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Question : 32 of 33
Marks: +1, -0
It is a well known fact that during a total solar eclipse the disc of the moon almost completely covers the disc of the Sun. From this fact determine the approximate diameter of the moon. Earth-Moon distance=3.84×108 mSun-Earth distance=1.496×1011 mSun’s diameter=1.39×109 m\begin{array}{l} \text{Earth-Moon distance}=3.84\times 10^{8}\,\text{m} \\ \text{Sun-Earth distance}=1.496\times 10^{11}\,\text{m} \\ \text{Sun's diameter}=1.39\times 10^{9}\,\text{m} \end{array}
Solution:  
Distance of moon from earth, ME=3.84×108 mME = 3.84 \times 10^{8}\,\text{m}
Distance of sun from earth,
SE=1.496×1011 m.SE = 1.496 \times 10^{11}\,\text{m}.
Diameter of sun AB=1.39×109 m.AB = 1.39 \times 10^{9}\,\text{m}.
The situation during total solar
eclipse is shown in figure
As Δs ABE\Delta s \, ABEand CDECDE are similar, therefore,
CD=AB×MESECD = AB \times \frac{ME}{SE}
=1.39×109×3.84×1081.496×1011= \frac{1.39 \times 10^{9} \times 3.84 \times 10^{8}}{1.496 \times 10^{11}}
=3.5679×106 m= 3.5679 \times 10^{6}\,\text{m}
=3567.9 km= 3567.9\,\text{km}
This is the diameter of the moon.
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