Identify the correct statements: [JEE Online 15-04-2018] | (A) The image executes periodic motion |
| (B) The image executes no n-periodic motion |
| (C) The turning points of the image are asymmetric w.r.t the image of the point at \[x=10cm\] |
| (D) The distance between the turning points of the oscillation of the image is\[\frac{100}{21}\] |
A) (B), (D)
B) (B), (C)
C) (A), (C), (D)
D) (A), (D)
Correct Answer: C
Solution :
For mean, \[\Rightarrow \frac{-1}{10}+\frac{1}{v}=-\frac{1}{5}\] \[\Rightarrow \frac{1}{v}=\frac{1}{10}-\frac{1}{5}\] \[\Rightarrow \frac{1}{v}=\frac{1}{5}\left[ \frac{1}{2}-1 \right]\] \[\Rightarrow \frac{1}{v}=\frac{-1}{10}\] \[\Rightarrow v=-10cm\] As image copies the time period of object (A) is right as well. It will be periodic motion. For one extreme \[\Rightarrow \frac{-1}{8}+\frac{1}{v}=-\frac{1}{5}\] \[\Rightarrow \frac{1}{v}=\frac{1}{8}-\frac{1}{5}\] \[\Rightarrow \frac{1}{v}=-\frac{3}{40}\] Right arrow\[=\frac{-40}{3}cm\] For other extreme \[\Rightarrow \frac{-1}{12}+\frac{1}{v}=-\frac{1}{5}\] \[\Rightarrow \frac{1}{v}=\frac{1}{12}-\frac{1}{5}\] \[\Rightarrow \frac{1}{v}=\frac{-7}{60}c{{m}^{-1}}\] \[\Rightarrow v=\frac{-60}{7}cm\] These points are asymmetric about \[{{x}_{0}}=10cm\] So, (c) is right. Amplitude of oscillation of image \[\Rightarrow \frac{40}{3}-\frac{60}{7}\] \[\Rightarrow 10\left[ \frac{4}{3}-\frac{6}{7} \right]\] \[10\times \frac{10}{21}\] \[\Rightarrow \frac{100}{21}cm\] (d) is right
You need to login to perform this action.
You will be redirected in
3 sec
