GATE CE 2017 Official Paper: Shift 2 (Previous Year Paper)

The sentence is in the perfect continuous tense which can be noticed because of the usage of "been". 

When you see "would have" in a sentence it means that the action didn't actually happen, because something else didn't happen first.

From the sentence we know that the "event" has already happened. 

Hence the correct answer is "had been" as it would convey the correct tense, which is the past perfect continuous.

Option 2 is correct.

Let us look at the meanings of all the words:

Through: having finished using or doing something.

Pretty: (of a person, especially a woman or a child) attractive in a delicate way without being truly beautiful.

Complete: finish making or doing.

Sloppy: careless and unsystematic; excessively casual.

Haphazard: lacking any obvious principle of organization.

Example sentence: I've got some work to do but I should be through in an hour if you can wait.

Option 2 is correct.

Faces available on the cards are: 2, 3, red, and blue.

Proposition: If a card has an even value on one side, then its opposite face is red.

We need to check the card with face 2 whether its opposite is red or not. The opposite side will be RED, if the given proposition is TRUE.

No need to check the card with face 3 as it is not even number.

No need to check the card with face red as it can have either even or odd number as its opposite number. (The converse of the given proposition may or may not be true).

We need to check the card with face blue whether its opposite number is even or odd. Its opposite number will be odd, if the given proposition is TRUE.

\(81 \times {\left( {\frac{{16}}{{25}}} \right)^{x + 2}} \div {\left( {\frac{3}{5}} \right)^{2x + 4}} = 144\)

\(\Rightarrow 81 \times \frac{{{{\left( {\frac{{16}}{{25}}} \right)}^{x + 2}}}}{{{{\left( {\frac{3}{5}} \right)}^{2x + 4}}}} = 144\)

\(\Rightarrow 81 \times \frac{{{{\left( {\frac{4}{5}} \right)}^{2x + 4}}}}{{{{\left( {\frac{3}{5}} \right)}^{2x + 4}}}} = 144\)

\(\begin{array}{l} \Rightarrow 81 \times {\left( {\frac{4}{3}} \right)^{2x + 4}} = 144\ \Rightarrow {81 \times\left( {\frac{4}{3}} \right)^{2x}} \times {\left( {\frac{4}{3}} \right)^4} = 144 \end{array} \)

\(\begin{array}{l} \Rightarrow {\left( {\frac{{16}}{9}} \right)^x} = \frac{{144}}{{256}}\ \Rightarrow {\left( {\frac{{16}}{9}} \right)^x} = \frac{9}{{16}} = {\left( {\frac{{16}}{9}} \right)^{ - 1}} \end{array}\)

⇒ x = -1

Two dice are thrown simultaneously.

Total number of possibilities = 6² = 36

The possible perfect squares = 1, 4, 9, 16, 25, 36

Possible outcomes to get the product as 1 = (1, 1)

Possible outcomes to get the product as 4 = (1, 4), (4, 1), (2, 2)

Possible outcomes to get the product as 9 = (3, 3)

Possible outcomes to get the product as 16 = (4, 4)

Possible outcomes to get the product as 25 = (5, 5)

Possible outcomes to get the product as 36 = (6, 6)

Total number of possibilities = 8

Probability = 8/36 = 2/9

Explanation:

Option 1 can be safely concluded from the given passage.

Option 2 is incorrect because it says 'Customers were served in an arbitrary order which is not given in the passage.

Option 3 is incorrect because it says 'the system is likely to operate on a last-come-first-served basis' which is not given in the passage.

Option 4 is incorrect because it says 'Entering the center early ensured that one would have shorter service times' which is not given in the passage.

Hence option 1 is correct.

A map shows the elevations of Darjeeling (D), Gangtok (G), Kalimpong (K), Pelling (P), and Siliguri (S).

Kalimpong is at a lower elevation than Gangtok.

⇒ K < G       ----(1)

Pelling is at a lower elevation than Gangtok.

⇒ P < G       ----(2)

Pelling is at a higher elevation than Siliguri.

⇒ S < P        ----(3)

Darjeeling is at a higher elevation than Gangtok.

⇒ G < D       ----(4)

From inequalities (1) and (4), we get

⇒ K < G < D

From inequalities (2) and (3), we get

⇒ S < P < G

i. Pelling is at a higher elevation than Kalimpong (K < P): There is no clear relation between K and P, so it may or may not TRUE.

ii. Kalimpong is at a lower elevation than Darjeeling (K < D): It is TRUE as K < G < D.

iii. Kalimpong is at a higher elevation than Siliguri (S < K): There is no clear relation between S and K, so it may or may not TRUE.

iv. Siliguri is at a lower elevation than Gangtok (S < G): It is TRUE as S < P < G.

P, Q, R, S, T and U are seated around a circular table.

R is seated two places to the right of Q.

