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: |
 |
| \[\frac{2}{3}\left( 2a \right)<2a\sin \theta <\frac{5}{6}\left( 2a \right)\] |
| \[\Rightarrow \]\[\frac{2}{3}<\sin \theta <\frac{5}{6}\Rightarrow \frac{\sqrt{11}}{6}<\cos <\frac{\sqrt{5}}{3}\] |
| \[\Rightarrow \]\[\frac{\sqrt{11}}{6}a<r<\frac{\sqrt{5}}{3}a\] |
| \[\therefore \]The given condition is satisfied if the midpoint of the chord lies within the region between the concentric 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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