
It takes 8 minutes and 20 seconds for sunlight to reach the earth, find the distance between the earth and the sun. The velocity of light is $3\times {{10}^{8}}{m}/{s}\;$.(a). $15\times {{10}^{10}}m$(b). $25\times {{10}^{10}}m$(c). $30\times {{10}^{10}}m$(d). $5\times {{10}^{10}}m$
Answer
243k+ views
Hint: We know that the relation between speed, distance and time can be given by using the formula,
Complete step by step answer:
Note: This question can also be solved by using the relation between velocity, time and distance, but in that considering the velocity of light constant is the key point each student should consider otherwise, they might make mistakes in that. The above solution is more preferable and less time consuming also.
$\text{speed}=\dfrac{\text{distance}}{\text{time}}$. But before using this formula we have to convert time which is 8 minutes into seconds because all our units should be in the MKS unit system.
Formula used: $\text{speed}=\dfrac{\text{distance}}{\text{time}}$
In the question we are given that sun rays take 8 minutes and 20 seconds to reach the Earth from the Sun and we have to find the distance between Earth and Sun. So, we will use the formula which is given as,
$\text{speed}=\dfrac{\text{distance}}{\text{time}}$
Or $\text{s}=\dfrac{\text{d}}{\text{t}}$
Where, s is speed in ${m}/{s}\;$, d is distance in metre, and t is time in seconds.
Now, we are given that time taken by rays is 8 minutes so, we will convert minutes into seconds which is given as,
$1\ \text{minute}=60\ \text{seconds}$
So, $8\ \text{minute and 20 seconds}=8\times 60\ \text{seconds+20}=480+20\ \sec $
$8\ \text{minute and 20 seconds}=500\sec $
Now, t is 500 sec, and s is $3\times {{10}^{8}}{m}/{s}\;$, substituting these values in main expression we will get,
$3\times 1{{0}^{8}}=\dfrac{\text{d}}{500}$
$d=500\times 3\times {{10}^{8}}=15\times {{10}^{10}}m$
Thus, the distance between Earth and Sun is $15\times {{10}^{10}}m$.
Hence, option (a) is correct.
Recently Updated Pages
WBJEE 2026 Registration Started: Important Dates Eligibility Syllabus Exam Pattern

Dimensions of Charge: Dimensional Formula, Derivation, SI Units & Examples

How to Calculate Moment of Inertia: Step-by-Step Guide & Formulas

Circuit Switching vs Packet Switching: Key Differences Explained

Dimensions of Pressure in Physics: Formula, Derivation & SI Unit

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2026: Session 1 Results Out and Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Electric Field of a Uniformly Charged Ring

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2025-26

CBSE Notes Class 11 Physics Chapter 4 - Laws of Motion - 2025-26

CBSE Notes Class 11 Physics Chapter 14 - Waves - 2025-26

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

CBSE Notes Class 11 Physics Chapter 9 - Mechanical Properties of Fluids - 2025-26

