Some Simple Physics JEE Questions 1

let at any instant of time distance of bird from water level is 'y"

so Distance of image of bird from lake is = $\mu$ . * y

now differentiate => ${ V }_{ { ball }_{ image } }$ .= $\mu$ . ${ V }_{ { ball }_{ } }$ .

${ V }_{ { ball }_{ } }$ .= $\sqrt { 2\times 10\times (20-12.8) }$ .

=> ${ V }_{ { ball }_{ } }$ .=12

=> ${ V }_{ { ball }_{ image } }$ .=16

velocity of ball when it is 12.8m above water level: $v^{2}$ = 2gh =>v =12m/s therefore velocity of ball as seen by fish = v/n(relative) =16m/s

fish will see the image of ball after reraction through water surface...

using

         n2/v - n1/u = n2-n1/R

R for plane surface is infinity & n2(refrative index of water) , n1(refractive index of air)

     4/3v - 1/u = 0

     v = 4u/3  ..........1

now differentiating eq 1 wrt t

(velocity of image)dv/dt = 4/3.(du/dt) (velocity of object)

now when ball has fallen 12.8m then its velocity is U..

so velocity of ball as seen by fish is (4/3) .U

    V = 4/3. (U) =4/3 (sqrt2g(H-h))                                    (change in height (H-h) = 7.2m)

                      =4*12/3=16m/s

Distance of ball seen by fish = actual distance * refractive index of observers medium

differentiate it , we get velocity of ball seen by fish = actual velocity * refractive index of observers medium Actual velocity= 2as= v^2 - u^2 S= 20-12.8=7.2 we get V = 12 now put in above equation 12*4/3=16