\sigma$^2 = \text{Var}(X_i) = mp(1-p) The Central Limit Theorem states that if $X_1, X_2, \dots, X_n$ are i.i.d. random variables with a finite mean $\mu$ and finite variance $\sigma^2$, then the distribution of their sum $Y = \sum_{i=1}^{n} X_i$ approaches a normal distribution as $n \to \infty$. The mean of $Y$ is:
E[Y] = E\left[\sum_{i=1}^{n} X_i\right] = \sum_{i=1}^{n} E[X_i] = n$\mu$ = nmp
Since the samples are independent, the variance of $Y$ is:
\text{Var}(Y) = \text{Var}\left(\sum_{i=1}^{n} X_i\right] = \sum_{i=1}^{n} \text{Var}(X_i) = n$\sigma$^2 = nmp(1-p) As $n \to \infty$, the cumulative distribution function of the standardized variable
Z = \frac{Y - n$\mu$}{$\sigma$ \sqrt{n}} converges to the standard normal distribution $N(0, 1)$. Thus, the distribution of $Y$ can be approximated as a Normal distribution:
Y $\sim$ N\left(nmp, \, nmp(1-p)\right) \quad \text{as } n \to $\infty$Correct Option: D

To find the elastic and plastic strains corresponding to a stress of $200\text{ MPa}$, we analyze the loading and unloading behavior shown in the stress-strain curve: During loading, the material is stressed up to $\sigma = 200\text{ MPa}$. At this peak stress, the total strain reached is:
$\varepsilon$_{\text{total}} = 0.03 When the material is fully unloaded from $200\text{ MPa}$ to $0\text{ MPa}$, the elastic strain is recovered, leaving only the permanent (plastic) deformation. From the figure, the residual strain after complete unloading is:
$\varepsilon$_p = 0.01 The total strain is the sum of the elastic and plastic strains:
$\varepsilon$_{\text{total}} = $\varepsilon$_e + $\varepsilon$_p Substituting the known values:
0.03 = $\varepsilon$_e + 0.01 $\varepsilon$_e = 0.03 - 0.01 = 0.02Thus, the elastic and plastic strains corresponding to $200\text{ MPa}$ stress are 0.02 and 0.01, respectively.Correct Option: B

A cylindrical surface is generated when a generatrix (in this case, a circle) moves along a directrix (a straight line) that is perpendicular to the plane of the generatrix. As the circular tool path moves along the straight line, it sweeps out a cylinder.

The standard sequence for creating and finishing a high-precision hole is:

Backlash is the play or clearance between the screw and the nut. In CNC machines, preloaded ballscrews are used to eliminate this. Preloading ensures that the ball bearings are in constant contact with both the screw and nut races, removing any axial play and thus eliminating backlash.

The center line average roughness ($R_a$) is defined as the arithmetic mean of the absolute values of the profile height deviations from the mean line over the sampling length $L$:$$R_a = \frac{1}{L} \int_{0}^{L} |y(x)| dx$$From the given roughness profile, the height $y(x)$ alternates between $+1\mu m$ and $-1\mu m$ over intervals of $L/4$. Therefore, the absolute value $|y(x)|$ is constant and equal to $1\mu m$ for the entire length $L$.$$R_a = \frac{1}{L} \int_{0}^{L} 1 dx = \frac{1}{L} [x]_0^L = \frac{L}{L} = 1\mu m$$

The Carnot cycle consists of two reversible isothermal processes and two reversible adiabatic (isentropic) processes. The Stirling cycle consists of two reversible isothermal processes and two reversible constant-volume (isochoric) processes with regeneration. Other cycles like Otto, Diesel, and Brayton do not contain isothermal processes in their ideal air-standard forms.

