$1N = Zkgf$ (approx.), then what is the value of $Z$?
(A) $0.1$
(B) $1$
(C) $10$
(D) $0$
Answer
Verified
122.7k+ views
Hint: To solve this question, we need to use the formula for the gravitational force on a body on the surface of the earth. Then we need to substitute the approximate value of the acceleration due to gravity on the surface of the earth. Also, we need to substitute the value of the force given in the question, after which we will get the required value of $Z$.
Formula used: The formula used to solve this question is given by
$F = mg$, here $F$ is the magnitude of the gravitational force acting on a body on a particle of mass $m$, and $g$ is the acceleration due to gravity on the surface of earth.
Complete step-by-step solution:
We know that the gravitational force acting on a body on the surface of the earth is given by the expression
$F = mg$
Now, we know that the value of the acceleration due to gravity on the surface of the earth is approximately equal to $10m/{s^2}$. The value of the force given in the question is equal to $1N$. Also, the mass of the body is given as $Zkg$. Therefore substituting $F = 1N$ $m = Zkg$ and $g = 10m/{s^2}$ in the above expression, we get
$1 = 10Z$
Dividing both sides by $10$, we get
$Z = \dfrac{1}{{10}}$
$ \Rightarrow Z = 0.1$
Thus, the value of $Z$ is equal to $0.1$.
Hence, the correct answer is option A.
Note: The unit of force given here is called one kilogram force. It is the non standard, gravitational metric unit of the force. It is a measure of the gravitational force exerted by the earth on an object having a mass equal to one kilogram.
Formula used: The formula used to solve this question is given by
$F = mg$, here $F$ is the magnitude of the gravitational force acting on a body on a particle of mass $m$, and $g$ is the acceleration due to gravity on the surface of earth.
Complete step-by-step solution:
We know that the gravitational force acting on a body on the surface of the earth is given by the expression
$F = mg$
Now, we know that the value of the acceleration due to gravity on the surface of the earth is approximately equal to $10m/{s^2}$. The value of the force given in the question is equal to $1N$. Also, the mass of the body is given as $Zkg$. Therefore substituting $F = 1N$ $m = Zkg$ and $g = 10m/{s^2}$ in the above expression, we get
$1 = 10Z$
Dividing both sides by $10$, we get
$Z = \dfrac{1}{{10}}$
$ \Rightarrow Z = 0.1$
Thus, the value of $Z$ is equal to $0.1$.
Hence, the correct answer is option A.
Note: The unit of force given here is called one kilogram force. It is the non standard, gravitational metric unit of the force. It is a measure of the gravitational force exerted by the earth on an object having a mass equal to one kilogram.
Recently Updated Pages
The ratio of the diameters of two metallic rods of class 11 physics JEE_Main
What is the difference between Conduction and conv class 11 physics JEE_Main
Mark the correct statements about the friction between class 11 physics JEE_Main
Find the acceleration of the wedge towards the right class 11 physics JEE_Main
A standing wave is formed by the superposition of two class 11 physics JEE_Main
Derive an expression for work done by the gas in an class 11 physics JEE_Main
Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More
JEE Main Login 2045: Step-by-Step Instructions and Details
Class 11 JEE Main Physics Mock Test 2025
JEE Main Chemistry Question Paper with Answer Keys and Solutions
JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking
JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions
Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs
NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids
Units and Measurements Class 11 Notes: CBSE Physics Chapter 1
JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More
NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line