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An object is initially at a distance of 100 cm from a plane mirror. If the mirror approaches the object at a speed of 5 cm/s. Then after 6 s the distance between the object and its image will be
- a)60 cm
- b)140 cm
- c)170 cm
- d)150 cm
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
An object is initially at a distance of 100 cm from a plane mirror. If...
Given that,
The distance of object from plane mirror D=100cm
So, initial distance of image and object
=100−(−100)
=200cm
Now, object approaches the mirror with the speed=v=5cm/s
So, distance travelled by object in 6 sec
D=5×6
D=30cm
At this instant,
Distance between mirror and object
D=100−30
D=70cm
Now, the distance between image and object is
D=70−(−70)
D=140cm
Hence, the distance between the object and its image is 140 cm.
The distance of object from plane mirror D=100cm
So, initial distance of image and object
=100−(−100)
=200cm
Now, object approaches the mirror with the speed=v=5cm/s
So, distance travelled by object in 6 sec
D=5×6
D=30cm
At this instant,
Distance between mirror and object
D=100−30
D=70cm
Now, the distance between image and object is
D=70−(−70)
D=140cm
Hence, the distance between the object and its image is 140 cm.
Most Upvoted Answer
An object is initially at a distance of 100 cm from a plane mirror. If...
Solution:
Given,
Initial distance between object and mirror, u = 100 cm
Speed of mirror, v = 5 cm/s
Time, t = 6 s
Let the final distance between the object and its image be v'.
Using mirror formula,
1/v' = 1/u + 1/v
=> v' = uv/(u+v)
As the mirror is approaching the object, the relative speed between them will be (v-u).
Using relative speed concept,
v' = (v-u)t + u
=> v' = (5-(-5))6 + 100
=> v' = 140 cm
Therefore, after 6 seconds, the distance between the object and its image will be 140 cm.
Hence, the correct answer is option B.
Given,
Initial distance between object and mirror, u = 100 cm
Speed of mirror, v = 5 cm/s
Time, t = 6 s
Let the final distance between the object and its image be v'.
Using mirror formula,
1/v' = 1/u + 1/v
=> v' = uv/(u+v)
As the mirror is approaching the object, the relative speed between them will be (v-u).
Using relative speed concept,
v' = (v-u)t + u
=> v' = (5-(-5))6 + 100
=> v' = 140 cm
Therefore, after 6 seconds, the distance between the object and its image will be 140 cm.
Hence, the correct answer is option B.
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An object is initially at a distance of 100 cm from a plane mirror. If...
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An object is initially at a distance of 100 cm from a plane mirror. If the mirror approaches the object at a speed of 5 cm/s. Then after 6 s the distance between the object and its image will bea)60 cmb)140 cmc)170 cmd)150 cmCorrect answer is option 'B'. Can you explain this answer?
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