Northern Electricity Supply PLC (NESCO)- 2026

Solution: The first law states that energy can neither be created nor destroyed; for a cycle ∮δQ = ∮δW and for a process Q − W = ΔU. It is essential because every process must satisfy this energy balance. However, it is not sufficient because:

• It does not tell the direction of a process. Heat flowing by itself from a cold body to a hot body satisfies the first law, but never happens.
• It treats heat and work as equivalent forms of energy, but does not say how much of the heat supplied can be converted into work. A heat engine converting all the heat supplied into work (η = 100%) does not violate the first law.
• It cannot predict whether a process is possible or reversible, and it gives no idea of the quality (availability) of energy.

These limitations are removed by the second law of thermodynamics, which introduces entropy and fixes the direction of processes and the maximum possible efficiency.

Solution: A heat engine receives heat Q_H from a high-temperature source, converts part of it into net work W_net and rejects the remaining heat Q_L to a low-temperature sink.

$${\eta }_{th}=\frac{W_{net}}{Q_H}=1-\frac{Q_L}{Q_H}$$

Solution: According to the Kelvin–Planck statement of the second law, it is impossible to construct a device operating in a cycle that receives heat from a single reservoir and produces an equal amount of work. A heat engine must reject some heat Q_L to a low-temperature sink, so W = Q_H − Q_L < Q_H. Even the ideal (reversible) Carnot engine has

$${\eta }_{Carnot}=1-\frac{T_L}{T_H}$$

which can be 100% only if T_L = 0 K or T_H = ∞, both of which are impossible. In actual engines, friction, heat losses and other irreversibilities reduce the efficiency further. Hence 100% efficiency is impossible.

Solution: Petrol (SI) engines are light, compact, smooth and cheap, and are used mainly for low and medium power:

• Motorcycles, scooters, three-wheelers and passenger cars
• Small portable generators and water pumps
• Lawn mowers, chain saws and other small agricultural machines
• Outboard motors for small boats
• Light aircraft and small racing engines

Solution:

Fuel is burnt in the boiler and the feed water is converted into superheated steam. The steam expands in the turbine, which drives the generator. The exhaust steam is condensed in the condenser by cooling water (cooled in the cooling tower), and the condensate is returned to the boiler by the feed pump through the feed water heaters and economiser.

Solution: Consider a pump lifting water from a sump (section 1) to an overhead tank (section 2). Applying the energy (extended Bernoulli) equation between the two sections:

$$\frac{p_1}{\rho g}+\frac{V_1^2}{2g}+z_1+H_p=\frac{p_2}{\rho g}+\frac{V_2^2}{2g}+z_2+h_L$$

where H_p is the head added by the pump and h_L is the total head loss (friction and minor losses) in the suction and delivery pipes. For open sump and tank, p₁ = p₂ = atmospheric and V₁ ≈ 0, so

$$H_p=\left(z_2-z_1\right)+h_L+\frac{V_2^2}{2g}=H_{static}+h_{fs}+h_{fd}+\frac{V_d^2}{2g}$$

The required pump power is

$$P=\frac{\rho gQ{H}_p}{{\eta }_o}$$

Solution: Such problems are solved with the following relations:

$$\text{Air-standard (Diesel cycle) efficiency:}\eta =1-\frac{1}{r^{\gamma -1}}\left[\frac{{\rho }^{\gamma }-1}{\gamma (\rho -1)}\right]$$

$$\text{Indicated power:}IP=\frac{p_{mi}LAnk}{60},n=\frac{N}{2}\text{(four-stroke)}$$

$${\eta }_{bth}=\frac{BP}{{\dot{m}}_f\times CV},{\eta }_{ith}=\frac{IP}{{\dot{m}}_f\times CV},{\eta }_m=\frac{BP}{IP}$$

where r = compression ratio, ρ = cut-off ratio, k = number of cylinders.

Solution: The normal stress in each wire is

$${\sigma }_1=\frac{P_1}{A_1},{\sigma }_2=\frac{P_2}{A_2},A=\frac{\pi d^2}{4}$$

If the load is shared by the two wires, the load in each wire is first found from equilibrium (ΣF = 0, ΣM = 0) and, if necessary, from compatibility of elongation (δ = PL/AE).

Solution: Stress: The internal resisting force per unit cross-sectional area developed in a body under external load, σ = P/A (unit: N/m² or Pa).

Strain: The deformation per unit original dimension, ε = δL/L. It is dimensionless.

Solution:

Up to A stress is proportional to strain (Hooke’s law). After the yield point the material deforms plastically; from D to E strain hardening occurs; beyond E necking starts and the specimen breaks at F.

Solution:

Solution: $$\text{Gear ratio, }G=\frac{T_G}{T_P},\frac{N_P}{N_G}=\frac{T_G}{T_P}\Rightarrow N_G=N_P\times \frac{T_P}{T_G}$$

where T_P, T_G are the numbers of teeth and N_P, N_G the speeds (rpm) of the pinion and the gear.

