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A fish in an aquarium approaches the left wall at a rate of $2.5 \;m/s$ observes a fly approaching it at $8 m/s$ If the refractive index of water is $(4/3)$ . Then the actual velocity of fly is

$(a)\;2.75\;m/s \\ (b)\;4.75\;m/s \\ (c)\;3.75\;m/s \\ (d)\;1.75\;m/s $

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For fish in an aquarium the apparent distance of fly from wall is $\mu x,y$. x is actual distance
Then apparent velocity will be $d \large \frac{(\mu x)}{dt}$
$(V_{app/fly})=\mu vf+y$
fish observes the velocity of fly to be 8 m/s
=> apparent relative velocity will be $=8 \;m/s$
$V_{fish}+(V_{app})fly=8 m/5$
=> $3+\mu V_{fly}=8$
$V_{fly} = 5 \times \large\frac{3}{7}$
$\qquad= 3.75 m/s$
Hence c is the correct answer.
answered Feb 11, 2014 by meena.p

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