P is seated three places to the left of R.

S is seated opposite U.

According to the above information, there are two possibilities:

Case 1: The arrangement is as shown below.

After switching the seats of P and U, the arrangement becomes

P is immediately right of Q.

Case 2:

The arrangement is as shown below.

After switching the seats of P and U, the arrangement becomes

T is immediately to the left of P.

So, T is immediately to the left of P or P is immediately to the right of Q.

Let Budhan cycles at a speed of x km/hr and walks at a speed of y km/hr.

Budhan covers a distance of 19 km in 2 hours by cycling one fourth of the time and walking the rest.

I.e. Budhan cycled for half an hour and had walk for remaining time.

⇒ 0.5x + 1.5y = 19       ----(1)

The next day he cycles (at the same speed as before) for half the time and walks the rest (at the same speed as before) and covers 26 km in 2 hours.

⇒ x + y = 26      ----(2)

By solving the equations (1) and (2), x = 20 km/hr and y = 6 km/hr

So, the speed at which Budhan walks is 6 km/hr.

At 28^th minute and at 39^th minute elevator directly moves form second floor to fifth floor.

So, statement (i) is not correct.

Elevator stays at ground floor for 21 minutes whereas it stays for 19 minutes at fourth floor.

So, statement (ii) is not correct.

3x₁ + 2x₂ = c₁

3x₁ + x₂ = c₂

\(\left[ {\begin{array}{*{20}{c}} 3&2\ 4&1 \end{array}} \right] = \left[ {\rm{A}} \right]\)

[A] – λ[I] = 0

\(\left[ {\begin{array}{*{20}{c}} {3 - {\rm{\lambda }}}&4\ 2&{1 - {\rm{\lambda }}} \end{array}} \right] = 0\)

3 + λ² – 4λ – 8 = 0

λ² – 4λ – 5 = 0

w = f (x, y)

Partial differentiation with respect to x and y gives:

\(\frac{{{\rm{dw}}}}{{{\rm{dt}}}} = \frac{{\partial {\rm{w}}}}{{\partial {\rm{x}}}}\frac{{{\rm{dx}}}}{{{\rm{dt}}}} + \frac{{\partial {\rm{w}}}}{{\partial {\rm{y}}}}\frac{{{\rm{dy}}}}{{{\rm{dt}}}}\)

If scope of construction work is well-defined with all its drawings, specification quantities and estimates, then lump sum contract is used.

Concept:

The dry density (γ_d) of soil is given by,

\({{\rm{γ }}_{\rm{d}}} = \frac{{\left( {1 - {{\rm{η }}_{\rm{a}}}} \right){{\rm{G}}_{\rm{}}}{{\rm{γ }}_{\rm{w}}}}}{{1 + {\rm{e}}}}\)

η_a - Percentage of air voids

G - Specific gravity of soil solids

e - Voids ratio

For zero air voids ⇒ volume of air is zero, hence η_a = 0, a_c = 0                       

Also, S + a_c = 1

Hence, S = 1      

So above equation becomes, \({{\rm{γ }}_{\rm{d}}} = \frac{{\left( {1 - 0} \right){{\rm{G}}_{\rm{}}}{{\rm{γ }}_{\rm{w}}}}}{{1 + \frac{{{\rm{w}}{{\rm{G}}_{\rm{}}}}}{{\rm{S}}}}}\)

⇒ \({{\rm{γ }}_{\rm{d}}} = \frac{{{{\rm{G}}_{\rm{}}}{{\rm{γ }}_{\rm{w}}}}}{{1 + {\rm{w}}{{\rm{G}}_{\rm{}}}}}\)

Explanation

Skempton’s Pore Pressure coefficient are A and B,

where,

A → depends upon the degree of saturation and over consolidation ratio

B → depends upon degree of saturation of soil

Value of B lies between 0 and 1. For completely saturated soil, B = 1 and for completely dry soil, B = 0.

Value of A may be as large as 2 to 3 for very loose saturated fine sand and it can be less than zero for over consolidated clay.

Pressure diagram for partially submerged soil with surcharge is shown below:

But no option is matching.

If the water table is at ground table then Option b is the answer.

Explanation:

The specific speed is of a fluid machine is a type number representative of its performance. To compare the performance of different pumps (or turbines), it is necessary to have some terms that will be common to all the pumps (or turbines), the term ordinarily used for this purpose is the specific speed. This can be used to predict one pump (or turbine) behavior based on tests of similar, but different sized pumps (or turbines).

