Official Paper

GATE Civil Engineering (CE) Official Paper (Held On: 04 Feb, 2024 Shift 2) (Previous Year Paper)

65 questions · 180 minutes · with answers · free

General Aptitude (10 questions)

If '→' denotes increasing order of intensity, then the meaning of the words [drizzle → rain → downpour] is analogous to [_________ → quarrel → feud]. Which one of the given options is appropriate to fill the blank?

((a))
dither
((b))
dodge
((c))
bog
((d))
bicker

Show Answer

Answer: ((d))

bicker

Explanation:

Drizzle means intermittant rain
Downpour means heavy rain
Quarrel means an angry argument
Feud means an angry and bitter argument
Bicker means to argue about things that are not important.

Now its very clear that

drizzle → rain → downpour is analogous to bicker → quarrel → feud

Five cubes of identical size and another smaller cube are assembled as shown in Figure A. If viewed from direction X, the planar image of the assembly appears as Figure B.

If viewed from direction Y, the planar image of the assembly (Figure A) will appear as

((a))
((b))
((c))
((d))

Show Answer

Answer: ((c))

Explanation:

In the given text, the blanks are numbered (i)-(iv).

Select the best match for all the blanks.

Yoko Roi stands (i)__ as an author for standing (ii) as an honorary fellow, after she stood ___(iii) her writings that stand

(iv) the freedom of speech.

((a))
(i) down (ii) out (iii) by (iv) in
((b))
(i) down (ii) out (iii) for (iv) in
((c))
(i) out (ii) down (iii) by (iv) (iv) for
((d))
(i) out (ii) down (iii) in (iv) for

Show Answer

Answer: ((c))

(i) out (ii) down (iii) by (iv) (iv) for

Explanation:

stands out
standing down
by her writtings
for the freedom of speech is the best filling

so option (c) is correct

A student was supposed to multiply a positive real number p with another positive real number q. Instead, the student divided p by q. If the percentage error in the student's answer is 80%, the value of q is

((a))
5
((b))
√5
((c))
√2
((d))
2

Show Answer

Answer: ((b))

√5

Explanation:

Actual result = pq

wrong result = p/q

$\frac{\text { Wrong result }}{\text { Actual result }}=$ = 0.2 (result reduced by 80%)

$\frac{p / q}{p q} =0.2=\frac{2}{10}=\frac{1}{5}$

$\frac{1}{q^2} =\frac{1}{5} \Rightarrow q=\sqrt{5}$

Statements:

All heroes are winners.
All winners are lucky people. Inferences:

I. All lucky people are heroes.

II. Some lucky people are heroes.

III. Some winners are heroes.

Which of the above inferences can be logically deduced from statements 1 and 2?

((a))
Only III
((b))
Only I and III
((c))
Only II and III
((d))
Only I and II

Show Answer

Answer: ((c))

Only II and III

Explanation:

From above Venn diagram its clear that (ii) and (iii) are valid deductions.

Seven identical cylindrical chalk-sticks are fitted tightly in a cylindrical container. The figure below shows the arrangement of the chalk-sticks inside the cylinder.

The length of the container is equal to the length of the chalk-sticks. The ratio of the occupied space to the empty space of the container is

((a))
5/2
((b))
7/2
((c))
9/2
((d))
3

Show Answer

Answer: ((b))

7/2

Explanation:

As per given data:

Let the radius of cylinder = R

and the radius of chalk = r

2r + 2r + 2r = 2R

r = (R/3)

Volume of cylinder = πR²h

Volume of chalk $=π\left(\frac{\mathrm{R}}{3}\right)^2 \mathrm{~h}$

$\frac{\text { Volume of occupied space }}{\text { Volume of empty space }} =\frac{7 π \frac{\mathrm{R}^2}{9} h}{π R^2 h-\frac{7 π R^2}{9}}$

$=\frac{7 / 9}{2 / 9}=\left(\frac{7}{2}\right)$

The plot below shows the relationship between the mortality risk of cardiovascular disease and the number of steps a person walks per day. Based on the data, which one of the following options is true?

((a))
The risk reduction on increasing the steps/ day from 0 to 5000 is less than the risk reduction on increasing the steps/day from 15000 to 20000.
((b))
For any 5000 increment in steps/day the largest risk reduction occurs on going from 0 to 5000 .
((c))
The risk reduction on increasing the steps/ day from 0 to 10000 is less than the risk reduction on increasing the steps/day from 10000 to 20000.
((d))
For any 5000 increment in steps/day the largest risk reduction occurs on going from 15000 to 20000.

Show Answer

Answer: ((b))

For any 5000 increment in steps/day the largest risk reduction occurs on going from 0 to 5000 .

Explanation:

For any 5000 increment in steps/day the largest risk reduction occurs on going from 0 to 5000.

Observing the graph, the curve shows a steep decline from 0 to 5000 steps/day, indicating a significant reduction in the mortality risk of cardiovascular disease.
As the number of steps increases beyond 5000, the curve flattens, indicating smaller reductions in risk for each additional 5000 steps/day increment.

Therefore, the largest reduction in risk happens when increasing steps from 0 to 5000 steps/day compared to any other 5000-step increment.

Visualize a cube that is held with one of the four body diagonals aligned to the vertical axis. Rotate the cube about this axis such that its view remains unchanged. The magnitude of the minimum angle of rotation is

((a))
60°
((b))
180°
((c))
120°
((d))
90°

Show Answer

Answer: ((c))

120°

Explanation:

When the cube is rotated that is held with one of the four body diagonals aligned to the vertical axis

If the sum of the first 20 consecutive positive odd numbers is divided by 20², the result is

((a))
1/2
((b))
20
((c))
2
((d))
1

Show Answer

Answer: ((d))

Explanation:

1 + 3 = 4 = 2²

1 + 3 + 5 = 9 = 3²

1 + 3 + 5 + 7 = 4²

(1 + 3 + 5 + 4_________ 2 times) = 20²

$\frac{20^2}{20^2}=1$

The ratio of the number of girls to boys in class VIII Is the same as the ratio of the number of boys to girls in class IX. The total number of students (boys and girls) in classes VIII and IX is 450 and 360 , respectively. If the number of girls in classes VIII and IX is the same, then the number of girls in each class is

((a))
150
((b))
175
((c))
250
((d))
200

Show Answer

Answer: ((d))

200

Explanation:

Let no. of girls in 8^th= no. of girls in 9^th

= x

$\frac{\text { No. of girls in } 8^{\text {th }} \text { class }}{\text { No. of girls in } 9^{\text {th }} \text { class }}=\frac{\text { No. of boys in } 9^{\text {th }} \text { class }}{\text { No. of girls in } 9^{\text {th }} \text { class }}$

$\frac{x}{450-x} =\frac{360-x}{x}$

x² = (450 - x) (360 - x)

x² = 16200 - 450x - 360x + x²

$x =\left(\frac{16200}{810}\right)$

= 200

Civil Engineering (55 questions)

Which one of the following products is NOT obtained in anaerobic decomposition of glucose?

((a))
H2S
((b))
CO2
((c))
H2O
((d))
CH4

Show Answer

Answer: ((a))

H2S

Explanation:

Anaerobic decomposition of glucose occurs in the absence of oxygen, leading to the production of:

Carbon dioxide (CO₂)
Methane (CH₄)
Water (H₂O) (as a byproduct in some cases)

However, Hydrogen sulfide (H₂S) is not a direct product of anaerobic glucose decomposition. H₂S is typically produced by sulfate-reducing bacteria when they break down sulfur-containing compounds, which is a different biochemical process.

The following figure shows a plot between shear stress and velocity gradient for materials/fluids Q, R, S and T.

Which one of the following options is CORRECT?

