Practice Test: Civil Engineering (CE)- 7 - Civil Engineering (CE)

The sentence talks about a past incident and a condition has been mentioned. Thus option 2 fits here correctly. 'Would have' is the correct tense to be used here. Option 1 is illogical. 'Would' must be used here and not 'will' as it is in the past tense. Past perfect tense is incorrect here.

The word that fits here is a noun thus options 3 and 4 are eliminated. The form should be singular as for abstract nouns we do not use the plural form. Option 1 is thus the correct answer.

The word powerful is an adjective and 'empowering' is a verb.

The radius of outer circle = r

∴ Diagonal of square = 2r

Side of square (a)=√2r

Now, the radius of the inner circle = r₁ = 

Average speed = Total distance/Total time

Total distance = a + a + a = 3a

Total time = 

Average speed = 

The sentence mentions a person doing something as per her wishes thus the first word must be a positive reaction. 'Restive' which means 'restless' and 'jinxed' which means 'cursed' are incorrect here. Between options 1 and 3,the word 'elated' which means 'too happy' and the word 'venture' which means 'undertake a risky or daring journey or course of action' fit here correctly. The word 'cease' means 'stop' and does not convey a proper meaning. Thus option 3 is the correct answer.

As, the faster runner is twice as fast as the slowest runner, the faster runner would have complete two rounds by the time the slowest runner completes one round.

Their first meeting takes place after the fastest runner takes 6 min to complete two rounds and the slowest runner completes one round.

Time taken to complete one round by fastest runner = 6/2 = 3 minutes

Rounds needed to complete the race = 1600/400 = 4

Time taken to complete the race = 4 × 3 = 12 min

The young one of a horse is called a foal. Similarly, the young one of a goose is known as gosling. Thus, option 2 contains the pair that best expresses a relationship similar to that expressed in the given pair. In all the other pairs, the first word expresses the male form of the latter.

Since teenagers here have tender age and commit crimes unknown to its consequences, clearly they should not be treated as any criminal. They need to be taught so that they can make difference between right and wrong and do not commit these crimes in future. Hence inference in option 1 is correct.

450 y = n³ implies that 450 y is the cube of an integer. 

When we prime-factorize the cube of an integer, we get 3 (or a multiple of 3) of every prime factor.

8 is the cube of an integer because 8 = 2³ = 2*2*2.

Thus, when we prime-factorize 450 y, we need to get at least 3 of every prime factor. 

Here's the prime factorization of 450 y:

450 y = 2 * 3² * 5² * y 

Since 450 provides only one 2, two 3's, and two 5's, and we need at least 3 of every prime factor, the missing prime factors must be provided by y. Thus, y must provide at at least two more 2's, one more 3, and one more 5. 

Smallest possible case: 

y = 2² * 3 * 5

In this problem, cumulative frequency is given, so converting into frequency

From the table, it is clear that option b, i.e.

60% of the scholars published at least 2 papers is correct.

A matrix is singular if and only if its determinant is zero.

x(3+1)−2(x²+1)=0

−2x²+4x−2 = 0

x²−2x+1 = 0

(x−1)² = 0

x = 1 

According to formula, L(tⁿ f(t)) =

Here, f(t) = sin (2t) ⇒ F(s) = 

∴ L(t²sin (2t)) =  

ii) (4 - x)_x=2 = 2

Hence at x = 2, f(x) is continuous

∴ f(x) is not differentiable.

The given limit is in 0/0 form, So

Applying L hospitals rule

⇒ log b - 1 = -1 - log b

⇒ 2 log b = 0

⇒ b = 1

y log y dx + (x – log y) dy = 0

It is a Leibnitz’s equation in x.

Integrating factor = 

In context of the Real and conjugate Beam:

(i) The fixed end of the real beam will become the free end of the conjugate beam

(ii) The diagram of the Real beam will be loading diagram for the conjugate beam

(iii) Hogging moment diagram (-ve) in the real beam will give downward loading in the conjugate beam

As end A is fixed in the real beam, so it will be free in conjugate beam. Similarly for end B in real beam will become fixed end in conjugate beam. The  diagram of the real beam will become loading diagram for the conjugate beam. 

Concept:

Using the relations/ formula

At surface y = y_max , ε = ε_max

ε_surface = 

Calculation:

ε_surface = 0.00084

L = 250 mm

y_max  = = 0.7mm

ε_surface =

ε_surface =

ε_surface = 0.00084

0.00084 = 

δ = 9.375 mm

As per IS 456 : 2000,

(i) The minimum stripping time for props to slab spanning up to 4.5 m and spanning over 4.5 m is 7 days and 14 days respectively.

