An object kept on the principle axis is moving in the same direction as that of a mirror as shown in the figure. Speed of the object and mirror is $10m{s^{ - 1}}$ and $\dfrac{{40}}{{13}}m{s^{ - 1}}$ . Radius of the curvature of the mirror is $20cm$ . If the distance of the object from the mirror at this instant is $5xcm$ , the velocity of the image at this instant is found to be zero. Find x.

Hint: In this question the values of the speed of the object, the mirror, and the radius of curvature are already given. We need to use the relation between the speed of the image and the object to find the required solution.


Formula used:
${\vec V_{im}} = - \dfrac{{{v^2}}}{{{u^2}}}{\vec v_{om}}$


Complete answer:
Before starting the solution of the question let us write all the given values which are present in the question,
The velocity of the object = $10m{s^{ - 1}}$
The velocity of the mirror = $\dfrac{{40}}{{13}}m{s^{ - 1}}$
The radius of curvature of the mirror = $20cm$
To find = the distance of the object from the mirror at an instant of $5xcm$ and also to find the value of x.
As we start the solution by using the below formula,
${\vec V_{im}} = - \dfrac{{{v^2}}}{{{u^2}}}{\vec v_{om}}$
By putting all the values in the above formula, we get the result as,
$0 - \dfrac{{40}}{{13}} = - {\left( {\dfrac{{ - 10}}{{ - 10 + x}}} \right)^2}\left( {10 - \dfrac{{40}}{{13}}} \right)$
By doing further solution of the above equation we get,
$\dfrac{{ - 10}}{{ - 10 + x}} = \pm \dfrac{2}{3}$
By using the cross multiplication method, we get the following values,
$5x = 25, - 5$
As we know that the object is in front of the mirror,
Hence, $x = 5$ .
Therefore, the correct answer of $x$ is $5$ .




Note: Choose the formula wisely for solving this problem. Some does the mistake of using the mirror formula. Also, it is mentioned that If the distance of the object from the mirror at that instant is 5xcm, the velocity of the image at this instant is found to be zero. Thus, we just can’t put the value of the speed of the object and the mirror directly in the formula.