\({N_{{s_T}}}\) (specific speed for turbines) = \(N{P^{\frac{1}{2}}}/{H^{\frac{5}{4}}}\)

\({N_{{s_P}}}\) (specific speed for the pump) = \(N{Q^{\frac{1}{2}}}/{H^{\frac{3}{4}}}\)

For a turbine, \({N_{{s_T}}}\) is the speed of a member of the same homologous series as the actual turbine, so reduced in size as to generate unit power under a unit head of the fluid. Similarly, for a pump, \({N_{{s_P}}}\) is the speed of a hypothetical pump with reduced size but representing a homologous series so that it delivers unit flow rate at a unit head. The specific speed (\({N_s}\)) is, therefore, not a dimensionless quantity.

As per Noise pollution (Regulation and control) Rules 2000 of India.

\(\begin{array}{l} {\rm{EL}}{{\rm{R}}_1} = \frac{{{\rm{\Delta T}}}}{{{\rm{\Delta h}}}} = \frac{{15.5 - 15.0}}{{60 - 10}} = 0.01^\circ \frac{{\rm{c}}}{{\rm{m}}} = ;10^\circ \frac{{\rm{c}}}{{{\rm{km}}}};\ {\rm{EL}}{{\rm{R}}_2} = \frac{{15.0 - 14.3}}{{130 - 60}} = 0.01^\circ \frac{{\rm{C}}}{{\rm{m}}} = ;10^\circ \frac{{\rm{c}}}{{{\rm{km}}}} \end{array}\)

Adiabatic lapse rate (ALR) is 9.8°C/km for dry adiabatic and 6.c/km for wet adiabatic condition.

∵ ELR > ALR

So, it is super adiabatic and the atmosphere is said to be highly unstable.

Heterotrophs cannot produce organic compounds from inorganic sources and therefore rely on consuming other organisms in the food chain. In activated sludge process, in the presence of food mass and oxygen heterotrophic bacteria converts food mass into biomass.

Concept:

Equilibrium cant \(= \frac{{{\rm{G}}{{\rm{V}}^2}}}{{127{\rm{R}}}}\)

V - Design speed (kmph)

R - Radius of curvature (m)

For B.G track G = 1.676

\(\begin{array}{l} {\rm{Cant}} = \frac{{1.676{\rm{;}}{{\rm{V}}^2}}}{{127{\rm{;R}}}}{\rm{m}}\ {\rm{e}} = \frac{{1.315{\rm{;}}{{\rm{V}}^2}}}{{\rm{R}}}{\rm{cm}} \end{array}\)

The essence of the intersection control is to resolve these conflicts at the intersection for the safe and efficient movement of both vehicular traffic and pedestrians.

The safety within a roundabout and the efficiency of a roundabout can be increased respectively by decreasing the entry radius and increasing the exit radius.

Note:

There are two methods of intersection controls 

1. time sharing and

2. space sharing.

The type of intersection control that has to be adopted depends on the traffic volume, road geometry, cost involved, importance of the road etc.

Resection method of orientation is employed when the plane table occupies a position not yet plotted on the drawing sheet.

Concept:

Total degree of freedom when axial and shear deformations in all members are allowed. 

D_k = 3 + 3 + 3 + 3 + 1 + 1 = 14

When axial and shear deformations are negligible,

The degree of freedom equals to the number of members would be deducted i.e. 6

∴ Kinematic indeterminacy for the frame if axial and shear deformation is considered negligible, (D_k) = 14 - 6 = 8.

Hence, reduction in degree of kinematical indeterminacy = 14 – 8 = 6

Mistake Point:

Kindly do read the Question again as it is asked about the reduction in the degree of kinematical indeterminacy i.e 14 – 8 = 6

Concept:

The correct expression as per IS 456: 2000 to find Target mean strength is:

fcm = fck + (k × σ)

fcm = Target mean strength

fck = Characteristic strength

σ = Standard deviation

 For not more than 50% of results are likely to fall k = 0.

Since characteristic strength has been defined as the value below which not more than 50% of results are likely to fall.

Hence,

Target mean strength = f_t = fck + (0× σ) = Characteristic strength (f_ck) = 25 N/mm²

Shear strain in an element is positive when the angle between two positive faces (or two negative faces) is reduced. The strain is negative when the angle between two positive (or two negative) faces increase.

Angle between 1 & 4 is increased by 0.0005 rad.

∴ γ_xy = – 0.0005 = 0.001 K

∴ K = – 0.5

In plate load test bearing capacity of clay does not depend upon size of footing.

q_up (for plate) = q_uf (for clay) = 180 kPa

\(\begin{array}{l} {\rm{Z}} = \frac{{{{\rm{T}}_{\rm{s}}} - {{\rm{T}}_{\rm{E}}}}}{\sigma }\ 1.64 = \frac{{{T_s} - 200}}{{6.1}} \end{array}\)

T_s = 210 days

\({\rm{V}} = {{\rm{x}}^2}{\rm{\hat i}} + 2{{\rm{y}}^3}{\rm{\hat j}} + {{\rm{z}}^4}{\rm{\hat k}}\)