Using the real gas equation of state involving the compressibility factor:$$Pv = ZRT$$where:
$P = 1500\text{ kPa} = 1500 \times 10^3\text{ Pa}$
$v = 2.75\text{ m}^3/\text{kmol}$ (molar specific volume)
$Z = 0.95$
$R = 8.314\text{ kJ/kmol}\cdot\text{K} = 8314\text{ J/kmol}\cdot\text{K}$ (Universal gas constant)Rearranging for temperature $T$ in Kelvin:$$T = \frac{Pv}{ZR} = \frac{1500 \times 10^3 \times 2.75}{0.95 \times 8314} \approx 522.26\text{ K}$$Converting to Celsius:$$T(^\circ\text{C}) = T(\text{K}) - 273.15 = 522.26 - 273.15 = 249.11^\circ\text{C}$$Rounding to the nearest integer, we get $249^\circ\text{C}$.

Heisler charts are used for transient heat conduction analysis in objects where the lumped capacitance method is not applicable. These charts provide the temperature distribution within an object as a function of time and position. The two primary dimensionless parameters used in Heisler charts are:Therefore, Biot number and Fourier number are required to determine the instantaneous center temperature of the ball using a Heisler chart.The final answer is $\boxed{A}$

For an infinitely long fin, the rate of heat transfer ($Q$) is given by the formula:
$Q = \sqrt{h P k A_c} (T_b - T_\infty)$
where:

• $h$ is the convective heat transfer coefficient
• $P$ is the perimeter of the fin
• $k$ is the thermal conductivity of the fin material
• $A_c$ is the cross-sectional area of the fin
• $T_b$ is the base temperature of the fin
• $T_\infty$ is the ambient air temperature

Let $Q_1$ be the initial heat transfer rate with thermal conductivity $k$. So,
$Q_1 = \sqrt{h P k A_c} (T_b - T_\infty)$Now, the thermal conductivity of the fin material is doubled, so the new thermal conductivity is $k' = 2k$. All other parameters ($h, P, A_c, T_b, T_\infty$) remain constant. Let the new heat transfer rate be $Q_2$.
$Q_2 = \sqrt{h P (2k) A_c} (T_b - T_\infty)$
$Q_2 = \sqrt{2} \sqrt{h P k A_c} (T_b - T_\infty)$Substitute the expression for $Q_1$ into the equation for $Q_2$:
$Q_2 = \sqrt{2} Q_1$We need to find the ratio $Q_2/Q_1$:
$\frac{Q_2}{Q_1} = \sqrt{2}$The final answer is $\boxed{A}$

Relative humidity ($\phi$) is defined as the ratio of the partial pressure of water vapor ($p_v$) to the saturation pressure of water vapor ($p_{sat}$) at the same temperature.Given:
Relative humidity ($\phi$) = 50% = 0.50
Partial pressure of water vapor = $p_v$
Saturation pressure of water at 300 K = $p_{sat}$The formula for relative humidity is:
$\phi = \frac{p_v}{p_{sat}}$Substitute the given value of relative humidity:
$0.50 = \frac{p_v}{p_{sat}}$Rearranging the equation to solve for $p_v$:
$p_v = 0.50 \times p_{sat}$Thus, the correct relation for the air-water mixture is $p_v = 0.5 p_{sat}$.The final answer is $\boxed{A}$

The secondary unbalance force in a reciprocating engine is given by:$$F_s = m R \omega^2 \frac{\cos(2\theta)}{n}$$where $n = L/R$. Substituting $n$:$$F_s = m R \omega^2 \frac{R}{L} \cos(2\theta) = m \frac{R^2}{L} \omega^2 \cos(2\theta)$$A virtual crank of radius $r_v$ rotating at twice the engine speed ($2\omega$) produces a primary unbalance force:$$F_{pv} = m r_v (2\omega)^2 \cos(2\theta) = 4 m r_v \omega^2 \cos(2\theta)$$Equating the two forces:$$m \frac{R^2}{L} \omega^2 \cos(2\theta) = 4 m r_v \omega^2 \cos(2\theta)$$$$\frac{R^2}{L} = 4 r_v \implies r_v = \frac{R^2}{4L}$$The speed is $2\omega$, which is twice the engine speed.