GATE CE 2017 Set 1

The sentence expresses a general scientific fact or a habitual action in the passive voice. For such statements, the simple present tense is used. The structure is 'it is + past participle'. Therefore, 'is' is the correct auxiliary verb.

"Tampering" means to interfere with something in a way that causes damage or is unauthorized. In the context of an email account, unauthorized access or modification is referred to as tampering.

Let's analyze the statements using set theory:Now evaluate the inferences:
i. Some beds are lamps: From (2), we know that any lamp that is a bulb cannot be a bed. However, there is no information regarding lamps that are not bulbs. They could overlap with beds or not. Thus, this is not a necessary inference.
ii. Some lamps are beds: This is logically equivalent to inference (i) and is also not necessarily true.Since neither inference follows logically from the given statements, the correct option is (D).

The volume of a right circular cone is given by the formula:$$V = \frac{1}{3} \pi r^2 h$$Let the initial radius be $r$ and the initial height be $h$. The initial volume is $V_1 = \frac{1}{3} \pi r^2 h$.If the radius is increased by $50\%$, the new radius $r'$ becomes:$$r' = r + 0.5r = 1.5r$$Assuming the height $h$ remains constant, the new volume $V_2$ is:$$V_2 = \frac{1}{3} \pi (1.5r)^2 h = \frac{1}{3} \pi (2.25r^2) h = 2.25 V_1$$The increase in volume is:$$\Delta V = V_2 - V_1 = 2.25 V_1 - V_1 = 1.25 V_1$$The percentage increase in volume is:$$\text{Percentage Increase} = \frac{\Delta V}{V_1} \times 100\% = 1.25 \times 100\% = 125\%$$Thus, the correct option is (C).

Given sequence in increasing order: $1, x, x, x, y, y, 9, 16, 18$. There are $n = 9$ terms.$$\text{Mean} = \frac{1 + 3x + 2y + 9 + 16 + 18}{9} = \frac{44 + 3x + 2y}{9}$$According to the problem:$$\text{Mean} = \text{Median} = 2 \times \text{Mode}$$$$y = 2x \implies x = \frac{y}{2}$$Also, Mean = Median:$$\frac{44 + 3x + 2y}{9} = y$$$$44 + 3x + 2y = 9y$$$$44 + 3x = 7y$$Substitute $x = \frac{y}{2}$ into the equation:$$44 + 3\left(\frac{y}{2}\right) = 7y$$$$44 + 1.5y = 7y$$$$44 = 5.5y$$$$y = \frac{44}{5.5} = 8$$Checking the conditions: If $y = 8$, then $x = 4$. The sequence is $1, 4, 4, 4, 8, 8, 9, 16, 18$. It is in increasing order. Mode $= 4$, Median $= 8$, Mean $= (1+12+16+9+16+18)/9 = 72/9 = 8$. All conditions are satisfied. The correct option is (D).

The passage describes the demolition of an old concert hall due to fears that its foundation would be affected by new metro line construction. It also mentions that despite using modern technology and safeguards to mitigate the impact of pressurized air pockets, there was still concern about the stability of the soil below the concert hall. This directly implies that metro construction requires careful consideration of its potential impact on the foundations of existing buildings.Let's evaluate the options:

• (A) the foundations of old buildings create pressurized air pockets underground, which are difficult to handle in metro construction. This statement is incorrect. The passage states that pressurized air pockets are created by the excavation for the metro line, not by the foundations of old buildings.
• (B) metro construction has to be done carefully considering its impact on the foundations of existing buildings. This aligns perfectly with the passage. The demolition of the concert hall due to fears for its foundation, despite safeguards, highlights the necessity of careful consideration.
• (C) old buildings in an area form an impossible hurdle to metro construction in that area. The passage states it was a fear that the soil would not be stable, leading to demolition. It doesn't necessarily imply an "impossible hurdle" for all old buildings, but rather a significant challenge requiring careful assessment.
• (D) pressurized air can be used to excavate large amounts of soil from underground areas. While the passage mentions "pressurized air pockets created by the excavation," it doesn't state that pressurized air is used for excavation. It refers to the impact of these pockets, not their utility in excavation.

