GATE CE 2017 Official Paper: Shift 1 (Previous Year Paper)

Correct answer: Option 3.

The sentence talks about a general truth: bacteria gets destroyed when milk gets heated to 80 degree Celsius.

Because it is a general truth, the tense should be simple present tense.

Correct answer: 3.

The sentence implies that somebody is trying to create problems regarding somebody else's email account.

Hence, the blank requires a word which means to interfere with (something) in order to cause damage or make unauthorized alterations.

For given condition ⇒ All benches are beds. No bed is a bulb. Some bulbs are lamps, the following Venn diagrams is possible.

According to statement ⇒ Some beds are lamps

According to statement ⇒ Some lamps are beds.

∴ None of the statement can be logically inferred from the given condition.

Concept:

The volume of a right circular cone (V) is given by,

 \(\rm V = \frac{1}{3} \times \pi \times {R^2} \times H\)

Where

R = radius of a cone

H = height of a cone

∴ V ∝ R²

Calculation:

∴ V₁ ∝ R² and V₂ ∝ (1.5)² × R²

∴ V₂ = 2.25 V₁

∴ Increase in volume  \(\Rightarrow {V_{\rm{\Delta }}} = \frac{{{V_2} - {V_1}}}{{{V_1}}} \times 100 = \frac{{2.25{V_1} - {V_1}}}{{{V_1}}} \times 100 = 1.25 \times 100 = 125% \)

Given increasing order of number sequence is 1, x, x, x, y, y, 9, 16, 18

Total number of numbers = 9

It is given that Mean = Median = 2 Mode

\({\rm{Mean}} = \frac{{{\rm{Sum;of;Numbers}}}}{{{\rm{Total;number;of;nubers}}}} = \frac{{1 + x + x + x + y + y + 9 + 16 + 18}}{9} = \frac{{3x + 2y + 44}}{9}..\left( i \right)\)

Median ⇒ As number of terms are odd i.e. 9 so, median will be middle value of the sequence = y

Mode = Number showing maximum frequency of repetition = x (three times)

From the given condition

y = 2x       …(ii)

From (i) and (ii)

\(\frac{{3x + 2y + 44}}{9} = 2x \Rightarrow \frac{{3x + 4x + 44}}{9} = 2x \Rightarrow 11x = 44\)

∴ x = 4 and y = 8

The paragraph implies that metro construction is affecting the foundation of an old concert hall which can be generalized to other existing buildings. Hence, option 2 is the correct answer.

Pressurized air pockets are created by the excavation of large amounts of soil during metro construction and not by the foundations of buildings. Therefore, option 1 is false.

Option 3 is false as there is no mention of hurdles faced by metro construction.

Option 4 cannot be inferred as pressurized air pockets are created by the excavation of large amounts of soil.

Amar is senior most guy and prefers room ‘R’ and hence R is allotted to him. Akbar is second most senior student and prefers room ‘R’ and ‘S’ only. Since, room ‘R’ is already allotted to Amar, room ‘S’ is provided to Akbar.

Anthony prefers room ‘P’, so it is allocated to him. Remaining only room ‘Q’ is allocated to Ajit.

Given equation ⇒ (2171)⁷ + (2172)⁹ + (2173)¹¹ + (2174)¹³

The last digit of the power value obtained for a particular number repeat in the cycle of the four number.

For e.g.

Since the cycle after very four interval, Last digit of 7⁸ would be 3 only.

All power values are divided with 4 then the possible remainders are 3, 1, 3 and 1

(1)³ + (2)¹ + (3)³ + (4)¹ = 1 + 2 + 27 + 4 = 34

∴ The last digit is ‘4’

If Machine M₁ executes jobs P and Q

For Job ‘P’ time taken 30 minutes with 2 objects = 2 × 30 = 60 min

For Job ‘Q’ is taken 20 minutes with 3 objects = 20 × 3 = 60 min

∴ Total time taken by Machine (M₁) for jobs P and Q = 60 min + 60 min = 2 hours

If Machine M₂ executes jobs R and S

For Job ‘R’ is taken 60 min with 1 object = 60 × 1 = 60 min

For Job ‘S’ is taken 15 min with 4 objects = 4 × 15 = 60 min 

∴ Total time taken by machine (M₂) for Jobs R and S = 60 min + 60 min = 2 hours

From the given bar graph,

i. The number of beds made by carpenter C₂ = 8 Nos

ii. The number tables made by carpenter C₃ = 8 Nos

∴ Statement (i) is true

i. The total number of chairs made by all carpenters

= C₁ + C₂ + C₃ +C₄ + C₅

= 2 + 10 + 5+ 2 + 4 = 23 Nos

ii. The total numbers of tables made by all carpenters

= C₁ + C₂ + C₃ + C₄ + C₅

= 7 + 2 + 8 + 3 + 9 = 29 Nos

The total number of chairs made by all carpenters (23 Nos) < The total numbers of tables made by all carpenters (29 Nos)

∴ Statement (ii) is also true

∴ Both the statements (i) and (ii) are true

Given, P = Q^-1

Post multiply by Q

PQ = Q^-1Q (we know Q^-1Q = I)

PQ = I

Similary, pre-multiply by Q

QP = QQ^-1

QP = I (QQ^-1 = I)

So, PQ = I and QP = I

Probability distribution function (PDF) of an exponential distribution is

\(f\left( {x,\lambda } \right) = \left{ {\begin{array}{*{20}{c}} {\lambda {e^{ - \lambda x}}}&{x \ge 0}\ 0&{x < 0} \end{array}} \right.\)

Cumulative distribution function of an exponential distribution is

\(f\left( {x,\lambda } \right) = \left{ {\begin{array}{*{20}{c}} {1 - {e^{ - \lambda x}}}&{x \ge 0}\ 0&{x < 0} \end{array}} \right.\)

where λ > 0 is the parameter of distribution. So only one parameter in exponential

distribution.

