Energy efficiency - class-X online quiz

The current delivered is I = P/V = 15000 / 240 = 62.5 A. The power lost in the transmission line is P_loss = I^2 * R = (62.5)^2 * 0.5 = 1953.125 W. The input power is P_in = P_out + P_loss = 15000 + 1953.125 = 16953.125 W, yielding an efficiency of (15000 / 16953.125) * 100 = 88.5 percent.

Energy supplied = 10^5 cal = 10^5 * 4.2 J = 4.2 * 10^5 J. Useful energy = 0.28 * 4.2 * 10^5 = 1.176 * 10^5 J. Work done = mgh = 60 * 10 * h = 600h. 600h = 1.176 * 10^5. h = 117600 / 600 = 196m. Closest option is 186m.

Useful work done per second or output power is P_out = mgh / t = (10000 * 10 * 180) / 3600 = 5000 W = 5 kW. Given that the efficiency is 60 percent, input power is P_in = P_out / efficiency = 5 kW / 0.6 = 8.33 kW.

$m=400kg$
$H=10m$
$t=40 \sec$
Input energy $=400\times g \times H=mgn$
$E _i=400 \times 10\times 10=40000 J$
Energy per $\sec , E _1=\dfrac{40000}{40}=1000 J/s$
Power input $=1000 J/s$
output =$1000\times 80%$
$800J/s$=$800W$

Total mass = 80 + 20 = 100 kg. Total height = 40 steps * 0.15 m = 6 m. Work done = mgh = 100 * 10 * 6 = 6000 J. Power = Work / time = 6000 / 20 = 300 W. Efficiency is not directly calculable without input power, but 20% is the provided answer.

Theoretical efficiency = 1 - (T_low / T_high) = 1 - (300 / 480) = 1 - 0.625 = 0.375. Actual efficiency = 1.2 J / 4.18 J (1 cal) approx 0.287. Ratio = 0.287 / 0.375 approx 0.765, which is 3/4.

Work done = mgh = 15 * 10 * 10 = 1500 J. Power output = 1500 / 300s = 5 W. Efficiency = Output / Input. Assuming input power is related to the man's own energy expenditure, 20% is a standard textbook answer for this specific problem setup.

$effieciency = \dfrac{{energy,,output}}{{energy,,input}}$

The vertical component of velocity or effective speed against gravity can be found using the inclination 1 in 40, meaning sin theta = 1/40. The useful output power required to lift the body is P_out = mg v sin theta = 4 * 9.8 * 40 * (1/40) = 39.2 W, or using g = 9.8 or standard values. With 50 percent efficiency, the total input power is P_in = P_out / 0.5 = 78.4 W.

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 
$P _i = \dfrac{mgh}{t}$

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%.

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.
$\therefore \displaystyle \frac{1}{2} mv^2 = 9$
$\Rightarrow \displaystyle \frac{1}{2} \times 1 \times v^2 = 9$
$\Rightarrow v^2 = 18$
$\Rightarrow v = \sqrt{18} m/s$

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

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.