Tag: physics

Questions Related to physics

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

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

In a stationary wave, 

physics superposition of waves-2: stationary (standing) waves: vibrations of air columns from moving to stationary stationary (or standing) waves formation of stationary waves

  1. Phase is same at all points in a loop

  2. Amplitude is same at all points

  3. Energy is constant at all points

  4. Temperature is same at all points

Show answer
Correct Option: A
Explanation:

Let two waves be $y _1=A \sin\ (wt-kx)$
$y _2=A \sin\ (wt+kx)$
$y=y _1+y _2$
$=(2A \cos\ kx)\sin\ wt.$ 
For all point in one loop i.e as $x$ varies in $2A \cos k x$, the phase is same. The phase changes only after crossing a node.

Standing waves can be produced in.

physics superposition of waves-2: stationary (standing) waves: vibrations of air columns from moving to stationary stationary (or standing) waves formation of stationary waves

  1. Solid only

  2. Liquid only

  3. Gases only

  4. All of the above

Show answer
Correct Option: A
Explanation:

Standing wave produces when two waves of identical frequency interfere with one another while travelling in opposite directions and this coincidence directions and this coincidence is not possible in fluids or gases.

In strings, the position of antinodes are obtained at

physics superposition of waves-2: stationary (standing) waves: vibrations of air columns from moving to stationary stationary (or standing) waves formation of stationary waves

  1. $\lambda,\space2\lambda, \space3\lambda$

  2. $0,\space\lambda,/2 \space\lambda$

  3. $2\lambda,\space4, \space6\lambda$

  4. $\lambda/4,\space3\lambda/4, \space5\lambda/4$

Show answer
Correct Option: D
Explanation:

In a string which is connected at both ends (similarity to sine wave), anti-nodes appear at odd multiples of $\dfrac{\lambda}{4}$.