Sonali Bank PLC- 2026

Solution: Q7, QB-197, Dhaka Power Distribution Company Limited (DPDC)-2020.

$$Q_L=5\times 3.5=17.5\text{kW},COP=\frac{17.5}{3.75}=4.67,\dot{m}=\frac{17.5}{56}=0.3125\text{kg/s}$$

Answer: Q4, QB-241, Karnaphuli Gas Distribution Company Limited (KGDCL)-2021.

Solution: Given, ρ = 920 kg/m³, Q = 0.055 m³/s, z = 10 m, d = 0.22 m, η_o = 0.60, f = 0.05. The pipe length is taken equal to the lift (L = 10 m).

$$A=\frac{\pi }{4}(0.22{)}^2=0.0380{\text{m}}^2,V=\frac{0.055}{0.0380}=1.447\text{m/s}$$

$$h_f=\frac{fL{V}^2}{2gd}=\frac{0.05\times 10\times {1.447}^2}{2\times 9.81\times 0.22}=0.243\text{m},H=10+0.243=10.24\text{m}$$

$P=\frac{\rho gQH}{{\eta }_o}=\frac{920\times 9.81\times 0.055\times 10.24}{0.60}=8.47\text{kW}$ (Answer)

Answer: $U=\frac{\pi DN}{60}=\frac{\pi \times 1.2\times 3000}{60}=188.5\text{m/s},C_1=\frac{U}{{\rho }_b}=\frac{188.5}{0.42}=448.8\text{m/s}$ (Answer)

Answer: $Nu=\frac{hD}{k}\Rightarrow h=\frac{Nu\cdot k}{D}=\frac{30\times 0.05}{0.12}=12.5{\text{W/m}}^2\text{K}$ (Answer)

Solution:

Solution: $$V=\frac{\pi DN}{1000}=\frac{\pi \times 100\times 800}{1000}=251.3\text{m/min},F=f\times N=0.25\times 800=200\text{mm/min}$$

∴ (i) Cutting speed = 251.3 m/min; (ii) Feed rate = 200 mm/min (Answer)

Solution:

Solution: A gear drive should be selected. A heavy-duty machine transmits high power and torque, often with shock loads, and requires an exact velocity ratio.

A belt drive is preferred only for light loads, long centre distances or where slip protection against overload is desired.

Solution: Option 2, the multistage pump, should be selected. A multistage pump has several impellers in series on one shaft; each stage adds head, so

$$H_{total}=n\times H_{stage}$$

A single-stage pump would need a very large impeller or a very high speed to produce a high head, which reduces efficiency and increases stress and vibration. For continuous operation, the higher efficiency of the multistage pump reduces the energy cost, which soon offsets its higher initial cost.

Solution:

Solution:

Solution: Span AD = 6 m; UDL w = 30 kN/m over BC (2 m). Total load = 60 kN acting 4 m from A.

$$R_D\times 6=60\times 4\Rightarrow R_D=40\text{kN},R_A=60-40=20\text{kN}$$

Shear force: AB: +20 kN; BC: V = 20 − 30(x − 3), so V_C = −40 kN; CD: −40 kN. Zero shear at x = 3 + 20/30 = 3.667 m from A.

Bending moment:

$$M_B=20\times 3=60\text{kNcdotpm},M_C=40\times 1=40\text{kNcdotpm}$$

$$M_{\max }=20\times 3.667-\frac{30\times {0.667}^2}{2}=73.33-6.67=66.67\text{kNcdotpm}\left(\text{at }x=3.667\text{m}\right)$$

Solution: Volute casing: A spiral casing surrounding the impeller whose area of flow increases gradually from the tongue to the delivery. The increasing area reduces the velocity of water and converts part of the kinetic energy into pressure energy. Eddies are formed, so its efficiency is somewhat lower.

Vortex casing: A circular chamber (whirlpool chamber) is provided between the impeller and the volute. Water first moves in a free vortex in this chamber, which converts a part of the velocity head into pressure head, and then enters the volute. Eddy losses are reduced and the efficiency is higher than that of the simple volute casing.

Solution: Supply chain management is the coordinated planning and control of the flow of materials, products, information and money from the suppliers of raw materials, through manufacturing and distribution, up to the final customer.

Role in customer satisfaction: right product available at the right time (no stock-out); shorter and reliable delivery lead time; lower cost through efficient purchasing, inventory and transport, allowing competitive prices; consistent quality through supplier selection and traceability; and quick response to changes in customer demand through information sharing.

Solution: SQC is the use of statistical methods to monitor, control and improve the quality of products and processes.

Principles: every process has variation — chance (common) causes and assignable (special) causes; decisions are based on data from samples, not guesswork; processes are monitored continuously to detect assignable causes early; defects are prevented rather than detected after production.

Contribution: uniform and better quality, less scrap and rework, lower inspection and production cost, higher productivity and customer satisfaction, and a basis for continuous improvement.

Tools with examples:

• Control charts (X̄–R, p, c charts): monitoring the fill volume of bottles in a beverage plant.
• Acceptance sampling: accepting or rejecting a lot of bolts from a supplier by inspecting a sample.
• Pareto chart: finding the few defect types causing most rejections in a garment line.
• Cause-and-effect (fishbone) diagram: analysing the causes of poor surface finish in machining.
• Histogram: studying the variation of package weights.
• Check sheet: recording types of defects on an assembly line.

Explanation: Since a = √(γRT), a ∝ √T: a₂ = 330 × √(327/300) m/s.

Answer: Raise the pressure of the feed water (condensate) to the boiler pressure.

Explanation: By Dalton’s law, p = 0.06 + 0.007 = 0.067 bar.

Explanation: Q_L = 120 − 30 = 90 kW; COP = 90/30 = 3.

Answer: Laminar flow.

Answer: Uniform flow.

Explanation: L₂ = L₁ × k₂/k₁ = 0.5 × 0.14/0.7 = 0.1 m.

Explanation: ε_axial = 60 × 10⁻⁵/0.3 = 2 × 10⁻³; E = 300/(2 × 10⁻³) = 150,000 MPa = 150 GPa.

Answer: The bending stress is the same at every section along its length.

Answer: Inner surface.

Explanation: The three masses and the rotation of the right pulley are independent; the rotation of the left pulley is fixed by the motion of mass A.

Explanation: N = n(n − 1)/2 = 4 × 3/2 = 6.

Answer: Perpendicular to the line joining the centre of the roller and the point of contact with the obstacle.