A) coming towards earth with velocity of light
B) moving away from earth with velocity \[1.5\times {{10}^{6}}m/sec\]
C) stationary with respect to earth
D) moving away from earth with velocity of light
Correct Answer: B
Solution :
Wavelength displacement \[\Delta \lambda =\frac{\upsilon }{c}\lambda \] ...(1) Suppose the wavelength coming from a source on the earth\[=\lambda \] and the wavelength coming from galaxy\[=\lambda '\] \[\lambda '=(1+0.5%)\lambda \] Or \[\frac{\lambda '}{\lambda }=1.005\] Or \[\frac{\Delta \lambda }{\lambda }=\frac{\lambda '-\lambda }{\lambda }\] Or \[\frac{\Delta \lambda }{\lambda }=\left( \frac{\lambda '}{\lambda }-1 \right)\] Or \[\frac{\Delta \lambda }{\lambda }=1.005-1\] Or \[\frac{\Delta \lambda }{\lambda }=0.005\] \[\therefore \] \[\frac{\Delta \lambda }{\lambda }=\frac{\upsilon }{c}\] \[\Rightarrow \] \[0.005=\frac{\upsilon }{3\times {{10}^{8}}}\] \[\Rightarrow \] \[\upsilon =0.005\times 3\times {{10}^{8}}\] \[=1.5\times {{10}^{6}}m/s\] Since, wavelength is increasing, therefore galaxy is moving away from the earth with a velocity of\[1.5\times {{10}^{6}}\text{ }m/s\].
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