Divergence \(\left( \nabla V \right) = \frac{\partial }{{\partial {\rm{x}}}}\left( {{{\rm{x}}^2}} \right){\rm{}} + \frac{\partial }{{\partial {\rm{y}}}}\left( {2{{\rm{y}}^3}} \right){{}} + \frac{\partial }{{\partial {\rm{x}}}}\left( {{{\rm{z}}^4}} \right){\rm{}}\)

= 2x + 6y² + 4z³

At x= 1, y=2 and z=3

[Divergence (V)]= 2 × 1 + 6 × 2² + 4 × 3³ = 134

Probability of getting (H.H.H) = ½ × ½ × ½ = 1/8

Probability of getting (H.H.H.H) = ½ × ½ × ½ × ½ = 1/16

Given condition is that (H.H.H) is already realized conditional probability of getting next H after

(H.H.H) \(= \frac{{\frac{1}{{16}}}}{{\frac{1}{8}}} = 0.5\)

Or we can say that getting head or tail on next trial is independent of previous trials. Hence probability of getting head for the fourth time given 3 heads on previous attempt = 0.5

Concept:

The discharge (Q) passing through a flow net is given by,

\({\rm{Q}} = \frac{{{\rm{k\times H\times}}{{\rm{N}}_{\rm{f}}}}}{{{{\rm{N}}_{\rm{D}}}}}\)

k - Coefficient of soil

H - Difference between upstream and downstream heads

N_f - Number of flow channels in flow net

N_d - Number of potential drops in flow net

Calculation:

Given:

k= 10-6 m/s

Nf = No. of flow lines – 1 = 5 – 1 = 4

Nd = No. of equipotential lines – 1 = 11 – 1 = 10

Q = 10 ^-6 m/s × 4m × 4/10 per meter width

⇒ 1.6 × 10^–6 m³/s/m width

⇒ 1.6 cm³/s/meter width

Important Points

For anisotropic soils,

\({\rm{Q}} = \frac{{{\rm{k'\times H\times}}{{\rm{N}}_{\rm{f}}}}}{{{{\rm{N}}_{\rm{D}}}}}\)

\(;{\rm{k'}} = \sqrt {{k_x} \times {k_y}} \)

k_x - Coefficient of permeability in x direction

k_y - Coefficient of permeability in y direction.

Explanation

For flat vertical curve we assume that l₁ = l₂

l = l₁ + l₂ = 100 + 100 = 200 m

Rainfall intensity = 3.5 cm/hr

ϕ – index = 1.5 cm/hr

Intensity of effective rainfall = 3.5 – 1.5 = 2 cm/hr

C₁ = f₀ (final infiltration capacity) = 25 mm/hr                      

C₂ = f₁ – f₀ = 200 – 25 = 175 mm/hr

f = 90 mm/hr after 1 hr

Hence, 90 = 25 + 175 × e^-k×1

\(\Rightarrow {{\rm{e}}^{ - {\rm{k}}}} = \frac{{90 - 25}}{{175}} = 0.3715\)

Taking log_e on both side

K = 0.99

R = 50

Grade compensation \(= \frac{{30 + {\rm{R}}}}{{\rm{R}}} = \frac{{30 + 50}}{{50}} = \frac{{80}}{{50}} = 1.6\)

(Grade compensation)_max \(= \frac{{75}}{{\rm{R}}} = \frac{{75}}{{50}} = 1.5\)

Grade compensation should not be more than (Grade compensation)max

Hence, (grade compensation)_max = 1.5

Target is having inclination of 45° with x-axis

\(\begin{array}{l} \Rightarrow \frac{{dy}}{{dx}} = \tan 45^\circ \ \Rightarrow \frac{{d\left( {xln;x} \right)}}{{dx}} = 1\ \Rightarrow \ln x + \frac{x}{x} = 1 \end{array}\)

∴ At x = 1, y = 1 × ln1 = 0

Explanation:

\(I = \mathop \smallint \limits_0^1 \frac{{{{\left( {{{\sin }^{ - 1}}x} \right)}^2}}}{{\sqrt {1 - {x^2}} }}dx\)

Put sin^-1 x = t

\(\begin{array}{l} \frac{{dx}}{{\sqrt {1 - {x^2}} }} = dt\ = \mathop \smallint \limits_0^{\frac{\pi }{2}} {t^2}dt\ = \left. {\frac{{{t^3}}}{3}} \right]_0^{\frac{\pi }{2}}\ = \frac{{{\pi ^3}}}{{24}} \end{array}\)

