
If the weight of the body is 1000 dyne, then the mass of the body will be
(A) \[{10^{ - 1}}kg\]
(B) \[{10^{ - 2}}kg\]
(C) \[{10^{ - 3}}kg\]
(D) \[{10^{ - 4}}kg\]
Answer
231.6k+ views
Hint Weight is actually the force generated by the body on earth. Quantitatively, it is the mass times the acceleration due to gravity of the planet on which the mass is placed. In this question we are given the force in terms of dynes. We first need to convert it into the SI units which is N and then proceed further by dividing the weight by the acceleration due to gravity. This will give us the mass of the body.
Complete step by step solution
We are given force in terms of dynes or in the CGS system. The SI unit of force is N which is kg \[\dfrac{m}{{{s^2}}}\] . We need to convert each unit to its counterpart in the CGS system. As we know that 1kg = 1000g and 1m = 100cm, Therefore.
\[\dfrac{{1kg \times m}}{{{s^2}}} = \dfrac{{1000g \times 100cm}}{{{s^2}}}\]
\[
\Rightarrow \dfrac{{1 kg \times m}}{{{s^2}}} = {10^5}\dfrac{{g \times cm}}{{{s^2}}} \\
\Rightarrow 1N = {10^5}dynes \\
\Rightarrow 1dyne = {10^{ - 5}}N \\
\]
Therefore 1000 dynes will be equal to \[{10^{ - 2}}N\] . Now we will find the mass of the body from this force.
\[
W = mg \\
\Rightarrow {10^{ - 2}} = m \times 10 \\
\Rightarrow {10^{ - 3}} = m \\
\]
Therefore, the answer with the correct option is option C.
Note
The weight of the body is subject to the acceleration due to gravity of that planet. This acceleration due to gravity is affected by the mass and radius of the planet on which the body is kept. However weight is a universal quantity and its value will remain the same whatever the external conditions be.
Complete step by step solution
We are given force in terms of dynes or in the CGS system. The SI unit of force is N which is kg \[\dfrac{m}{{{s^2}}}\] . We need to convert each unit to its counterpart in the CGS system. As we know that 1kg = 1000g and 1m = 100cm, Therefore.
\[\dfrac{{1kg \times m}}{{{s^2}}} = \dfrac{{1000g \times 100cm}}{{{s^2}}}\]
\[
\Rightarrow \dfrac{{1 kg \times m}}{{{s^2}}} = {10^5}\dfrac{{g \times cm}}{{{s^2}}} \\
\Rightarrow 1N = {10^5}dynes \\
\Rightarrow 1dyne = {10^{ - 5}}N \\
\]
Therefore 1000 dynes will be equal to \[{10^{ - 2}}N\] . Now we will find the mass of the body from this force.
\[
W = mg \\
\Rightarrow {10^{ - 2}} = m \times 10 \\
\Rightarrow {10^{ - 3}} = m \\
\]
Therefore, the answer with the correct option is option C.
Note
The weight of the body is subject to the acceleration due to gravity of that planet. This acceleration due to gravity is affected by the mass and radius of the planet on which the body is kept. However weight is a universal quantity and its value will remain the same whatever the external conditions be.
Recently Updated Pages
Dimensions of Pressure in Physics: Formula, Derivation & SI Unit

Circuit Switching vs Packet Switching: Key Differences Explained

JEE General Topics in Chemistry Important Concepts and Tips

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Electricity and Magnetism Explained: Key Concepts & Applications

Trending doubts
JEE Main 2026: Admit Card Out, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

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

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

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

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

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

