A) \[\frac{5}{16}\]
B) \[\frac{1}{16}\]
C) \[\frac{1}{4}\]
D) \[\frac{5}{12}\]
Correct Answer: C
Solution :
Let \[l\] be the length of the chord \[AB\] of the given circle of radius a and r be the distance of the midpoint D of the chord from the Centre c, then \[r=a\cos \theta \,\]and \[l=2a\sin \theta \] according to given condition: |
\[\frac{2}{3}(2a)<2a\,sin\theta <\frac{5}{6}(2a)\] \[\Rightarrow \frac{2}{3}<\sin \theta <\frac{5}{6}\Rightarrow \frac{\sqrt{11}}{6}<\cos \theta <\frac{\sqrt{5}}{3}\]\[\Rightarrow \frac{\sqrt{11}}{6}a<r<\frac{\sqrt{5}}{3}a\] |
\[\therefore \]The given condition is satisfied if the midpoint of the chord lies within the region between the concentric circles of radii\[\frac{\sqrt{11}}{6}\]a and \[\frac{\sqrt{5}}{3}a\] |
Hence, the required probability \[=\frac{\pi {{\left( \frac{\sqrt{5}}{3}a \right)}^{2}}-\pi {{\left( \frac{\sqrt{11}}{6}a \right)}^{2}}}{\pi {{a}^{2}}}=\frac{1}{4}\] |
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