### Efficiency

**49.** Formula for efficiency is

(a) (Energy input / energy output) * 100 (b) (energy output / energy input) * 100

(c) (Useful energy input
/ energy output) * 100 **(d) (useful energy output / total energy
input) * 100 **

**Explanation: **Any machine capable of
doing work is using energy; using part of it to useful work and the remaining
energy is lost. Energy can’t be created or destroyed. Energy lost is also
transferred to some other form of energy, for example, heat, sound etc.
However, this conversion is not required, therefore, we use the term useful
energy output.

**50.** A car uses total energy of 2500 J and output is 750 J. Efficiency
of the car is

(a) 50% **(b) 30%** (c)
25% (d) 80%

Solution

Formula Efficiency = (useful energy output)/(total energy input)*100 = (750/2500) * 100 = 0.3 * 100 = 30%

**51.** A machine uses total
energy of 5000 J and within process one-fourth of energy is lost to surroundings.
Efficiency of machine is

(a) 25% (b) 50% **(c) 75%** (d) 90%

**Explanation:**
If one-fourth of the energy is lost to the surrounding, this means 25% energy
is wasted and 75% of it is the useful output. (Can you solve it
quantitatively)?

**52.** Efficiency of a machine is 80% and total energy input is
2000 J. Output of the machine is

(a) 2500 J **(b) 1600 J** (c)
1000 J (d) 2080 J

% Efficiency = (output/input) * 100

Put the values 80 = (output/2000) * 100 OR (80 * 2000)/100 = 1600 J

**53.** An elevator motor is used to lift 1000 N load to a height of
0.25 km. If total amount of energy input is 1MJ, which of following pairs of
values of energy wasted and efficiency are correct?

(a) Energy wasted = 750 Kj, Efficiiency = 75% (b) Energy wasted = 500 KJ, Efficie = ncy = 75%

(c) Energy wasted = 500
KJ, Efficiency = 25% **(d) Energy wasted = 750 KJ, Efficiency =
25%**

**Solution: ** Convert km to meters first. Height = d = 0.25
* 1000 m = 250 m, Weight = F = 1000 N

Useful work = Fd = 1000
* 250 = 250000 J = output Input = 1
MJ = 1 * 10^{6} J

As total energy input = useful output + energy lost

Therefore, 1000000 =
250000 + energy lost OR Energy lost = 1000000 – 250000 = 750000 =
750 * 10^{3} J = 750 KJ

%Efficiency = (250000/1000000) * 100 = (25/100) * 100 = 25%