GATE CE 2016 Official Paper: Shift 2 (Previous Year Paper)

The correct answer is 'wouldn't buy'.

Key Points

• The given sentence is an example of a second conditional sentence.
• The second conditional sentence is a structure used for talking about unreal situations in the present or in the future.
• Structure:
• If+past simple, would+ infinitive.
• Example: 
• If I won the lottery, I would buy a big house.
• Thus, from the above-given explanation, it is clear that 'wouldn't buy' will be used in the blank.

Therefore, the correct sentence is: 'If I were you, I wouldn't buy that laptop. It's much too expensive.'

The correct answer is 'Ignored'.

Key Points

• Let's explore the meaning of the given word and the marked option:
• ‘Turning a deaf ear' is a phrase which means to pretend to have not heard what was said.
• ​Example: You can plead all day but he's turning a deaf ear to everyone.
• Ignored: to avoid or pretend to have no knowledge about something.
• Example: Ignore those who try to discourage you.
• Thus, from the above-given explanation, we can say that the correct answer is option 1.

Additional Information

• Let's explore the meaning of the other words:
• Appreciated: to like and admire.
• Twisted: physically turn something or to understand something differently from what is meant i.e to get a negative meaning.
• Returned: to give back.

The correct answer is 'Where, there, there'.

Key Points

• 'Where there is a will there is a way' is a proverb which means if one really wants to achieve something then he/she will keep on getting ways to it and will be successful in the end. 
• Example: The man drove through eight countries in his car. It is true, where there is a will, there is a way.
• The famous saying highlights the ‘importance of willpower or strong determination in life’.
• Therefore, the correct answer is option 3.

The correct sentence is: Where there is a will, there is a way.

(x % of y) + (y % of x) 

$= \frac{x}{{100}} \times y + \frac{y}{{100}} \times x$

= 2xy/100

= 2% of xy

Let the ones digit be a and tens digit be b.

Given,

sum of the digits of a two digit number is 12.

∴ a + b = 12 ---------eqn 1

Given,

New number formed by reversing the digits is greater than the original number by 54.

(b + 10a) = (a + 10b) + 54

9a – 9b = 54

∴ a – b = 6 -------- eqn 2

From equation 1 and 2

a = 9 and b = 3

∴ The orginal number is = a + 10b  = 39

Concept:

SI = Simple interest = (P × r × t)

where,

P = principal amount

R = rate of interest in %

t = time for which money is lend (in days or months or years)

Calculation:

Rate of interest provided by company P in 2006 = r_P = 6%

Rate of interest provided by company Q in 2006 = r_Q = 4%

Time for which money was lend = t = t_P = t_Q = 1 year

Given, amounts invested in the companies, P and Q, in 2006 are in the ratio 8 : 9.

∴ $\frac{{{P_P}}}{{{P_Q}}} = \frac{8}{9};$

$\frac{{{{\left( {S.I} \right)}_P}}}{{{{\left( {S.I} \right)}_Q}}} = \frac{{{P_P} \times {r_P} \times {t_P}}}{{{P_Q} \times {r_Q} \times {t_Q}}};$

$\frac{{{{\left( {S.I} \right)}_P}}}{{{{\left( {S.I} \right)}_Q}}} = \frac{{8 \times 6 \times 1}}{{9 \times 4 \times 1}} = \frac{4}{3}$

Amounts received after one year as interests from companies P and Q is in the ratio 4 : 3.

Explanation:

In the given statement, it has been told that the availability of evidence helps the historians identify greatness of the king at their time.

Statement (A) In this statement it has been said that historians ‘tend’ to correlate the greatness of a king at his time with the availability of evidence today. This means that sculpted evidence is only one of the means to know about the greatness of a king. Hence this statement is wrong.

Statement (B) is incorrect as Ashoka, who lived thousands of years ago, must not have produced stone-carved edicts so that later historians will respect him but for the as a socio-cultural token for the contemporary time.

Statement (C) is incorrect as the statement not at all mentions ‘Statues of kings’.

Statement (D) is correct as it is the past evidence that historians take into account to conclude about their greatness. So it is through the historical remains that historians interpret about the kings’ greatness and it is through their interpretation that we know them.

Let us consider each statement separately.

I. All mammals build houses.

This is clearly false as only some mammals (humans) are engineers.

II. Engineers are mammals.

All engineers are definitely humans, and all humans are mammals, therefore this is true.

III. Some humans are not engineers.

The given information states that ‘some humans are engineers’. Therefore, we cannot determine how many humans are engineers. There is the possibility of all humans being engineers as well, so this is definitely false.

Therefore, only II is true.

Mistake Points

1) If the statements are positive, then the negative conclusions will be false in definite cases and vice versa.

2. "Some" does not say anything certain about "Some not".

3) The conclusion must be 100% true. The conclusions which are 99% true will be considered false.

For Example:

Statement: Some A are B  

Conclusion: some A are not B, It is possible, not definite. We can't say anything about the negative conclusion.

