question_answer

A object is moving with velocity v (w.r.t. earth) parallel to plane mirror\[{{M}_{2}}\]. Another plane mirror \[{{M}_{1}}\] makes an angle \[\beta \] with the vertical as shown in figure. Then velocity of image in mirror \[{{M}_{2}}\] w.r.t. the image in \[{{M}_{1}}\] is -

A) \[2\text{ }v\text{ }sin\text{ }\beta \]                    

B) \[v\text{ }cos\text{ }2\text{ }\beta \]

C) \[2v\,\,sin\,\,\beta ~\]     

D) \[v\sqrt{2}\]

Correct Answer: C

Solution :

[C]

From image diagram

\[={{\upsilon }_{{{I}_{2}}{{I}_{1}}}}=\sqrt{{{(v-v\cos 2\beta )}^{2}}+{{(v\sin 2\beta )}^{2}}}\]

\[=2\text{v}\,\,\sin \beta \]