\(\begin{array}{l} A = \left[ {\begin{array}{*{20}{c}} 1&5\ 6&2 \end{array}} \right];\left[ B \right] = \left[ {\begin{array}{*{20}{c}} 3&7\ 8&4 \end{array}} \right]\ A{B^T} = \left[ {\begin{array}{*{20}{c}} 1&5\ 6&2 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 3&8\ 7&4 \end{array}} \right]\ = \left[ {\begin{array}{*{20}{c}} {3 + 35}&{8 + 20}\ {18 + 14}&{48 + 8} \end{array}} \right]\ = \left[ {\begin{array}{*{20}{c}} {38}&{28}\ {32}&{56} \end{array}} \right] \end{array}\)

Explanation:

y” – 4y’ + 3y = 2t – 3t²

f(D) = D² – 4D + 3

\(\begin{array}{l} P.I = \frac{I}{{f\left( D \right)}}\left( {2t - 3{t^2}} \right)\ = \frac{1}{{{D^2} - 4D + 3}}\left( {2t - 3{t^2}} \right)\ = \left[ {\frac{1}{{1 - D}} - \frac{1}{{3\left( {1 - \frac{D}{3}} \right)}}} \right]\left( {t - \frac{{3{t^2}}}{2}} \right)\ \left( {1 + D + {D^2} + \ldots } \right)\left( {t - \frac{{3{t^2}}}{2}} \right) - \frac{1}{3}\left( {t - \frac{D}{3} + \frac{{{D^2}}}{9} + \ldots } \right) \times \left( {t - \frac{{32{t^2}}}{2}} \right)\ = \left( {t - \frac{{3{t^2}}}{2} + 1 - 3t - 3} \right) - \frac{1}{3}\left( {1 - \frac{{3{t^2}}}{2} + \frac{1}{3} - 8 - \frac{1}{3}} \right) \end{array}\)

= – 2 – 2t – t²

1. Le chatelier Test: This test is used to measure the soundness of OPC due to lime. Lime & Magnesia are two primary compounds responsible for soundness of cement.

2. Vee Bee Test: It is one of the methods of measuring the workability of concrete.

3. Blaine Air Permeability: It is used to measure fineness of cement.

4. The Vicat Apparatus: It is used to measure setting time and consistency of concrete.

\(\begin{array}{l} {{\rm{\delta }}_1} = \frac{5}{{384}}\frac{{w{L^4}}}{{EI}}\ = \frac{5}{{384}} \times \frac{{6 \times {4^4}}}{{10000;m}} = 20;mm \end{array}\)

\(\begin{array}{l} {{\rm{\delta }}_{\rm{L}}} = \frac{{P{L^3}}}{{48EI}}\ = \frac{{120 \times {2^3}}}{{48 \times 10000;m}} \end{array}\)

= 20 mm

∴ δ₁ = δ₂

Explanation:

For P:

Σf_v = 0

⇒ V = 80 N

∴ Shear force at support = S_P = 80 N.

Moment at support = Pl ⇒ 80 × 8 = 640 Nm.

∴ M_P = 640 Nm.

For Q:

Σf_v = 0

⇒ V = wx ⇒ 20 × 8 = 160 N

∴ Shear force at support = S_Q = 160 N.

Moment at support = M_Q ⇒ \(\frac{wx^2}{2}\) 

\(\therefore \frac{20;\times;8^2}{2}⇒640;Nm\)

For R:

Σf_v = 0

⇒ V = 0 = S_R

Moment at support = M_R = 640 Nm

Hence,

S_P < S_Q > S_R & M_P = M_Q = M_R

Explanation:

As per IS 1200 (part 3), 1976, Cl No- 4, brick work is measured in the following way:

No deduction is made for following.

• Opening each up to 0.1 m².
• Ends of beams rafters up to 0.05 m² in section.

When the ground is not uniform levels shall be taken before the start, after site clearance and after the completion of the work and the quantity of excavation in cutting computed from these levels.

Concept:

End Bearing Resistance = Effective Stress at Base × N_q

End Bearing Resistance = qNq

Q_P1 = qN_q 

Q_P2 = qN_q 

Find \(\frac{{{Q_{{P_1}}}}}{{{Q_P}_2}} \)

Calculation:

Q_P1= 18 × 20 × N_q  

Q_P1= 18 × (20N_q ) = 360 N_q

Q_P2 = qN_q 

= (19 – 10) × 20 × N_q 

= 9 × (20 N_q) = 180 N_q

(Assuming γ_w = 10 kN/m³)

\(\frac{{{Q_{{P_1}}}}}{{{Q_P}_2}} = 2\)

Hence, by 100%.

When σ₁ = σ₃

q = 0, \(p = \frac{{2{\sigma _1}}}{2} = {\sigma _1}\)

So stress point lies on p-axis of p-q plot

If σ₁ > σ₃

\(\begin{array}{l} q = \frac{{{\sigma _1} - {\sigma _3}}}{2};so;q > 0\ p = \frac{{{\sigma _1} + {\sigma _3}}}{2};so;p > 0 \end{array}\)

So, stress point p & q are positive.