(ii) The minimum stripping time for props to aches spanning up to 6 m and spanning over 6 m is 14 days and 21 days respectively.

(iii) When bar of two different diameter are to be spliced the lap length shall be calculated on the basis of bar of larger diameter.

(iv) A random sampling procedure shall be adopted to ensure that each concrete batch shall have a reasonable chance of being tested that is, the sampling should be spread over the entire period of concreting and cover all mixing units.

Member AB and AG are co-linear,

So F_AB = F_AG = 0 kN

Now removing the member AB and AG. Now member BC and BG are also collinear. So, F_BC = F_BG = 0 kN

Consider Joint G

∑F_x = 0

F_GC Sin 63.43° = 0

F_GC = 0 kN

∑F_V = 0

20 + F_GF + F_GC Cos 63.43° = 0

F_GF = - 20 kN 

Concept:

Tensile strength of a section, (T_dn)

where A_n = net section area, mm²

f_u = ultimate strength of the material

γ_m = pantial safety factor.

Calculation:

For Fe 410 grade of steel

f_u = 410 N/mm², f_y = 250 N/mm²

γ_m = 1.25

T_dn = 826.56 kN.

Concept:

Vertical stress increment is given by:

r = radius of circular area

z = depth of soil stata

q = uniformly distributed circular load

Calculation:

z = 6 m

q = 200 kN/m²

σ_z = 29.2 kN/m²

Concept:

Using Peck, Hanson equation:

The net allowable bearing pressure (kN/m²) is given by

q_a(net) = 0.41 N S C_W

N = Final corrected SPT Number

C_w = Water table correction factor

S = Permissible Settlement of foundation (mm)

Calculation:

N = 28

S = 40 mm

C_w = 0.8

q_a(net) = 0.41 × 28 × 40 × 0.8

q_a(net) = 367.36 kN/m²

Consistency Index (I_C) of the soil is:

I_P = W_L – W_P

W_n = 30%

W_L = 55%

W_P = 25%

Liquidity Index (I_L) will be:

W_n = 30%

W_P = 25%

W_L = 55%

As the soil is in Plastic stage of consistency because both liquidity index and consistency index lies in between 0 to 1.

Concept:

Height of capillary rise is given by:

Where,

σ = Surface tension of fluid at 20° C

ρ = Density of fluid (kg/m³)

D = Diameter of the glass tube

h = height of capillary rise/fall

Calculation:

σ = 0.0825 N/m

θ = 0°

ρ = 1000 × 9.6 = 960 kg/m³

Diameter (D) = 5.60 mm

For 30% increase in “h”

h_I = 8.125 mm

D_I = 4.31 mm

% Decrease 

As per IS10500:2012, the total concentration of manganese (as Mn) and iron (as Fe) for drinking water purpose shall not exceed 0.3 mg/L.

Concept:

Carbonate hardness = minimum of {Alkalinity, Total Hardness}

Total hardness = Carbonate Hardness + Non-carbonate Hardness.

Calculation:

Carbonate Hardness= minimum of {200 mg/L , 250 mg/L}

Carbonate Hardness = 200 mg/L

Non – Carbonate Hardness = Total Hardness – Carbonate Hardness

= 250 – 200

= 50 mg/L

Concept:

Ultimate BOD = L_o

BOD_5­ = 5day BOD, of water/waste water

BOD at any time ‘t’ (L_t)

L_t = L_o(1−e^−kDt) 

Where L_o = ultimate BOD

K_D = Rate constant

Calculation:

L_t = 300 × (1 – e^{-0.3585 × t})

L₅ = BOD₅ = 300 × (1 – e^{-0.3585 × 5})

= 250 mg/L

Capacity of different types of roads in rural areas.

Concept:

We know that by Martin’s formula for a Broad Gauge track in a transition curve. For speed greater than 100 kmph.

Speed. 

Where V is speed in kmph

R is the radius of curvature in m.