Concept:

For square pyramid

Lateral surface area of the pyramid = 4 × area of one of the triangles of the pyramid.

perimeter of a square  = 4 × Side of a square

Calculation:

Given,

Square pyramid has a base perimeter = X

perimeter of a square  = 4 × Side of a square

∴ Side of the square base = X/4

The slant height  = half of the perimeter.= X/2

we know that

Perpendicular height of the triangles = Slant height of the pyramid = X/2

Base of the triangle = Side of the square base = X/4

Lateral surface area of the pyramid = 4 × area of one of the triangles of the pyramid.

Lateral surface area of the pyramid = 4 × ½ × base × height

Lateral surface area of the pyramid = 2 × (x/4) × (x/2) = 0.25x²

Let the number of pages of the book be n.

Given,

Time needed To read a book

Ananth takes 6 hours

Bharath takes 4 hours 

∴ Number of pages read in 1 hour

Ananth reads n/6 pages

Bharath reads n/4 pages 

Number of pages remaining to be read

Number of pages remaining to be read by Ananth = $n - \frac{{nt}}{6}$

Number of pages remaining to read by Bharath = $n - \frac{{nt}}{4}$

Let after ‘t’ hours the number of pages remaining to be read by Ananth is twice that number of pages remaining to read by Bharath.

(Number of pages remaining to be read by Ananth) = 2 × (Number of pages remaining to read by Bharath)

$\therefore n - \frac{{nt}}{6} = 2\left( {n - \frac{{nt}}{4}} \right)$

⇒ 1 – t/6 = 2 – t/2

⇒ t/3 = 1

⇒ t = 3 hours

Concept:

The method of finding Maxima and Minima of Y = f(x)

1. Find f’(x) and f”(x) for given function Y = f(x)

2. Equate f’(x) to zero to obtain stationary points x = a

3. calculate f”(x) at each stationary points x = a (i.e f”(a))

4. we obtain following three conditions

(i) If f”(a) > 0 then f(x) has a minimum at x = a and minimum value will be f(a)

(ii) If f”(a) < 0 then f(x) has a maximum at x = a and maximum value will be f(a)

(iii) If f”(a) = 0 then f(x) may or may not have a maximum or a minimum at x = a 

Calculation:

Given function f(x) = x2 – 4x + 2

1) f’(x) = 2x - 4, f”(x) = 2

2) f’(x) = 0, ∴ 2x - 4 = 0, x = 2(stationary point)

3. f"(x) at stationary point x = 2, f”(x) = 2

4. f"(x) > 0

∴ f(x) is minimum at x = 2

i.e.,f(2)= (2)² – 4(2) + 2 = – 2 

∴ The optimum value of f(x) is – 2 (minimum)

The given fourier series is,

$f\left( x \right) = \frac{\pi }{4} + \frac{2}{\pi }\left[ {\frac{{\cos x}}{1^2} + \frac{{\cos 3x}}{{{3^2}}} \ldots \ldots \ldots } \right] + \left[ {\frac{{\sin x}}{1} + \frac{{\sin 2x}}{2} + \frac{{\sin 3x}}{3} + \ldots \ldots \ldots } \right]$

$f\left( 0 \right) = \frac{\pi }{4} + \frac{2}{\pi }\left[ {\frac{{1}}{1^2} + \frac{{1}}{{{3^2}}} + \frac{{1}}{{{5^2}}} \ldots \ldots \ldots } \right]$

The convergence of f(x) at x = 0 is valid if

$f\left( 0 \right) = \frac{{f\left( {{0^ - }} \right) + f\left( {{0^ + }} \right)}}{2}$

$f\left( 0 \right) = \frac{{f\left( {{0^ - }} \right) + f\left( {{0^ + }} \right)}}{2}$

where $f\left( {{0^ - }} \right) = \mathop {\lim }\limits_{x \to 0;} \left( {\pi - x} \right) = \pi$

$f\left( 0 \right) = \frac{\pi}{2}$

$\frac{\pi }{2} = \frac{\pi }{4} + \frac{2 }{\pi} \left[ {\frac{1}{{{1^2}}} + \frac{1}{{{3^2}}} + \ldots } \right]$

$\frac{1}{1} + \frac{1}{{{3^2}}} + \frac{1}{{{5^2}}} + \ldots =\frac{{{\pi^2}}}{8}$

\(\mathop \sum \limits_{n = 1}^\infty \frac{1}{{{{\left( {2n - 1} \right)}^2}}} = \frac{\pi^2 }{8}\)

P(X ∪ Y^C) = 0.7 

P(X ∪ YC) = P(X) + P(Y^C) – P(X) P(Y^C)

∴ P(X) + P(YC) – P(X) P(YC)= 0.7

(Since C, Y are independent events)

P(X) + 1 – P(Y) – P(X) {1 – P(Y)} = 0.7

P(Y) – P(X ∩ Y) = 0.3    

Now we know that

P(X ∪ Y) = P(X) + P(Y) – P(X ∩ Y)

∴ P(X ∪ Y) = 0.4 + 0.3 = 0.7

Put x = 0 and then y = 0,

i) $\mathop {\lim }\limits_{x \to 0 } \frac{{xy}}{{{x^2} + {y^2}}}\mathop {\lim }\limits_{y \to0 } \left( {\frac{0}{{{0^2} + {y^2}}}} \right) = 0$

