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The speed of light in vacuum is taken as unity. If light takes 6 min 40 s to reach the Earth from the Sun, the distance between the Sun and the Earth in new unit is :

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Solution written and verified by Roshan, science educator with 5 years of experience teaching NEET and JEE aspirants. Last reviewed September 2026.
Options
1
3 × 10⁸
2
3 × 10¹⁰
3
400
4
500
Correct Answer
Option 3 : 400
Step-by-Step Solution
1

Given: Speed of light in vacuum c = 1 new unit/s (taken as unity)

Time taken by light to travel from Sun to Earth: t = 6 min 40 s

2

Convert time to seconds:

t = 6 × 60 + 40 = 360 + 40 = 400 s

3

Calculate distance:

d = c × t = 1 × 400 = 400 new units

d = c × t = 1 (new unit/s) × 400 s = 400 new units

Theory: Units, Dimensions & Measurements
1. System of Units

A unit is a standard quantity used to express and compare physical quantities. The SI (Système International) system is the internationally accepted system. Its seven base units are: metre (length), kilogram (mass), second (time), ampere (current), kelvin (temperature), mole (amount of substance), candela (luminous intensity).

In this question, a natural unit system is used where c = 1. This is common in relativistic physics and astrophysics. When c = 1, distances can be measured in light-seconds, light-minutes, or light-years — the distance light travels in that time unit.

2. Distance = Speed × Time

d = v × t

When v = c = 1 new unit/s and t = 400 s → d = 400 new units

This distance is called 400 light-seconds

3. Actual Sun-Earth Distance

📌 Actual speed of light: c = 3 × 10⁸ m/s

📌 Time for light Sun → Earth: ≈ 8 min 20 s = 500 s (actual average)

📌 In this problem: 6 min 40 s = 400 s is given

📌 Actual Sun-Earth distance: ≈ 1.5 × 10¹¹ m = 1 AU (Astronomical Unit)

📌 In natural units (c=1): distance = 500 light-seconds (actual), 400 in this problem

4. Dimensional Analysis

Dimensional analysis is used to: (1) Check the correctness of equations — both sides must have same dimensions. (2) Derive relationships between physical quantities. (3) Convert units from one system to another.

📌 [Distance] = [Speed] × [Time] = LT⁻¹ × T = L ✓

📌 In new unit system: [c] = 1 new unit/s, [t] = s → [d] = new units ✓

📌 Principle of homogeneity: all terms in an equation must have same dimensions

5. Significant Figures

Significant figures represent the precision of a measurement. Rules: all non-zero digits are significant; zeros between non-zero digits are significant; leading zeros are NOT significant; trailing zeros after decimal point are significant. In calculations: addition/subtraction — result has same decimal places as least precise measurement; multiplication/division — result has same significant figures as least precise factor.

6. Natural Units in Physics

In natural unit systems, fundamental constants are set equal to 1 to simplify equations. Examples: Planck units (c = 1, ℏ = 1, G = 1), Atomic units (e = 1, mₑ = 1, ℏ = 1). When c = 1, energy and mass become equivalent (E = mc² becomes E = m), and distances are measured in time units (light-seconds, light-years). This is widely used in particle physics and astrophysics.

Frequently Asked Questions
1. Why is option 3 (400) correct and not option 4 (500)? ⌄
The problem states light takes 6 min 40 s = 6×60+40 = 360+40 = 400 s. With c = 1 new unit/s, distance = 1 × 400 = 400 new units. Option 4 (500) would correspond to 8 min 20 s = 500 s, which is the actual average Sun-Earth light travel time — but the question specifies 6 min 40 s.
2. Why are options 3×10⁸ and 3×10¹⁰ wrong? ⌄
3×10⁸ is the speed of light in m/s — not a distance. 3×10¹⁰ is c in cm/s — also not the answer. These options are distractors designed to confuse students who try to use c = 3×10⁸ m/s. In this new unit system, c = 1 (not 3×10⁸), so the distance is simply c×t = 1×400 = 400.
3. What is 6 min 40 s in seconds? ⌄
6 min 40 s = (6 × 60) + 40 = 360 + 40 = 400 seconds. Key step: multiply minutes by 60 to convert to seconds, then add the remaining seconds. This unit conversion is the entire calculation in this problem.
4. What is an Astronomical Unit (AU)? ⌄
1 AU = average distance from Earth to Sun = 1.496 × 10¹¹ m ≈ 1.5 × 10¹¹ m. Other useful units: 1 light-year = 9.46 × 10¹⁵ m (distance light travels in 1 year). 1 parsec = 3.086 × 10¹⁶ m = 3.26 light-years. 1 parsec is the distance at which 1 AU subtends an angle of 1 arcsecond.
5. What are natural units? ⌄
Natural units set fundamental physical constants equal to 1 to simplify equations. In this problem's system: c = 1 new unit/s. So distances are automatically measured in light-seconds. 1 light-second = 3×10⁸ m (the distance light travels in 1 second). The Sun is 400 light-seconds away (as per this problem's data).
6. How do you convert between unit systems? ⌄
Using dimensional analysis: n₁u₁ = n₂u₂, where n = numerical value and u = unit. Example: 1 km = 1000 m. For this problem: d in new units = d in metres / (3×10⁸) = (1.5×10¹¹) / (3×10⁸) ≈ 500 light-seconds. But the problem gives t = 400s, so d = 400 in those units.
7. What is the significance of c in physics? ⌄
The speed of light c = 3×10⁸ m/s is the universal speed limit — nothing with mass can travel at or beyond c. It appears in: E = mc² (mass-energy equivalence), Maxwell's equations (electromagnetic wave speed = c), special relativity (time dilation, length contraction). Setting c = 1 is not just mathematical convenience — it reveals deep connections between space and time.
8. What are the rules for significant figures in multiplication? ⌄
In multiplication and division, the result should have the same number of significant figures as the measurement with the fewest significant figures. Example: 2.5 × 3.14 = 7.85 → round to 7.9 (2 sig figs, same as 2.5). In this problem: c = 1 (exact) × t = 400 s (3 sig figs) → d = 400 new units (3 sig figs).
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