((a))
P → Real solid; Q → Ideal Bingham plastic S → Newtonian fluid; T → Ideal Fluid
((b))
P → Real solid; Q → Newtonian Fluid R → Ideal Bingham Plastic; T → Ideal Fluid
((c))
P → Ideal Fluid; Q → Ideal Bingham Plastic R → Non-Newtonian Fluid; S → Newtonian Fluid
((d))
P → Ideal Fluid; Q → Ideal Bingham Plastic R → Non-Newtonian Fluid; T → Real solid

Show Answer

Answer: ((a))

P → Real solid; Q → Ideal Bingham plastic S → Newtonian fluid; T → Ideal Fluid

Explanation:

Various types of newtonian & non-newtonian fluids are shown in the figure.
Fluids which obeys Newton's law of viscosity $\left(\tau=\mu \frac{d u}{d y}\right)$ are called Newtonian Fluids and those fluids which do not obey this rule are called Non-Newtonian Fluids.
General relationship between shear stress and velocity gradient is given by $\tau=A\left(\frac{d u}{d y}\right)^n+B$
In the figures shown above, slope of the curve is called apparent viscosity.
Fluid for which apparent viscosity increases with du/dy are called Dilatant.
Dilatant fluids are also called shear thickening fluids. Examples of dilatant fluids are solution with suspended starch or sand, sugar in water.
Fluids for which apparent viscosity decreases with du/dy are called Pseudo Plastic.
Pseudo plastic fluid are also called shear thinning fluid. Examples are paints, polymer solutions, blood, paper pulp, syrup, molasses, milk, gelatine.
Bingham Plastic (ideal plastic) fluids require a certain minimum shear stress t_y (yield stress) before they start flowing. Examples: tooth paste, sewage sludge, drilling mud have time dependent Newtonian Behaviour.

The second derivative of a function f is computed using the fourth-order Central Divided Difference method with a step length h.

The CORRECT expression for the second derivative is

((a))
$\frac{1}{12 h^2}\left[-f_{i+2}+16 f_{i+1}-30 f_i+16 f_{i-1}-f_{i-2}\right]$
((b))
$\frac{1}{12 h^2}\left[-f_{i+2}+16 f_{i+1}-30 f_i+16 f_{i-1}+f_{i-2}\right]$
((c))
$\frac{1}{12 h^2}\left[-f_{i+2}-16 f_{i+1}+30 f_i-16 f_{i-1}-f_{i-2}\right]$
((d))
$\frac{1}{12 h^2}\left[\mathrm{f}{\mathrm{i}+2}+16 \mathrm{f}{\mathrm{i}+1}-30 \mathrm{f}{\mathrm{i}}+16 \mathrm{f}{\mathrm{i}-1}-\mathrm{f}_{\mathrm{i}-2}\right]$

Show Answer

Answer: ((a))

$\frac{1}{12 h^2}\left[-f_{i+2}+16 f_{i+1}-30 f_i+16 f_{i-1}-f_{i-2}\right]$

Explanation:

The fourth-order Central Divided Difference formula for the second derivative of a function $f(x)$ is given by:

$f''(x) = \frac{1}{12h^2} \left[ -f_{i+2} + 16f_{i+1} - 30f_i + 16f_{i-1} - f_{i-2} \right]$

Now, comparing this with the given options:

$(a) \quad \frac{1}{12h^2} \left[ f_{i+2} + 16f_{i+1} - 30f_i + 16f_{i-1} - f_{i-2} \right]$ (Incorrect sign for $f_{i+2}$ )
$(b) \quad \frac{1}{12h^2} \left[ -f_{i+2} - 16f_{i+1} + 30f_i - 16f_{i-1} - f_{i-2} \right]$ (Incorrect sign for several terms)
$(c) \quad \frac{1}{12h^2} \left[ -f_{i+2} + 16f_{i+1} - 30f_i + 16f_{i-1} + f_{i-2} \right]$ (Incorrect sign for $f_{i-2}$ )
$(d) \quad \frac{1}{12h^2} \left[ -f_{i+2} + 16f_{i+1} - 30f_i + 16f_{i-1} - f_{i-2} \right]$ (Correct expression)

Thus, the correct answer is option (d).

The longitudinal sections of a runway have gradients as shown in the table.

End of end for sections or runway (m)	Graident (%)
0 to 200	+1.0
200 to 600	-1.0
600 to 1200	+0.8
1200 to 1600	+0.2
1600 to 2000	-0.5

Consider the reduced level (RL) at the starting point of the runway as 100 m. The effective gradient of the runway is

((a))
0.18%
((b))
0.02%
((c))
0.35%
((d))
0.28%

Show Answer

Answer: ((d))

0.28%

Explanation:

End-to-end section of runway (m)	Gradient (%)	RL (m)
0		100 (given)
200	+1	$100+200 \times \frac{1}{100}=102$
600	-1	$102-400 \times \frac{1}{100}=98$
1200	+0.8	$98+600 \times \frac{0.8}{100}=102.8$
1600	+0.2	$102.8+400 \times \frac{0.2}{100}=103.6$
2000	-0.5	$103.6-400 \times \frac{0.5}{100}=101.6$

The effective gradient is given by

$=\frac{103.6-98}{2000} \times 100=0.28 \%$

The steel angle section shwon in the figure has elastic section modulus of 150.92 cm3 about the horizontal X-X axis, which passes through the centroid of the section.

The shape factor of the section is ________- (rounded off to 2 decimal places)

To finalize the direction of a survey, four surveyors set up a theodolite at a station P and performed all the temporary adjustments. From the station P, each of the surveyors observed the bearing to a tower located at station Q with the same instrument without shifting it. The bearing observed by the surveyors. are 30°30'00'', 30°29'40'', 30°30'20'' and 30°31'20''. Assuming that each measurement is taken with equal precision, the most probable value of the bearing is

((a))
30°31'20''
((b))
30°29'40''
((c))
30°30'20''
((d))
30°30'00''

Show Answer

Answer: ((c))

30°30'20''

Explanation:

The most probable value with equal weightage is given by:

= $30°30'00''+30°29'40''+30°30'20''+30°31'20'' \over 4$ = 30°30'20''

Consider two ordinary differential equations (ODEs):

$P: \frac{d y}{d x}=\frac{x^4+3 x^2 y^2+2 y^4}{x^3 y}$

$Q: \frac{d y}{d x}=\frac{-y^2}{x^2}$

Which one of the following options is CORRECT?

((a))
P is homogeneous ODE and Q is an exact ODE.
((b))
P is a nonhomogenous ODE and Q is not an exact ODE
((c))
P is homogeneous ODE and Q is not an exact ODE
((d))
P is a non homogeneous ODE and Q is an exact ODE

Show Answer

Answer: ((c))

P is homogeneous ODE and Q is not an exact ODE

Explanation:

As per given data:

$P: \frac{d y}{d x}=\frac{x^4+3 x^2 y^2+2 y^4}{x^3 y}$

$\frac{d y}{d x}=\left(\frac{x^4}{x^3 y}\right)+\left(\frac{3 x^2 y^2}{x^3 y}\right)+\left(\frac{2 y^4}{x^3 y}\right)$

$\frac{d y}{d x}=\left(\frac{x}{y}\right)+3\left(\frac{y}{x}\right)+2\left(\frac{y}{x}\right)^3$

$\frac{d y}{d x}=\left(\frac{1}{y / x}\right)+3\left(\frac{y}{x}\right)+2\left(\frac{y}{x}\right)^3$

$\frac{d y}{d x}=f\left(\frac{y}{x}\right)$

So P is a homogenous ODE

$Q: \frac{d y}{d x}=\frac{-y^2}{x^2}$

x²dy = -y²dx

y²dx + x²dy = 0

$M=y^2 \quad \frac{d M}{\partial y}=2 y$

$N=x^2 \quad \frac{\partial N}{\partial x}=2x$

So Q is non exact ODE

∴ The correct answer is option (3).