The normal (or Gaussion) distribution is a very common continuous probability distribution.

The probability density of normal distribution is

\(\delta \left( {\frac{x}{\mu },{\sigma ^2}} \right) = \frac{1}{{\sqrt {2\pi \sigma } }}{e^{ - \frac{{{{\left( {x - \mu } \right)}^2}}}{{2{\sigma ^2}}}}}\)

So there are two parameters i.e. (μ and σ₂) in gaussian distribution.

\(\begin{array}{l} g\left( x \right) = \smallint f\left( x \right)dx\ f\left( x \right) = {e^{ - x - {e^{ - x}}}}\ g\left( x \right) = \smallint {e^{ - x - {e^{ - x}}}}dx\ = \smallint \frac{{{e^{ - x}}}}{{{e^{{e^{ - x}}}}}}dx \end{array}\)

Substitude e^-x = t

-e^-x dx = dt

\(\begin{array}{l} g\left( x \right) = \smallint - \frac{{dt}}{{{e^t}}}\ = \smallint - {e^{ - t}}dt\ g\left( x \right) = {e^{ - t}};\ g\left( x \right) = {e^{ - {e^{ - x}}}} \end{array}\)

Concept:

From compatibility

δ_T - δ_R = 0

δT = δR

\(\Rightarrow L \alpha {\rm{\Delta }}T = \frac{{RL}}{{AE}}\)

\(\Rightarrow \sigma = \frac{R}{A} = E\alpha \Delta T\)

Calculation:

\( \sigma = E\alpha \Delta T\)

Hence stress is independent of length of bar.

To Revise the concept of Thermal Stress, click here.

Explanation:

Statement i:

• Maximum or minimum shear force occurs where the radius of curvature is zero.
• This statement is incorrect. The radius of curvature relates to bending moments and the curvature of the beam, not shear force. Shear force is maximum at the support for a simply supported beam under a uniform load.

Statement ii:

• Maximum or minimum bending moment occurs where the shear force is zero.
• This statement is correct. For a simply supported beam, the bending moment is maximum at the point where the shear force is zero. This typically occurs at the center of the beam under a uniform load.

Statement iii:

• Maximum or minimum bending moment occurs where the radius of curvature is zero.
• This statement is incorrect. The maximum or minimum bending moment occurs where the shear force is zero (as mentioned above), not where the radius of curvature is zero.

Statement iv:

• Maximum bending moment and maximum shear force occur at the same section.
• This statement is incorrect. For a simply supported beam under uniform loading, the maximum bending moment occurs at the center of the beam, while the maximum shear force occurs at the supports.

For limiting or balanced depth of neutral axis.

For the recommendation that,

\({ \in _{st}} \ge \frac{{.87{f_y}}}{{{E_s}}} + .002\)

And strain in concrete at collapse, should be max of .0035

\(\begin{array}{l} \frac{{0.0035\left( {d - {x_u}} \right)}}{{{x_u}}} \ge \frac{{0.87{f_y}}}{{{E_s}}} + 0.002\ \left( {\frac{d}{{{x_u}}} - 1} \right) \ge \frac{{\frac{{0.87{f_y}}}{{{E_s}}} + 0.002}}{{0.0035}}\ \frac{d}{{{x_u}}} \ge \frac{{\frac{{0.87{f_y}}}{{{E_s}}} + 0.0055}}{{0.0035}}\ \frac{{{x_u}}}{d} \le \frac{{0.0035}}{{\frac{{0.87{f_y}}}{{{E_s}}} + 0.0055}} \end{array}\)

Thus limiting value of neutral axis depth is given by.

\({\left( {\frac{{{x_u}}}{d}} \right)_{lim}} = \frac{{0.0035}}{{0.0055 + \frac{{0.87{f_y}}}{{{E_s}}}}}\)

Hence, balanced or limiting depth of neutral axis is dependent on grade of steel only.

Concept:

If two-hinged parabolic is subjected to uniformly distributed load of intensity q per unit length.

The bending moment at anywhere in the arch is zero. So, the maximum bending moment in the arch is equal to zero.

Thixotropy is that property of soil due to which loss of strength on remoulding can be regained if left undisturbed for some time.

If rate of loading is larger and soil is saturated positive pore water pressure will develop. This reduces effective stress and hence shear strength. If effective stress reduces to zero. The soil will lose all its shear strength. This is known as liquefaction. It occurs during dynamic loading such as pile driving vibration of machine, explore blasting, earthquake etc.

When upward flow is taking place at critical hydraulic gradient a soil such as sand losses all its shearing strength. This condition is called quick sand condition or boiling of sand.

Degree of disturbance achieved on remoulding is expressed by sensitivity.

