Question:
When an object is kept at a distance of $30 \mathrm{~cm}$ from a concave mirror, the image is formed at a distance of $10 \mathrm{~cm}$ from the mirror. If the object is moved with a speed of $9 \mathrm{cms}^{-1}$, the speed (in $\mathrm{cms}^{-1}$ ) with which image moves at that instant is_______.
Solution:
$\left|\left(\frac{\mathrm{dv}}{\mathrm{dt}}\right)\right|=\left|\frac{\mathrm{v}^{2}}{4^{2}}\right| \frac{\mathrm{du}}{\mathrm{dt}} \mid$
$=\left(\frac{10}{30}\right) 2 \times 9=1 \mathrm{~m} / \mathrm{s}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.