The apparent wavelength of light from a star moving away from the earth is 0.02% more than actual wavelength. What is the velocity of the star?
(A) $30km{s^{ - 1}}$
(B) $60km{s^{ - 1}}$
(C) $90km{s^{ - 1}}$
(D) None of these
Answer
668.4k+ views
Hint: To answer this question we need to find the change in wavelength from the formula. Then we have to find the expression for velocity from the formula and then we have to put the values from the question in the formula. This will give us the answer to the required question.
Complete step by step answer:
We know that wavelength is expressed as $\lambda $. So now the change in the wavelength is given as follows:
$\dfrac{{\Delta \lambda }}{\lambda } = \dfrac{v}{C}$
Here v is the velocity and C represents the speed of light.
Hence we can express V as:
$V = \dfrac{{\Delta \lambda }}{\lambda }C$
Now we have to put the values in the above expression, to get:
$
V = \dfrac{{0.02}}{{100}} \times 3 \times {10^8}m{s^{ - 1}} \\
V = 60km{s^{ - 1}} \\
$
Hence the velocity of the star is 60 km/s.
So the correct answer is option B.
Note: We should know that wavelength is defined as the distance which is calculated between identical points, which are also known as adjacent crests, in the corresponding cycles of a waveform signal. This waveform signal is propagated in the space or it can also propagate along a wire. Usually in the wireless systems, the length is calculated in meters, centimetres or also in millimetres.
It should always be remembered that for solving questions which involve the speed of light, the default value of light not necessarily should be mentioned. So in case the value of the speed of light is not mentioned then we have to assume the value as $3 \times {10^8}m{s^{ - 1}}$, as in this question.
Complete step by step answer:
We know that wavelength is expressed as $\lambda $. So now the change in the wavelength is given as follows:
$\dfrac{{\Delta \lambda }}{\lambda } = \dfrac{v}{C}$
Here v is the velocity and C represents the speed of light.
Hence we can express V as:
$V = \dfrac{{\Delta \lambda }}{\lambda }C$
Now we have to put the values in the above expression, to get:
$
V = \dfrac{{0.02}}{{100}} \times 3 \times {10^8}m{s^{ - 1}} \\
V = 60km{s^{ - 1}} \\
$
Hence the velocity of the star is 60 km/s.
So the correct answer is option B.
Note: We should know that wavelength is defined as the distance which is calculated between identical points, which are also known as adjacent crests, in the corresponding cycles of a waveform signal. This waveform signal is propagated in the space or it can also propagate along a wire. Usually in the wireless systems, the length is calculated in meters, centimetres or also in millimetres.
It should always be remembered that for solving questions which involve the speed of light, the default value of light not necessarily should be mentioned. So in case the value of the speed of light is not mentioned then we have to assume the value as $3 \times {10^8}m{s^{ - 1}}$, as in this question.
Recently Updated Pages
Explain the structure of megasporangium class 12 biology CBSE

Why is chloroform kept in dark coloured bottles class 12 chemistry CBSE

Derive the balancing condition of a Wheatstone bri class 12 physics CBSE

Draw VI characteristics of a pn junction diode Explain class 12 physics CBSE

Is it possible to carry pride in ones language too class 12 english CBSE

In a population at Hardy Weinberg equilibrium the allele class 12 biology CBSE

Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE

Which are the Top 10 Largest Countries of the World?

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

One megawatt is equal to how many units of electri class 12 physics CBSE

Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE

Give 10 examples of unisexual and bisexual flowers