The contact presure distribution shown in the figure belongs to a

((a))
flexible footing resting on a cohesive soil
((b))
rigid footing resting on a cohesive soil
((c))
rigid footing resting on a cohesionless soil
((d))
flexible footing resting on a cohesionless soil

Show Answer

Answer: ((b))

rigid footing resting on a cohesive soil

Explanation:

1. Flexible footing over clayey soil:

In flexible footing, the contact pressure at the interface between footing and soil is uniformly distributed producing dish-shape pattern in clayey soil.

2. Flexible footing over Granular soil:

In granular soil, modulus of elasticty (E_s) varies across the width being maximum at the centre and minimum at edge. As E is maximum at centre, defflection is less at centre. As E is less at edge deflection is more at edge.

3. Rigid footing on Clayey soil:

In case of flexible footing, deflection is more at centre. Hence pressure developed at centre is less. Deflection is less in flexible footing at edge, hence in rigid footing pressure developed is more at edge.

4. Rigid footing on Granular soil

Various stresses in jointed plain concrete pavement with slab size of 3.5m × 4.5m are denoted as follows:

Wheel load stress at interior = $S_{w l}^i$

Wheel load stress at edge = $S_{\mathrm{wl} }^{\mathrm{e}}$

Wheel load stress at corner = $\mathrm{S}_{\mathrm{wl} }^{\mathrm{C}}$

Warping stress at interior = $\mathrm{S}_{\mathrm{t}}^{\mathrm{i}}$

Warping stress at edge = $\mathrm{S}_{\mathrm{t}}^{\mathrm{e}}$

Warping stress at corner = $\mathrm{S}_{\mathrm{t}}^{\mathrm{c}}$

Frictional stress between slab and supporting layer = S_f

The critical stress combination in the concrete slab during a summer midnight is

((a))
$\mathrm{S}{\mathrm{wl} }^{\mathrm{C}}+\mathrm{S}{\mathrm{t}}^{\mathrm{C}}$
((b))
$\mathrm{S^c}{\mathrm{wl} }+\mathrm{S}{\mathrm{t}}^{\mathrm{c}}+\mathrm{S}_{\mathrm{f}}$
((c))
$\mathrm{S}{\mathrm{wl} }^{\mathrm{e}}+\mathrm{S}{\mathrm{t}}^{\mathrm{e}}+\mathrm{S}_{\mathrm{f}}$
((d))
$\mathrm{S}{\mathrm{wl} }^{\mathrm{e}}-\mathrm{S}{\mathrm{t}}^{\mathrm{e}}+\mathrm{S}_{\mathrm{f}}$

Show Answer

Answer: ((a))

$\mathrm{S}{\mathrm{wl} }^{\mathrm{C}}+\mathrm{S}{\mathrm{t}}^{\mathrm{C}}$

E×planation:

Given Information:

The pavement is jointed plain concrete with a slab size of 3.5m × 4.5m.
Different stresses acting on the slab:
Wheel load stress at different locations:
Interior: SiwlSwli S^i_{wl}

Edge: SewlSwle S^e_{wl}

**Corner: SCwlSwlC S^C_{wl}

Warping stress at different locations:
Interior: SitSti S^i_t

Edge: SetSte S^e_t

**Corner: SCtStC S^C_t

Frictional stress: SfSf S_f

Understanding Critical Stress Condition at Summer Midnight

At midnight during summer, the pavement slab cools down, leading to negative temperature gradients:
Warping stress at the corner (SCtStC S^C_t

) is negative (due to contraction).

Wheel load stress at the corner (SCwlSwlC S^C_{wl}

) is always positive.

**Frictional stress (SfSf S_f

) does not significantly affect corner stresses**.

Since the warping stress is negative, the critical stress at the corner is simply the sum of the wheel load stress and the warping stress:

$\mathrm{S}{\mathrm{wl} }^{\mathrm{C}}+\mathrm{S}{\mathrm{t}}^{\mathrm{C}}$

A 3 m long, horizontal, rigid, uniform beam PQ has negligible mass. The beam is subjected to a 3 kN concentrated vertically downward force at 1 m from P, as shown in the figure. The beam is resting on vertical linear springs at the ends P and Q. For the spring at the end P, the spring constant K_p = 100 kN/m.

If the beam does not rotate under the application of the force and displaces only vertically, the value of the spring constant K_Q (in kN/m) for the spring at the end Q is

((a))
150
((b))
50
((c))
100
((d))
200

Show Answer

Answer: ((b))

Explanation:

ΣM_Q↓= 0 ⇒ R_P × 3 - 3 × 2 = 0

RP = 2kN & R_Q= 1 kN

As beam PQ is rigid & for no rotation settlement at both P & Q should be same.

ΔP = ΔQ

$\frac{\mathrm{R}{\mathrm{P}}}{\mathrm{~K}{\mathrm{P}}} =\frac{\mathrm{R}{\mathrm{Q}}}{\mathrm{~K}{\mathrm{Q}}} $

$⇒ \mathrm{~K}{\mathrm{Q}} =\frac{\mathrm{R}{\mathrm{Q}}}{\mathrm{R}{\mathrm{P}}} × \mathrm{K}{\mathrm{P}}=\frac{1}{2} × 100=50 \mathrm{kN} / \mathrm{m}$

The value of the spring constant KQ (in kN/m) for the spring at the end Q is 50 kN/m.

The function f(x) = x³- 27x + 4, 1 ≤ x ≤ 6 has

((a))
Inflection point
((b))
Saddle point
((c))
Minima point
((d))
Maxima point

Show Answer

Answer: ((c))

Minima point

Explanation:

As peer given data:

f(x) = x3 - 27x + 4

f'(x) = 3x2 - 27

f'(x) = 0

3x2 - 27 = 0

x = ±3

x = 3 ∈ (1, 6)

So at x = 3 function has point at local minima.

The structural design method that DOES NOT take into account the safety factors on the design load is

((a))
working stress method
((b))
Ultimate load method
((c))
load factor method
((d))
limit state method

Show Answer

Answer: ((a))

working stress method

Explanation:

The Working Stress Method (WSM) is a traditional structural design approach that uses elastic theory and ensures that stresses in materials remain within their permissible limits under working loads. It does not incorporate safety factors on loads, but instead applies safety factors to material strength.

Additional Information

Ultimate Load Method (ULM) applies a factor of safety on loads to ensure failure does not occur at the ultimate stage.
Load Factor Method is another term for ULM, incorporating load factors for safety.
Limit State Method (LSM) considers safety factors on both loads and material strengths, making it the most advanced and widely used approach.

A partial differential equation

$\frac{\partial^2 T}{\partial x^2}+\frac{\partial^2 T}{\partial y^2}=0$

is defined for the two-dimensional field T : T (x, y), inside a planar square domain of size 2m × 2m. Three boundary edges of the square domain are maintained at value T = 50, whereas the fourth boundary edge is maintained at T = 100.

The value of T at the center of the domain is

((a))
75.0
((b))
50.0
((c))
87.5
((d))
62.5

Show Answer

Answer: ((d))

62.5

Explanation:

The given partial differential equation is the Laplace equation:

$\frac{\partial^2 T}{\partial x^2}+\frac{\partial^2 T}{\partial y^2}=0$

This equation describes steady-state heat conduction in a 2D square domain.

Given Boundary Conditions:

The square domain is 2m × 2m.
Three boundaries are kept at T = 50.
The fourth boundary is maintained at T = 100.
We need to determine the temperature at the center of the domain.

