A motor of a corn mill machine operates on electrical energy of \(1.44 \times 10^{5}\)J. If one-quarter of the energy is wasted on sound and heat, determine the efficiency of the machine.
25%
36%
64%
75%
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Ans=75%
E=(workoutput/workinput)×100
1/4 of work input was lost to heat=1/4 ×(1.44×10^5)=36000
workoutput=144000-36000=108000
E=(108000/144000)×100=75%
Answer D
irs simple really
efficiency is work done by machine/work done on machine × 100
since 1.44×10^5 j of energy was put into the machine (work done on machine)
but the machine wasted one quarter of this energy on heat and other stuffs
e.g ¼*1.44×10^5 of the initial energy was wasted = 0.36×10^5j remains
This means that 1.44*10^5 j - 0.36*10^5 j was the work done by the machine = 1.08*10^5j
therefore
eff = 1.44*10^5/1.08*10^5 × 100 = 0.75 * 100 = 75%
E =workoutput/output
workinput=1/4*1.44×10^5=36000
workoutput=1.44×10^5-36000=108000
E=(108000/36000)×100
E=75%
E=workoutput/output
workinput=1/4*1.44×10^5=36000
workoutput=1.44×10^5-36000=108000
E=(108000/36000)×100
E=75%
take note of the word operate 'on' meaning the machine work with initial energy 1.44 raise to power 5
it's 25%
Remember work input is more than work output because of energy lost
workinput = 1.44×10^5 J
workoutput = 1/4 ×1.44×10^5 J= 36000 J
Efficiency =36000 × 100 / 1.44×10^5
= 25%
The answer should be A. Reason:
Efficiency = workoutput/workinput×100
workinput = 1.44×10^5J
workoutput = 1/4 ×1.44×10^5J = 3600J
Efficiency =3600/1.44×10^5×100
= 25%