Therefore, the most accurate inference is that metro construction must be done carefully, considering its impact on existing buildings' foundations.The final answer is $\boxed{B}$

Let's trace the room allocation process based on seniority, from highest (1) to lowest (4).Initial State:

• Room P: Amar (Seniority 1)
• Room Q: Anthony (Seniority 3)
• Room S: Ajit (Seniority 4)
• Akbar (Seniority 2): No room

Allocation Step 1: Amar (Seniority 1)

• Amar is currently in Room P. His preferences are R, S, Q.
• Check his preferences in order:
• R: Room R is currently empty. Amar moves to Room R.
• Result after Amar's move:
• Room R: Amar
• Room P: Empty
• Room Q: Anthony
• Room S: Ajit
• Akbar: No room

Allocation Step 2: Akbar (Seniority 2)

• Akbar has no current room. His preferences are R, S.
• Check his preferences in order:
• R: Room R is occupied by Amar (Seniority 1). Akbar cannot displace Amar as Amar is more senior.
• S: Room S is occupied by Ajit (Seniority 4). Akbar (Seniority 2) is more senior than Ajit (Seniority 4). According to the rule "Students already staying in a room will move if they get a room in their preferred list" and "Preferences of lower ranked applicants are ignored during allocation", Akbar can take Room S from Ajit.
• Result after Akbar's move:
• Room R: Amar
• Room P: Empty
• Room Q: Anthony
• Room S: Akbar
• Ajit: No room (displaced from S)

Allocation Step 3: Anthony (Seniority 3)

• Anthony is currently in Room Q. His preference is P.
• Check his preference:
• P: Room P is currently empty. Anthony moves to Room P.
• Result after Anthony's move:
• Room R: Amar
• Room P: Anthony
• Room Q: Empty
• Room S: Akbar
• Ajit: No room

Allocation Step 4: Ajit (Seniority 4)

• Ajit was displaced from Room S and currently has no room. His preferences are Q, P, R.
• Check his preferences in order:
• Q: Room Q is currently empty. Ajit moves to Room Q.
• Result after Ajit's move:
• Room R: Amar
• Room P: Anthony
• Room Q: Ajit
• Room S: Akbar

Therefore, Ajit will stay in Room Q.The final answer is $\boxed{B}$

To find the last digit of the sum, we need to find the last digit of each term and then sum their last digits. The last digit of $2171$ is $1$. Any positive integer power of a number ending in $1$ will also end in $1$.
So, the last digit of $(2171)^7$ is $1$. The last digit of $2172$ is $2$. We look at the cyclicity of the last digits of powers of $2$:
$2^1 = 2$
$2^2 = 4$
$2^3 = 8$
$2^4 = 16 \rightarrow 6$
$2^5 = 32 \rightarrow 2$
The cycle of last digits for powers of $2$ is $(2, 4, 8, 6)$, with a length of $4$.
To find the last digit of $2^9$, we divide the exponent $9$ by the cycle length $4$: $9 \div 4 = 2$ with a remainder of $1$.
The last digit is the $1^{st}$ digit in the cycle, which is $2$.
So, the last digit of $(2172)^9$ is $2$. The last digit of $2173$ is $3$. We look at the cyclicity of the last digits of powers of $3$:
$3^1 = 3$
$3^2 = 9$
$3^3 = 27 \rightarrow 7$
$3^4 = 81 \rightarrow 1$
$3^5 = 243 \rightarrow 3$
The cycle of last digits for powers of $3$ is $(3, 9, 7, 1)$, with a length of $4$.
To find the last digit of $3^{11}$, we divide the exponent $11$ by the cycle length $4$: $11 \div 4 = 2$ with a remainder of $3$.
The last digit is the $3^{rd}$ digit in the cycle, which is $7$.
So, the last digit of $(2173)^{11}$ is $7$. The last digit of $2174$ is $4$. We look at the cyclicity of the last digits of powers of $4$:
$4^1 = 4$
$4^2 = 16 \rightarrow 6$
$4^3 = 64 \rightarrow 4$
The cycle of last digits for powers of $4$ is $(4, 6)$, with a length of $2$.
For odd exponents, the last digit is $4$. For even exponents, the last digit is $6$.
Since the exponent $13$ is an odd number, the last digit is $4$.
So, the last digit of $(2174)^{13}$ is $4$.Now, we sum the last digits we found:
Last digit of sum = Last digit of $(1 + 2 + 7 + 4)$
Last digit of sum = Last digit of $(14)$
The last digit of $14$ is $4$.Therefore, the last digit of the given expression is $4$.The final answer is $\boxed{B}$

To find the minimum time, we first calculate the total workload for each job type:

• Job $P$: $2 \text{ objects} \times 30 \text{ min/object} = 60 \text{ minutes}$
• Job $Q$: $3 \text{ objects} \times 20 \text{ min/object} = 60 \text{ minutes}$
• Job $R$: $1 \text{ object} \times 60 \text{ min/object} = 60 \text{ minutes}$
• Job $S$: $4 \text{ objects} \times 15 \text{ min/object} = 60 \text{ minutes}$

Total workload = $60 + 60 + 60 + 60 = 240 \text{ minutes}$.Since there are two machines working simultaneously, the theoretical minimum time is $\frac{240}{2} = 120 \text{ minutes} = 2 \text{ hours}$.We can achieve this balance by assigning:

• Machine $M_1$: Job $R$ (60 min) + Job $P$ (2 objects $\times$ 30 min = 60 min) = 120 minutes.
• Machine $M_2$: Job $Q$ (3 objects $\times$ 20 min = 60 min) + Job $S$ (4 objects $\times$ 15 min = 60 min) = 120 minutes.

Thus, the minimum time is 2 hours.

Let's extract the data from the stacked bar graph for each carpenter:Statement i: Number of beds by C2 = 8. Number of tables by C3 = 8. Since $8 = 8$, statement i is true.Statement ii:
Total chairs = $2 + 10 + 5 + 2 + 4 = 23$.
Total tables = $7 + 2 + 8 + 3 + 10 = 30$.
Since $23 < 30$, statement ii is true.Both statements are true, so the correct option is (C).

By the definition of the inverse of a matrix, if $P$ is the inverse of $Q$, then $P = Q^{-1}$. This relationship implies that $PQ = Q^{-1}Q = I$ and $QP = QQ^{-1} = I$. Therefore, both $PQ = I$ and $QP = I$ must be true.

Thus, the number of parameters are 1 and 2 respectively.

To evaluate the integral $g(x) = \int e^{-x-e^{-x}} dx$, we can rewrite the integrand as:$$f(x) = e^{-e^{-x}} \cdot e^{-x}$$Let $u = -e^{-x}$. Then, the derivative is $du = e^{-x} dx$.
Substituting these into the integral:$$g(x) = \int e^u du = e^u + C$$Substituting back for $u$:$$g(x) = e^{-e^{-x}} + C$$Comparing with the options, the correct form is $e^{-e^{-x}}$.

Thermal stress in a bar fixed at both ends is given by the formula $\sigma = E \alpha \Delta T$. This expression is independent of the length $L$ of the bar. When the temperature increases, the bar tries to expand by $\Delta L_{th} = L \alpha T$. Since the ends are fixed, a compressive force $P$ develops to counteract this expansion, such that $\Delta L_{mech} = \frac{PL}{AE} = L \alpha T$. Solving for stress $\sigma = \frac{P}{A}$, we get $\sigma = E \alpha T$. Since $L$ cancels out, doubling the length has no effect on the axial stress.

For a beam subjected to transverse loading, the relationship between the shear force ($V$) and the bending moment ($M$) is given by:$$V = \frac{dM}{dx}$$This mathematical relationship implies that the bending moment reaches a local maximum or minimum at points where its first derivative, the shear force, is zero. Therefore, statement (B) is true.According to the Euler-Bernoulli beam theory, the curvature ($\kappa$) of the beam's neutral axis is proportional to the bending moment:$$M = EI\kappa$$where $E$ is the Young's modulus and $I$ is the area moment of inertia. This relationship indicates that the curvature is zero at sections where the bending moment is zero. For a simply supported beam under a uniformly distributed load (UDL), the bending moment is zero at the supports ($x = 0$ and $x = L$). At these same support locations, the shear force is at its maximum absolute value ($|V_{max}| = \frac{wL}{2}$). Consequently, the maximum shear force occurs at the sections where the curvature is zero. Thus, statement (A) is also true.Since both statements (A) and (B) are technically correct for the given conditions, the official GATE answer key accepts both options as correct.

According to IS 456:2000, the limiting (balanced) depth of the neutral axis $x_{u,max}$ (or $x_{u,bal}$) is derived from the strain compatibility condition at failure. It is given by:$$\frac{x_{u,max}}{d} = \frac{0.0035}{0.0035 + 0.87\frac{f_y}{E_s} + 0.002}$$where $f_y$ is the yield strength of steel and $E_s$ is the modulus of elasticity of steel ($2 \times 10^5$ MPa). This formula shows that the depth depends only on the grade of steel ($f_y$) and is independent of the grade of concrete.

