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Explanation:

Basic Runway length:

It is the length of runway under the following assumed conditions at the aircraft

(i) Airport altitude at sea level.

(ii) Temperature at the airport is standard (15° C)

(iii) Runway is leveled in the longitudinal direction.

(iv) No wind is blowing on runway.

(v) Aircraft is loaded to its full loading capacity.

(vi) There is no wind blowing enroute to the destination.

(vii) Enroute temperature is standard.

Correction for Elevation, Temperature and Gradient:

(i) Correction for Elevation: Basic runway length is increased at the rate of 7% per 300 m rise in elevation above the mean sea level.

(ii) Correction for Temperature: 

Airport reference temperature = \({T_a} + \frac{{{T_m} - {T_a}}}{3}\)

Where, Ta = Monthly mean of average daily temperature

Tm = Monthly mean of the maximum daily temmperature for the same month of the year

Total correction for elevation plus temperature ⇒ 35% of basic runway length

(c) Correction for Gradient:

(i) Steeper gradient results in greater consumption of energy and as such longer length of runway is required to attain the desired ground speed.

(ii) After having been corrected for elevation and temperature should be further increased at the rate of 20% for every 1% of effective gradient.

Explanation:

Given data:

The radius of well (r_w) = 15 cm

Hydraulic conductivity (k) = 10 m/day

(r₁) distance of first observation well = 10m

(r₂) distance of second observation well = 100m

(S₁) drawdown in first observation well = 5m

(S₂) drawdown in second observation well = 1m

(H) thickness of unconfined aquifer/initial water table height = 20m

(Q) puming rate/discharge (m³/day) = ?

(h₁) height of water table at r₁ = H – S₁ = 20 – 5 = 15 m

(h₂) height of water table at r₂ = H – S₂ = 20 – 1 = 19m

The corresponding pumping rate (in m³/day) from the well is

\(\mathrm{Q}=\frac{\pi \mathrm{k}\left(\mathrm{~h}_{2}^{2}-\mathrm{h}_{1}^{2}\right)}{\ln \left(\frac{\mathrm{r}_{2}}{r_{1}}\right)}\)

\(\mathrm{Q}=\frac{\pi \times 10\left(19^{2}-15^{2}\right)}{\ln \frac{100}{10}}=1855.55\)

Explanation:

Ferrous metals (P) → Magnetic separator (iv)

• Ferrous metals (like iron and steel) are separated using a magnetic separator.

Aluminum and copper (Q) → Eddy current separator (iii)

• Non-ferrous metals (like aluminum and copper) are separated using eddy current separators.

Food waste (R) → Rapid composting (ii)

• Organic waste like food is treated using rapid composting.

Cardboard (S) → Incineration (i)

• Cardboard can be incinerated as part of waste-to-energy processes.

Explanation:

Given data:

Characteristic strength = 30 MPa

No. of cubes = 40

Hence, Characteristic strength is the strength below which not more than 5% of the test results are expected to fall.

95% of the test results are having strength ≥ characteristic strength.

0.95 × 40 = 38

38 = 38

The minimum number of specimens having at least 30 MPa strength is = 38

Explanation:

1. Understanding Under-Reinforced Sections (Reference: Limit State Design of Reinforced Concrete by Varghese & IS 456:2000)

• A balanced section is when both concrete in compression and steel in tension reach their limit state at the same time.
• A section becomes under-reinforced when steel in tension yields before the concrete in compression fails. This makes the failure ductile, which is desirable for safety.

2. Effect of Changes in Reinforcement Area

• Decreasing the area of tension reinforcement

Correct (Under-Reinforced) → Less steel means it reaches yield stress earlier, leading to ductile failure.

• Decreasing the area of compression reinforcement

Incorrect (Not necessarily Under-Reinforced) → This affects compression strength, but the section may still remain balanced or shift towards an over-reinforced state.

• Increasing the area of tension reinforcement

Incorrect (Over-Reinforced) → More steel means higher resistance, making the section over-reinforced.

• Increasing the area of compression reinforcement

Correct (Under-Reinforced) → Adding compression reinforcement improves the compression zone, making the steel in tension yield earlier, shifting failure towards a ductile mechanism.

