Tag: constellations and galaxies

Questions Related to constellations and galaxies

Which of the following is different from others?

physics constellations and galaxies light year evolution and end stages of stars in the world of stars

  1. light year

  2. parsec

  3. astronomical unit

  4. micron

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Correct Option: D

One light year is equal to

physics constellations and galaxies light year evolution and end stages of stars in the world of stars

  1. 3.26 parsec

  2. 3.26km

  3. 3.26 A.U.

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

Show answer
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?

physics constellations and galaxies light year evolution and end stages of stars in the world of stars

  1. light year

  2. parsec

  3. KM

  4. astronomical unit

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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

physics constellations and galaxies light year evolution and end stages of stars in the world of stars

  1. $10^5$

  2. $10^4$

  3. $10^3$

  4. 10

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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?

physics constellations and galaxies light year evolution and end stages of stars in the world of stars

  1. light year

  2. parsec

  3. astronomical unit

  4. km

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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

physics constellations and galaxies light year evolution and end stages of stars in the world of stars

  1. 4.3 light years

  2. 3.26 light years

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

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

Show answer
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 _________.

physics constellations and galaxies light year evolution and end stages of stars in the world of stars

  1. $3.26$ parsec

  2. $3.26$ km

  3. $3.26$ AU

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

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Correct Option: D
Explanation:

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

Velocity of light is

physics constellations and galaxies light year evolution and end stages of stars in the world of stars

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

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

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

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

Show answer
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?

physics constellations and galaxies light year evolution and end stages of stars in the world of stars

  1. $3\ min$

  2. $498.66\ s$

  3. $8\ min$ $30\ s$ 

  4. $554\ s$

Show answer
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 $=9.461\times {10}^{12}\ km$)

physics constellations and galaxies light year evolution and end stages of stars in the world of stars

  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

Show answer
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.