Physics

Power, Energy, and Efficiency

443 Questions

Power, energy, and efficiency are fundamental physics concepts that quantify work and the rate of doing work. Questions cover electrical energy consumption in kilowatt hours, the efficiency of thermodynamic cycles like the Brayton cycle, and mechanical power lifting loads. This topic is essential for general science and engineering exam preparation.

Electrical energy consumptionBrayton cycle efficiencyPower calculation formulasCommercial electricity unitsHydraulic lift mechanics

Power, Energy, and Efficiency Questions

Multiple choice
  1. 15.2

  2. 18.2

  3. 30.4

  4. 45.6

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\text{Given} : t_\gamma=20m m, t_f=18mm, b=100 mm, \\ R=250mm,N=10rpm.\sigma_0=300MPa\\ \text {We know, Roll strip contact length is given by,} $

Multiple choice
  1. 8 minutes

  2. 12 minutes

  3. 16 minutes

  4. 20 minutes

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Multiple choice
  1. 45

  2. 50

  3. 55

  4. 60

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\text{Hence, standard production rate of the weld joint}\\ \hspace{6cm}=\frac{8\times60}{10.67}\\ \hspace{6cm}=45units$

Multiple choice physics electric current and its effects coil and fuse heating effect of electric current- uses applications of heating effect of electric current

A coal based thermal power plant producing electricity operates between the temperatures $27^o C$ and $227^o C$ The plant works at 80% of its maximum theoretical efficiency. Complete burring of 1 kg of coal yields 3600 KJ of heat. A house needs 10 units of electricity each day. Coal used for supplying the amount of energy for the house in one year is 

  1. 1141 kg

  2. 580 kg

  3. 605 kg

  4. 765 kg

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Carnot efficiency: η_max = 1 - T_c/T_h = 1 - 300/500 = 0.4. Actual efficiency at 80%: η = 0.4 × 0.8 = 0.32. Yearly electricity needed: 10 × 365 = 3650 units. Each unit requires energy. With 3600 kJ/kg coal and 32% efficiency, usable energy per kg = 3600 × 0.32 = 1152 kJ/kg. Converting units to kJ and calculating: Coal needed ≈ 3650 × (energy per unit in kJ)/(1152) ≈ 580 kg after unit conversions.