
If a boy of mass 30kg goes from earth to the moon. What will be his:
(A) Mass on the moon?
(B) Weight on the moon in N? $\left( {1\,kgf = 10N} \right)$
Answer
219.9k+ views
Hint: To find the answers of both the parts of the question, you first need to know that the mass does not vary for any condition unless the body is moving with very high velocity, comparable to the velocity of light. As for the weight, you must know that the acceleration due to gravity is one-sixth of that on earth.
Complete step by step solution:
We will approach the solution to the question exactly the same way as explained in the hint section of the solution to the question. The answer to the first part is pretty simple and straight-forward, while for the solution of the second part, you must know that the acceleration due to gravity on the moon is one-sixth of the acceleration due to gravity on earth. This can be represented as:
${g_m} = \dfrac{{{g_e}}}{6}$
The value of acceleration due to gravity on earth is given as following in the question:
${g_e} = 10\,m{s^{ - 2}}$
Also, let us have a look at what else is given in the question:
Mass of the boy is given to be $m = 30\,kg$
Since mass is independent of everything, the answer to the first part of the question is rather simple:
The mass of the boy on the moon will still be the same as the mass of the boy on earth since mass does not change when you go from one place to another in the universe.
As for the second part, we need to find the weight of the boy. Weight can be found as:
$W = mg$
Where, $W$ is the weight of the object,
\[g\] is the acceleration due to gravity and,
$m$ is the mass of the object
In our case, we can write it as:
${W_m} = m{g_m}$
Where, ${g_m}$ is the acceleration due to gravity on moon
If we substitute the values as we found them to be, we get:
${W_m} = \left( {30} \right)\left( {\dfrac{{10}}{6}} \right)$
Upon solving, we get the value of weight on the moon as:
${W_m} = 50\,N$
Or, ${W_m} = 5\,kgf$
Note: A common mistake that many people make is that do not interpret the given relation of $\left( {1\,kgf = 10N} \right)$ as the given value of acceleration due to gravity and thus, get confused about what they should take as the acceleration due to gravity on earth. Also, you must always remember that the acceleration due to gravity on the moon is exactly the one-sixth of that on the earth.
Complete step by step solution:
We will approach the solution to the question exactly the same way as explained in the hint section of the solution to the question. The answer to the first part is pretty simple and straight-forward, while for the solution of the second part, you must know that the acceleration due to gravity on the moon is one-sixth of the acceleration due to gravity on earth. This can be represented as:
${g_m} = \dfrac{{{g_e}}}{6}$
The value of acceleration due to gravity on earth is given as following in the question:
${g_e} = 10\,m{s^{ - 2}}$
Also, let us have a look at what else is given in the question:
Mass of the boy is given to be $m = 30\,kg$
Since mass is independent of everything, the answer to the first part of the question is rather simple:
The mass of the boy on the moon will still be the same as the mass of the boy on earth since mass does not change when you go from one place to another in the universe.
As for the second part, we need to find the weight of the boy. Weight can be found as:
$W = mg$
Where, $W$ is the weight of the object,
\[g\] is the acceleration due to gravity and,
$m$ is the mass of the object
In our case, we can write it as:
${W_m} = m{g_m}$
Where, ${g_m}$ is the acceleration due to gravity on moon
If we substitute the values as we found them to be, we get:
${W_m} = \left( {30} \right)\left( {\dfrac{{10}}{6}} \right)$
Upon solving, we get the value of weight on the moon as:
${W_m} = 50\,N$
Or, ${W_m} = 5\,kgf$
Note: A common mistake that many people make is that do not interpret the given relation of $\left( {1\,kgf = 10N} \right)$ as the given value of acceleration due to gravity and thus, get confused about what they should take as the acceleration due to gravity on earth. Also, you must always remember that the acceleration due to gravity on the moon is exactly the one-sixth of that on the earth.
Recently Updated Pages
Electricity and Magnetism Explained: Key Concepts & Applications

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

States of Matter Chapter For JEE Main Chemistry

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Understanding Uniform Acceleration in Physics

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

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

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

NCERT Solutions For Class 11 Physics Chapter 8 Mechanical Properties Of Solids

Motion in a Straight Line Class 11 Physics Chapter 2 CBSE Notes - 2025-26

NCERT Solutions for Class 11 Physics Chapter 7 Gravitation 2025-26