Explanation:

Given data:

• 20 trains travel at a speed of 40 km/hr
• 15 trains travel at a speed of 50 km/hr
• 12 trains travel at a speed of 60 km/hr
• 8 trains travel at a speed of 70 km/hr
• 3 trains travel at a speed of 80 km/hr

The velocity is given by:

\(\rm V_{e q}=\frac{20 \times 40+15 \times 50+12 \times 60+8 \times 70+3 \times 80}{20+15+12+8+3}\) = 52.93 km/hr

The equilibrium cant is given by

= \(\rm \frac{GV_{e q}^{2}}{127 R}\) = \(\rm \frac{1.75 \times 52.93^{2}}{127 \times 437}\)= 0.08834 m ≈ 88.3 mm

Closest integer = 88 mm < 165 mm

Explanation:

Step 1: Understanding the Given Stress States

• P: Pure shear stress (equal and opposite shear forces) → Mohr’s circle should be symmetric about the τ-axis.
• Q: Equal biaxial tension (same normal stress in both directions) → Mohr’s circle should be centered at the same positive stress value.
• R: Uniaxial tensile stress in one direction → Mohr’s circle should have one normal stress component.
• S: Equal biaxial compression (same normal stress in both directions but negative) → Mohr’s circle should be a point on the σ-axis.

Step 2: Matching with Mohr’s Circles

• P (Pure shear) → III (Circle symmetric about τ-axis).
• Q (Equal biaxial tension) → IV (Circle centered at positive σ with equal principal stresses).
• R (Uniaxial tension) → I (Circle with normal stress +10).
• S (Equal biaxial compression) → II (Single point on σ-axis at -10).

Explanation:

As per given data:

Using equilibrium equations

∑F_V = 0 ⇒ R_A + R_L – 10 = 0 …(i) 

∑FH = 0 ⇒ H_A + H_L - R_K + 10 = 0 …(ii) 

 …(iii)

BM at D = 0 ⇒ R_A × 2 – H_A × 2 = 0 …(iv)

BM at J = 0 ⇒ H_L × 2 = 0 …(v)

From (v), H_L = 0

From (iv), R_A = H_A

From (ii), R_A + 0 – R_K + 10 = 0 ⇒ R_K = R_A + 10

From (iii), 70 – 6R_L – 2(R_A + 10) = 0

⇒ 50 – 6RL – 2RA = 0 …(α)

From (i),  2R_A + 2R_L – 20 = 0 …(β) 

30 – 4R_L = 0 ⇒ R_L = 7.5 kN

Explanation:

As per given data:

\(\rm \frac{T_{P}}{T_{R}}=4\ &\ \frac{J_{1}}{J_{2}}=2\)

As P & R are fixed supports,

Hence, 

|ϕ_QP| = |ϕ_RQ|

\(\rm \frac{T_{\mathrm{P}} \times \mathrm{L}_{1}}{J_{1} \times G_{1}}=\frac{T_{R} \times L_{2}}{G_{2} \times J_{2}}\)

\(\rm \Rightarrow \frac{L_{1}}{L_{2}}=\frac{T_{R}}{T_{P}} \times \frac{J_{1}}{J_{2}} \times\left(\frac{G_{P}}{G_{2}}\right)^{1}\)

\(\rm \frac{L_{1}}{L_{2}}=\frac{1}{4} \times 2=\frac{1}{2}=0.50\)

CONCEPT:

The relationship between speed ( u )  and density ( k ) is given by the

\({\bf{u}} = {{\bf{u}}_{\bf{f}}} - \left( {\frac{{{{\bf{v}}_{\bf{f}}}}}{{{{\bf{k}}_{\bf{j}}}}}} \right) \times {\bf{k}}\)

u - Mean speed at density k ( m/s )

k- Density of stream ( veh/km )

uf - Free mean speed

kj – Jam density

Relation between flow, speed, and density is given by,

q = k × u

q – Flow given in veh/hr

Space headway is defined as the distance between corresponding points of two successive vehicles at any given time. It can also be taken as the average space occupied by each vehicle.

\({\bf{k}} = \frac{{1000}}{{{\bf{Space}};{\bf{headway}}}}\)

At capacity or maximum flow,

\({{\rm{k}}_0} = \frac{{{{\rm{k}}_{\rm{j}}}}}{2}\)

\({u_0} = \frac{{{{\rm{u}}_{\rm{f}}}}}{2}\)

\({{\bf{q}}_{{\bf{max}}}} = {{\bf{k}}_0} \times {{\bf{u}}_0} = \frac{{{{\bf{k}}_{\bf{j}}} \times {{\bf{u}}_{\bf{f}}}}}{4}\)

qmax – Maximum flow or flow at capacity

k0, u0 – Density and speed at capacity

Calculation:

\(\rm q_{\max }=\frac{V_{f} K_{j}}{4}=\frac{60 \times \frac{1000}{8}}{4}=1875 \mathrm{veh} / \mathrm{hr}\)