Calculation:

Bearing of PQ = 60°

Bearing of RQ = N50° W (WCB)

Bearing of RQ = 360° - 50° = 310° (QBS)

Bearing of line QR = Bearing (RQ) – 180°

Bearing of line QR = 310° - 180° = 130°

Bearing of line QP = 180° + Bearing of PQ = 180° + 60° = 240°

∠PQR = (Bearing)_QP = - (Bearing)_QR

∠PQR = 240° - 130°

∠PQR = 110°

Relative precision = 

where p = perimeter of traverse

e = closing error 

Relative precision = 

Relative precision = 

IS 6403 – 1981: Indian standard code of Practice for determination for Bearing Capacity of shallow foundation

IS 4640 – 1980: Indian standard specification for split spoon sampler

IS 1892 – 1979: Code of Practice for site investigations for foundations

IS 2132 – 1972: Code of Practice for Thin walled tube sampling for soils

Concept:

Metacentric height of the rectangular block is given by:

I_min = Moment of inertia of the plan about X – X or Y – Y (minimum value should be taken)

V_Dis = Volume of the displaced fluid

B G = Distance between centre of Buoyancy and centre of gravity of the block

Calculation:

Let the Depth of immersion of the block will be ‘x’

GM = 1.4 m

I_min = I_yy = 32 m⁴

BG = PG – PB

PG = 1 m

V_Dis = x × 4 × 6 = 24x

3x² – 14.4x + 8 = 0

X₁ = 4.16 m

X₂ = 0.641 m

X₁ = 4.16 m is not possible

So, X₂ = x = 0.641 m

So, depth of immersion will be 0.641 m.

Concept:

The empirical method to evaluate evaporation based on Dalton’s law in developed by Meyer’s and the equation is as follow.

E = Rate of evaporation in mm/day

e_w = saturated vapour pressure in mm of Hg

e_a = vapour pressure of air in mm in Hg

V₉ = Mean wind velocity in km/hr at a height of 9 m from the ground.

k_m = constant, 0.36 for large deep water and 0.50 for small shallow water.

Calculation:

(e_s - e_a) = 15 mm of Hg

V₉ = 10 m/s

= 36 km/hr

E = 17.55 mm/day

Concept:

The optimum signal cycle is given by as per Webster’s Method

Where L = total lost time per cycle, in sec

= 2n + R (n = number of lanes and sum of R is all red-time)

Y = Sum of Ratio of normal flow to the saturation flow.

= y₁ + y₂

y₁, y₂ are the ratio of normal flow to saturation flow in direction 1 and 2.

Where q₁ = Normal flow on each approach

q₂ = Normal flow on each approach

S = saturation flow per unit time.

Calculation:

Given:

L = 11 sec

y_p = 0.3 sec

y_Q = 0.25 sec

∴ Y = y_p + y_Q = 0.3 + 0.25

Y = 0.55

The optimum cycle length

C_o = 47.77 sec

Given that,    is an Eigen vector of A.

⇒ AX = λ₁X

For all values of a and b, is an Eigen vector of A.

Given that,  is an Eigen vector of B

⇒ BX = λ₂X

being an eigen vector of matrix B

a + b = λ₂ a      ----(1)

2a = λ₂ b      ----(2)

From (1) and (2)

⇒ ab + b² = 2a²

⇒ 2a² - ab - b² = 0

⇒ 2a² - 2ab + ab - b² = 0

⇒ 2a (a - b) + b (a - b) = 0

⇒ b = a, b = -2a

For b = a, (or) b = -2a, X is an eigen vector of matrix B.

Since the above function f(x) satisfy the assumption of Rolle’s theorem in interval [-1, 1]

∴ f(x) is continuous in [-1, 1]

F(x) is differentiable in (-1, 1)

F(-1) = f(1)

∴ for continuity

LHL = RHL = Functional value at 0 i.e f(0)

= 1

= q

∴ LHL = RHL

⇒ q = 1

Also for differentiability

LHD = RHD

at x = 0

at x = 0

p – q = 1

⇒ p = 1 + q

⇒ p = 1 + 1

⇒ p = 2

∴ ordered pair (p, q) ≡ (2, 1)

Integrating on both sides:

Now given y (0) = 0

Thus,  

C = 0

ln (1 – 2y) = -x²

1−2y = e − x²

2y =1 − e ^{- x2}

Assuming poisson distribution

​

Alternate:

• Probability a particular cashew is in particular biscuit is 
• Probability a particular cashew is not in a particular biscuit = 
• Probability that no cashew in = 

For f(x) = 1, x, x² there is no error in the formula

f(x) = 1

= a(1) + b(1) + c(1)

2h = a + b + c

f(x) = x

= a(−h) + b(0) + c(h)  

a – c = 0

f(x) = x²

= a(h²) + b(0) + c(h²)

⇒ a = c =

Concept:

The bending strength or the plastic moment capacity of a laterally unsupported beam is given by:

Md = β_b Z_p f_bd

Where

β_b = 1 → for plastic and concept section

 for semi - compact section. 