Put y = 0 and then x = 0)

ii) \(\mathop {\lim }\limits_{\begin{array}{*{20}{c}} {x \to 0 } \end{array}} \frac{{xy}}{{{x^2} + {y^2}}}\mathop {\lim }\limits_{x \to 0} \left( {\frac{0}{{{x^2} + 0}}} \right) = 0\)

Put y = mx,

iii) \(\mathop {\lim }\limits_{\begin{array}{*{20}{c}} {x \to 0 } \end{array}} \frac{{xy}}{{{x^2} + {y^2}}}\mathop {\lim }\limits_{x \to 0} \frac{{x\left( {mx} \right)}}{{{x^2} + {m^2}{x^2}}}\)

$\mathop {\lim }\limits_{x \to 0 } \left( {\frac{m}{{1 + {m^2}}}} \right) = \frac{m}{{1 + {m^2}}}$

It does not depend on limits, it depends on m. 

∴ Limits do not exist

Concept:

The following table shows the kinematic indeterminacy for pin jointed and rigid frames

Where J = number of joints in the frame,  r = number of support reactions

Calculations:

J = 7

r = 3

Kinetic indeterminacy

D_K = 2j – r_e

D_K= 2 × 7 – 3 = 11

Explanation:

If shear stress is greater than **τ_cmax **concrete will fail due to the crushing of concrete at the web in the diagonal direction prior to the yielding of stirrups, resulting in a brittle failure. This type of failure of concrete is known as diagonal compression. **τ_cmax **depends on grade of concrete only.

Table 20 of IS 456

Stipulates the maximum shear stress of reinforced concrete in beams τcmax as given below in Table. Under no circumstances, the nominal shear stress in beams τv shall exceed τcmax given in the table for different grades of concrete.

Important Points

 When the shear stress in concrete exceeds these values it leads to brittle failure due to diagonal compression.

Taking moment about point B

R_D × 2L – P × L = 0

R_D = P/2

At joint D:

ΣF_y = 0

$\begin{array}{l} T;cos45^\circ = \frac{P}{2}\ \therefore T = \frac{P}{{\sqrt 2 }} \end{array}$

As per IS 1077: 1992 clause 4.1,

• The  acceptable minimum strength of burnt clay bricks is 3.5 MPa.
• The maximum strength of burnt clay brick goes upto 35 MPa

Explanation:

Total float can be determined once the activity times i.e. EST, EFT, LST, and LFT are known. Total float

F_T = LST – EST= LFT – EFT

For activity 5 – 7,

EST = 38

EFT = 63

LFT = 63

LST = 38

F_T = LST – EST = 38 - 38 = 0

For activity 11 – 12,

EST = 80

EFT = 81

LFT = 81

LST = 80

F_T = LST – EST = 80 - 80 = 0

Net ultimate bearing capacity of a footing embedded in a clay stratum is given by: 

q_nu = cN_c

Generally, net ultimate bearing capacity of a footing is independent of depth and size of footing as shown in the above formula.

${q_{u}} = c{N_c} \ +\ \gamma D_f$

Hence, Ultimate bearing capacity of soil is also independent of width of footing and remains same if width is changed.

Concept:

The relation between porosity (n) and voids ratio (e) is given by,

$n = \frac{e}{{1 + e}}$

The critical hydraulic gradient is given by,

$i; = \frac{{G - 1}}{{1 + e}}$

Calculation:

n = 0.3

\(\begin{array}{l} e = \frac{n}{{1 - n}} = \frac{{0.3}}{{1 - 0.3}} = \frac{3}{7}\We \ know \ that\ i = \frac{{G - 1}}{{1 + e}}\ \therefore i = \frac{{G - 1}}{{1 + \frac{3}{7}}} \end{array}\)

G - 1 = 1.43i

G = 1.43i + 1

Concept:

Over Consolidation Ratios (O.C.R)

It is defined as the ratio of ‘maximum applied effective stress in the past’ to the ‘maximum applied effective stress in the present’

${\rm{O}}.{\rm{C}}.{\rm{R}} = \frac{{{\rm{Maximum;applied;effective;stress;in;the;past}}}}{{{\rm{Maximum;applied;effective;stress;in;the;present}}}} = \frac{{Past;{σ _{max}}}}{{Present;{σ _{max}}}}$

For over consolidated soils, OCR > 1

Calculation:

γ_sat = 16 kN/m³ 

Past σ_max = 90 kN/m² 

\(\begin{array}{l} OCR = \frac{{{σ _{max;}}Past}}{{{ σ _{max}}; Present;}}\ {Present\ Effective\ σ _{max}}\ =({\gamma _{sat}} \times h) - \left( {{\gamma _w} \times h} \right) \{Present\ Effective\ σ _{max}}\ =16 \times 10 - 10 \times 10 = 60;kN/{m^2}\ OCR = \frac{{90}}{{60}} = 1.5 \end{array}\)

Concept:

The percentage of uplift pressure of the weir is given by,

$\phi = \frac{h}{H} \times 100$

Where,

h = uplift pressure heads

H = operating head of weir

Calculation:

h_Q = uplift pressure head at point Q = 3.14 m

hR = uplift pressure head at point R = 2.75 m

hS = uplift pressure head at point S = 0 m

we know that

$\phi = \frac{h}{H} \times 100$

\(\begin{array}{l} {\phi _R} = \frac{{2.75}}{4} \times 100 = 68.75% \ {\phi _P} = 100 - {\phi _R} = 31.25% \{\phi _P} = \frac{{Pressure;head;at;point;P}}{{Total;head}} \times 100\ 31.25 = \frac{h}{4} \times 100 \end{array}\)

h = 1.25 m

∴ Uplift Pressure head at point P = 1.25 m

Concept:

According to Newton’s law of viscosity,

${\rm{\tau ;\alpha }}\frac{{{\rm{du}}}}{{{\rm{dy}}}}$

${\rm{\tau }} = {\rm{\mu }} \times \frac{{{\rm{du}}}}{{{\rm{dy}}}}{\rm{;}}$

μ = Dynamic viscosity in Pa.sec

According to Power-law model

${\rm{\tau }} = {\rm{m}} \times {\left( {\frac{{{\rm{du}}}}{{{\rm{dy}}}}} \right)^{\rm{n}}}$

where, where n = flow behavior index, m = consistency index.Explanation:

The lowest and closest layer of the atmosphere is called the troposphere as shown in the figure above.

Note:

1. The sound pressure of the faintest sound that a normal healthy individual can hear is 20 μPa.
2. It is taken as reference sound pressure level.
3. A 20 μPa pressure is 0 dB on the sound pressure level scale.

Concept:

Radius of relative stiffness

$l = {\left[ {\frac{{E{h^3}}}{{12k\left( {1 - {μ ^2}} \right)}}} \right]^{\frac{1}{4}}}$

where,

E = modulus of elasticity of pavement in kg/cm²

μ = poisson's ratio of concrete = 0.15

h = slab thickness in cm

K = modulus of subgrade reaction in kg/cm³

Explanation:

Statement 1: Radius of relative stiffness is directly related to modulus of elasticity of concrete and inversely

related to Poisson's ratio.

• As E increases, Relative thickness also increases because l α E^1/4
• But As μ increases then 1 – μ² decreases and thus Relative stiffness also increases, because $l;α \ \frac{1}{{{{\left( {1 - {\mu ^2}} \right)}^{\frac{1}{4}}}}}$

∴ Statements 1 is false

Statement 2: Radius of relative stiffness is directly related to thickness of slab and modulus of subgrade

reaction.

• As thickness of slab increases , relative thickness also increases, because l α h^3/4
• As modulus of subgrade reaction increases, relative thickness decreases, because $l;α \ \frac{1}{{{{\left( {K} \right)}^{\frac{1}{4}}}}}$

∴ Statement 2 is also false

Sample size for axle load survey:

Minimum percentage is 15 % for axle load determination when commercial vehicles are between 3000-6000

1.Side overlap (Lateral overlap):

1. The overlap or common coverage between the photograph of two adjacent flights is called the side lap.
2. It is required to prevent the gaps between flight strips.
3. It also prevents us from using extreme edges of the photographs.

2.End overlap (Longitudinal overlap):

1. The overlap or common coverage by two successive photos of the same flight in the direction of flight is called end overlap.
2. It is done to prevent the gaps due to crib, tilt, flying height variations, terrain variations, etc

Hence the optimal flight planning is the one in which both the side overlap and end overlap are taken, to eliminate all the errors.

Concept:

L = latitude is the projection on North-South meridian

D = departure is the projection on East-west meridian

θ = bearing angle

l = length of the line

L = l × Cos θ 

D = l × Sin θ 

$Line;closure = L.C = \sqrt {{{\left( {{\bf{\Sigma }}L} \right)}^2} + {{\left( {{\bf{\Sigma }}D} \right)}^2}}$

Calculation:

Given,

l = 10 m

θ = 30° 

The departure of the line

D = l × sin θ

D = 10 sin 30° = 5 m

Concept:

The tangent length of the circular curve is given by,

$V{T_1} = R\tan \frac{{\rm{Δ }}}{2}$

Where,

R = radius of the circular curve

Δ = Deflection angle = 180° - Intersection angle

Calculation:

Tangent length,

$V{T_1} = R\tan \frac{{\rm{Δ }}}{2}$

$V{T_1} = R\tan \frac{{\rm{60^{} }}}{2}$= 0.577 R

Concept:

The system AX = B has

1. A unique solution if and only if Rank of A = Rank of [A|B] = Number of variables
2. Infinitely many solutions if Rank of A = Rank of [A|B] < Number of variables
3. No solution (inconsistent) if Rank of A ≠ Rank of [A|B], i.e Rank of A < Rank of [A|B]

​Calculation:

The system of equations can be written in matrix form as

\(\left[ {\begin{array}{{20}{c}} 1&2&{ - 3}\ 2&3&3\ 5&9&{ - 6} \end{array};\left| {;\begin{array}{{20}{c}} a\ b\ c \end{array}} \right.} \right]\)