Radius of Mohr circle \(= \sqrt {{{\left( {\frac{{{\sigma _x} - {\sigma _y}}}{2}} \right)}^2} + {\tau ^2}} \)

Given: σ_x = σ_y = σ_z = σ     τ = 0

\(\begin{array}{l} R = \sqrt {{{\left( {\frac{{\sigma - \sigma }}{2}} \right)}^2} + {0^2}} = 0\ Centre = \frac{{{\sigma _x} + {\sigma _y}}}{2} = \frac{{\sigma + \sigma }}{2} = \sigma \end{array}\)

Hence Mohr circle reduces to a point at a distance of σ from origin.

Ultimate bearing capacity for a strip footing is

\({q_u} = C{N_c} + \gamma {D_f}{N_q} + 0.5\gamma B{N_\gamma }\)

For pure clay, N_c = 5.14, N_q = 1 and Nγ = 0 (∵ assuming smooth footing)

Footing is on the ground surface i.e. D = 0

q_u = c_uN_c

qu = 5.14 c_u

qu = (π + 2)c_u

Due to uniformly distributed line load vertical stress

\(= \frac{{2q}}{{\pi z}}{\left[ {\frac{1}{{1 + {{\left( {\frac{x}{z}} \right)}^2}}}} \right]^2}\)

Vertically below line load x = 0

\(\begin{array}{l} {\sigma _z} = \frac{{2q}}{{\pi z}}\ \frac{{{\sigma _1}}}{{{\sigma _2}}} = \frac{{{z_2}}}{{{z_1}}} = \frac{4}{2} = 2 \end{array}\)

\(v = \frac{1}{{2\mu }}\left( {\frac{{ - \partial p}}{{\partial x}}} \right)\left( {By - {y^2}} \right)\)

Shear stress

\(\begin{array}{l} \tau = \mu \frac{{du}}{{dy}}\ = \frac{1}{2} \times \left( { - \frac{{\partial p}}{{\partial x}}} \right)\left( {B - 2y} \right)\ y = \frac{B}{2},;\tau = 0 \end{array}\)

Y = 0, τ = τ_max

The order of reaction with respect to a given substance (reactant, catalyst or product) is defined as the index or exponent to which its concentration term in the rate equation is raised. For typical rate equation of form

r = k[A]^x [B]^y

Overall reaction order = x + y

So first order reaction α + β = 1

Explanation:

At (A) rate of reoxygenation is equal to the rate of deoxygenation.

Before (A) rate of reoxygenation is less than the rate of deoxygenation. The DO continuously decreases when rate of deoxygenation > Rate of reoxygenation

Acid rain results due to conversion of atmospheric gases into acidic compounds. NO_x, SO_x and CO₂ present in the atmospheric reacts with the water vapour and sunlight forming HNO₃, H₂SO₄ and H₂CO₃ which are present in the acid rain.

Concept:

Let the weight of vehicle = W in stopping condition the friction force F

should be greater than w sin θ to prevent the sliding F > w sin θ

From Figure,

F = w sinθ 

for a smaller value of θ 

tan θ = e = sin θ

F = w × e ...... (1)

We can also written,

F = w cosθ  × f

F = w × f (As θ is low so Cos θ = 1) .........(2)

From equation 1 & 2

w × f ≥ w × e

f > e

Explanation:

Correction for elevation: 7% increase per 300 m

So, correction \(= \frac{7}{{100}} \times \frac{{535}}{{300}} \times 2000\)

= 249.66 m

Corrected length = 2000 + 249.66 = 2249.66 m

Correction for temperature:

Standard atmospheric temperature = 15 – 0.0065 × 535 = 11.5225°C

Rise of temp = 22.65°C – 11.523°C = 11.127°C

\(Correction = \frac{{2249.66}}{{100}} \times 11.127 = 250.320;m\)

Correct length = 2249.66 + 250.320 = 2499.98 m

Check for total correction for elevation plus temperature

\(Total;correction;% = \frac{{2499.98 - 2000}}{{2000}} \times 100 = 24.99%\)

According to ICAO, this should not exceed by 35%.

Accuracy of EDMI is generally stated in terms of constants instruments error and measuring

Error proportional to the distance being measured.

± (a mm + b ppm)

The first part ‘a’ in this expression indicates a constant instrument error that is independent of the length of line measured.

Second component ‘b’ is distance related error.

Classification of EDM instruments based on the precision obtainable:

Less precise → Standard deviation of one measurement greater than ± (5 mm + 5 ppm)

Moderately precise → Standard deviation of one measurement = ± (5 mm + 1 ppm)

Highly precise → Instruments having a standard deviation of one measurement = ± (1 mm + 1 ppm)

Landsat enhanced Thematic Mapper sensor on the remote sensing satellite Landsat-7 has

8 number of spectral bands which are as follows:

Band 1 – Blue

Band 2 – Green

Band 3 – Red

Band 4 – Near Infrared (NIR)

Band 5 – Shortwave Infrared (SWIR)₁

Band 6 – Thermal

Band 7 – Shortwave Infrared (SWIR)₂

Band 8 – Panchromatic

\(3\frac{{{\partial ^2}\phi }}{{\partial {x^2}}} + B\frac{{{\partial ^2}\phi }}{{\partial x\partial y}} + 3\frac{{{\partial ^2}\phi }}{{\partial {y^2}}} + 4\phi = 0\)

Compare above equation with \(A\frac{{{\partial ^2}\phi }}{{\partial {x^2}}} + B\frac{{{\partial ^2}\phi }}{{\partial x\partial y}} + C\frac{{{\partial ^2}\phi }}{{\partial {y^2}}} + D\phi = 0\)