For a two-hinged parabolic arch subjected to a uniformly distributed load over the entire span, the horizontal thrust $H$ is given by:$$H = \frac{qL^2}{8h}$$where $h$ is the rise of the arch. The bending moment at any section $x$ from the support is:$$M_x = \mu_x - H \cdot y$$where $\mu_x$ is the beam bending moment and $y$ is the vertical coordinate of the arch at $x$. For a parabolic arch:$$y = \frac{4h}{L^2}x(L-x)$$The beam bending moment for a UDL is:$$\mu_x = \frac{q}{2}x(L-x)$$Substituting $H$ and $y$ into the bending moment equation:$$M_x = \frac{q}{2}x(L-x) - \left(\frac{qL^2}{8h}\right) \left(\frac{4h}{L^2}x(L-x)\right)$$$$M_x = \frac{q}{2}x(L-x) - \frac{q}{2}x(L-x) = 0$$Thus, the bending moment is zero at every section of the arch.

The correct matches are:

• P. Regain of strength with time: This is known as Thixotropy (3). It is the property of some soils to regain a part of their lost strength over time at a constant water content.
• Q. Loss of strength due to cyclic loading: This leads to Liquefaction (2), where saturated soil loses its strength and stiffness in response to an applied stress, such as earthquake shaking.
• R. Loss of strength due to upward seepage: This causes Boiling (1) or the quicksand condition, where the effective stress becomes zero due to upward seepage pressure.
• S. Loss of strength due to remolding: This is related to Sensitivity (4), which is the ratio of the unconfined compressive strength of an undisturbed soil to that of the same soil in a remolded state.

Hydrostatic pressure implies that the normal stress is the same in all directions, i.e., $\sigma_1 = \sigma_2 = \sigma_3 = \sigma$.
The radius of the Mohr circle is given by:$$R = \frac{\sigma_1 - \sigma_3}{2} = \frac{\sigma - \sigma}{2} = 0$$The center of the Mohr circle is at:$$C = \frac{\sigma_1 + \sigma_3}{2} = \frac{\sigma + \sigma}{2} = \sigma$$Since the radius is zero, the Mohr circle collapses to a single point located at a distance $\sigma$ from the origin on the normal stress axis.

According to Terzaghi's bearing capacity theory for a strip footing on pure clay (where the internal friction angle $\phi = 0$): $$q_u = c N_c + q N_q + 0.5 \gamma B N_\gamma$$For $\phi = 0$, the bearing capacity factors are $N_c = 5.14$, $N_q = 1$, and $N_\gamma = 0$.
For a footing resting on the ground surface, the depth of embedment $D_f = 0$, which implies the surcharge $q = \gamma D_f = 0$.
Thus, the ultimate bearing capacity simplifies to:$$q_u = c_u N_c$$The value $5.14$ is the plasticity theory solution (Prandtl's solution) which is exactly $\pi + 2 \approx 3.1416 + 2 = 5.1416$.
Therefore, $q_u = (\pi + 2) c_u$.

The vertical stress $\sigma_z$ at a depth $z$ directly below a line load $q$ is given by Boussinesq's equation for a line load:$$\sigma_z = \frac{2q}{\pi z}$$Since the load $q$ is constant, the vertical stress is inversely proportional to the depth: $\sigma_z \propto \frac{1}{z}$.
The ratio of the vertical stress at depth $z_1 = 2\text{ m}$ to the vertical stress at depth $z_2 = 4\text{ m}$ is calculated as:$$\frac{\sigma_{z1}}{\sigma_{z2}} = \frac{z_2}{z_1} = \frac{4}{2} = 2.0$$

In a steady incompressible laminar flow between two parallel stationary plates (Plane Poiseuille flow), the velocity profile $u(y)$ is parabolic. The shear stress $\tau$ is related to the velocity gradient by Newton's law of viscosity: $\tau = \mu \frac{du}{dy}$.
For this specific flow, the shear stress distribution is linear across the gap. It varies from zero at the center line (where the velocity is at its maximum and the gradient is zero) to a maximum value at the stationary plates (where the velocity is zero due to the no-slip condition).