Applying ${R_2} \to {R_2} - 2{R_1}$

\(\begin{array}{{20}{l}} {{R_3} \to {R_3} - 5{R_1}}\ {\left[ {\begin{array}{{20}{c}} 1&2&{ - 3}\ 0&{ - 1}&9\ 0&{ - 1}&9 \end{array};\left| {;\begin{array}{{20}{c}} a\ {b - 2a}\ {c - 5a} \end{array}} \right.} \right]}\ {{R_3} \to {R_3} - {R_2}}\ {\left[ {\begin{array}{{20}{c}} 1&2&{ - 3}\ 0&{ - 1}&9\ 0&0&0 \end{array};\left| {;\begin{array}{*{20}{c}} a\ {b - 2a}\ {c - 3a - b} \end{array}} \right.} \right]} \end{array}\)

For system to be consistent,

Rank of [A] = Rank of [A|B]

∴  c – 3a – b = 0

It can be written as 3a + b – c = 0

$E_1(x)=\int^{\infty}_{-\infty}xf(x)dx$

$⇒ \int^0_{-a}xf(x)dx+\int^a_0af(x)dx$

$⇒ \int^0_{-a}x\left(\frac{x}{a}+1\right)+\int^a_{0}x\left(-\frac{x}{a}+1\right)$

$⇒ \int^0_{-a}\frac{x^2}{a}dx+\int^0_{-a}xds+\int^a_{0}\frac{-x^2}{a}dx+\int^a_{0}xdx$

⇒ 0

∴ $\left[\int^0_{-a}\frac{x^2}{a}=-\int^a_0-\frac{x^2}{a}\right]$

∴ $\left[\int^0_{-a}xdx=-\int^a_0xdx\right]$

$E_2(x)=\int^{\infty}_{-\infty}xg(x)dx$

⇒ $\int^0_{-a}x\left(-\frac{x}{a}\right)dx+\int^0_{a}x\left(\frac{x}{a}\right)dx$

⇒ $\int^0_{-a}-\frac{x^2}{a}dx+\int^0_a\frac{x^2}{a}dx$

⇒ 0

∴ $\left[\int^0_{-a}-\frac{x^2}{a}dx=-\int^0_a\frac{x^2}{a}dx\right]$

Variance

E(x²) - {E(x)}²

$E_1(x^2)=\int^{\infty}_{-\infty}x^2f(x)$

⇒ $\int^0_{-a}x^2\left(\frac{x}{a}+1\right)dx+\int^a_{0}x^2\left(-\frac{x}{a}+1\right)dx$

$\Rightarrow \int^0_{-a}\frac{x^3}{a}dx+\int^0_{-a}x^2dx+\int^a_0-\frac{x^3}{a}dx+\int^a_0x^2dx$

$\Rightarrow -\frac{a^3}{4}+\frac{a^3}{3}-\frac{a^3}{4}+\frac{a^3}{3}=\frac{a^3}{6}$

$E_2(x^2)=\int^{\infty}_{-\infty}x^2g(x)$

$\int^0_{-a}x^2\left(-\frac{x}{a}\right)dx+\int^a_0x^2\left(\frac{x}{a}\right)dx$

$\int^0_{-a}\frac{-x^3}{a}dx+\int^a_0\frac{x^3}{a}dx$

$\left[-\frac{x^4}{4a}\right]^0_{-a}+\left[\frac{x^4}{4a}\right]^a_{0}$

\(\left{0-\left[\frac{-(a)^4}{4a}\right]\right}+\left{\frac{a^4}{4a}-0\right}\)

$\frac{a^3}{4}+\frac{a^3}{4}=\frac{a^3}{2}$

Mean of f(x) is E(x):

\(\begin{array}{l} = \mathop \smallint \limits_{ - a}^0 X \left( {\frac{X}{a} + 1} \right)dx + \mathop \smallint \limits_0^a X \left( {\frac{{ - X}}{a} + 1} \right)dx\ = \left( {\frac{{{X^3}}}{{3a}} + \frac{{{X^2}}}{2}} \right)_{ - a}^0 + \left( {\frac{{ - {X^3}}}{{3a}} + \frac{{{X^3}}}{3}} \right)_0^a = 0\end{array}\)

\(\begin{array}{l} = \mathop \smallint \limits_{ - a}^0 X^2 \left( {\frac{X}{a} + 1} \right)dx + \mathop \smallint \limits_0^a X^2 \left( {\frac{{ - X}}{a} + 1} \right)dx\ \end {array}\)

$\left( {\frac{{{X^4}}}{{4a}} + \frac{{{X^3}}}{3}} \right)_{ - a}^0 + \left( {\frac{{ - {X^4}}}{{4a}} + \frac{{{X^3}}}{3}} \right)_0^a = \frac{a^3}{6}$

⇒ Variance is $\frac{{{a^3}}}{6}$

Next, mean of g(x) is E(x)

\(= \mathop \smallint \limits_{-a}^0 x\left( {\frac{{ - x}}{a}} \right)dx + \mathop \smallint \limits_0^a x\times \left( {\frac{X}{a}} \right)dx = 0\)