A = 3      B = ?      C = 3

For equation to be parabolic

B² – 4AC = 0

B² – 4 × 3 × 3 =

B² = 36

\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\tan x}}{{{x^2} - x}}} \right)\)

0/0 form so we can apply L-Hospital’s Rule

\(= \mathop {\lim }\limits_{x \to 0} \frac{{{{\sec }^2}x}}{{2x - 1}} = \frac{1}{{ - 1}} = - 1\)

Concept:

The time required for any degree of consolidation is given by Taylor's formula

\(\left( {{{\rm{T}}_{\rm{v}}}} \right) = \frac{{{{\rm{C}}_{\rm{v}}}{\rm{t}}}}{{{{\rm{d}}^2}}}\)

Tv - Time factor

Cv - Coefficient of consolidation

t - Time taken for consolidation

d - Drainage path

Drainage path represents the maximum distance the water particles have to travel to reach the free drainage layer.

Calculation:

Given:

Since the bottom is impermeable water can drain only through top sand  layer

It is one way drainage case so

H = 3 m = 3000 mm

\(\begin{array}{l} {T_v} = \frac{{{c_v}t}}{{{H^2}}}\ t = \frac{{0.85 \times {{3000}^2}}}{3} = 2250000;minutes \end{array}\)

 ⇒ 1770.83 days ≈ 1771 days

Double drainage condition refers to the situation when both faces are permeable and pore water can dissipate through both faces but in single drainage only one face is permeable and hence it takes a longer time in single drainage.

Single drainage d = H

Double drainage d = H/2

Concept:

Hardy cross method:

This method is applicable to closed-loop pipe networks. The outflows from the system are assumed to occur at the nodes. 

The Hardy-Cross analysis is based on the principles that:

1. ​At each junction, the total inflow must be equal to the total outflow.
2. The algebraic sum of the head losses around any closed-loop is zero.

Calculation:

Given:

If the discharge supplied at point A equally divided so Q_AB = 50 m³/s

Q_AC = 50 m³/s

Head loss in closed loop is zero

∑rQⁿ = 0

∑rQ^1.8 = 0

1 × 50^1.8 – 2 × 20^1.8 - r × 50^1.8 = 0

r × 50^1.8 = 703.838

r = 0.615 ≈ 0.62

Explanation:

Lagging 2-hr ordinate unit hydrograph by 2-hr and adding it with the 2-hr unit hydrograph

Hence ordinate of the resulting hydrograph at 3hrs would be = 15 m³/s

Probability of time headway being greater than 8 sec.

\(P\left( {h \ge 8} \right) = {e^{ - 8\lambda }} = {e^{ - 8 \times \frac{{900}}{{3600}}}}\)

= e^-2 = 0.1353

Concept:

For a given probability density function f(x) to be valid, the integration of it over the complete range must equal 1.

Mathematically, it must satisfy:

\(\mathop \smallint \limits_{ - \infty }^\infty f\left( x \right) = 1\)

Application:

Given f(x) = a + bx, 0 ≤ x ≤ 1

For the given density function to be valid, it must satisfy:

\(\mathop \smallint \limits_{ - \infty }^\infty a + bx = 1\)

\(\mathop \smallint \limits_0^1 \left( {a + bx} \right)dx = 1\)

\(\begin{array}{l} ax + \left. {\frac{{b{x^2}}}{2}} \right]_0^1 = 1\ a + \frac{b}{2} = 1 \end{array}\)

To satisfy the above equation:

a = 0.5, b = 1

Concept:

\(\begin{array}{l} \frac{{dQ}}{{dt}} + Q = 1\ I.F = {e^{\smallint \rho dt}} = {e^{\smallint 1.dt}} = {e^t}\ Solution ⇒ Q\left( t \right){e^t} = \smallint {e^t}dt \end{array}\)

⇒ Q(t) e^t = e^t + C

At t = 0 Q = 0

0 = 1 + C

C = -1

Q(t) = 1 – e^-t

\(\left[ A \right] = \left[ {\begin{array}{*{20}{c}} 5&{ - 1}\ 4&1 \end{array}} \right]\)

For eigen value

\(\left[ {A - \lambda I} \right] = \left[ {\begin{array}{*{20}{c}} {5 - \lambda }&{ - 1}\ 4&{1 - \lambda } \end{array}} \right] = 0\)

(5 - λ) (1 - λ) + 4 = 0

5 – 5λ - λ + λ² + 4 = 0

λ² - 6λ + 9 = 0

(λ - 3)² = 0

λ = 3

∴ eigenvalue 3 has multiplicity 2

for eigen vector

\(\begin{array}{l} \left[ {A - \lambda I} \right]\left[ {\begin{array}{*{20}{c}} x\ y \end{array}} \right] = 0\ \left[ {\begin{array}{*{20}{c}} {5 - 3}&{ - 1}\ 4&{1 - 3} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x\ y \end{array}} \right] = 0 \end{array}\)

2x – y = 0

\(\left[ {\begin{array}{*{20}{c}} x\ y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1\ 2 \end{array}} \right]\)

And 4x – 2y = 0

\(\left[ {\begin{array}{*{20}{c}} x\ y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1\ 2 \end{array}} \right]\)

So, only one independent eigen vector.

For the given truss,

 members (m) = 15, joints (j) = 8

External Indeterminacy = r – 3

r = no of support reactions = 4

Hence, External Indeterminacy = 4 – 3 = 1

Internal Indeterminacy = m – (2j – 3) = 15 – ( 2 x 8 – 3) = 2

Non modular outlet: These are outlets through which the discharge depends upon the difference of head between the distributary and the water course.