Concept of Temperature Distribution:

The Laplace equation ensures that the temperature field follows a smooth variation.
In a square plate with three sides at a lower temperature and one side at a higher temperature, the center temperature will be an average of the boundary conditions due to symmetry.

Calculation:

Using approximate averaging techniques for steady-state conduction in such problems, the center temperature TcenterT_{\text{center}}Tcenter is roughly estimated as:

T_center ≈ [3(50)+100]/4 = [150 + 100]/4 = 250/4 = 62.5

For a thin-walled section shown in the figure, points P, Q and R are located on the major bending axis X-X of the section. Point Q is located on the web whereas point S is located at the intersection of the web and the top flange of the section.

Qualitatively, the shear center of the section lies at

((a))
Q
((b))
P
((c))
R
((d))
S

Show Answer

Answer: ((c))

Explanation:

The shear center is the point through which the applied transverse shear force must pass to avoid any twisting of the section.

Key Observations:

The section consists of a vertical web and two unequal flanges.
The top flange is asymmetric, extending further to the right than the bottom flange.
The shear center shifts towards the more extended flange to balance the torque effects.

Why is the Shear Center at Point R?

For thin-walled open sections like C-sections or asymmetric I-sections, the shear center is generally located towards the larger flange.
Since the top flange extends more to the right, the shear center must shift rightward along the X-X axis.
Point R is located towards the extended top flange, making it the correct location for the shear center.

A reinforced concrete pile of 10 m length and 0.7 m diameter is embedded in a saturated pure clay with unit cohesion of 50 kPa. If the adhesion factor is 0.5, the net ultimate uplift pullout capacity (in kN) of the pile is ______. (rounded off to the nearest integer).

In general, the outer edge is raised above the inner edge in horizontal curves for

((a))
Highways and Railways only
((b))
Highways only
((c))
Railways and Taxiways only
((d))
Highways, Railways and Taxiways

Show Answer

Answer: ((a))

Highways and Railways only

Explanation:

The concept of raising the outer edge above the inner edge in horizontal curves is known as super-elevation (cant or banking) and is primarily used in:

Highways – To counteract centrifugal force on vehicles, reducing the risk of skidding and overturning.
Railways – To provide a smooth transition for trains on curves, reducing lateral forces and rail wear.

However, taxiways generally do not require super-elevation because aircraft move at lower speeds on taxiways, and the main concern is ensuring smooth, level surfaces for ground maneuvering rather than counteracting lateral forces.

Thus, the correct choice is Highways and Railways only

What is the CORRECT match between the air pollutants and treatment techniques given in the table?

	Air pollutants		Treatment techniques
P.	NO₂	i.	Flaring
Q.	SO₂	ii.	Cyclonic separator
R.	CO	iii.	Lime scrubbing
S.	Particles	iv.	NH₃injection

((a))
P - iv, Q - iii, R - i, S - ii
((b))
P - i, Q - ii, R - iii, S - iv
((c))
P - ii, Q - i, R - iv, S - iii
((d))
P - ii, Q - iii, R - iv, S - i

Show Answer

Answer: ((a))

P - iv, Q - iii, R - i, S - ii

Explanation:

P. NO₂ → iv. NH₃ injection

(Selective catalytic reduction (SCR) or selective non-catalytic reduction (SNCR) involves NH₃ injection to reduce NO₂ emissions.)

Q. SO₂ → iii. Lime scrubbing

(Lime scrubbing is used in flue gas desulfurization (FGD) to remove SO₂ from emissions.)

R. CO → i. Flaring

(Flaring is used to burn off excess CO and other hydrocarbons in industrial processes.)

S. Particles → ii. Cyclonic separator

(Cyclonic separators help in removing particulate matter from air streams.)

The statements P and Q are related to matrices A and B , which are conformable for both addition and multiplication.

P : (A + B)^T= A^T+ B^T

Q : (AB)^T= A^TB^T

Which one of the following options is CORRECT?

((a))
Both P and Q are false
((b))
Both P and Q are true
((c))
P is false and Q is true
((d))
P is true and Q is false

Show Answer

Answer: ((d))

P is true and Q is false

Explanation:

According to the properties of a matrix

(i) (A + B)T = AT + BT

The sum of the transpose of matrices is equal to the transpose of the sum of two matrices.

(ii) (AB)T = AT BT

The product of the transpose of two matrices in reverse order is equal to the transpose of the product of them.

What is the CORRECT match between the survey instruments/parts of instruements shown in the table and the operations carried out with them?

	Instruments/Parts of instruments		Operations
P.	Bubble tube	i.	Tacheometry
Q.	Plumb bob	ii.	Minor movements
R.	Tangent screw	iii.	Centering
S.	Stadia cross-wire	iv.	Levelling

((a))
P - ii, Q - iii, R - iv, S - i
((b))
P - iv, Q - iii, R - ii, S - i
((c))
P - iii, Q - iv, R - i, S - ii
((d))
P - i, Q - iii, R - ii, S - iv

Show Answer

Answer: ((b))

P - iv, Q - iii, R - ii, S - i

Explanation:

P. Bubble tube → iv. Levelling

(The bubble tube is used in leveling instruments to check and adjust the horizontal level.)

Q. Plumb bob → iii. Centering

(A plumb bob helps in centering the instrument precisely over a survey point.)

R. Tangent screw → ii. Minor movements

(Tangent screws allow fine adjustments and minor movements for precise alignment.)

S. Stadia cross-wire → i. Tacheometry

(Stadia cross-wires are used in tacheometry for measuring distances using a theodolite or level.)

Consider the statements P and Q.

P: In a Pure project organization, the project manager maintains complete authroity and has maximum control over the project.

Q: A matrix organization structure facilitates quick response to changes, conflicts, and project needs.

Which one of the following options is CORRECT?

((a))
P is false and Q is true
((b))
Both P and Q are false
((c))
Both P and Q are true
((d))
P is true and Q is false

Show Answer

Answer: ((c))

Both P and Q are true

Explanation:

Statement P: "In a Pure project organization, the project manager maintains complete authority and has maximum control over the project."

True

A pure project organization (also called a projectized organization) gives the project manager full authority over project execution, decision-making, and resources.
The project team reports directly to the project manager, as confirmed by standard project management literature.

Statement Q: "A matrix organization structure facilitates quick response to changes, conflicts, and project needs."

True

As per the matrix organization structure description in the images, one of its advantages is:
"The structure facilitates quick response to changes, conflicts, and project needs."
The matrix structure combines functional and project-based management, ensuring efficient adaptability to project demands and conflicts.

A 2 m wide rectangular channel is carrying a discharge of 30 m³/s at a bed slope of 1 in 300. Assuming the energy correction factor as 1.1 and acceleration due to gravity as 10 m/s², the critical depth of flow (in meters) is _________ (rounded off) to 2 decimal places)

Which one of the following saturated fine-grained soils can attain a negative Skempton's pore pressure coefficient (A)?

((a))
Quick clays
((b))
Lightly-consolidated clays
((c))
Normally-consolidated clays
((d))
Over-consolidated clays

Show Answer

Answer: ((d))

Over-consolidated clays

Explanation:

For normally consolidated clays A = 0.5 - 1

For OC clays, A = f (OCR), for heavily over consolidated clays, A < 0.

Additional Information

Pore Pressure Parameters:

(i) Sometimes it is not possible to determine pore pressure practically, then the theoretical approach given by skempton can be adopted.

(ii) Pore pressure parameters A and B are empirical coefficients that are used to express the response of pore pressure to changes in vertical pressure and lateral pressure under "undrained condition".