Z_e = Elastic section modulus

Z_p = Plastic section modulus

f_bd = design bending compressive strength

y_mo = Partial safety factor for material = 1.1

X_LT = bending stress reduction factor to account for lateral torsional buckling

X_LT ≤ 1.0

Calculation:

For a semi-compact section of a laterally unsupported steel bean plastic moment capacity

X_LT = 1.0

γ_mo = 1.1

f_y = 250 Mpa for grade fe 410

Md = 200 =  454.54 kNm

For No tension failure

Where

B_min = Minimum width of the dam

H = Height of the dam

G = specific gravity of the dam

C = coefficient of uplift pressure.

For critical condition

c = 0

Calculation:

Minimum width

= 30.69

= 31m

Concept:

Minimum height of the hump

Where y₁ = depth of flow

F₁ = Froude numbers at section depth y₁.

Calculation:

ΔZ_m = 0.186 m

ΔZ_m = 18.6 cm

Alternate solution:

= 1.371 m

y_c = 0.79

E_c = 1.5 × 0.79 = 1.185

Δz_m = E₁ – E_c = 1.371 – 1.185 = 0.186 m = 18.6 cm

Δz_m = 0.186, Δz_m = 18.6 cmZ

Concept:

Primary consolidation settlement is given by

C_C = compression index

e₀ = initial void ratio

H₀ = thickness of clay layer.

 initial effective stress within the clay layer.

increase in effective stress

Δ H = primary consolidation settlement

Calculation:

C_C = 0.40

e₀ = 1.345

Δ H = 20 mm

H₀ = 1.5 m

Concept:

To find the cumulative standard axle load repetition using Modified CBR method

Where A = Number of commercial vehicle per day when construction is completed.

r = Rate of growth of traffic

n = design life of pavement.

LDF = Lane distribution factor

VDF = Vehicle damage factor.

Calculation:

N_s = 110.27 million standard axles.

Concept:

Practical capacity of Rotary

Where W = width of weaving section.

e = Average width at entry and exit = 

P = proportion of weaving traffic

L = length of the weaving section.

The above expression will be valid only when these four conditions will be satisfied

A. 6m ≤ w ≤ 18m

D. 0.4 ≤ P ≤ 1

Calculation:

Given:

W = 15m

L = 54m

e = 8m

P = 60%

Practical capacity of Rotary

Qp = 3665.5 vehicle/hr

Concept:

The distance AB on ground is given by:

Calculation:

h_a = 502 m, h_b = 525 m

H = 1670 m

x_a = +15.50 mm

x_b = = +109.20 mm

y_a = -71.20 mm

y_b = -25.20 mm

X_b = 791.35 m

X_a = 114.58 m

Y_b = -182.62 m

Y_a = -526.34 m

AB = 759.05 m

Concept:

Maximum height of vertical cut is given by:

γ = γ_sat

For pure clay ɸ = 0°

c_u = undrained cohesion in clay

Where,

τ_u = undrained shear strength of clay

C_u = undrained cohesion in clay

ɸ_u = undrained friction angle

Calculation:

As, ɸ_u = 0°

τ_u = C_u + 0

150 = C_u

H_c = 31.58 m

H_o = Actual height of vertical cut

H_o = 7.895 m

Equating the moments

P = Prestressing force = 2000 kN

l = span of the beam = 9 m

w = load on the beam

e = eccentricity of the tendon = 0.18 m

w = 35.55 kN/m

Calculation:

Let us assume the weight of cement will be w.