Semi modular outlet: These are outlets through which the discharge is independent of the distributary so long as a minimum working head is available and depends upon distributary water surface level.

Modular outlet: These are outlets through which discharge is constant and fixed within limits irrespective of the fluctuations of the water levels of either the distributary or of the water course or of both.

Concept:

Discharge of channel by Manning's Equation is

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

Where,

Q – Discharge through the channel, n – Manning’s Constant

R – Hydraulic Radius, S –Slope of the channel

Calculation:

Given - Normal depth  (y_n) = 0.5 m

​\(Q = \frac{1}{0.04} \times 1\times0.5\times {0.5^{\frac{2}{3}}} \times {0.0016^{\frac{1}{2}}}\)​

= 0.198 m³/s

\(q = \frac{Q}{b} = \frac{{0.198}}{1} = 0.198\) m³/s/m

\({y_c} = {\left( {\frac{{{q^2}}}{g}} \right)^{\frac{1}{3}}} = {\left( {\frac{{0.198}}{{9.81}}} \right)^{\frac{1}{3}}} = 0.16;m\)

Measured flow depth is 0.6 m, so y > y_n > y_c

Hence, M₁ profile.

Mass of oven dry aggregate, A = 1000 gm

Mass of saturated surface dry aggregate, B = 1025 gm

Mass of saturated surface dry aggregate under water, C = 625 gm

Bulk Specific gravity

\(\Rightarrow \frac{{Wt.;of;solids}}{{Wt.;of;equivalent;vol.;of;water}}\)

\(\Rightarrow \frac{A}{{B - C}} = \frac{{1000}}{{1025 - 625}}\) = 2.5

Water absorption \(\Rightarrow \frac{{1025 - 1000}}{{1000}} \times 100\) = 2.5%

Vehicle arrived upto 60 sec (Red time) = 10

\(Arrival;rate = \frac{{10}}{{60}} \times 3600 = 600\frac{V}{h}\)

V = 600 V/h

Between = 60 second to 90 second total vehicle departed

= Vehicle arrived upto 60 second + Vehicle arriving between 60 sec to 90 sec

\(= 10 + \frac{{600}}{{3600}} \times 30\)

= 10 + 5

= 15

So, departure rate = Saturation flow

\(= S = \frac{{15}}{{30}} \times 3600\)

S = 1800 v/h

Average delay time is given by

\({t_d} = \frac{{\frac{C}{2}{{\left( {1 - \frac{g}{c}} \right)}^2}}}{{1 - \frac{V}{S}}}\)

C = 96 seconds

g = 96 – 60 = 36 seconds

V = 600 Vph

S = 1800 Vph

\({t_d} = \frac{{\frac{{96}}{2}{{\left( {1 - \frac{{36}}{{96}}} \right)}^2}}}{{1 - \frac{{600}}{{1800}}}}\)

t_d = 28.125 seconds

Given ⇒ R = 120 m

\({V_{design}} = 60\frac{{km}}{{hr}}\)

f = 0.15

\(e + f = \frac{{{v^2}}}{{gR}}\)

If full lateral friction is assumed to develop, then

\(\begin{array}{l} \Rightarrow e + 0.15 = \frac{{{{\left( {60 \times \frac{5}{{18}}} \right)}^2}}}{{9.81 \times 120}}\ e = \frac{{{{\left( {60 \times \frac{5}{{18}}} \right)}^2}}}{{9.81 \times 120}} - 0.15 = \frac{1}{{11.6}}\ e + f = \frac{{{v^2}}}{{gR}} \end{array}\)

If, e = 0

\(\Rightarrow f = \frac{{{{\left( {60 \times \frac{5}{{18}}} \right)}^2}}}{{9.81 \times 120}}\)

f= 0.24

Concept:

Local attraction is a term used to denote any influence which prevents the needle from pointing to Magnetic north.

If the difference between fore and back bearing is 180^°, it may be taken that both the stations are free from the local attraction.

Calculation:

Fore bearing and back bearing difference for PQ and ST is 180°. So, P, Q, S, T are free from any local attraction.

∴ R is affected by local attraction.

Given ⇒ W_LL = 60% = 0.6

W_PL = 25% = 0.25

% of clay sized fraction 25% = 0.25, Natural moisture content w_n = 18% = 0.18

\(\begin{array}{l} {I_L} = \frac{{{W_n} - {W_{PL}}}}{{{W_{LL}} - {W_{PL}}}}\ = \frac{{0.18 - 0.25}}{{0.6 - 0.25}} = - 0.2\ Activity = \frac{{{I_P}}}{{% ;of;clay;sized;fraction}}\ = \frac{{{W_{LL}} - {W_{PL}}}}{{0.25}} = \frac{{0.6 - 0.25}}{{0.25}} = 1.4 \end{array}\)

\(\frac{{du}}{{dt}} = 3{t^2} + 1\)

Forward Euler Method

y₁ = y₀ + hf(t)

at t = 0, u = 0

y₁ = 0 + 2(3 × 0² + 1)

u₁ = 2

\(\begin{array}{l} \frac{{du}}{{dt}} = 3{t^2} + 1\ du = \mathop \smallint \limits_0^2 \left( {3{t^2} + 1} \right)dt\ {u_1} = \left. {{t^3} + t} \right]_0^2 \end{array}\)

u₁ = 10

Absolute error = 10 – 2 = 8

Concept:

Loss of prestress due to elastic deformation = m × f_c

m - Modular ratio

f_c - stress in concrete at the level of steel

\({f_c} = \frac{P}{A} + \frac{{P{e^2}}}{I}\)

P - Prestressing force

A - Area of cross-section of beam

Calculation:

Given:

A = 150 × 300, P = 1200 × 3 × 50 N, m = 6

\({f_c} = \frac{P}{A} + \frac{{P{e^2}}}{I}\)

e = 50 mm

\({f_c} = \frac{1200\times 3\times 50}{150\times 300} + \frac{{1200\times 3\times 50\times{50^2}}}{3375\times10^5}\)

⇒ \(4 + \frac{4}{3} = \frac{{16}}{3} kN/m^2\)

\(Loss = \frac{{16}}{3} \times 6=32 ~kN/m^2\)

Loss in force = 32 × 3 × 50 = 4800 N = 4.8 kN

\(U = \sum \frac{{{P^2}{L_1}}}{{2{A_1}{E_1}}}\)

\(= \frac{{{{\left( {1000} \right)}^2} \times 400}}{{2 \times 100 \times 2 \times {{10}^5}}} + \frac{{{{\left( {1000} \right)}^2} \times 900}}{{2 \times 60 \times 3 \times {{10}^5}}}\) = 10 + 25 = 35 N.mm

Concept:

The slope deflection equation is given as

\({M_{BA}} = {M_{FBA}} + \frac{{2EI}}{l}\left( {2{\theta _B} + {\theta _A} - \frac{{3{\rm{\Delta }}}}{l}} \right)\)

Calculation:

Applying slope deflection equation for span AB 

\({\theta _B} = 0\), \({\theta _A} = 0,;{\rm{\Delta }} = 0\)

\({M_{FBA}} = \frac{{w{l^2}}}{{12}} = \frac{{30 \times {4^2}}}{{12}} = 40;kNm\)

\(\begin{array}{l} {M_{BC}} = {M_{FBC}} + \frac{{3EI}}{6}\left( {{\theta _B} - \frac{{\rm{\Delta }}}{l}} \right),\ {\rm{\Delta }} = 0,;{\theta _B} = 0,;{M_{FBC}} = 0,;{M_{BC}} = 0 \end{array}\)

M = M_BA + M_BC = 40

Now as per compatibility, \(\Rightarrow \frac{{w{l^4}}}{{8EI}} + \frac{{M{l^2}}}{{2EI}} = \frac{{{R_B} \times {l^3}}}{{3EI}}\)

\(\begin{array}{l} \frac{{30 \times 4}}{8} + \frac{{40}}{{2 \times 4}} = \frac{{{R_B}}}{3}\ 15 + 5 = \frac{{{R_B}}}{3} \end{array}\)

R_B = 60 kN

⇒ According to Muller Breslau Principle ILD for moment at C

θ₁ = θ₂ = tan45°

for maximum bending moment

Hence, M_max = 50 × 4 = 200 kNm

Concept:

Crippling load:

\(P = \frac{{{\pi ^2}EI}}{{L_e^2}}\)

Calculation:

1 - One end fixed other free: L_e1 = 2L

2 - One end fixed and other hinged: Le2 = L/√2

\(\frac{{{P_2}}}{{{P_1}}} = \left( {\frac{{{L_1}}}{{{L_2}}}} \right)_{eff}^2 = {\left( {\frac{{2l}}{{\frac{l}{{\sqrt 2 }}}}} \right)^2} = 8\)

P = 10 kN, e = 15 cm

r₁ = r₂ = r₃ = r₄ = 5 cm

Direct load to bolt \(\left( 1 \right) = \frac{P}{4}\)

\({F_1} = \frac{{10}}{4} = 2.5kN\)

Force in bolt (1) due to moment

\(\begin{array}{l} = \frac{{Pe{r_1}}}{{\sum r_1^2}}\ {F_2} = \frac{{10 \times 15 \times 5}}{{4 \times {5^2}}} = 7.5;KN \end{array}\)

Angle between force F₁ and F₂ = 135°

\(\begin{array}{l} R = \sqrt {F_1^2 + F_2^2 + 2{F_1}{F_2}\cos \left( {135^\circ } \right)} \ = \sqrt {{{2.5}^2} + {{7.5}^2} + 2 \times 2.5 \times 7.5 \times \cos \left( {135^\circ } \right)} \end{array}\)

= 5.99 kN

Concept:

The relationship between force (F) and mass (m) is given by

F = m × a

\(F = m \times \frac{dv}{dt}\)

Calculation:

Given:

f(t) = 3t², m = 2 kg 

\(\begin{array}{l} m\frac{{dv}}{{dt}} = 3{t^2}\ 2\mathop \smallint \limits_{1.5}^v dv = \mathop \smallint \limits_0^2 3{t^2}dt \end{array}\)

2(v – 1.5) = 8

v = 5.5 m/s

Explanation:

Activity on Arrow (AOA) diagram:

Time along path 1 – 2 – 4 – 5 – 7 = 6 + 15 + 12 + 16 = 49 days

Time along path 1 – 2 – 3 – 4 – 6 – 7 = 6 + 10 + 14 + 16 = 46 days

Time along path 1 – 2 – 4 – 6 – 7 = 6 + 15 + 14 + 16 = 51 days

The minimum time for the completion of the project will be = 51

Additional Information Dummy activity:

• A dummy activity is an activity added to a project schedule as a placeholder. It has no activity time associated with it.
• A dummy activity is intended to show a path of action in a project activity diagram and is employed when a logical relationship between two activities cannot be linked by showing the use of arrows linking one activity to another. for example activity (3-4), in above solution is a dummy activity.