Parameter B:

(i) This parameter is defined under cell pressure stage and represents the ratio of change in pore pressure to the change in cell pressure.

B = ΔUc/Δσc

(ii) B varies from o to 1 depending on the degree of saturation. "B" is zero for dry soil and is equals unity for fully saturated soil.

Parameter A:

(i) This parameter is valid in deviator stage and is defined in terms of another parameter A̅, such that

A̅ = A.B

(ii) The parameter A̅ represents the ratio of change in pore pressure to change in deviator stress during shear stage.

$\bar A = \dfrac{{\Delta {U_d}}}{{\Delta {\sigma _d}}} = \dfrac{{\Delta {U_d}}}{{\Delta ({\sigma _1} - {\sigma _3})}} = \dfrac{{\Delta {U_d}}}{{\Delta {\sigma _1} - \Delta {\sigma _3}}}$

(iii) The parameter depends upon strain in soil, degree of saturation, over consolidated ratio, stratification of soil, sample disturbance, etc. Its value may be as low as -0.5 for over consolidated soil with high O.C.R. to as high as 3 or loose saturated sand.

For a reconnaisssance survey, it is necessary to obtain vertical aerial photographs of a terrain at an average scale of 1 : 13000 using a camera. If the permissible flying height is assumed as 3000 m above a datum and the average terrain elevation is 1050 m above the datrum, the required focal length (in mm ) of the camera is

((a))
125
((b))
100
((c))
150
((d))
200

Show Answer

Answer: ((c))

150

Explanation:

Given Data:

Required scale of aerial photograph = 1 : 13,000
Permissible flying height above datum = 3000 m
Average terrain elevation above datum = 1050 m

Now,

Scale $=\frac{1}{13000}$

Hence,

$\mathrm{S} =\frac{\mathrm{f}}{\mathrm{H}-\mathrm{h}}$

$\frac{1}{13000} =\frac{\mathrm{f}}{3000-1050}$

∴ f = 0.15 m = 150 mm

Consider teh following data for a project of 300 days duration.

Budgeted cost of work scheduled (BCWS) = Rs. 200

Budgeted cost of work performed (BCWP) = Rs. 150

Actual cost of work performed (ACWP) = Rs. 190 The 'schedule variance' for the project is

((a))
(-)Rs. 50
((b))
(+)Rs. 50
((c))
(+)50 days
((d))
(-)50 days

Show Answer

Answer: ((a))

(-)Rs. 50

Explanation:

As per given data:

Budgeted Cost of Work Scheduled (BCWS) = Rs. 200
Budgeted Cost of Work Performed (BCWP) = Rs. 150
Actual Cost of Work Performed (ACWP) = Rs. 190

Schedule Variance is given by

= BCWP - BCWS

= 150 - 200

= -50

A simply supported, uniformly loaded, two-way slab panel is torsionally unrestrained. The effective span lengths along the short span ( x ) and long span ( y ) directions of the panel are I_x and I_y respectively. The design moments for the reinforcements along the x and y directions are M_ux and M_uy respectively. By using Rankine-Grashoff method, the ratio M_ux/M_uy is proportional to

((a))
Iy/Ix
((b))
Ix/Iy
((c))
(Iy/Ix)²
((d))
(Ix/Iy)2

Show Answer

Answer: ((c))

(Iy/Ix)²

Explanation:

Grashoff method is used to determine the bending moment in two way RCC slab. It is used for load distribution when the slab is simply supported on all four edges and the corners are not held down.

As per Rankine Grashoff method

$M_{u x}=\frac{r^4}{8\left(1+r^4\right)} w \epsilon_x^2$

$M_{u y}=\frac{r^2}{8\left(1+r^4\right)} w \epsilon_x^2$

Here $r=\frac{\ell_y}{\ell_x}$

$\Rightarrow \quad \frac{\mathrm{M}{\mathrm{ux}}}{\mathrm{M}{\mathrm{uy}}}=\mathrm{r}^2=\left(\frac{\ell_{\mathrm{y}}}{\ell_{\mathrm{x}}}\right)^2$

The expression for computing the effective interest rate (i_eff) using continous compounding for a nominal interest rate of 5% is

$i_{\text {eff }}=\lim _{m \rightarrow \infty}\left(1+\frac{0.05}{m}\right)^m-1$

The effective interest rate (in percentage) is _________ (rounded off to 2 decimal places).

In a sample of 100 heart partients, each patients has 80% chance of having a heart attack without medicine X. It clinically known that medicine X reduces the probability of having a heart attack by 50%. Medicine X is taken by 50 of these 100 patients. The probability that a randomly selected patient, out of the 100 patients, takes medicine X and has a heart attack is

((a))
40%
((b))
30%
((c))
20%
((d))
60%

Show Answer

Answer: ((c))

20%

Explanation:

Given Data:

Total patients = 100
Probability of a heart attack without medicine X = 80% = 0.8
Medicine X reduces the probability of a heart attack by 50%.
Number of patients taking medicine X = 50
Number of patients not taking medicine X = 50

Step 1: Probability of a heart attack for patients taking medicine X

Without medicine X, probability of a heart attack = 0.8
With medicine X, probability of a heart attack = 0.8 × 0.5 = 0.4

Step 2: Probability of selecting a patient who takes medicine X and has a heart attack

Probability that a randomly selected patient takes medicine X = 50/100 = 0.5
Probability that a patient who takes medicine X has a heart attack = 0.4

Thus, the required probability:

P(Takes medicine X and has a heart attack)=P(Takes medicine X)×P(Heart attack | Takes medicine X)

= 0.5 × 0.4 = 0.2 (or 20%)

A 2 m × 1.5 m tank of 6 m height is provided with a 100 mm diameter orifice at the center of its base. The orifice is plugged and the tank is filled up to 5 m height. Consider the average value of discharge coefficient as 0.6 and acceleration due to gravity (g) as 10 m/s². After unplugging the orifice, the time (in seconds) taken for the water level to drop from 5 m to 3.5 m under free discharge condition is ___________. (rounded off to 2 decimal places).

A round-bottom trianglular lined canal is to be liad at a slope of 1 m in 1500, to carry a discharge of 25 m³/s. The side slopes of the canal cross-section are to be kept at 1.25 H: 1V. If Manning's roughtness coefficient is 0.013, the flow depth (in meters) will be in the range of

((a))
1.94 to 1.97
((b))
2.61 to 2.64
((c))
2.24 to 2.27
((d))
2.39 to 2.42

Show Answer

Answer: ((d))

2.39 to 2.42

Explanation:

cotθ = 1.25

θ = 0.675 rad

n = 0.013

$\mathrm{Q} =25 \mathrm{~m}^3 / \mathrm{s}, \mathrm{~S}=\frac{1}{1500}$

Using manning's equation:

Area = y² (θ + cot θ) \

A = 1.925 y²

$R =\frac{y}{2}$

$Q =\frac{A}{n} R^{2 / 3}(S)^{1 / 2}$

$25 =\frac{1.925 y^2}{0.013}\left(\frac{y}{2}\right)^{2 / 3}\left(\frac{1}{1500}\right)^{1 / 2}$

⇒ y = 2.40 m

The flow depth (in meters) is 2.40.

Differential levelling is carried out from point P(BN +200.000 m) to point R.

The reading taken are given in the table.

Points	Staff readings (m)	Remarks
Back Sight	Fore Sight
P	(-)2.050		BM: +200.000 m
Q	1.050	0.950	Q is a change point
R		(-)1.655

educed level (in meters) of the point R is _________ (rounded off to 3 decimal places)

The horizontal beam PQRS shown in the figure has a fixed support at point P, an internal hinge at point Q, and a pin support at point R. A concentrated vertically downward load (V) of 10 kN can act at any point over the entire length of the beam.

