MCQ
The wavelength of light observed on the earth, from a moving star is found to decrease by $0.05\%$. Relative to the earth the star is
  • A
    Moving away with a velocity of $1.5 \times {10^5}m/s$
  • Coming closer with a velocity of $1.5 \times {10^5}m/s$
  • C
    Moving away with a velocity of $1.5 \times {10^4}m/s$
  • D
    Coming closer with a velocity of $1.5 \times {10^4}m/s$

Answer

Correct option: B.
Coming closer with a velocity of $1.5 \times {10^5}m/s$
b
(b)$\frac{{\Delta \lambda }}{\lambda } = \frac{v}{c}\, \Rightarrow \,\frac{{0.05}}{{100}} = \frac{v}{{3 \times {{10}^8}}}$

$ ==> v = 1.5 × 10^{5} m/s$
(Since wavelength is decreasing, so star coming closer)

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