Tag: energy efficiency

Quizzes (0) Questions (17)

Questions Related to energy efficiency

Multiple choice energy efficiency energy transformations and energy transfers physics

The power of a heart which pumps $5\times{10}^{3}cc$ of blood per minute at a pressure of $120mm$ of mercury ($g=10m{s}^{-2}$ and density of $Hg=13.6\times{10}^{3}kg/m$) is

  1. $1.36W$
  2. $13.6W$
  3. $0.136W$
  4. $136W$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Pressure P = h * rho * g = 0.12m * 13600 * 10 = 16320 Pa. Volume V = 5 * 10^3 cc = 5 * 10^-3 m^3. Work = P * V = 16320 * 5 * 10^-3 = 81.6 J. Power = Work / time = 81.6 J / 60s = 1.36 W.

Multiple choice energy efficiency energy transformations and energy transfers physics

The work done during the process when 1 mole of gas is allowed to expand freely into vacuum is:

  1. zero

  2. +ve

  3. -ve

  4. either of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Free expansion into a vacuum means there is no external pressure (P_ext = 0). Since work done W = integral(P_ext * dV), the work done is zero.

Multiple choice energy efficiency energy transformations and energy transfers physics

A man weighing 60 kg lifts a body of 15 kg to the top of a building 10 m high in 30 minutes. His efficiency is:

  1. 10%

  2. 20%

  3. 30%

  4. 40%
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Here, a man weighing 60 kg lifts a body of 15 kg to the top of a building 10 m high in 30 minutes. initially, while lifting the body(doing work) the power is 
$P _i = \dfrac{mgh}{t}$

$P _i = \dfrac{(60+15)(10)(10)}{(30)(60)}$

$P _i = \dfrac{7500}{1800}$

After reaching to the top the power is

$P _f = \dfrac{mgh}{t}$

$P _f = \dfrac{(15)(10)(10)}{(30)(60)}$

$P _f = \dfrac{1500}{1800}$

The efficiency of the man is the ratio of final power to the initial one
$\eta = \dfrac{P _f}{P _i}$

$\eta = \dfrac{\dfrac{1500}{1800}}{\dfrac{7500}{1800}} \times 100$ 

$\eta = \dfrac{1}{5} \times 100$ 

$\eta = 20$%

Multiple choice energy efficiency energy transformations and energy transfers physics

A motor has an electrical input of $30 kJ$ and is used to raise $100 kg$ load to a height of $25 m$ when fired to a crane winch. What is the efficient of winch ? ($g = 10 \ ms^{-1}$)

  1. $0.75 %$
  2. $83.3 %$
  3. $75 %$
  4. $17.5 %$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Efficiency is calculated as (useful work output / total energy input) * 100. The useful work is potential energy mgh = 100 * 10 * 25 = 25,000 J or 25 kJ. Efficiency = (25 kJ / 30 kJ) * 100 = 83.33%.

Multiple choice energy efficiency energy transformations and energy transfers physics

A machine which is 75% efficient, uses 12 J of energy in lifting 1 kg mass through a certain distance. The mass is then allowed to fall through the same distance. The velocity at the end of its fall is:

  1. $\sqrt{12} $ m/s
  2. $\sqrt{18} $ m/s
  3. $\sqrt{24} $ m/s
  4. $\sqrt{32} $ m/s
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Efficiency = 75%, Input energy = 12 J
$\therefore \displaystyle \frac{75}{100} = \frac{\text{Output energy}}{\text{Input energy}}$
$\Rightarrow $ Output energy $= \displaystyle \frac{75}{100} \times 12 = 9 J$
$\therefore$ P.E. of the mass = 9 J
At the end of the fall it will be converted to K.E.
$\therefore \displaystyle \frac{1}{2} mv^2 = 9$
$\Rightarrow \displaystyle \frac{1}{2} \times 1 \times v^2 = 9$
$\Rightarrow v^2 = 18$
$\Rightarrow v = \sqrt{18} m/s$

Multiple choice energy efficiency energy transformations and energy transfers physics

An installation consisting of an electric motor driving a water pump left $75 L$ of water per second to a height of $4.7 m$. If the motor consumes a power of $5 kW$, then the efficiency of the installation is

  1. $39$%
  2. $69$%
  3. $93$%
  4. $96$%
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Power consumed by motor$5kW= 5 \times 10^3 W= 5000W$

power used i lifting water = $\dfrac{mgh}{t}=7.5 \times 9.8 \times 4.7= 3454.5 W$
Efficiency = $\dfrac{\text{Power used}}{\text{Power consumed}} \times 100$% = $\dfrac{3454.5}{5000} \times 100$% = 69 %

Multiple choice energy efficiency energy transformations and energy transfers physics

A man of 60 kg gains 1000 cal of heat by eating 5 mangoes. His efficiency is 56%. To what height he can jump by using this energy?

  1. 4m

  2. 20 m

  3. 078 m

  4. 0.2 m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total heat energy gained is 1000 cal = 1000 * 4.184 J approx 4184 J. Effective mechanical energy available is 56 percent of this, which is 0.56 * 4184 = 2343 J. Equating this to gravitational potential energy mgh yields h = 2343 / (60 * 9.8) = 3.98 m, which rounds to 4 m.