The maximum magnitude of the moment reaction (in $\mathrm{kN} . \mathrm{m}$ ) that can act at the support P due to V is __________ (in integer).

A rectangular channel is 4.0 m wide and carries a discharge of 2.0 m³/s with a depth of 0.4 m. The channel transitions to a maximum width contraction at a downstream location, without influencing the upstream flow condtions. The width (in meters) at the maximum contraction is _________(rounded off to 2 decimal places)

The consoliated data of a spot study for a certain stretch of a highway is given in the table.

Speed range (kmph)	Number of observations
0-10	7
10-20	31
20-30	76
30-40	129
40-50	104
50-60	78
60-70	29
70-80	24
80-90	13
90-100	9

The "upper speed limit" (in kmph) for the traffic sign is

((a))
70
((b))
55
((c))
50
((d))
65

Show Answer

Answer: ((b))

Explanation:

As per given data:

Mid speed (kmph)	% of vehicles	Cumulative %
5	1.4	1.4
15	6.2	7.6
25	15.2	22.8
35	25.8	48.6
45	20.8	69.4
55	15.6	85
65	5.8	90.8
75	4.8	95.6
85	2.6	98.2
95	1.8	100

Hence 85^th percentile speed or safe speed = 55 km/hr

A linearly elalstic beam of length 2/ with flexural rigidity El has neglitible mass. A massless spring with a spring constant k and a rigid block of mass m are attached to the beam as shown in the figure.

The natural frequency of this system is

((a))
$\sqrt{\frac{\mathrm{k} \beta^3+6 \mathrm{EI}}{\mathrm{m} \beta^3}}$
((b))
$\sqrt{\frac{k \beta^3+48 E I}{m \beta^3}}$
((c))
$\sqrt{\frac{6 \mathrm{Elk}}{\left(\mathrm{k} \beta^3+6 \mathrm{EI}\right)_{\mathrm{m}}}}$
((d))
$\sqrt{\frac{48 \mathrm{Elk}}{\left(\mathrm{k} \beta^3+48 \mathrm{EI}\right) \mathrm{m}}}$

Show Answer

Answer: ((a))

$\sqrt{\frac{\mathrm{k} \beta^3+6 \mathrm{EI}}{\mathrm{m} \beta^3}}$

Explanation:

As per given data:

Let us consider the stiffness of the beam as k_b.

Here both the stiffness elements are in parallel.

k_eq= k₁+ k₂

$=\mathrm{k}+\frac{48 \mathrm{EI}}{(2 \ell)^3}=\mathrm{k}+\frac{6 \mathrm{EI}}{\ell^3}$

Natural frequency is given by

$\omega_{\mathrm{n}}=\sqrt{\frac{\mathrm{k}_{\mathrm{eq}}}{\mathrm{~m}}}$

$=\sqrt{\frac{\left(\mathrm{k}+\frac{6 \mathrm{EI}}{\ell^3}\right)}{\mathrm{m}}}=\sqrt{\frac{\mathrm{k} \ell^3+6 \mathrm{EI}}{\mathrm{~m} \ell^3}}$

Consider the statements P and Q related to the analysis/design of retaining walls.

P: When a rough retaining wall moves toward the backfill, the wall friction force/resistance mobilizes in upward direction along the wall.

Q: Most of the earth pressure theories calculate the earth pressure due to surcharge by neglecting the actual distribution of stresses due to surcharge.

Which of the following options is CORRECT?

((a))
Both P and Q are true
((b))
P is true and Q is false
((c))
P is false and Q is true
((d))
Both P and Q are false

Show Answer

Answer: ((a))

Both P and Q are true

Explanation:

Statement P:

"When a rough retaining wall moves toward the backfill, the wall friction force/resistance mobilizes in the upward direction along the wall."

True

When a rough retaining wall moves toward the backfill, it activates passive earth pressure conditions.
In this scenario, wall friction mobilizes in an upward direction due to the interaction between the soil and the wall.

Statement Q:

"Most of the earth pressure theories calculate the earth pressure due to surcharge by neglecting the actual distribution of stresses due to surcharge."

True

Earth pressure theories, such as Rankine’s and Coulomb’s theories, generally assume uniform stress distribution due to surcharge load.
The actual variation of stresses due to surcharge is neglected for simplicity in calculations.

A circular settling tank is to be desinged for primary treatment of sewage of a flow rate of 10 million liters/day. Assume a detention period of 2.0 hours and surface loading rate of 40000 liters/m²/day. The height (in meters) of the water column in the tank is __________ (rounded off to 2 decimal places)

Consider two matrices

$A=\left[\begin{array}{lll}2 & 1 & 4 \ 1 & 0 & 3\end{array}\right]$ and $B=\left[\begin{array}{cc}-1 & 0 \ 2 & 3 \ 1 & 4\end{array}\right]$

The determinant of the matrix AB is __________ (in integer).

A concrete column section of size 300 mm × 500 mm as shown in the figure is subjected to both axial compression and bending along the major axis. The depth of the neutral axis (x_u) is 1.1 times the depth of the column, as shown.

The maximum compressive strain (ε_c) at highly compressive extreme fiber in concrete, where there is no tension in the section, is _________ × 10^-3 (rounded off to 2 decimal places)

A vertical summit curve on a freight corridor is formed at the intersection of two gradients, +3.0% and -5.0%.

Assume the following:

Only large-sized trucks are allowed on this corridor.

Design speed =80 kmph

Eye height of truck drivers above the road surface = 2.30 m

Height of object above the road surface for which trucks need to stop = 0.35 m

Total reaction time of the truck drivers = 2.0 s

Coefficient of longitudinal friction of the road = 0.36

Stopping sight distance gets compesnated on the gradient.

The design length of the summit curve (in meters) to accommodate the stopping sight distance is ____________ (rounded off to 2 decimal places).

An organic waste is represented as C₂₄₀O₂₀₀H₁₈₀N₅S

(Atomic weights: S-32, H-1, C-12, O-16, N-14)

Assume complete conversion of S to SO₂ while burning.

SO₂ generated (in grams) per kg of this waste is __________ (rounded off to 1 decimal place).

In the context of pavement material characterization, the CORRECT statement(s) is/are

((a))
In compacted bituminous mix. voids in the mineral aggregate (VMA) is equal to the sum of total volume of air voids (VV) and total volume of bitumen (Vb).
((b))
The toughness and hardness of road aggregates are determined by Los Angeles abrasion test and aggregate impact test, respectively.
((c))
The load penetration curve of CBR test may need origin correction due to the non-vertical penetrating plunger of the loading machine.
((d))
Grading of normal (unmodified) bitumen binders is done based on viscosity test results.

Show Answer

Answer: ((a))

In compacted bituminous mix. voids in the mineral aggregate (VMA) is equal to the sum of total volume of air voids (VV) and total volume of bitumen (Vb).

Explanation:

The correct statements regarding pavement material characterization are:

"In compacted bituminous mix, voids in the mineral aggregate (VMA) is equal to the sum of total volume of air voids (Vᵥ) and total volume of bitumen (Vᵦ)."

Correct - VMA represents the total void space within the aggregate structure, including both air voids and bitumen. 2. "The toughness and hardness of road aggregates are determined by Los Angeles abrasion test and aggregate impact test, respectively."

Incorrect -

The Los Angeles Abrasion Test measures the abrasion resistance (not toughness).
The Aggregate Impact Test measures toughness (resistance to sudden impact), whereas hardness is generally evaluated using crushing strength tests.

"The load penetration curve of the CBR test may need origin correction due to the non-vertical penetrating plunger of the loading machine."