Stress at B = 16 × 2 = 32 kN.m²

Average stress in AB = 32/2 = 16 kN/m²

In part AB:

Axial frictional resistance in AB = (Kσ_avg tanδ)A_AB

A_AB = Area of AB = 3 × 2 = 6 m²

Frictional Resistance = (Kσ_avg tanδ) × 6

= (1 × 16 × tan23°) × 6

= 40.75 kN

In part BC:

Stress at C = 16 × 2 + (19 - 9.81) x 5 = 77.95 kN.m²

Effective vertical stress variation

\({\sigma _{avg}} = \frac{{32 + 77.95}}{2}\) = 54.975 kN/m²

A_BC = 3 × 5 = 15 m²

Axial frictional resistance in part BC

= (Kσ_avgtanδ)A_BC

= (1 × 54.975 × tan23°) × 15

= 350.03 kN

Total axial frictional resistance

= 350.03 + 40.75 = 390.78 kN

\(\begin{array}{l} F.O.S = \frac{{Effective;stress}}{{Total;stress}}\ F.O.S = \frac{{{\gamma _{sub}}Hco{s^2}\beta \tan \phi }}{{{\gamma _{sat}}H\cos \beta \sin \beta }}\ 1 = \frac{{{\gamma _{sub}}\tan \phi }}{{{\gamma _{sat}}\tan \beta }}\ 1 = \frac{{\left( {21 - 9.81} \right)\tan \phi }}{{21\tan 20^\circ }}\ \tan \phi = \frac{{21 \times \tan 20^\circ }}{{11.19}} \end{array}\)

ϕ = 34.33

There is no energy loss so

E₁ = E₂

\(\begin{array}{l} {y_1} + \frac{{v_1^2}}{{2g}} = {y_2} + \frac{{v_2^2}}{{2g}}\ 1 + \frac{Q}{{2gA_1^2}} = 0.2 + \frac{Q}{{2gA_2^2}}\ \frac{{{Q^2}}}{{2g}}\left[ {\frac{1}{{A_2^2}} - \frac{1}{{A_1^2}}} \right] = 1 - 0.2\ \frac{{{Q^2}}}{{2 \times 10}}\left[ {\frac{1}{{{{0.2}^2}}} - \frac{1}{{{1^2}}}} \right] = 0.8\ {Q^2} = \frac{{16}}{{24}}\ {Q^2} = \frac{2}{3} \end{array}\)

Q = 0.82 m³/s

Concept:

Pressure at the exit of nozzle is taken as zero because water at the outlet of nozzle will be discharging to atmosphere.

\(\frac{{{P_1}}}{{{\gamma _w}}} + {Z_1} + \frac{{V_1^2}}{{2g}} = \frac{{{P_2}}}{{{\gamma _w}}} + {Z_2} + \frac{{V_2^2}}{{2g}}\)

Calculation:

Given:

z₁ = z₂ and P₂ = 0, so above equation reduces to

\(\begin{array}{l} \frac{{140 \times {{10}^3}}}{{10 \times 1000}} + 0 = 0 + 0 + \frac{{{Q_2}}}{{2g}}\left( {\frac{1}{{A_2^2}} - \frac{1}{{A_1^2}}} \right)\ 280 = {Q^2}\left( {\frac{1}{{{{\left( {\frac{\pi }{4} \times \frac{{{{0.14}^2}}}{\pi }} \right)}^2}}} - \frac{1}{{{{\left( {\frac{\pi }{4} \times \frac{{{{0.27}^2}}}{\pi }} \right)}^2}}}} \right) \end{array}\)

Q = 0.085 m³/s

\(\begin{array}{l} {v_1} = ;\frac{Q}{{{A_1}}} = \frac{{0.085}}{{\frac{\pi }{4} \times {{\left( {\frac{{0.27}}{{\sqrt \pi }}} \right)}^2}}} = 4.664;m/sec\ {v_2} = ;\frac{Q}{{{A_2}}} = \frac{{0.085}}{{\frac{\pi }{4} \times {{\left( {\frac{{0.14}}{{\sqrt \pi }}} \right)}^2}}} = 17.347;m/sec \end{array}\)

In x-direction momentum equation is given by:

\(\begin{array}{l} {P_1}{A_1} - {F_x} = \rho Q\left( {{V_2} - {V_1}} \right),;{V_2} = 0;in;x;direction,;{V_1} = \frac{Q}{{{A_1}}}\ {F_x} = 140 \times {10^3} \times \frac{\pi }{4} \times \frac{{{{0.27}^2}}}{\pi } + ;1000 \times 0.085 \times 4.664 \end{array}\)

= 2.947 kN

In y-direction momentum:

F_y = ρQV₂

\(= 1000 \times 0.085 \times 17.347\)

= 1.47 kN'

\(Resultant;force = \sqrt {{{2.947}^2} + {{1.47}^2}} = 3.29;KN\)

Confining pressure = σ₃ = 100 kPa

Deviator stress = σ₁ – σ₃ = 80 kPa

σ₁ = 80 + σ₃ = 180 kPa

Pore pressure = u = 50 kPa

\({\bar \sigma _1} = {\sigma _1} - u\)

