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Light year - class-XI

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The distance of a galaxy from the earth is of the order of $10^{25}$ m. The time taken by light to reach the earth from the galaxy is

  1. $3 \times 10^{14}$ s

  2. $3 \times 10^{16}$ s

  3. $3 \times 10^{18}$ s

  4. $3\times10^{20}$ s


Correct Option: B
Explanation:

Speed of light, c = 3 x 10$^8$ m s$^{-1}$
Time taken by light to reach the earth from the galaxy is
$\displaystyle t = \frac {10^{25}m}{3\times 10^8m\,s^{-1}}\,=\,3\times 10^{16}s$

The $6563 A^0 H _\alpha$ line emitted by hydrogen in a star is found to be red-shifted by $15 A^0$. The speed with which the star is receding from the earth is

  1. $3.2 \times 10^5m s^{-1}$

  2. $6.87 \times 10^5m s^{-1}$

  3. $2 \times 10^5m s^{-1}$

  4. $12.74 \times 10^5m s^{-1}$


Correct Option: B
Explanation:

Wavelength of $H _\alpha$ line, 
$\lambda=6563A^o$
$=6563\times10^{-10}$


Redshift observed in star $(\acute{\lambda-\lambda})=15A^o=15\times 10^{-10}$

Let the velocity of the star with which it is receding away from the earth be v.

Red shift relation, $\acute{\lambda}-\lambda$=$\dfrac{v}{c}\lambda$

$=\dfrac{c}{\lambda}\times (\acute{\lambda}-\lambda)$

$=\dfrac{3\times 10^8\times 15\times 10^{-10}}{6563\times 10^{-10}}$

$=6.87 \times 10^5m s^{-1}$

The Milky Way Galaxy is about

  1. 150000 light years

  2. 200000 light years

  3. 250000 light years

  4. 170000 light years


Correct Option: A
Explanation:

Milky way is about 100,000 light years or 30kpc.

The Milky Way Galaxy is about 150000 light years

What is known as Astronomical Unit (AU)?

  1. Average distance between Earth and Sun

  2. Average distance between Sun and Moon

  3. Average distance between Moon and Earth

  4. Average distance between Earth and Star


Correct Option: A
Explanation:

Average distance from the centre of the Earth to the centre of the Sun forms the astronomical unit (AU). It equals to around 149.6 million kilometres.

Which method is used by astronomers to calculate how far the star is?

  1. Parallax

  2. Equillax

  3. Purplex

  4. None


Correct Option: A
Explanation:

Parallax is used by astronomers to measure the distance between the stars. Parallax is also known as trigonometric parallax.

1 Parsec is:

  1. 2.3 light years

  2. 3.3 light years

  3. 4.3 light years

  4. 5.3 light years


Correct Option: B
Explanation:

Parsec is a unit of length used in astronomy to measure distances outside our solar system. One parsec equals to around 3.26156 light years.

The method of measuring distance to stars beyond 100 light-years is

  1. Capheid variable stars

  2. Heid variable stars

  3. Lowheid variable stars

  4. None


Correct Option: A
Explanation:

The brightness of the stars changes over time. The difference in the brightness over time allows calculating the distance to the star. This is the cepheid variable stars method that is used to measure the distance to stars beyond 100 light years.

The method of measuring the distance to stars beyond 100 light-years is  Cepheid variable stars

Which is the another unit used by astronomers for measuring distance to other parts of Milky Way Galaxy?

  1. Km

  2. Light year

  3. Parsec

  4. Meter


Correct Option: C
Explanation:

Units used in astronomy by astronomers are an astronomical unit, light year and parsec.

another unit used by astronomers for measuring the distance to other parts of Milky Way Galaxy is Parsec

How can Cepheid variable stars helpful to figure out true brightness?

  1. These stars change their brightness over time

  2. These stars lower their brightness over time

  3. These stars raise their brightness over time

  4. None


Correct Option: A
Explanation:

The brightness of the stars changes over time. The difference in the brightness over time allows calculating the distance to the star. This is the cepheid variable stars method that is used to measure the distance to stars beyond 100 light years.

These stars change their brightness over time


Distance from Earth to Proxima Centauri, the next nearest star is:

  1. 4.24 light years

  2. 5.5 light years

  3. 4.0 light years

  4. 3.5 light years


Correct Option: A
Explanation:

Proxima Centauri, a part of a triple star system named Alpha Centauri, is the closest star to our solar system. It is about 4.24 light years away from the Earth.