P: d = 60t

Q: d = 60t²

Space Headway (S) = 60t² – 60t

For space headway to be max,

\(\frac{{ds}}{{dt}} = 0\)

120t – 60 = 0

⇒ 60(2t – 1) = 0

⇒ t = ½ hr

Maximum space Headway

= 60 × ½ - 60 × (½)²

= 60 × ½ - 60 × ¼

= 30 – 15 = 15 km

Case-I: Undrained condition

\({\rm{F}}.{\rm{O}}.{\rm{S}} = \frac{{{\rm{resisting;shear;strength}}}}{{{\rm{Acting;shear;stress}}}} = ;\frac{{80}}{{50}} = 1.6\)

Case-II: Drained condition

\(\begin{array}{l} {\rm{F}}.{\rm{O}}.{\rm{S}} = \frac{{{\rm{\bar \sigma }}\tan \phi ' + c'}}{{{\rm{Acting;shear;stress}}}}\ = \frac{{\left[ {2 \times 18 + 4\left( {20 - 9.81} \right)} \right] \times \tan 18^\circ + 20}}{{50}} = 0.9 \end{array}\)

Calculation:

Given:

For Tower B,

Radial distance of top of tower = r = 6cm

Length of image = d = 2cm

Height of tower (h₂) = 80m

∴ Flying height above datum = H – h₁

\(\begin{array}{l} d = \frac{{r{h_2}}}{{H - {h_1}}}\ \Rightarrow H - {h_1} = \frac{{r{h_2}}}{d}\ = \frac{{6 \times 80}}{2} = 240;m \end{array}\)

For Tower A,

Both on same plane

\(d = \frac{{r{h_2}}}{{H - {h_1}}}\)

r = 4 cm

d = 1.5 cm

H – h₁ = 240 m

\({h_2} = \frac{{d \times \left( {H - {h_1}} \right)}}{r} = \frac{{1.5 \times 240}}{4};\)

h₂ = 90 m

Concept:

The torsion equation is given as,

\(\frac{T}{{{I_P}}} = \frac{τ }{r} = \frac{{G\theta }}{l}\)

T - Torque acting on the shaft, I_P - Polar moment of inertia

τ - Shear stress, C - Shear modulus of shaft material, l - Length of the shaft

Calculation:

Given:

D_o = 100 mm

D_i = 50 mm

τ_max = 125 MPa

\(\begin{array}{l} {I_P} = \frac{\pi }{{32}}\left( {D_0^4 - D_2^4} \right) = \frac{\pi }{{32}}\left( {{{100}^4} - {{50}^4}} \right)\ T \Rightarrow \frac{{τ {I_P}}}{r}\ \Rightarrow \frac{{125}}{{50}} \times \frac{\pi }{{32}}\left( {{{100}^4} - {{50}^4}} \right) \times {10^{ - 6}} \end{array}\)

⇒ 23 KN.m

Prestressing force = 1600 kN

\(\begin{array}{l} {M_{max}} = \frac{{w{l^2}}}{8}\ = \frac{{30 \times {8^2}}}{8} = 240;kNm \end{array}\)

Shape of tendon profile will follow the shape of bending moment diagram and for equilibrium of the section having maximum Bending Moment.

M_max = Pe

240 = 1600 × e

\(\Rightarrow \frac{{240 \times {{10}^3}mm}}{{1600;}} = e\)

e = 150 mm = 0.15 m

Concept:

The load taken by the weld (P) is given by,

P = tt × leff × τ

 t_t - Throat thickness = 0.7 ×  size of weld

l_eff - Effective length of weld  

τ - Permissible shear stress in weld

Load taken by plate = Permissible stress in plate × Area of the plate

Calculation:

Given:

l_eff = 100 + 50 + 100 = 250 mm

Permissible stress in weld = 110 N/mm²

Permissible stress in plate = 150 N/mm2

Area of the plate = 50 × 8 = 400 mm²

Maximum load taken by plate = 150 × 400 = 60 kN

Maximum load taken by the weld = t_t × l_eff × τ

= 0.7 × 6 × 250 × 110 = 115.5 kN

∴ Permissible load = Minimum of above two = 60 kN

Moment at \({\rm{Q}} = {{\rm{M}}_{\rm{Q}}} = 2000 \times 2 - 1650 \times \frac{{{2^2}}}{2}\)

= 700 kN.m (anticlockwise)

\(\begin{array}{l} \begin{array}{*{20}{c}} {{{\rm{K}}_{{\rm{QR}}}} = \frac{{4{\rm{EI}}}}{4} = \frac{{{\rm{EI}}}}{{1{\rm{m}}}}}&{{K_{QS}} = \frac{{4EI}}{4} = \frac{{EI}}{{1m}}} \end{array}\ \therefore {K_Q} = {K_{QR}} + {K_{QS}} = \frac{{2EI}}{{1m}} \end{array}\)

∴ M_Q = K_Q θ_Q

\(\begin{array}{l} \therefore {\theta _Q} = \frac{{{M_Q}}}{{{K_Q}}} = \frac{{700 \times {{10}^6} \times 1000}}{{2 \times 2.5 \times {{10}^4} \times 8 \times {{10}^8}}} = 0.0175;rad\ = 0.0175 \times \frac{{180}}{\pi } = 1;degree \end{array}\)

The area enclosed by the inclined line and the vertical axis is called complementary strain energy. For linearly elastic material, the complementary strain energy and strain energy is same.