= 180 – 50 = 130 kPa

\({\bar \sigma _3} = {\sigma _3} - u\)

= 100 – 50 = 50 kPa

For NC soil C’ = 0

\(\begin{array}{l} {{\bar \sigma }_1} = {{\bar \sigma }_3}\left( {\frac{{1 + \sin \phi }}{{1 - \sin \phi }}} \right)\ 130 = 50\left( {\frac{{1 + \sin \phi }}{{1 - \sin \phi }}} \right)\ \sin \phi = \frac{8}{{18}} \end{array}\)

ϕ = 26.4°

Flood hydrograph peak = 200 m³/s

Base flow = 20 m³/s

Rainfall = 2 cm

φ = 0.4 cm/hr

Effective rainfall = 2 – 0.4 × 2 = 1.2 cm

Peak of DRH = 200 – 20 = 180 m³/s

Peak of 2-h unit hydrograph

\(\begin{array}{l} = \frac{{Peak;DRH}}{{Effective;rainfall}}\ = \frac{{180}}{{1.2}} = 150\frac{{{m^3}}}{s} \end{array}\)

Concept:

It is the statistical value of sound pressure level that can be equated to any fluctuating noise. Thus, It is defined as the constant noise level, which over a given time, expands the same amount of energy, as it expanded by the fluctuating levels over the same time.

The value  can be expressed as

\({L_{eq}} = 10\log \mathop \sum \limits_{{\rm{i}} = 1}^{{\rm{i}} = {\rm{n}}} {\left( {10} \right)^{\frac{{{{\rm{l}}_{\rm{i}}}}}{{10}}}} \times {{\rm{t}}_{\rm{i}}}\)

Where n = Total number of sound samples

Li = The noise level of any ith sample

ti = Time duration of the ith sample expressed as a fraction of total sample time.

Calculation

\Given,

noise levels 60, 69, 70, and 79 dB

 As time is not mentioned assume all noise lasts from start to end of duration T

 ∴ ti = T/T = 1

Now, 

\({L_{eq}} = 10\log \mathop \sum \limits_{{\rm{i}} = 1}^{{\rm{i}} = {\rm{n}}} {\left( {10} \right)^{\frac{{{{\rm{l}}_{\rm{i}}}}}{{10}}}} \times {{\rm{t}}_{\rm{i}}}\)

\({L_{eq}} = 10\log[(10^{6}+10^{6.9}+10^{7}+10^{7.9})\times 1]\)

 = 79.9 dB

Note: The equivalent sound level for the group of sound levels will be near the highest sound level ( here its 79 dB )

Concept:

The settling velocity of particles is given by stokes law as

\(\begin{array}{l} Settling;velocity(V_s) = \frac{{\left ( {{\ G} - \ 1} \right){\times}{g}{\times}{d^2}}}{{18~ν }}\ \frac{\mu }{\rho } = {10^{ - 6}}\frac{{{m^2}}}{s} \end{array}\)

ν  = 10^-3 N-s/m²

Calculation:

Given - G = 3, r = 0.01 mm

Diameter of particles should be taken twice the value of radius

d = 0.01 × 2 = 0.02 mm = 2 × 10^-5 m

\({V_s} = \frac{{\left( {3 × 9810 - 9810} \right) × {{\left( {2 × {{10}^{ - 5}}} \right)}^2}}}{{18 × {{10}^{ - 3}}}}\)

⇒ 4.36 × 10^-4 m/s

⇒ 0.436 mm/s

Concept:

Yt represents the total amount of organic matter oxidized in 't' days, BOD of t days is given by,

Y_t = L × (1 - e^{- (k_D × t)})

Yt - BOD of t days

t – Time in days

k_{D (T)} – Deoxygentaion constant at T^°C ( to base 10)

Relation between k_D,20 and deoxygenation constant at any temperature T

k_D (T) = k_{D (}20) × (1.135)(T-20)

Calculation:

Given:

For stream A:

Y₅ = L × (1 - 10- ( 0.115 × 5 ))

⇒ L = 68.13 mg/l

k_{D (10)} = k_{D (}20) × (1.135)(10-20)

⇒ 0.115 × (1.135)^-10 = 0.0324

Ultimate BOD remains same irrespective of temperature, hence (L)_A = (L)_B = 68.13 mg/l

For stream C:

Y5 = 68.13 × (1 - 10- ( 0.0324 × 5 ))

⇒ 68.13 [1 – 10^{-0.0324 × 5}] = 21.21 mg/l

Concept:

For Complete mixing in the detention tank, effluent concentration

\(L ={ L_o \over (1+K_D \times t)}\)

\(Detention;Time;\left( {{t_d}} \right) = \frac{{volume}}{{Discharge}} \)

Note that if decay constant is assumed to have been given at base ‘e’

Calculation:

Given ⇒ Initial concentration \(\left( {{L_0}} \right) = 54\frac{{mg}}{l}\)

Q = 0.8 m³/day

V = 2 × 4 × 2 = 16 m³

\(Detention;Time;\left( {{t_d}} \right) = \frac{{16}}{{0.8}} = 20;day\)

For Complete mixing in the detention tank, effluent concentration

\(L ={ L_o \over (1+K_D \times t)}\)

\(L ={ 54 \over (1+0.1 \times 20)}\) = 18 mg/l