A) 15 m/s
B) \[\frac{1}{10}m/s\]
C) \[\frac{1}{15}m/s\]
D) 10 m/s
Correct Answer: C
Solution :
For mirror \[\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\] \[\frac{-1}{{{v}^{2}}}\frac{dv}{dt}-\frac{1}{{{u}^{2}}}\frac{du}{dt}=0\] \[\frac{dv}{dt}=\frac{-{{v}^{2}}}{{{u}^{2}}}\frac{du}{dt}={{\left( \frac{f}{f-u} \right)}^{2}}\frac{du}{dt}\] \[=-{{\left( \frac{20}{20+280} \right)}^{2}}\frac{du}{dt}\] \[={{\left( \frac{1}{15} \right)}^{2}}\frac{du}{dt}=\frac{1}{15}m/s\]
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