Therefore, For 1m³ of concrete

(1 × w + 3 × w + 6 × w) + 0.5w = 2250 kg

10.5 w = 2250

w = 214.29 kg

Therefore, the amount of course aggregate for 1m³ of concrete will be

= 3 × 214.29

= 642.87 kg

For 50m³ of concrete

= 32.14 tonnes

Concept: Power developed in the shaft will be

Where, P = power developed / transmitted

T = Applied Torsion

N = Speed of rotation of shaft (rpm)

Torsion Equation:

I_P = Polar section modulus of the shaft

L = Length of the shaft

θ = Angle of rotation of the shaft

G = shear modulus of the shaft material

τ = shear stress developed in the shaft material

r = radial distance of the shaft

Calculation:

P = 500 kW

T = 13.642 kN-m

D_ex = 112.33 mm

D_int = 0.75 × 112.33

D_int = 84.25 mm

θ_max = 0.0554 radian 

Concept:

Major (σ₁) and Minor (σ₂) principal stresses are given by:

Calculation:

σ_x = 15 MPa

σ_y = 10 MPa

τ_xy = 5 MPa

σ₁ = 18.10 MPa

σ₂ = 12.5 – 5.59

σ₂ = 6.91 MPa

Concept:

Temporal Mean Velocity Gradient (G^’)

μ = dynamic viscosity of water,   

V = Volume of the raw water to be treated, m³

P = Power dissipated, watts

Calculation:

Given:

G’ = 106/second

V = 2500 m³

νν = 8.9 × 10^-7 m²/s

∴ dynamic viscosity (μ) = ρν 

μ = 1000 kg/m3 × 8.9 × 10^-7 m²/s

⇒ P = μVG^’2

⇒ P = 8.9 × 10^-4 × 2500 × 106 × 10^-3

⇒ P = 0.23585 kW

Possible mechanism for the beam is

By Principle of Virtual work

P × 5 × Q = MP × θ + QM_P × 2θ

⇒ P = 0.6 M_P

Concept:

∴ T₂ = P = T₁ e^μθ ​

If the tension in the one side of the belt is known, then the tension in the other side of the belt can be calculated as

T₂ = T₁ e^μθ​

Where, T₁ and T₂ are tension in the rope

μ = coefficient of friction between rope and shaft material.

q = contact angle in radian.

Calculation:

T₂ = P and T₁ = 50 kN

 q = 1.5 × 2π

q = 3π (q must be in radian)

∴ T₂ = P = T₁ e^μθ​

P = 50 × (e)^0.15×3π

P = 205.6 kN

P = 20.56 x 10⁴ N

Concept:

Probability factor

Where

T_S = Scheduled time required to complete the project

T_E = expected completion time of the project

σ = standard deviation

Calculation:

Expected completion time of the project, T_E = 50 weeks

variance = 25 weeks

Scheduled Completion time of project = 63weeks

∴  standard deviation

Probability factor

z = 2.6

Probability = 97.72 + (2.6 - 2) 

Probability = 99.01%

Concept:

From Mohr Coulomb failure criterion:

ɸ^I = effective angle of friction

c^I = effective cohesion

 = major effective principal stress at failure

 = Effective cell pressure

Calculation:

C^I = 25 kN/m²

ɸ^I = 23°

Because in unconsolidated undrained test, total stress parameters are used

350 - u_f = (200 - u_f) × 2.283 + 75.54

350 - u_f = 456.6 - 2.283 u_f + 75.54

(2.283-1) u_f = 456.6 + 75.54 – 350

u_f = 141.96 kN/m²

Concept:

F.O.S against boiling is giving by

i_er = critical hydraulic gradient

i_per = permissible hydraulic gradient

η = percentage air voids

a_c = air content

η = porosity

calculation:

i_cr = i_per x FOS

i_cr = 0.315 × 3

i_cr = 0.945

e = 0.746

η = 0.427 = 42.7%

a_c = 15%

η_a = 0.427 × 0.15

η_a = 0.06405

η_a = 6.405%

Concept:

Discharge through the trapezoidal section is given by

n → manning’s roughness coefficient

R → Hydraulic mean depth

For efficient cross-section,

y₀ = depth of flow

A = Area of cross-section

For efficient cross-section,

S₀ = Bed slope of the channel

Calculation:

Q = 210 m³/sec

F_r = 0.48

D = Hydraulic depth

For efficient cross-section,

y₀ = 6.132 m [depth of flow]

Concept:

To check the possibility of flow:

Stream function (Ψ)

Velocity potential function (ϕ):

Calculation:

Hence flow is possible

Angular velocity (W_Z)

W_Z ≠ 0

Flow is rotational

Also, For rotational floes velocity potential (ϕ) function does not exist.

Proof:

But W_Z ≠ 0

So, ∇ ² ϕ ≠ 0

ϕ not exist

check for stream function (Ψ)

For stream function, if there is a flow, stream function (Ψ) always exist.