The order of a chemical reaction is defined as the sum of the exponents of the concentration terms in the rate law expression. Given the rate law: $$\text{Rate} = -k[A]^\alpha [B]^\beta$$The overall order of the reaction ($n$) is given by:$$n = \alpha + \beta$$For a reaction to be a first-order reaction, the overall order must be 1 ($n = 1$).Evaluating the given options:

• (A) $\alpha = 0, \beta = 0 \implies n = 0 + 0 = 0$ (Zero-order reaction)
• (B) $\alpha = 1, \beta = 0 \implies n = 1 + 0 = 1$ (First-order reaction)
• (C) $\alpha = 1, \beta = 1 \implies n = 1 + 1 = 2$ (Second-order reaction)
• (D) $\alpha = 1, \beta = 2 \implies n = 1 + 2 = 3$ (Third-order reaction)

Therefore, option (B) is the only valid statement for a first-order reaction.

This scenario is described by the Streeter-Phelps oxygen sag curve. When organic waste is discharged into a river at location $S$, the dissolved oxygen (DO) level begins to drop because the rate of deoxygenation (due to microbial decomposition of organics) is initially higher than the rate of reaeration (replenishment of oxygen from the atmosphere). The DO level continues to decrease until it reaches a minimum point, known as the critical DO point. This critical point occurs at some distance downstream from the point of discharge $S$. After this point, as the organic matter is depleted, the reaeration rate begins to exceed the deoxygenation rate, and the DO level starts to recover.

Acid rain is primarily caused by the presence of strong mineral acids like sulfuric acid ($H_2SO_4$) and nitric acid ($HNO_3$), which are formed when sulfur dioxide ($SO_2$) and nitrogen oxides ($NO_x$) react with water, oxygen, and other chemicals in the atmosphere. Therefore, acetic acid is the one not typically associated with the formation of acid rain.

To prevent a stationary vehicle ($v=0$) from sliding down the slope of a super-elevated curve, the component of gravity pulling it down the slope must be balanced by the frictional force.
$W \sin \theta \leq f_{max} W \cos \theta$
$\tan \theta \leq f_{max}$
Since $e = \tan \theta$, the condition is $e \leq f_{max}$.

$L_1 = 2000 \times \left(1 + \frac{7}{100} \times \frac{535}{300}\right) = 2000 \times (1 + 0.12483) = 2249.67\text{ m}$.Standard temperature at elevation $h$: $T_s = 15 - 0.0065 \times h = 15 - 0.0065 \times 535 = 11.5225^\circ\text{C}$.
Airport reference temperature $T_a = 22.65^\circ\text{C}$.
Rise in temperature $\Delta T = T_a - T_s = 22.65 - 11.5225 = 11.1275^\circ\text{C}$.
Correction: $1\%$ increase for every $1^\circ\text{C}$ rise above standard temperature.
$L_2 = L_1 \times \left(1 + \frac{1}{100} \times 11.1275\right) = 2249.67 \times 1.111275 = 2500.00\text{ m}$.

In the EDMI accuracy specification $\pm (a\text{ mm} + b\text{ ppm})$, '$a$' represents the constant instrumental error (independent of distance), and '$b\text{ ppm}$' (parts per million) represents the distance-dependent error. The total error component from the second term is $b \times 10^{-6} \times D$, which varies in proportion to the distance $D$ being measured.

The Enhanced Thematic Mapper Plus (ETM+) sensor on Landsat-7 has 8 spectral bands: Bands 1-5 and 7 are multispectral (30m resolution), Band 6 is thermal (60m resolution), and Band 8 is panchromatic (15m resolution).

A second-order linear partial differential equation of the form $A \frac{\partial^2 \phi}{\partial x^2} + B \frac{\partial^2 \phi}{\partial x \partial y} + C \frac{\partial^2 \phi}{\partial y^2} + f(x, y, \phi, \phi_x, \phi_y) = 0$ is classified based on the discriminant $D = B^2 - 4AC$:Given the equation:$$3 \frac{\partial^2 \phi}{\partial x^2} + B \frac{\partial^2 \phi}{\partial x \partial y} + 3 \frac{\partial^2 \phi}{\partial y^2} + 4 \phi = 0$$Comparing this with the standard form, we identify the coefficients of the second-order terms:
$A = 3$
$B = B$
$C = 3$For the equation to be classified as parabolic, the discriminant must be equal to zero:
$B^2 - 4AC = 0$
$B^2 - 4(3)(3) = 0$
$B^2 - 36 = 0$
$B^2 = 36$Thus, the value of $B^2$ must be 36.