Solution: A good tariff should:

• Recover the total cost of generation, transmission and distribution, with a reasonable return on investment.
• Be fair to the different types of consumers (domestic, commercial, industrial, agricultural).
• Be simple and easily understood by the consumers.
• Encourage the consumers to improve power factor and load factor and to use energy at off-peak hours.
• Charge the consumer according to both maximum demand and energy consumed.
• Be reasonably stable, and attractive enough to encourage the use of electricity.

Answer: Lathe, drilling machine, milling machine, shaper, planer, slotter, grinding machine, boring machine, broaching machine, power hacksaw and CNC machining centre.

Solution: Facing, plain (straight) turning, step turning, taper turning, chamfering, knurling, thread cutting, drilling, boring, reaming, grooving, parting-off and form turning.

Solution: For a thin cylindrical shell with riveted/welded joints:

$${\sigma }_h=\frac{pd}{2t{\eta }_l}\Rightarrow d=\frac{2t{\sigma }_h{\eta }_l}{p}$$

$${\sigma }_l=\frac{pd}{4t{\eta }_c}\Rightarrow d=\frac{4t{\sigma }_l{\eta }_c}{p}$$

where η_l and η_c are the efficiencies of the longitudinal and circumferential joints. The smaller of the two diameters is the safe diameter.

Answer: For a two-part tariff:

Monthly bill = (Demand charge per kW × Maximum demand in kW) + (Energy rate per kWh × Energy consumed in kWh)

উত্তর: কাপলান টারবাইন (Kaplan turbine)।

উত্তর: 50 Hz।

উত্তর: তটিনী (এছাড়া তরঙ্গিণী, স্রোতস্বিনী, প্রবাহিনী)।

উত্তর: কপট বা ছদ্মবেশী; যে নিজের প্রকৃত স্বরূপ গোপন রাখে।

উত্তর: ২০২৩ সালে (সাইবার নিরাপত্তা আইন, ২০২৩)। পরবর্তীতে এটি রহিত করে ‘সাইবার সুরক্ষা অধ্যাদেশ, ২০২৫’ জারি করা হয়।

উত্তর: ষট্ + থ।

উত্তর: 0

x⁴ − x² + 1 = 0 ⇒ x² + 1/x² = 1 ⇒ (x + 1/x)² = 3 ⇒ x + 1/x = √3

x³ + 1/x³ = (x + 1/x)³ − 3(x + 1/x) = 3√3 − 3√3 = 0

উত্তর: বাস্তব সংখ্যায় অসংখ্য সমাধান; তবে স্বাভাবিক (ধনাত্মক পূর্ণ) সংখ্যায় ২টি — (x, y) = (3, 3) ও (6, 1)।

উত্তর: ল্যান্স নায়েক মুন্সী আব্দুর রউফ (৮ এপ্রিল ১৯৭১)।

উত্তর: জার্মানি (সদর দপ্তর: বার্লিন)।

Answer: Synonym: Frighten; Antonym: Reassure.

Innings.

উত্তর: 255/128 (≈ 1.992)

$$S_8=\frac{a\left(1-r^8\right)}{1-r}=\frac{1-(1/2{)}^8}{1-1/2}=2\left(1-\frac{1}{256}\right)=\frac{255}{128}$$

উত্তর: Neither … nor দ্বারা যুক্ত subject-এর ক্ষেত্রে verb নিকটবর্তী (দ্বিতীয়) subject অনুযায়ী বসে। যেমন: Neither the manager nor the workers are present; Neither the workers nor the manager is present.

উত্তর: ২৫ বর্গ সে.মি.

প্রতিটি বাহু = 10/√2 সে.মি.; ক্ষেত্রফল = ½ × (10/√2)² = 25 বর্গ সে.মি.

উত্তর: x = 1

2^x−1 = 2^4(x−1) ⇒ x − 1 = 4x − 4 ⇒ x = 1

উত্তর: কয়লা (আমদানিকৃত)।

উত্তর: কর্মকারকে শূন্য বিভক্তি।

উত্তর: গাইবান্ধার সুন্দরগঞ্জে (তিস্তা সোলার পার্ক, ২০০ মেগাওয়াট)।

Answer: Serious / Sedate / Temperate.

$$f^{\prime }(x)=\frac{\ln x-1}{{\left(\ln x\right)}^2}=0\Rightarrow \ln x=1\Rightarrow x=e$$

উত্তর: ৫ জুন।

উত্তর: বৃহস্পতিবার।

২ অক্টোবর ২০২৬ থেকে ২ অক্টোবর ২০২৭ পর্যন্ত ৩৬৫ দিন (৫২ সপ্তাহ + ১ দিন), তাই ২ অক্টোবর ২০২৭ শনিবার। ২ সেপ্টেম্বর ২০২৭ তার ৩০ দিন আগে (৪ সপ্তাহ + ২ দিন), অর্থাৎ শনিবারের ২ দিন আগে = বৃহস্পতিবার।

উত্তর: 4/29

$${\cos }^{2}x=\frac{1}{1+{\tan }^{2}x}=\frac{1}{1+25/4}=\frac{4}{29}$$

উত্তর: মুমূর্ষু।