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(B) Initial distance of the object from the mirror is $u = 100 \; cm$. Since the mirror moves towards the object at a speed of $v = 10 \; cm/s$,the di

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$80$

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(B) Initial distance of the object from the mirror is $u = 100 \; cm$. Since the mirror moves towards the object at a speed of $v = 10 \; cm/s$,the distance covered by the mirror in $t = 6 \; s$ is $d = v 	imes t = 10 	imes 6 = 60 \; cm$. After $6 \; s$,the new distance between the object and the mirror is $u' = 100 - 60 = 40 \; cm$. In a plane mirror,the image is formed at the same distance behind the mirror as the object is in front of it. Thus,the distance of the image from the mirror is also $40 \; cm$. The total distance between the object and its image is the sum of the distance of the object from the mirror and the distance of the image from the mirror: $40 \; cm + 40 \; cm = 80 \; cm$.

An object is initially at a distance of $100 \; cm$ from a plane mirror. If the mirror approaches the object at a speed of $10 \; cm/s$,then after $6 \; s$,the distance between the object and its image will be: (in $; cm$)

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