Correct - A non-vertical plunger can cause errors in the initial part of the CBR curve, requiring origin correction. 4. "Grading of normal (unmodified) bitumen binders is done based on viscosity test results."

Correct - Bitumen binders are graded based on viscosity tests, which determine their performance in different temperature conditions.

A hypothetical multimedia filter, consisting of anthracite particles (specific gravity: 1.50), silica sand (specific gravity: 2.60), and ilmenite sand (specific gravity: 4.20 ), is to be designed for treating water/wastewater. After backwashing, the particles should settle forming three layers: coarse anthracite particles at the top of the bed, silica sand in the middle, and small ilmenite sand particles at the bottom of the bed.

Assume

(i) Slow discrete settling (Stoke's law is applicable)

(ii) All particles are spherical

(iii) Diameter of silica sand particles is 0.20 mm

The correct option fulfilling the diameter requirements for this filter media is

(a)

((a))
diameter of anthracite particles is slightly less than 0.64 mm and diameter of ilmenite particles is slightly less than 0.10 mm.
((b))
diameter of anthracite is slightly greater than 0.64 mm and diameter of ilmenite particles is slightly than 0.10 mm
((c))
diameter of anthracite is slightly greater than 0.35 mm and diameter of ilmenite particles is slightly less than 0.141 mm.
((d))
diameter of anthracite particles is slightly less than 0.35 mm and diameter of ilmenite particles is slightly greater than 0.141 mm

Show Answer

Answer: ((d))

diameter of anthracite particles is slightly less than 0.35 mm and diameter of ilmenite particles is slightly greater than 0.141 mm

Explanation:

Anthracite Gs = 1.50
Silica sand Gs = 2.60 D = 0.2 mm
IImenite sand Gs = 4.20

Let the settling velocity of the anthracite pacticle at top, silica pacticle at middle & ilmenite pacticle at bottom, after back washing be (V_S)_T,(VS), (VS)_B respectively.

For Middle silica sand

$v_S =\frac{\left(G_{s_1}-1\right) \gamma_w D_1^2}{18 \mu}$

vS = K(2.6 - 1)(0.2)2

v_S = K(1.6)(0.2)² where $\mathrm{K}=\frac{\gamma_{\mathrm{w}}}{18 \mu}$

For Top Anthracite

(V_S)_T= K $\left(\mathrm{G}_{\mathrm{s}_2}-1\right)$ (D₂)²

(VS)T = K(0.5) (D2)2

As anthracite lies above the silica layer

So, anthracite particles should have a settling velocity less than silica particles

(V_S)_T< V_S

So, K(0.5) (D2)2 < K(1.6) (0.2)²

D₂ or $D_{\text {Top }}<\sqrt{\frac{1.6 \times(0.2)^2}{0.5}}$

D₂ or D_Top < 0.357 mm

For bottom ilmenite sand

(V_s)_B= K $\left(G_{s_3}-1\right)\left(D_3^2\right)$

(V_S)_B= K(4.2-1) $\left(D_3^2\right)$

As, ilmenite layer lies below silica layers

So, ilemite particles should have a settling velocity greater than silica

i.e., (V_S)_B> V_S

$K(3.2)\left(D_3^2\right)>K(1.6)(0.2)^2$

$D_3 \text { or } D_{\text {bottom }}>\sqrt{\frac{1.6 \times(0.2)^2}{3.2}}$

D₃ or D_bottom > 0.141 mm

Hence, diameter of anthracite particle is slightly greater than 0.35 mm & diameter of ilmenite particle is slightly greater than 0.141 mm

So (d) is the correct option.

A storm with a recorded precipitation of 11.0 cm , as shown in the table, produced a direct run off of 6.0 cm

Time from start (hours)	1	2	3	4	5	6	7	8
Recorded cumulative precipitation (cm)	0.5	1.5	3.1	5.5	7.3	8.9	10.2	11.0

The ϕ-index of this storm is ________ cm/hr (rounded off to 2 decimal places)

A homogeneous earth dam has a maximum water head difference of 15 m between the upstream and downstream sides. A flownet was drawn with the number of potential drops as 10 and the average length of the element as 3 m . Specific gravity of the soil is 2.65 . For a factor of safety of 2.0 against piping failure, void ratio of the soil is _____________ (rounded off to 2 decimal places).

Three vectors $\overrightarrow{\mathrm{p}}, \overrightarrow{\mathrm{q}}$ and $\overrightarrow{\mathrm{r}}$ are given as

$\overrightarrow{\mathrm{p}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$

$\overrightarrow{\mathrm{q}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$

$\overrightarrow{\mathrm{r}}=2 \hat{i}+3 \hat{j}+4 \hat{k}$

Which of the following is/are CORRECT?

((a))
$\overrightarrow{\mathrm{p}} \times(\overrightarrow{\mathrm{q}} \times \overrightarrow{\mathrm{r}})=(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{r}}) \overrightarrow{\mathrm{q}}-(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}}) \overrightarrow{\mathrm{r}}$
((b))
$\overrightarrow{\mathrm{r}} \cdot(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})=(\overrightarrow{\mathrm{q}} \times \overrightarrow{\mathrm{p}}) \cdot \overrightarrow{\mathrm{r}}$
((c))
$\vec{p} \times(\vec{q} \times \vec{r})=(\vec{p} \times \vec{q}) \times \vec{r}$
((d))
$\vec{p} \times(\vec{q} \times \vec{r})+\vec{q} \times(\vec{r} \times \vec{p})+\vec{r} \times(\vec{p} \times \vec{q})=\overrightarrow{0}$

Show Answer

Answer: ((a))

$\overrightarrow{\mathrm{p}} \times(\overrightarrow{\mathrm{q}} \times \overrightarrow{\mathrm{r}})=(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{r}}) \overrightarrow{\mathrm{q}}-(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}}) \overrightarrow{\mathrm{r}}$

Explanation:

(a) $\vec{p} \times(\vec{q} \times \vec{r})=(\vec{p} \cdot \vec{r}) \vec{q}-(\vec{p} \cdot \vec{q}) \vec{r}$

(This is always true for any three given vectors)

(b) We know that $\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}$ is always true but $\vec{a} \times \vec{b} \neq \vec{b} \times \vec{a}$ because $\vec{a} \times \vec{b}=-\vec{b} \times \vec{a}$

This can be true only when $\vec{a} \times \vec{b}=0$

So, $\quad \vec{r} \cdot(\vec{p} \times \vec{q})=\vec{r} .(\vec{q} \times \vec{p})$

$\vec{r} \cdot(\vec{p} \times \vec{q})=-\vec{r} \cdot(\vec{p} \times \vec{q})$

This can be true if

$\vec{r} \cdot(\vec{p} \times q) =0$

$\vec{p} \times \vec{q} =\hat{i}-2 \hat{j}+j$

$(\vec{r} \cdot \vec{p} \times \vec{q})=0$ this is true in this case

(c) $\overrightarrow{\mathrm{p}} \times(\overrightarrow{\mathrm{q}} \times \overrightarrow{\mathrm{r}})$ can't be equal to $(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}) \times \overrightarrow{\mathrm{r}}$ because $\vec{p} \times(\vec{q} \times \vec{r}) \perp \vec{p}$ and $(\vec{p} \times \vec{q}) \times \vec{r}) \perp \vec{r}$

So, $(\vec{p} \times(\vec{q} \times \vec{r}) \neq(\vec{p} \times \vec{q})) \times \vec{r}$

(d) $\overrightarrow{\mathrm{p}} \times(\overrightarrow{\mathrm{q}} \times \overrightarrow{\mathrm{r}})+\overrightarrow{\mathrm{q}} \times(\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{p}})+\overrightarrow{\mathrm{r}} \times(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})=\overrightarrow{0}$

$(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{r}}) \overrightarrow{\mathrm{q}}-(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}}) \overrightarrow{\mathrm{r}}+(\overrightarrow{\mathrm{q}} \cdot \overrightarrow{\mathrm{p}}) \overrightarrow{\mathrm{p}}-(\overrightarrow{\mathrm{q}} \cdot \overrightarrow{\mathrm{r}}) \overrightarrow{\mathrm{p}}+(\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{q}}) \overrightarrow{\mathrm{p}}-(\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{q}}) \overrightarrow{\mathrm{q}}$

0 = 0

$\because \quad \vec{p} \cdot \vec{r} =\vec{r} \cdot \vec{p}$

$\vec{p} \cdot \vec{q} =\vec{q} \cdot \vec{p}$

$\vec{q} \cdot \vec{r} =\vec{r} \cdot \vec{q}$

(Hence this is proved)

A critical activity in a project is estimated to take 15 days to complete at a cost of Rs. 30,000. The activity can be expedited to complete in 12 days by spending a total amount of Rs. 54,000. Consider the statements P and Q.