\(\epsilon= \frac{\delta }{l} = \frac{{2.5}}{{2000}} = \frac{1}{{800}}\)

σ_y = 250 MPa

Complementary strain energy = strain energy

\( = \frac{1}{2} \times {f_y} \times \epsilon \times vol.;of;bar\)

\(= \frac{1}{2} \times 250 \times \frac{1}{{800}} \times \frac{\pi }{4} \times {8^2} \times 2000 = 15708;N.mm\)

As per Boussinesq’s equation, stress due to one corner load

\(\begin{array}{l} {\sigma _z} = \frac{{3P}}{{2\pi {z^2}}}{\left( {\frac{1}{{1 + {{\left( {\frac{r}{z}} \right)}^2}}}} \right)^{\frac{5}{2}}}\ r = \sqrt {{2^2} + {z^2}} = 2\sqrt 2 m \end{array}\)

z = 5 m

P = 5000 kN

\({\sigma _z} = \frac{{3 \times 5000}}{{2\pi \times {5^2}}}{\left[ {\frac{1}{{1 + {{\left( {\frac{{2\sqrt 2 }}{5}} \right)}^2}}}} \right]^{\frac{5}{2}}}\)

= 47.71 kN/m²

So due to four load vertical stress = 4σ_z

= 4 × 47.71 = 190.84 kN/m²

Height of water column added \(= \frac{{5;c{m^3} \times 1000}}{{5;mm \times 5;mm}} = 20;cm\)

P_A = P_A = P_atm

P_A + 20ρ_wg – 2xρ_Mg = P_B

\(x = \frac{{10}}{{13.6}} = 0.735;cm\)

New height = 20 + x = 20.735 cm

Let, Subcritical depth = y

then, super critical depth \(= \frac{y}{2}\)

Specific energy is constant at both depths

\(\begin{array}{l} \Rightarrow y + \frac{{{Q^2}}}{{2g{B^2}{y^2}}} = \frac{y}{2} + \frac{{{Q^2}}}{{2g{B^2}{{\left( {\frac{y}{2}} \right)}^2}}}\ \Rightarrow \left( {y - \frac{y}{2}} \right) = \frac{{{Q^2}}}{{2g{B^2}}}\left( {\frac{{ - 1}}{{{y^2}}} + \frac{4}{{{y^2}}}} \right)\ \Rightarrow \frac{y}{2} = \frac{{{2^2}}}{{2 \times 9.81 \times {1^2}}} \times \frac{3}{{{y^2}}} \end{array}\)

y = 1.07

As per Theisson Polygon method,

Average Precipitation \(= \frac{{\frac{{{a^2}}}{8}\left( {{G_1} + {G_2} + {G_3} + {G_4}} \right) + \frac{{{a^2}}}{2} \times {G_5}}}{{{a^2}}}\)

\(\begin{array}{l} \Rightarrow \frac{{{G_1} + {G_2} + {G_3} + {G_4}}}{8} + \frac{{{G_5}}}{2}\ \Rightarrow \frac{{300 + 285 + 272 + 290}}{8} + \frac{{288}}{2} \end{array}\)

⇒ 287.37 mm

Calculation:

Given:

For Rice,

CCA = 10000 ha

B= 150 days

Δ = 130cm = 1.3 m

\(\begin{array}{l} {\rm{D}} = \frac{{{\rm{B}} ~\times~ 8.64}}{{\rm{\Delta }}}\ = \frac{{150~ \times~ 8.64}}{{1.3}} = 996.92;ha/cumecs \end{array}\)

Area irrigated = 2500 ha

∴ Design Discharge

\(= \frac{{2500}}{{996.923}} = 2.507;cumecs\)

For, wheat

B= 120 days

Δ = 50 cm

\(D = \frac{{8.64 \times 120}}{{0.5}} = 2073.6;ha/cumecs\)

Area irrigated = Duty × Design Discharge (Q)

= 2073.6 × 2.507 = 5198.5 ha.

