Combined Two Banks (Sonali Bank PLC & Rajshahi Krishi Unnayan Bank-RAKUB)- 2026

Solution: Given, Q = 0.055 m³/s, H_s = 16 m, D = 0.15 m, L = 25 m, η_o = 0.75, f = 0.05 (Darcy).

$$V=\frac{Q}{\frac{\pi }{4}{D}^2}=\frac{0.055}{0.01767}=3.112\text{m/s},h_f=\frac{fL{V}^2}{2gD}=\frac{0.05\times 25\times {3.112}^2}{2\times 9.81\times 0.15}=4.11\text{m}$$

The velocity head at the pipe outlet, V²/2g = 0.49 m

$$H=\frac{V^2}{2g}+H_s+h_f=0.49+16+4.11=20.6\text{m}$$

$P=\frac{\rho gQH}{{\eta }_o}=\frac{1000\times 9.81\times 0.055\times 20.6}{0.75}=14.8\text{kW}$ (Answer)

Performance characteristic curves (at constant speed): head H, power P and efficiency η plotted against discharge Q.

Solution: Reasons for uneven cooling:

• Unbalanced air distribution — wrongly set dampers or badly sized ducts
• Leaking, crushed or blocked ducts and closed vents
• Dirty filters, coils or blower, reducing airflow and capacity
• Low refrigerant charge or a weak compressor
• Thermostat located in a wrong place (in sunlight or near a supply diffuser)
• Unequal heat loads — sun-facing or crowded rooms, poor insulation
• Inadequate return air path

Remedies:

• Measure room temperatures and airflow at each diffuser and compare with design values
• Clean or replace filters; clean coils, fan and drain lines
• Check refrigerant charge, compressor, fan motors and controls
• Seal and repair ducts; remove obstructions; insulate ducts
• Carry out air balancing by adjusting dampers and diffusers
• Relocate the thermostat; provide zoning (VAV boxes), extra return grilles or supplementary units for high-load rooms

Solution:

The helical gear drive is preferred where low cost, quiet running and easy maintenance are required with parallel shafts. The planetary gear drive is preferred where high torque in a small space with coaxial shafts is required.

Solution: Total Quality Management (TQM): An organisation-wide approach in which every employee continuously improves the quality of products, services and processes to achieve customer satisfaction. Principles: customer focus, total employee involvement, process approach, continuous improvement and fact-based decisions. Tools: PDCA cycle, Pareto chart, cause-and-effect diagram, control charts and check sheets.

Kaizen: A Japanese term meaning “change for the better” — continuous small improvements by everyone, every day. Methods: 5S (sort, set in order, shine, standardise, sustain), PDCA, suggestion schemes, quality circles and elimination of waste (muda).

Just-in-Time (JIT): A production system (developed by Toyota) that produces and delivers only the required items, in the required quantity, at the required time. Features: pull system with kanban, small lots, quick set-up, reliable suppliers and zero defects. Benefit: minimum inventory and waste; limitation: a supplier delay can stop the line.

Solution: The slope of the shear diagram gives the load intensity and each jump gives a concentrated load:

• A–B (0–4 ft): shear rises from 0 to 900 lb → upward uniform load of 900/4 = 225 lb/ft
• B–C (4–6 ft): constant shear → no load
• At C (6 ft): drop of 900 lb → concentrated load of 900 lb downward
• C–D (6–10 ft): zero shear → no load
• At D (10 ft): drop of 900 lb → 900 lb downward
• D–E and E–F (10–18 ft): slope −480/4 = −120 lb/ft → downward uniform load of 120 lb/ft
• At E (14 ft): jump from −1380 to +480 lb → reaction of 1860 lb upward

Moments from the areas of the shear diagram: M_B = ½ × 4 × 900 = 1800 lb·ft; M_C = 1800 + 900 × 2 = 3600 lb·ft; M_D = 3600 lb·ft; M_E = 3600 − ½(900 + 1380) × 4 = −960 lb·ft; M_F = −960 + ½ × 4 × 480 = 0. The moment is zero in DE at 3.28 ft from D (x = 13.28 ft).