Variance of g(x) is E(x²) – {E(X)}², where

\(E\left( {{X^2}} \right) = \mathop \smallint \limits_{ - a}^0 {X^2}\left( {\frac{{ - X}}{a}} \right)dX + \mathop \smallint \limits_0^a {X^2}\left( {\frac{X}{a}} \right)dx = \frac{{{a^3}}}{2}\)

⇒ Variance is $\frac{{{a^3}}}{2}$

∴ Mean of f(x) and g(x) are same but variance of f(x) and g(x) are different

Given curve

x² = 4y                                   ……(i)

y² = 4x                                   ……(ii)

$2x = 4\frac{{dy}}{{dx}}$

At point (0,0)

${\left( {\frac{{dy}}{{dx}}} \right)_{\left( {0,0} \right)}} = \infty = {m_2}$

Let ${m_2} = \frac{1}{{m'}}$ 

where m’ = 0

$\tan \theta = \left| {\frac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right| = \left| {\frac{{{m_1}m' - 1}}{{m' + {m_1}}}} \right| = \left| {\frac{{0 - 1}}{{0 + 0}}} \right|$

$\Rightarrow \theta = \frac{\pi }{2} = 90^\circ ;$

Concept:

The Taylor's series expansion of f(x) about origin (i.e x = 0) is given by

$f\left( x \right) = f\left( 0 \right) + x × f'\left( 0 \right) + \frac{{{x^2}}}{{2!}} × f"\left( 0 \right) + .....$

It is also called Maclaurin's series.

Calculation:

f(x) = x³ - 3x²-5

f(0) = 0³ - 3 × 0² - 5 = - 5

f'(0) = 3x² - 6x = 0

f"(0) = 6x - 6 = - 6

The quadratic approximation of f(x) at the point x = 0 is

\(\begin{array}{l} f\left( x \right) = f\left( 0 \right) + \frac{x}{{1!}}f'\left( 0 \right) + \frac{{{x^2}}}{{2!}}f''\left( 0 \right);\ f(x)= \left( { - 5} \right) + x\left{ 0 \right} + \frac{{{x^2}}}{2}\left{ { - 6} \right}\ f(x) = - 3{x^2} - 5 \end{array}\)

Given,

The elastic body is in hydrostatic state of stress

∴ According to Pascal's law the body has equal stresses on the body

i.e σ_x = σ_y = σ_z

We know that the Volumetric strain is given by

$ϵ{_v} = \left( {\frac{{{σ _x} + {σ _y} + {σ _z}}}{E}} \right)\left( {1 - 2\mu } \right)$

ϵ_v = volumetric strain = δV/V  

$\therefore \frac{{δ V}}{V} = \left( {\frac{{{σ _x} + {σ _y} + {σ _z}}}{E}} \right)\left( {1 - 2\mu } \right)$

As no change in the volume is given

δV  = 0

Either ${σ _x} + {σ _y} + {σ _z} = 0\ or;1 - 2\mu = 0$

As stresses is not zero

1 – 2μ = 0

μ = 0.5

Concept:

Principle stress equation is given by,

${{\rm{σ }}_1},;{{\rm{σ }}_2} = \frac{{{{\rm{σ }}_x} + ;{{\rm{σ }}_y}}}{2} \pm \sqrt {{{\left( {\frac{{{{\rm{σ }}_x} - ;{{\rm{σ }}_y}}}{2}} \right)}^2} + {{\left( {{τ _{xy}}} \right)}^2}}$

Where,

σ₁ = maximum principal stress

σ₂ = minimum principal stress

σ_x = stress in x direction

σ_y = stress in Y direction

τ_xy = shear stress in xy plane 

Calculation:

Given:

σ_x = 5 MPa

σy = 5 MPa

From the above figure

σ_x  + σ_y = σ₁ + σ₂

5 + 5 = σ₁ + σ₂

we know thatσ₂ = 0

∴ σ₁ = 10 MPa 

Now,

$\begin{array}{l} {σ _{1}/σ_2} = \frac{{{σ _x} + {σ _y}}}{2} \pm \sqrt {{{\left( {\frac{{{σ _x} - {σ _y}}}{2}} \right)}^2} + τ _{xy}^2} \ ∴ {σ _1} = \frac{{5 + 5}}{2} + \sqrt {{{\left( {\frac{{5 - 5}}{2}} \right)}^2} + τ _{xy}^2} \end{array}$

⇒ 10 = 5 + τ _xy

∴ τ_xy = 5 MPa

As there is no horizontal force

Hence H_P = H_S = 0

Vp + Vs = wL

Symmetric structure V_p = wL/2 , V_S = wL/2

${\rm{\Sigma }}{M_Q} = - {H_P} \times \frac{L}{2} + {H_S} \times \frac{L}{2} - wL \times \frac{L}{2} + \frac{{wL}}{2} \times L$

∴ BM at Q = 0

Note: This beam acts similarly to a simply supported beam. 

∴ BM at supports is zero.

