A) 2
B) 6
C) 4
D) 5
Correct Answer: B
Solution :
Acceleration of block \[AB=\frac{3mg}{3m+m}=\frac{3}{4}g\] Acceleration of block \[CD=\frac{2mg}{2m+m}=\frac{2g}{3}\] Acceleration of image in mirror \[AB=2\times \] acceleration of mirror \[=2\left( \frac{-3g}{4} \right)=\frac{-3}{2}g\] Acceleration of image in mirror \[CD=2\left( \frac{2g}{3} \right)=\frac{4g}{3}\] \[\therefore \] Acceleration of the two images w.r.t. each other \[=\frac{4g}{3}-\left( \frac{-3g}{2} \right)=\frac{17g}{6}\]You need to login to perform this action.
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