The Andromeda Galaxy is

  1. 2.3 million light years

  2. 3.5 million light years

  3. 2 million light years

  4. 5.6 million light years


Correct Option: A
Explanation:

Andromeda galaxy is the nearest galaxy to our Milky Way. It is about 2.537 million light-years away.

 The Andromeda Galaxy is 2.3 million light years


Light moves at velocity of roughly __________ kms in every second.

  1. 300,000

  2. 500000

  3. 100000

  4. 400000


Correct Option: A
Explanation:

The velocity of the light is precisely 299792458 metres per second or 3 x 10$^8$ m/s.

Astronomers use light years to measure the speed of light.

  1. True

  2. False


Correct Option: B
Explanation:

Heavenly bodies are very far away from each other. It is impossible to measure their distances by using conventional units like miles and kilometres. Hence light year is used to measure the distance of heavenly bodies from each other.

All the stars are at the same distance from us. This statement is

  1. True

  2. False


Correct Option: B
Explanation:

Our Sun is the nearest star to the Earth. After the Sun, the next nearest star is 4.3 light years away. A light year is the distance light travels in one year. One light year is 10 trillion kilometers. Hence, all the stars are not the same distance from our Earth. They are at different distances. Hence, given statement is false.

The $6563 \mathring {A}$ line emitted by hydrogen atom in a star is found to be red shifted by $5\mathring{A}$. the speed with which the star is receding from the earth is:

  1. $17.3\times 10^3 $ m/s

  2. $4.29 \times 10^7 $m/s

  3. $3.39 \times 10^5 $m/s

  4. $ 2.29 \times 10^5$ m/s


Correct Option: D
Explanation:

$\dfrac{\triangle v}{v}=\dfrac{V _{radiant}}{C}$
$v=\dfrac{C}{\lambda}$
$d.v=\dfrac{C}{\lambda^2}.dv=\dfrac{\dfrac{x}{\lambda^2}}{\dfrac{C}{\lambda}}=\dfrac{d\lambda}{\lambda}$
$\dfrac{d\lambda}{\lambda }=\dfrac{V _{radial }}{C}$
$\dfrac{5 \times 10^{-10}}{6563 \times 10^{-10}}=\dfrac{C _{radial}}{3\times 10^8}$
Calculate $V _radial=\dfrac{5}{6563}\times 3\times 10^8$
$=\dfrac{15}{6563}\times 10^8$
$\dfrac{15}{6.56}\times 10^5$
$2.29\times 10^5m/s$

A star is seen to be rising on the eastern horizon at $23:00$ hrs. At what time the star will rise 2 days later

  1. $00:00$ hrs

  2. $21:40$ hrs

  3. $23:20$ hrs

  4. $23:00$ hrs


Correct Option: D
Explanation:
The stores appear to be static when seen from earth thus they only appear to be moving due to earth's rotation about its axis. It takes exactly $24$ hors to complete one rotation. So after days $(2\times 24\ hrs)$, the far will be seen rising at the same time
$(i.e., 23:00\ hrs)$

Which of the following is different from others?

  1. light year

  2. parsec

  3. astronomical unit

  4. micron


Correct Option: D

One light year is equal to

  1. 3.26 parsec

  2. 3.26km

  3. 3.26 A.U.

  4. $\displaystyle \frac{1}{3.26}$ parsec


Correct Option: D
Explanation:

One light year is the distance travelled by the light in one year. So,

$1$Parse $=3.08\times { 10 }^{ 16 }$m
$1$ light year$9.46\times { 10 }^{ 15 }$m
So
$1$ light year $=9.46\times { 10 }^{ 15 }$ parsec
                       $=3.08\times { 10 }^{ 16 }$
$1$ light year $=0.306$ parsec
                     $=\cfrac { 1 }{ 3.26 } $ parsec

Which of the following is the largest astronomical unit?

  1. light year

  2. parsec

  3. KM

  4. astronomical unit


Correct Option: B

We live on the outer edge of a spiral type of galaxy called the milky way, which is about ........light years in diameter

  1. $10^5$

  2. $10^4$

  3. $10^3$

  4. 10


Correct Option: A
Explanation:

The diameter of milky way is about ${ 10 }^{ 5 }$ light years


Which of the following is the smallest unit of distance?