L = 1m, A = 0.05 m²,

E = 30 × 106 N/m²

Consider joint S

F_SQ = F_SR

2F_SQ cos 45° = W

$\Rightarrow {F_{SQ}} = \frac{W}{2} \times \sqrt 2 = \frac{W}{{\sqrt 2 }}$

$\therefore {F_{SQ}} = \frac{W}{{\sqrt 2 }}$

As the truss is symmetrical

$\therefore {F_{QP}} = {F_{PR}} = \frac{W}{{\sqrt 2 }}$    (Tensile)

Now consider joint ‘Q’

$\begin{array}{l} {F_{QP}} = {F_{QS}} = \frac{W}{{\sqrt 2 }}\ {\rm{\Sigma }}{F_x} = 0 \end{array}$

$\Rightarrow \frac{{{F_{QP}}}}{{\sqrt 2 }} + \frac{{{F_{QS}}}}{{\sqrt 2 }} + {F_{QR}} = 0$

⇒ F_QR = W                           (Compressive)

∴ F_QR = 100 kN                   (Compressive)

Stress in member QR

\(\begin{array}{l} {\sigma {QR}} = \frac{{{F{QR}}}}{{2A}}\ \Rightarrow {\sigma _{QR}} = \frac{{100}}{{2 \times 0.05}}\ = \frac{{100 \times 100}}{{2 \times 5}} = 1000;kN/{m^2} \end{array}\)

∴ σ_QR = 1000 kN/m²

As the member QR consist compressive load so it will go under shortening

$\begin{array}{l} {\rm{\Delta }} = \frac{{PL}}{{2AE}} = \frac{{{F_{QR}}.{L_{QR}}}}{{2AE}}\ {L_{QR}} = \sqrt {{L^2} + {L^2}} = \sqrt 2 L\ \therefore {\rm{\Delta }} = \frac{{100 \times {{10}^3} \times \sqrt 2 \times 1}}{{2 \times 0.05 \times 30 \times {{10}^6}}}\ = \frac{{100 \times {{10}^3} \times \sqrt 2 }}{{0.1 \times 30 \times {{10}^6}}} = \frac{{\sqrt 2 }}{{30}} \end{array}$

∴ Δ = 0.471 m

Concept:

The design tensile strength of bolt based on net section rupture is given by,

$T_{db}={0.9;{A_{net}}\frac{{{f_u}}}{{{\gamma _{m1}}}}}$

Where,

A_net = net tensile area of bolt = 0.78 A_sb

Asb = Nominal shank area of bolt

Calculation:

Given,

Tensile strength of plate in figure (2) will be greater than in figure (1)

\(\begin{array}{l} {\left( {0.9;{A_{net}}\frac{{{f_u}}}{{{\gamma _{m1}}}}} \right)2} > {\left( {0.9;{A{net}}\frac{{{f_u}}}{{{\gamma _{m1}}}}} \right)_1}\ \left( {B - 2d + \frac{{{p^2}}}{{4g}}} \right)t > \left( {B - d} \right)t\ {p^2} > 4gd \end{array}\)

D_s = 2

∴ Number of plastic hinge required for complete collapse= D_s + 1 = 2 + 1 = 3

Mechanism 1:

${\rm{\Delta }} = \frac{{2L}}{{3\theta }} = \frac{{4L}}{3}$

⇒ θ = 2θ

For principal of virtual work done

$\begin{array}{l} - 2{M_P}\theta - 2{M_P}\theta - 2{M_P}\phi - {M_P}\phi + P\left( {\frac{{2L}}{3}\theta } \right) = 0\ \Rightarrow 4{M_P}\theta + 3{M_P}\phi = \frac{{2PL}}{3}\theta \ \Rightarrow 8{M_P}\phi + 3{M_P}\phi = \frac{{4PL}}{3}\phi \ \Rightarrow 11{M_P} = \frac{{4{P_u}L}}{3}\ \Rightarrow {P_u} = \frac{{33}}{4}\frac{{{M_P}}}{L}\ \therefore {P_u} = 8.25\frac{{{M_P}}}{L} \end{array}$

Mechanism 2 :

$\begin{array}{l} 2{M_P}\theta + {M_P}\theta + {M_P}\theta + {M_P}\theta = P\left( {\frac{{2L}}{3}} \right)\ \Rightarrow 5{M_P} = \frac{{2PL}}{3}\ \Rightarrow {P_u} = \frac{{15{M_P}}}{{2L}}\ \therefore {P_u} = 7.5\frac{{{M_P}}}{L} \end{array}$

Concept:

TiE  = Early start time 

T^jE = Early Finish time 

Tⁱ_L  = Latest start time 

T^jL = Latest Finish time - 

tij = Duration of the activity

T^j_E = Tⁱ_E +​ t_ij  

TjL  = TiL + tij   

Calculation:

The critical duration is 15 days which is observed along the path P – Q – T – W – X which is the critical path of the project.

