Q.

For a thin non-uniform disc of mass M and radius R the area density of mass, σ as a function of distance r from its centre is given as  σ=σ01rR . This disc is placed on a rough horizontal surface (coefficient of friction =μ ) and a constant horizontal force F = Mg, applied along the center of mass. The disc rolls without slipping on the surface, then choose the correct statement(s).[Moment of inertia of disc about its own axis is 3MR210

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!
An Intiative by Sri Chaitanya

a

The angular acceleration of the disc is 5g13R

b

The value of μ may be 0.3 

c

The linear acceleration of the disc is 5g13

d

The linear acceleration of the disc is  10g13

answer is A, C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Question Image

 

 

 

 

 

M=0Rσ01rR2πrdr=2πσ0r22r33R0R=2πσ0R26=πσ0R23=MI=0Rσ01rR2πr3dr=2πσ0r44r5R0R=2πσ0R420=πσ0R410I=3MR2R410=3MR210Ff=Ma

fR=310MR2,aRf=310MaF=1+310Ma=1310Maa=10F13M=10g13f=310M×10F13=3Mg13μMg>3Mg13μ>313

 

 

Watch 3-min video & get full concept clarity

tricks from toppers of Infinity Learn

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon