
A coin of mass 10 g is placed over a book of length 50cm. The coin is on the verge of sliding when one end of the book is lifted 10 cm up. The coefficient of static friction between the book and the coin is
A. 1.0
B. 0.4
C. 0.3
D. 0.2
Answer
219.6k+ views
Hint: When two surfaces come in contact with each other or are rubbed, then a force known as frictional force. There are different types of friction.
Formula used:
The relation between normal and frictional force is given by
\[F = \mu N\]
Where ‘F’ is the frictional force, ‘N’ is the normal force and \[\mu \] is the coefficient of static friction.
Complete step by step solution:
Given that the mass of the coin is, m=10g
Length of the book is, l=50cm
Height of the book is =10 cm
The relation between normal and frictional force is given by
\[F = \mu N\]
The coin will not slide if all the forces in the system are balanced. Therefore, it can be written that,
\[mg\sin \theta = \mu R\]
Substituting the value in above equation and solving,
\[mg\sin \theta = \mu mg\]
\[\Rightarrow \sin \theta = \mu \]
\[\Rightarrow \sin \theta = \dfrac{P}{H}\]
Or in this case, it can be written that
\[\sin \theta = \dfrac{\text{Length of the book}}{\text{Height}}\]
Substituting the values given in the question and solving, we get
\[\mu = \dfrac{{10}}{{50}}\]
\[\therefore \mu = 0.2\]
Therefore, the coefficient of static friction between the book and the coin will be 0.2.
Hence, option D is the correct answer.
Note: Now since it is known that the frictional force exists when two surfaces are in contact, the object tries to be in its state of rest and hence applies a resistive force which is known as coefficient of static friction. Basically the coefficient of static friction is used to determine whether the two surfaces in contact are rough or smooth or to what extent they are rough or smooth. Different materials have different coefficients of friction and when they are brought in contact the amount of force that is to be applied also varies.
Formula used:
The relation between normal and frictional force is given by
\[F = \mu N\]
Where ‘F’ is the frictional force, ‘N’ is the normal force and \[\mu \] is the coefficient of static friction.
Complete step by step solution:
Given that the mass of the coin is, m=10g
Length of the book is, l=50cm
Height of the book is =10 cm
The relation between normal and frictional force is given by
\[F = \mu N\]
The coin will not slide if all the forces in the system are balanced. Therefore, it can be written that,
\[mg\sin \theta = \mu R\]
Substituting the value in above equation and solving,
\[mg\sin \theta = \mu mg\]
\[\Rightarrow \sin \theta = \mu \]
\[\Rightarrow \sin \theta = \dfrac{P}{H}\]
Or in this case, it can be written that
\[\sin \theta = \dfrac{\text{Length of the book}}{\text{Height}}\]
Substituting the values given in the question and solving, we get
\[\mu = \dfrac{{10}}{{50}}\]
\[\therefore \mu = 0.2\]
Therefore, the coefficient of static friction between the book and the coin will be 0.2.
Hence, option D is the correct answer.
Note: Now since it is known that the frictional force exists when two surfaces are in contact, the object tries to be in its state of rest and hence applies a resistive force which is known as coefficient of static friction. Basically the coefficient of static friction is used to determine whether the two surfaces in contact are rough or smooth or to what extent they are rough or smooth. Different materials have different coefficients of friction and when they are brought in contact the amount of force that is to be applied also varies.
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

Understanding Collisions: Types and Examples for Students

Other Pages
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

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

