Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
31.3k views
in Physics by (34.4k points)

A solid cylinder of mass m is wrapped with an inextensible light string and, is placed on a rough inclined plane as shown in the figure. The frictional force acting between the cylinder and the inclined plane is :

[The coefficient of static friction, μs, is 0.4]

(1) \(\frac 72\) mg

(2) 5 mg

(3) \(\frac{mg}5\)

(4) 0

Please log in or register to answer this question.

1 Answer

+1 vote
by (34.9k points)

Correct option is (3) \(\frac{mg}5\)

Let's take solid cylinder is in equilibrium

T + f = mg sin60° ....(i)

TR – fR = 0 ....(ii)

Solving we get

T = freq = \(\frac{mgsin\,\theta}2\)

But limiting friction < required friction

μmgcos60° < \(\frac{mg\,sin\,60^o}2\)

\(\therefore\)  Hence cylinder will not remain in equilibrium

Hence f = kinetic

= μkN

= μkmgcos 60°

\(\frac{mg}5\)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...