Q = 0.01 m³/s

d = 100 mm

R_e = 800

g = 9.81 m/s²

\({h_f} = \frac{{fl{V^2}}}{{2gd}}\)

Since Re is less than 2000 the flow will be laminar flow

Hence, \(f = \frac{{64}}{{{R_e}}} = 0.08\)

\(\begin{array}{l} {h_f} = \frac{{fl{Q^2}}}{{12.1{d^5}}}\ = \frac{{0.08 \times 1000 \times \left( {0.01} \right)2}}{{12.1 \times {{\left( {0.1} \right)}^5}}} = 66.10;m \end{array}\)

Energy on dry basis: It will be the total energy when whole mass is dry.

In current situation only 60% mass is dry, so energy corresponding to 60% dry mass = 5805 kJ/kg

Using unitary method, energy corresponding to 100% dry mass \(= \frac{{5805}}{{60}} \times 100\) = 9675 kJ/kg

Energy based on as discarded basis = 5805 kJ/kg

So, difference = 9675 – 5805 = 3870 kJ/kg

Concept:

The BOD consumed in ‘ t ’ days is given by,

Yt = L₀ × [1 - e- (k × t)]

Yt – BOD consumed at t days

L₀ – Ultimate BOD

t - Time in days

k - Rate constant.

Calculation:

\(\therefore \frac{{{L_0}\left( {1 - {e^{ - 5k}}} \right)}}{{{L_0}}} = 0.68\)

⇒ 1- e^{-5 × k} = 0.68

⇒ e^{- 5 × k} = 0.32

\(\therefore k = \frac{1}{5}\ln \left( {\frac{1}{{0.32}}} \right)\)

= 0.2279 = 0.23

\(Q = 1000\frac{{{m^3}}}{{day}}\)

Overflow Rate = 20 m/day

H = 3m

Surface Area of circular Tank

\(= \frac{{1000}}{{20}} = 50{m^2}\)

Assuming diameter to be ‘d’

Hence, \(\frac{\pi }{4} \times {d^2} = 50\)

\(d = \sqrt {\frac{{50 \times 4}}{\pi }} = 7.98 = ;8;m\)

Explanation:

Hardness will be due to multivalent cations only

Ca²⁺ = 60 mg/L → equivalent weight = 20 (given)

Mg²⁺ = 36.6 mg/L → equivalent weight = 12.2 (given)

Miliequivalent of calcium and magnesium ions are as follows:

\(\begin{array}{l} C{a^2} + ; = \frac{{60}}{{20}} = 3\ M{g^{2 + }} = \frac{{36.6}}{{12.2}} = 3 \end{array}\)

Total hardness (mg/L as CaCO₃)

= (miliequivalent of Ca²⁺ + miliequivalent of Mg²⁺) × equivalent weight of CaCO₃

= (3 + 3) × 100/2

= 300 mg/L as CaCO₃

Concept

The radius of relative stiffness is given by

\(l = {\left( {\frac{{E; \times ;{h^3}}}{{12; \times; K ;\times; \left( {1 - {\mu ^2}} \right)}}} \right)^{\frac{1}{4}}}\)

Where

 E is the modulus of elasticity

 K is the modulus of subgrade

 μ is the Poisson ratio

h is the thickness of pavement

Calculation

Given,

 Data of two pavements P & Q is as follows 

\(\begin{array}{l} l = {\left[ {\frac{{E{h^3}}}{{12\left( {1 - {\mu ^2}} \right)k}}} \right]^{\frac{1}{4}}}\ \frac{{{l_P}}}{{{l_Q}}} = {\left[ {\frac{{{E_P}}}{{{E_Q}}} \times \frac{{h_P^3}}{{h_Q^3}} \times \frac{{{K_Q}}}{{{K_P}}}} \right]^{\frac{1}{4}}}\ = {\left[ {\frac{E}{E} \times \frac{{{h^3}}}{{{{\left( {0.5h} \right)}^3}}} \times \frac{{2K}}{K}} \right]^{\frac{1}{4}}} \end{array}\)

= [16]^0.25 = 2

∴ Ratio of relative stiffness is 2

Concept:

Combined Error due to curvature and refraction is given by

Ecomb = Ecurvature + Erefraction

Distance (km), \(d = 3.855\sqrt h\)

Rearranging the above equation can also be written as, 

h (m) = 0.06735 × d2

Where

d = distance traveled on the surface of the earth (km)

h = height of object (m)

Case - Let d1 be the distance of the observer from the horizon and d2 distance of the lighthouse from the horizon.

d1 = 3.855\(\sqrt{h_1}\)  km

d2 = 3.855\(\sqrt{h_2}\)  km

Where h1 = the height of the observer's eye above the sea level, h2 = The top of the lighthouse above the sea level

Calculation:

Let d1 is the distance of the observer from the horizon and d₂ distance of the lighthouse from the horizon.

\(0.0673;d_1^2 = 5;and,;0.0673;d_2^2 = 40\)

d₁ = 8.62 km, d₂ = 24.38 km

∴ Distance of observer from the lighthouse d = d₁ + d₂ = 33 km