Tag: constellations and galaxies

Quizzes (4) Questions (52)

Questions Related to constellations and galaxies

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

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
Reveal answer Fill a bubble to check yourself
D Correct answer
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

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

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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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. $554\ s$
Reveal answer Fill a bubble to check yourself
B Correct answer
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,  $Time=\cfrac{1.496\times {10}^{11}\ m}{3\times {10}^{8}\ m/s}$ = $\cfrac{1496}{3}s$ $=498.66\ s$ 

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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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.