A bird in air is diving vertically over a tank with speed `5 "cm//s"` , base of tank is silvered. A fish in the tank is rising upward abong the same line with speed 2cm/s. Water level is falling at rate of 2cm/s [Take :`mu_(water) = 4//3`]

`{:("Column"I(cm//s), "Column"II), ((A) "Speed of the image of fish as seen by the bird directly", (P) 8), ((B) "Speed of the image of fish formed after reflection in the mirror as seen by the bird", (Q) 6), ((C) "Speed of image of bird relative to the fish looking upwards" , (R) 3), ((D) "Speed of image of bird relative to the fish looking downwards in the mirror" , (S) 4):}`

Answer

A, B, C, D

Explanation

Distance of fish asseen by bird `X_(fB) = H + (d)/(mu)`
Distance of bird as seen by fish `X_(Bf) = d + muh`
By differentiating `(d(X_(fB))/(dt) = (dh)/(dt) + (1)/(mu) (d(d))/(dt)`
`v_(FB) implies 3 + (4)/(mu) = 6 cm//s " "[(dh)/(dt) = 5-2 = 3 cm//s , (d(d))/(dt) = 2 - (-2) = 4"cm"/s, mu= (4)/(3)]`
`v_(BF) = ((d(d))/(dt)) + mu((dh)/(dt)) implies 4 + ((4)/(3)) (3) implies 8 cm//s`
`(d(d))/(dt)` (for fish image after reflection = 0) `implies 3 + (1)/(mu)(0) = 3 cm//s`
Similarly speed of image of bird `implies 4 cm//s`