JEE MAIN - Physics (2022 - 28th July Evening Shift - No. 25)
An object 'O' is placed at a distance of $$100 \mathrm{~cm}$$ in front of a concave mirror of radius of curvature $$200 \mathrm{~cm}$$ as shown in the figure. The object starts moving towards the mirror at a speed $$2 \mathrm{~cm} / \mathrm{s}$$. The position of the image from the mirror after $$10 \mathrm{~s}$$ will be at _________ $$\mathrm{cm}$$.
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Answer
400
Explanation
The object after 10 second will be at $$u = - 80$$ cm.
So $${1 \over v} - {1 \over {80}} = - {1 \over {100}} $$
$$\Rightarrow v = {{8000} \over { + 20}} = 400$$ cm
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