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WAEC Physics 2017 Paper

Question : 20

Total: 50

1.44 \times 10^{5}

Solution:

Concept:

Efficiency is the fraction of input energy that is converted into useful output energy.

Explanation:

Given electrical energy input,

E_{\text{input}} = 1.44 \times 10^{5}\ \mathrm{J}

Energy wasted on sound and heat is

\frac{1}{4}

of the input energy.

So, useful energy

= E_{\text{input}} - \frac{1}{4}E_{\text{input}} = \frac{3}{4}E_{\text{input}}

Efficiency,

\eta = \frac{E_{\text{useful}}}{E_{\text{input}}} \times 100\%

Substituting the fraction,

\eta = \frac{\frac{3}{4}E_{\text{input}}}{E_{\text{input}}} \times 100\%

Thus,

\eta = \frac{3}{4} \times 100\% = 75\%

Answer:

D. 75%

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