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PhysicsUnits & Measurements
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 :
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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