Tag: physics

Quizzes (598) Questions (13441)

Questions Related to physics

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.

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

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. $36\times {10}^{9}km$
Reveal answer Fill a bubble to check yourself
A Correct answer
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
$Speed=\cfrac{Distance}{Time}$
$Distance=Speed \times time$ $=3\times {10}^{8}m/s\times 900 sec$ $=2700\times {10}^{8}m$ $=2.7\times {10}^{8}km$

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

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

In a stationary wave, 

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

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

Standing waves can be produced in.

  1. Solid only

  2. Liquid only

  3. Gases only

  4. All of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

In strings, the position of antinodes are obtained at

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

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

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

  4. $\lambda/4,\space3\lambda/4, \space5\lambda/4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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