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Q.

The periodic time of rotation of a certain star is 22 days and its radius is 7 $\times$ 10⁸ meters. If the wavelength of light emitted by its surface be 4320 $\overset{0}{A}$ , the Doppler shift will be (1 day = 86400 sec)

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a

$0 .33 \overset{0}{A}$

b

$3 .3 \overset{0}{A}$

c

$0 .033 \overset{0}{A}$

d

$33 \overset{0}{A}$

answer is A.

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Detailed Solution

$ω = \frac{2 π}{T} = \frac{2 \times 22}{7 \times 22 \times 86400} v = r ω = 7 \times 10^{8} \frac{2 \times 22}{7 \times 22 \times 86400} = \frac{10^{6}}{432} ∆ λ = \frac{v}{C} λ = \frac{10^{6}}{3 \times 10^{8}} \times \frac{1}{432} \times 4320 = 0.033 \overset{0}{A}$

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