  1. light year

  2. parsec

  3. astronomical unit

  4. km


Correct Option: D
Explanation:

Out of these four units kilometer is the smallest unit.

The nearest star to the Earth (apart from the Sun) is 'alpha centauri' which is about .......... away form the Earth

  1. 4.3 light years

  2. 3.26 light years

  3. $4.3 \times 10^{12}$ km

  4. $3.26 \times 10^{15}$ km


Correct Option: A
Explanation:

Alpha century is a star system closest to earth other than sun. Its distance from the earth is about $4.367$ light years.

One light year is equal to _________.

  1. $3.26$ parsec

  2. $3.26$ km

  3. $3.26$ AU

  4. $\cfrac{1}{3.26}$ parsec


Correct Option: D
Explanation:

1 light year = $\dfrac { 1 }{ 3.26 } $ parsec

Velocity of light is

  1. $3\times 10^4km/s$

  2. $3\times 10^6km/s$

  3. $3\times 10^5km/s$

  4. $3\times 10^3km/s$


Correct Option: C
Explanation:

Ans : $3\times { 10 }^{ 5 }km/s$

The average distance between Earth and the Sun is $1.496\times {10}^{8}\ km$ and the speed of light coming from the Sun is $3\times {10}^{8}\ m/s$. How much time will it take for Sun's rays to reach Earth?

  1. $3\ min$

  2. $498.66\ s$

  3. $8\ min$ $30\ s$ 

  4. $554\ s$


Correct Option: B
Explanation:

Distance between Earth and Sun$=1.496\times {10}^{8}\ km=1.496\times {10}^{11}\ m$
Speed of light $=3\times {10}^{8}\ m/s$
By using the formula:
$speed=\cfrac{Distance}{Time}$
or $3\times {10}^{8}\ m/s=\cfrac{1.496\times {10}^{11}m}{Time}$
So,  $Time=\cfrac{1.496\times {10}^{11}\ m}{3\times {10}^{8}\ m/s}$ = $\cfrac{1496}{3}s$ $=498.66\ s$ 

If light travelling from the Sun at the speed of $3\times {10}^{8}\ m/s$, reach a planet $A$ in $25\ min\  30\ sec$. Then what is the distance between the Sun and the planet? 

(1 light year $=9.461\times {10}^{12}\ km$)

  1. $3$ light minutes

  2. $0.48\times {10}^{-4}$ light year

  3. $1.96\times {10}^{4} $light year

  4. $2.5$ light years


Correct Option: B
Explanation:

Speed of light $=3\times {10}^{8}\ m/s$
Time taken $=25\ min\ 30\ sec = 1530\ sec$
By using the formula, 
$Speed=\cfrac{Distance}{Time}$
or 

$Distance=Speed \times Time$ $=3\times {10}^{8}\times 1530$ $=4590\times {10}^{8}\ m$ $=4590\times {10}^{5}\ km$
Distance (in light year) $=\cfrac{4590\times {10}^{5}}{9.461\times {10}^{12}}$ $=0.48\times {10}^{-4}$ light year.

If the light from star $A$ takes $15$ min to reach star $B$ and the speed of light is $3\times {10}^{8}m/s$, then what is the distance between the stars?

  1. $2.7\times {10}^{8}km$

  2. $2,9\times {10}^{11}km$

  3. $2.7\times {10}^{8}m$

  4. $36\times {10}^{9}km$


Correct Option: A
Explanation:

Given $t=15$min
Speed of light $=3\times {10}^{8}m/s$
$\therefore$ $t=15min=15\times 60=900 sec$
By using the formula
$Speed=\cfrac{Distance}{Time}$
$Distance=Speed \times time$ $=3\times {10}^{8}m/s\times 900 sec$ $=2700\times {10}^{8}m$ $=2.7\times {10}^{8}km$

How does astronomer calculate distance of star?

  1. Comparing apparent brightness of star to true brightness

  2. Comparing true brightness of star to apparent brightness

  3. Comparing true brightness of star to true brightness

  4. None


Correct Option: A
Explanation:

The brightness of the stars changes over time. The difference in the brightness (difference in the apparent brightness to the true brightness) over the time allows calculating the distance to the star. This is the cepheid variable stars method that is used to measure the distance to stars beyond 100 light years.

Comparing the apparent brightness of the star to true brightness

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