OMC-SP > OMC-MP and MDD-SP < MDD-MP

Concept:

According to Darcy’s law,

For laminar flow Vα i

V = K × i

Where,

V = velocity of water flowing in soil, K = coefficient of permeability

${\rm{i}} = hydraulic~ gradient= \frac{{{{\rm{h}}_{\rm{L}}}}}{{\rm{L}}}$

h_L = head difference, L = seepage length

Seepage velocity is given by,

V_S = V/n

Where,

 n = porosity of the soil

Calculation:

T = Time = 100 days

D  = distance = 100 m

h_L = head difference = 3 m

n = 15 % = 0.15

i = h_L/L = 3/100

V_S = D/T = 100/100 = 1 m/day

V = K × i

V = V_S × n

∴ V_S × n = K × i

$K = \frac{{{V_s} \times n}}{i} = \frac{{1 \times 0.15}}{{\frac{3}{{100}}}} = 5;m/day$

Concept:

The total hardness of water is given by,

$TH = \left[ {C{a^{2 + }}\left( {in;mg/lit} \right) \times ;\frac{{eq;wt;of;CaC{O_3}}}{{eq;wt;of;C{a^{2 + }}}}} \right] + \left[ {m{g^{2 + }}\left( {in;mg/lit} \right) \times \frac{{eq;wt;of;CaC{O_3}}}{{eq;wt;of;m{g^{2 + }}}}} \right];$

The total alkalinity of water due to HCO­₃ is given by,

$TA = HC{O_3}\left( {in;mg/lit} \right) \times \frac{{eq;wt;of;CaC{O_3}}}{{eq;wt;of;HC{O_3}}}$

We know that

$Quantity;of;ions;\left( {meq/lit} \right) = \frac{{quantity;of;ions;in\left( {mg/lit} \right)}}{{eq;wt;of;the;ions}}$

$\therefore TH = \left[ {C{a^{2 + }}\left( {in;meq/lit} \right) \times eq;wt;of;CaC{O_3};} \right] + \left[ {m{g^{2 + }}\left( {in;meq/lit} \right) \times eq;wt;of;CaC{O_3}} \right]$

$\therefore TA = HC{O_3}\left( {in;meq/lit} \right) \times eq;wt;of;CaC{O_3}$

Based on the relation between TH (total hardness) and TA (total alkalinity) estimation of Carbonaceous hardness (CH) and Non-carbonaceous hardness (NCH) is made.

Carbonate hardness = minimum of {total hardness, alkalinity}

When TH > TA then CH = TA, NCH = TH – CH

When TH ≤ TA then CH =TH, NCH = 0

Calculation:

Given:

Ca²⁺ = 2.65 meq/I, Mg²⁺ = 1.45 meq/I, Na⁺ = 2.25 meq/I

HCO^-₃ = 3.3 meq/I, SO^2-₄ = 0.6 meq/I, CI^- = 2.85 meq/I

We know that

The equivalent weight of CaCO₃ = 50

$\therefore TH = \left[ {C{a^{2 + }}\left( {in;meq/lit} \right) \times eq;wt;of;CaC{O_3};} \right] + \left[ {m{g^{2 + }}\left( {in;meq/lit} \right) \times eq;wt;of;CaC{O_3}} \right]$

$\therefore TH = \left[ {2.65 \times 50;} \right] + \left[ {1.45 \times 50} \right] = 205;mg/lit;as;CaC{O_3}$

$\therefore TA = HC{O_3}\left( {in;meq/lit} \right) \times eq;wt;of;CaC{O_3}$

$\therefore TA = 3.3 \times 50 = 165;mg/lit;as;CaC{O_3}$

As we can see TH > TA

∴ CH = TA = 165 mg/lit as CaCO₃

NCH = TH – CH = 205 – 165 = 40 mg/lit as CaCO₃

Concept:

Webster’s method for an optimum cycle time of a signal is given by

${C_o} = \frac{{1.5L + 5}}{{1 - Y}}$

Effective green time on i^th lane

${G_i} = \frac{{{Y_i}}}{Y} \times \left( {{C_o} - L} \right)$

L = lost time per cycle

Y = Sum of the ratios of normal to saturated flows

Calculation:

Given,

Y = y₁ + y₂ + y₃

Y= 0.30 + 0.25 + 0.25 = 0.80

L = 10 sec (given)

∴ Optimum cycle time

$\begin{array}{l} {C_0} = \frac{{1.5L + 5}}{{1 - Y}}\ {C_0} = \frac{{\left( {1.5 \times 10} \right) + 5}}{{1 - 0.80}} = \frac{{15 + 5}}{{0.20}} = \frac{{20}}{{0.2}} = 100;sec \end{array}$

Now green times are calculated by,

${G_1} = \frac{{{y_1}}}{y}\left( {{C_0} - L} \right) = \frac{{0.30}}{{0.80}}\left( {100 - 10} \right)$ = 33.75 ≈  34 sec

${G_2} = \frac{{{y_2}}}{y}\left( {{C_0} - L} \right) = \frac{{0.25}}{{0.80}}\left( {100 - 10} \right)$ = 28.11 ≈  28 sec

${G_3} = \frac{{{y_3}}}{y}\left( {{C_0} - L} \right) = \frac{{0.25}}{{0.80}}\left( {100 - 10} \right)$ = 28.11 ≈  28 sec