wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

An object is placed in front of a convex mirror at 60 m from the mirror. Its radius of curvature is 40 m. The object starts moving towards the mirror at 20 m s−1. Find the average speed of the image after a time interval of 2 seconds?

A
10 m s1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
20 m s1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2.5 m s1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
5 m s1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 2.5 m s1
Given:
The distance of object u from the pole is 60 m,
Radius of curvature (r)=40 m
Let f be the focal length and v be the image distance
We know that,
f=r2=12×40=20 m.
Applying the sign conventions, u=60 m and f=+20 m
We know, 1v + 1u = 1f
1v - 160 = 120
v=15 m
Positive sign shows that the image is virtual. Speed of the object =20 ms1 and time taken =2 s
Distance = Speed × time
20×2=40 m
Therefore, now the position of object will be u=6040=20 m
Let v be the new position of the image
Applying it in the mirror formula we get,
1v + 1u = 1f
1v - 120 = 120
1v = 4020×20
v=10 m
So, the image moves from 15 m to 10 m in 2 s. Hence, its average speed = 15102=2.5 m s1

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Mirror Equations and Focal Length Tackle
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon