
A rolling body is kept on a plank B. There is sufficient friction between A and B and no friction between B and the inclined plane. Then body

A. A Rolls
B. A doesn’t experience any friction
C. A and B have equal acceleration and unequal velocities.
D. A rolls depending upon the angle of inclination
Answer
219.9k+ views
Hint:when a body is moving under the influence on only gravitational force then the motion is said to be free fall. The body is at equilibrium along the direction perpendicular to the surface in contact.
Formula used:
\[{F_{net}} = ma\]
here \[{F_{net}}\] is the net force acting on the body of mass m and a is the acceleration produced.
Complete step by step solution:
As there is no friction between body B and the inclined surface, there will be no resistive force acting on it to oppose the motion. Using vector resolution for the weight of the plank B, we get the force on the plank B as,
\[{F_{B\parallel }} = {m_B}g\sin \theta \]
If acceleration of B on the inclined surface is \[{a_B}\] then using Newton’s 2nd law of motion,
\[{F_{B\parallel }} = m{a_B}\]
\[\Rightarrow {a_B} = \dfrac{{{m_B}g\sin \theta }}{{{m_B}}} = g\sin \theta \]
Similarly, for rolling sphere A, using vector resolution for the weight of the sphere A, we get the force on the sphere as,
\[{F_{A\parallel }} = m{a_A}\]
\[\Rightarrow {a_A} = \dfrac{{{m_A}g\sin \theta }}{{{m_A}}} = g\sin \theta \]
The magnitude of plank B and the sphere A along the inclined surface is same and also the direction is down the inclined surface. Hence, the relative acceleration of the sphere with respect to the plank is zero.
Using Newton’s 2nd law of motion,
\[{F_{net}} = ma\]
The force between the surfaces of the sphere and the plank is zero as the relative acceleration is zero. As we know that the frictional force is the resistive force which opposes the relative motion between the surfaces in contact, so there will be no friction between the surfaces of the sphere and the plank.
Therefore, the correct option is B.
Note: If the relative acceleration between the sphere and the plank was non-zero then frictional force will act on the sphere which will try to rotate the sphere and hence it may roll or slip based on the relative acceleration between A and B.
Formula used:
\[{F_{net}} = ma\]
here \[{F_{net}}\] is the net force acting on the body of mass m and a is the acceleration produced.
Complete step by step solution:
As there is no friction between body B and the inclined surface, there will be no resistive force acting on it to oppose the motion. Using vector resolution for the weight of the plank B, we get the force on the plank B as,
\[{F_{B\parallel }} = {m_B}g\sin \theta \]
If acceleration of B on the inclined surface is \[{a_B}\] then using Newton’s 2nd law of motion,
\[{F_{B\parallel }} = m{a_B}\]
\[\Rightarrow {a_B} = \dfrac{{{m_B}g\sin \theta }}{{{m_B}}} = g\sin \theta \]
Similarly, for rolling sphere A, using vector resolution for the weight of the sphere A, we get the force on the sphere as,
\[{F_{A\parallel }} = m{a_A}\]
\[\Rightarrow {a_A} = \dfrac{{{m_A}g\sin \theta }}{{{m_A}}} = g\sin \theta \]
The magnitude of plank B and the sphere A along the inclined surface is same and also the direction is down the inclined surface. Hence, the relative acceleration of the sphere with respect to the plank is zero.
Using Newton’s 2nd law of motion,
\[{F_{net}} = ma\]
The force between the surfaces of the sphere and the plank is zero as the relative acceleration is zero. As we know that the frictional force is the resistive force which opposes the relative motion between the surfaces in contact, so there will be no friction between the surfaces of the sphere and the plank.
Therefore, the correct option is B.
Note: If the relative acceleration between the sphere and the plank was non-zero then frictional force will act on the sphere which will try to rotate the sphere and hence it may roll or slip based on the relative acceleration between A and B.
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

