Q 12-09-169JEE MainJEE Main 2021 (26 Aug, Shift 1)Medium
Car $B$ overtakes another car $A$ at a relative speed of $40$ m s$^{-1}$. How fast will the image of car $B$ appear to move in the mirror of focal length $10$ cm fitted in car $A$, when the car $B$ is $1.9$ m away from the car $A$?
Answer: (A) $0.1$ m s$^{-1}$
The rear-view mirror is convex, $f = +10$ cm, and $u = -190$ cm.
$$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{10} + \frac{1}{190} = \frac{20}{190} \Rightarrow v = 9.5\ \text{cm}$$
Image speed $= \left(\dfrac{v}{u}\right)^2\times$ object speed $= \left(\dfrac{9.5}{190}\right)^2\times40 = \dfrac{40}{400} = 0.1$ m s$^{-1}$.
Solution by Sreeraj P, M.Sc Physics