P: It is economically advisable to complete the activity early by crashing if the indirect cost of the project is Rs. 8,500 per day.

Q: It is economically advisable to complete the activity early by crashing, if the indirect cost of the project is Rs. 10,000 per day.

Which one of the following options is CORRECT?

((a))
P is true and Q is false
((b))
Both P and Q is false
((c))
P is false and Q is true
((d))
Both P and Q are true

Show Answer

Answer: ((d))

Both P and Q are true

Explanation:

Given Data:

Normal duration: 15 days
Normal cost: Rs. 30,000
Crashed duration: 12 days
Crashed cost: Rs. 54,000
Indirect cost: Rs. 8,500 per day (for P) and Rs. 10,000 per day (for Q)

Cost slope (Crashing cost per day) is given by:

$\mathrm{C} / \mathrm{s}=\frac{54000-30000}{15-12}=8000 \mathrm{Rs} / \text { day }$

Crashing is economically beneficial if the savings in indirect cost per day exceed the direct cost of crashing per day.

If the indirect cost per day is Rs. 8,500 (Statement P):

Since Rs. 8,500 > Rs. 8,000, it is economically advisable to crash. So, P is true.

If the indirect cost per day is Rs. 10,000 (Statement Q):

Since Rs. 10,000 > Rs. 8,000, it is also economically advisable to crash. So, Q is true.

A child walks on a level surface from point P to point Q at a bearing of 30°, from point Q to point R at a bearing of 90° and then-directly returns to the starting point P at a bearing of 240°. The straight line paths PQ and QR are 4m each. Assuming that all bearings are measured from the magnetic north in degrees, the straight-line path length RP (in meters) is __________ (rounded off to the nearest integer)

A homogenous, prismatic, linearly elastic steel bar fixed at both the ends has a slenderness ratio (//r) of 105 , where / is the bar length and r is the radius of gyration. The coefficient of thermal expansion of steel is 12 × 10^-6/°C. Consider the effective length of the steel bar as 0.5 / and neglect the self-wieght of the bar.

The differential increase in temperature (rounded off to the nearest integer) at which the bar buckles is

((a))
85°C
((b))
250°C
((c))
400°C
((d))
298°C

Show Answer

Answer: ((d))

298°C

Explanation:

As per given data:

σ_temp = EαΔT

$σ_{\mathrm{cr}} =\frac{\pi^2 \mathrm{E}}{\lambda_{\text {eff }}^2}=\frac{\pi^2 \mathrm{E}}{\left(\ell_{\text {eff }} / r\right)^2}$

For given support $\rightarrow \ell_{\text {eff }}=\frac{1}{2}$

Hence,

$\mathrm{E} α Δ \mathrm{~T} =\frac{\pi^2 \mathrm{E}}{\left(\frac{I}{2 r}\right)^2}$

$Δ \mathrm{~T} =\frac{\pi^2 \times 4}{α 5\left(\frac{I}{2}\right)^2}=\frac{\pi^2 \times 4}{12 \times 10^{-6} \times 105^2}$

The differential increase in temperature (rounded off to the nearest integer) at which the bar buckles is

ΔT = 298°C

A 500 m long water distribution pipeline P with diameter 1.0 m, is used to convey 0.1 m³/s of flow. A new pipeline Q, with the same length and flow rate, is to replace P. The friction factors for P and Q are 0.04 and 0.01 , respectively. The diameter of the pipeline Q (in meters) is __________ (rounded off to 2 decimal places)

A horizontal curve of radius 1080 m (with transition curves on either side) in a Broad Gauge railway track is designed and constructed for an equilibrium speed of 70 kmph. However, a few years after construction, the Railway Authorities decided to run express trains on this track. The maximum allowable cant deficiency is 10 cm.

The maximum restricted speed (in kmph) of the express trains running on this track is ________ (rounded off to the nearest integer)

A drained triaxial test was conducted on a saturated sand specimen using a stress-path triaxial testing system. The specimen failed when the axial stress reached a value of 100 kN/m² from an initial confining pressure of 300 kN.

The angle of shearing plane (in degrees) with respect to horizontal is ____________ (rounded off to the nearest integer).

Consider the statements P, Q and R.

P: Compacted fine-grained soils with flocculated structure have isotropic permeability.

Q: Phreatic surface/line is the line along which the pore water pressure is always maximum.

R : The piping phenomenon occuring below the dam foundation is typically known as blowout piping.

Which of the following option(s) is/are CORRECT?

((a))
Both P and R are true
((b))
Both Q and R are false
((c))
P is false and Q is true
((d))
P is true and R is false

Show Answer

Answer: ((a))

Both P and R are true

Explanation:

Statement P:

"Compacted fine-grained soils with flocculated structure have isotropic permeability."

True - Fine-grained soils compacted on the dry side of optimum moisture content tend to have a flocculated structure (random arrangement). This structure provides approximately equal permeability in all directions (isotropic permeability), as confirmed by the provided explanation.

Statement Q:

"Phreatic surface/line is the line along which the pore water pressure is always maximum."

False - The phreatic line represents the topmost flow line below which seepage occurs in a dam body. However, pore water pressure is not maximum along the phreatic line; rather, it is atmospheric at this level. The maximum pore water pressure occurs below the phreatic line, not along it.

Statement R:

"The piping phenomenon occurring below the dam foundation is typically known as blowout piping."

False -

Piping phenomenon refers to the erosion of soil due to excessive seepage forces.
The correct term for piping failure below the dam foundation is "heave" or "boiling", not blowout piping. Blowout typically refers to failure due to excessive water pressure in confined aquifers rather than dam foundation piping.

The table shows the activities and their durations and dependencies in a project.

Activity	Duration(days)	Depends on
A	8	-
B	4	A
C	4	B
D	4	C, L
F	4	A
G	4	F
H	6	G, L
K	10	A
L	6	F, K

The total duration (in days) of the project is ___________ (in integer)

The in-situ percentage of voids of a sand deposit is 50%. The maximum and minimum densities of sand determined from the laboratory tests are 1.8 g/cm³ and 1.3 g/cm3, respectively. Assume the specific gravity of sand as 2.7.

The relative density index of the in-situ sand is ___________ (rounded off to 2 decimal places)

For the 6 m long horizontal cantilever beam PQR shown in the figure. Q is the mid-point. Segment PQ of the beam has flexural rigidity El = 2 × 10⁵kN, m² whereas the segment QR has infinite flexural rigidity. Segment QR is subjected to uniformly distributed, vertically downward load of 5 kN/m

The magnitude of the vertical displacement (in mm) at point Q is _________ (rounded off to 3 decimal places)

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