Tag: evolution and end stages of stars

Questions Related to evolution and end stages of stars

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

Show answer
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.

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?

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

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

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

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

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

Show answer
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?

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

  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

Show answer
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

Match the columns

A B
(1) Speed of light (a) 4.3 light years away from earth
(2) Light year                (b) 300,000 km/s
(3) Sun (c) Nearest star
(4) Alpha Centauri (d) distance travelled by light in one year
(e) 18 light minutes away from the Earth
physics life cycle of stars evolution and end stages of stars the stars stars

  1. 1-d 2-d, 3-a, 4-e

  2. 1-a, 2-b, 3-c, 4-e

  3. 1-d, 2-b, 3-e, 4-c

  4. 1-b, 2-d, 3-c, 4-a

Show answer
Correct Option: D
Explanation:

The speed of light is 300,000 km/s.
Light year is the distance travelled by light in one year.
Sun is the nearest star from Earth.
Alpha Centauri is 4.3 light years away from Earth.

The observed wavelength of light coming from a distant galaxy is found to be increased by $0.5\%$ as compared with that coming from a terrestrial source. The galaxy is.

physics life cycle of stars evolution and end stages of stars the stars stars

  1. Stationary with respect t to the earth

  2. Approching the earth with velocity of light

  3. Receding from the earth with velocity of light

  4. Receding from the earth with a velocity equal to $1.5\times 10^{6}m/s$

Show answer
Correct Option: D
Explanation:

$\displaystyle\frac{\Delta \lambda}{\lambda}=\frac{v}{c}$
Now, $\Delta\lambda=\displaystyle\frac{0.5}{100} \lambda$
$\Rightarrow \displaystyle\frac{\Delta\lambda}{\lambda}=\frac{0.5}{100}$
$\therefore v=\displaystyle\frac{0.5}{100}\times c=\frac{0.5}{100}\times 3\times 10^8$
$=1.5\times 10^6m/s$
increase in $\lambda$ indicates that the star is receding.

$1$ parsec equals to:

physics life cycle of stars evolution and end stages of stars the stars stars

  1. $3.26$ light years

  2. $3.1 \times 10^{13}\ km$

  3. $1.92 \times 10^{13}\ miles$

  4. all of the above

Show answer
Correct Option: D
Explanation:

$1$ parsec = $3.26$ light years

$= 3.6 \times 9.46 \times 10^{12}\ km$
$\approx 3.1 \times 10^{13}\ km$
$=\dfrac{3.1 \times 10^{13}}{1.6}\ miles$

             
 $\approx 1.92 \times 10^{13}\ miles$

The evidence for the rotation of stars comes mainly from the.

physics life cycle of stars evolution and end stages of stars the stars stars

  1. Stark effect

  2. Photoelectric effect

  3. Doppler effect

  4. Zeeman effect

Show answer
Correct Option: C