Integrating

From equation (i) and (ii)

Concept:

Settlement of foundation on sands is given by:

S_f = settlement of foundation (mm)

S_p = settlement of plate (mm)

B_f = width of foundation (m)

B_p = width of plate (m)

Calculation:

Load intensity on the foundation = 

From the chart,

S_p = 13.5 mm

B_f = 2.5m

B_p = 0.40 m

Settlement of foundation at 3m depth is:

S_f = 32.96 mm

Now correcting it for the depth

So, final settlement of foundation at 4m below the ground surface is

S_f = (Depth correction factor) × S_f

S_f = 0.88 × 32.96

S_f = 29.00 mm

Thiessen Polygon Method:

This method is also known as weighted area method. The rainfall data at every station is given a weightage on the basis of an area close to the station

Where Pavg = Average Precipitation

P1, P2, ........Pn are precipitation at station 1, 2, ....., n respectively

A1, A2, ........An is the area for station 1, 2, ....., n respectively

This method of fining average rainfall is suitable

→ when the area is large

→ rainfall varies from place to place

1) In order to find the representative area, all the stations are joined, so as to form a network, of the non-intersection triangle.

2) Perpendicular bisector is then drawer from the sides of the triangle in such a way that a representative area is obtained for each of the rain gauge stations.​

Calculation:

Catchment diameter = 24 km

Catchment area for station A

= (12 × 12) km²

= 144 km²

∴ Total area of catchment

= 452.4 km²

∴ Catchment area for B, C, D, and E will be equal and

= 77.1 km²

∴ Mean Rainfall

Mean Rainfall =228.7 cm = 0.2287 m

Concept:

Hydraulic Head Loss and Expansion of the Filter bed during Backwash:

Backwashing is accomplished by reversing the flow in filter media to force clean water to move upward through the filter media. In order to clean the filter bed, it is required to expand the bed, so that granular media filters are no longer in contact with each other.

To hydraulically expand a porous bed, the head loss (h_Le) must be at least equal to the buoyant weight of the filter bed. Even when the bed gets expanded to depth D_e, the head loss through the expanded bed essentially remains unchanged, because the total buoyant weight of the bed is constant.

Head loss, h_Le = D (1 – n) (G – 1)        …..(i)

where D = undisturbed depth of filter bed in m.

n = porosity

G = specific gravity

Also, h_Le = D_e (1 - n­_e)(G – 1)        ..…(ii)

where D_e = the depth of expanded bed in m.

n_e = porosity of expanded bed

∴ from equation (i) and (ii) we get:

Calculation:

⇒ n_e = 0.478

Therefore,

h_Le = D_e (1 – n_e) (G – 1)

h_Le = 0.75 (1 – 0.478) (2.72 – 1)

h_Le = 0.67338m

∴ h_Le = 673.4 mm

Alternate Solution:

Head loss in both the condition will be same i.e.

h_Le = D (1 – n) (G – 1) = De (1 – n_e)(G – 1)

h_Le = 0.675 (1 – 0.42) (2.72 – 1)

h_Le = 0.67338 m

h_Le = 673.38 mm

1) Amount of oxygen required for the complete combustion of solid waste, it is necessary to compute the oxygen requirement for the oxidation of carbon, hydrogen, and sulfur contained in the waste.

The basic reactions involved are:

For carbon

C + O₂ → CO₂

C - (12), O₂ - (32) and CO₂ - (44)

For Hydrogen

2H₂ + O₂ → 2H₂O

H - (4), O₂ - (32) and 2H₂O(64)

The amount of air required for the complete oxidation of 1 kg of carbon would be

= 11.52 kg.

2) Pyrolysis is a highly endothermic process and known as destructive distillation whereas combustion process is a highly exothermic process.

3) For proper development of anaerobic digestion, C/N ratio of the digestive material should be between 30 to 50 for optimum digestion.

4) Pulverization refers to the action of crushing and grinding, whereas shredding refers to the action of cutting and tearing.

5)  ​An electrostatic precipitator is also installed in the incineration plant to reduce air pollution, particularly due to the emission of dioxin.

Concept:

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

The value of L_eq can be expressed as

Where n = Total number of sound samples

L_i = The noise level of any i^th sample

t_i = Time duration of i^th sample, expressed as fraction of total sample time.

Calculation:

Total time for which observation has been done

= 40 min + 20 min

= 60 min

L_eq= 10 × 9.531 dB

L_eq = 95.31 dB