Solution: [Repeated question: MPL-2025] In a cascade system, two (or more) vapour compression cycles are connected in series through a cascade heat exchanger, which acts as the condenser of the low-temperature (bottoming) cycle and the evaporator of the high-temperature (topping) cycle. It is used for very low temperatures, where a single cycle would have a large pressure ratio and poor COP.

Solution: For a vapour compression cycle (1: evaporator exit, 2: compressor exit, 3: condenser exit, 4 = 3 after throttling):

$$COP=\frac{h_1-h_4}{h_2-h_1}$$

Solution: From the moment diagram, M_max = 5 kN·m; the section is 80 mm wide.

$$I=\frac{b{h}^3}{12}=\frac{80h^3}{12},c=\frac{h}{2}$$

$$\sigma =\frac{Mc}{I}\Rightarrow 20=\frac{5\times 10^6\times (h/2)}{80h^3/12}=\frac{375,000}{h^2}\Rightarrow h^2=18,750$$

∴ Minimum height, h = 137 mm (Answer)

Answer: Feed rate F = f × N (mm/min); cutting speed V = πDN/1000 (m/min); machining time T = L/(fN) (min); material removal rate MRR = πDNfd (mm³/min).

Solution: AB = 1.5 m (vertical), G is 2 m and the crate is 6 m horizontally from A.

$$W_c=1000\times 9.81=9.81\text{kN},W_L=2400\times 9.81=23.5\text{kN}$$

$$\Sigma M_A=0:B(1.5)-9.81(2)-23.5(6)=0\Rightarrow B=107.1\text{kN}(\to )$$

$$\Sigma F_x=0:A_x+B=0\Rightarrow A_x=-107.1\text{kN}(\leftarrow )$$

$$\Sigma F_y=0:A_y-9.81-23.5=0\Rightarrow A_y=33.3\text{kN}(\uparrow )$$

∴ B = 107.1 kN →; A_x = 107.1 kN ←; A_y = 33.3 kN ↑ (Answer)

Solution: QB-544, Prasangik Technical (BPSC).

Clearance: The difference between the hole size and the shaft size when the hole is larger; it permits relative motion. Clearance = Hole size − Shaft size.

Allowance: The intentional difference between the maximum material limits of mating parts (minimum hole − maximum shaft); it is the minimum clearance (positive allowance) or maximum interference (negative allowance).

Tolerance: The permissible variation in a dimension; Tolerance = Upper limit − Lower limit.

Solution: $$\Delta T_1=30-14=16^{\circ }\text{C},\Delta T_2=30-22=8^{\circ }\text{C},\Delta T_{lm}=\frac{16-8}{\ln (16/8)}={11.54}^{\circ }\text{C}$$

$$\dot{Q}=UA\Delta T_{lm}=2100\times 45\times 11.54=1.091\times 10^6\text{W}$$

$${\dot{m}}_w=\frac{\dot{Q}}{c_p\Delta T}=\frac{1.091\times 10^6}{4184\times 8}=32.6\text{kg/s},{\dot{m}}_s=\frac{\dot{Q}}{h_{fg}}=\frac{1091}{2431}=0.449\text{kg/s}$$

∴ Cooling water = 32.6 kg/s; Rate of condensation = 0.449 kg/s (Answer)

Solution: D₂ = 2 cm; 1 gal = 3.785 × 10⁻³ m³.

$$Q_1=\frac{20\times 3.785\times 10^{-3}}{60}=1.262\times 10^{-3}{\text{m}}^3/\text{s},Q_2=2.5\times \frac{\pi }{4}(0.02{)}^2=7.85\times 10^{-4}{\text{m}}^3/\text{s}$$

$$Q_3=Q_1-Q_2=4.77\times 10^{-4}{\text{m}}^3/\text{s},V_{jet}=\frac{Q_3}{100\times \frac{\pi }{4}(0.001{)}^2}=6.07\text{m/s}$$

∴ Jet velocity ≈ 6.1 m/s (Answer)