If light travelling from the Sun at the speed of $3 \times 10^{8} m / s$ , reach a planet $A$ in $25 m i n 30 s e c$ . Then what is the distance between the Sun and the planet?
(1 light year $= 9.461 \times 10^{12} k m$ )

3

light minutes

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0.48 \times 10^{- 4}

light year

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1.96 \times 10^{4}

light year

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2.5

light years

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Solution

The correct option is B $0.48 \times 10^{- 4}$ light year
Speed of light $= 3 \times 10^{8} m / s$
Time taken $= 25 m i n 30 s e c = 1530 s e c$
By using the formula,
$S p e e d = \frac{D i s t a n c e}{T i m e}$
or
$D i s t a n c e = S p e e d \times T i m e$ $= 3 \times 10^{8} \times 1530$ $= 4590 \times 10^{8} m$ $= 4590 \times 10^{5} k m$
Distance (in light year) $= \frac{4590 \times 10^{5}}{9.461 \times 10^{12}}$ $= 0.48 \times 10^{- 4}$ light year.

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Similar questions

Q. 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

A lightyear is a unit of distance. It is the distance travelled by light in one year. If the speed of light is equal to 3 x $10^{8}$ m/s, how many kilometres will make up 1 light year?

Q. Assertion :A light-year is a distance that light travels with velocity of

3 \times 10^{8}

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in one year. Reason: Light year is the unit for measuring time.

A light year is the distance that light can travel in one year.
$1$ light year $= 9, 460, 000, 000, 000 k m .$
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(b) The average distance between Earth and Sun is $1.496 \times 10^{8} k m .$ Is the distance between earth and Sun greater than, less than or equal to one light year?

Q. Correct the mistakes if any, in the following statements:
i) Astronomical unit is the mean distance of the surface of the sun from the surface of the earth.
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If light travelling from the Sun at the speed of 3×108 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 =9.461×1012 km)

The correct option is B 0.48×10−4 light yearSpeed of light =3×108 m/sTime taken =25 min 30 sec=1530 secBy using the formula, Speed=DistanceTimeor Distance=Speed×Time =3×108×1530 =4590×108 m =4590×105 kmDistance (in light year) =4590×1059.461×1012 =0.48×10−4 light year.

If light travelling from the Sun at the speed of $3 \times 10^{8} m / s$ , reach a planet $A$ in $25 m i n 30 s e c$ . Then what is the distance between the Sun and the planet?
(1 light year $= 9.461 \times 10^{12} k m$ )