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```
A horizontal disc rotating freely about a vertical axis makes 100rpm.A small piece of wax of mass 10kg falls vertically on the disc and adheres to it at a distance of 9cm from the axis.If the number of revolutions per minute is thereby reduced to 90.Calculate the moment of inertia of disc.

```
10 years ago

509 Points
```							taking wax & disk as a system ....
then there is no external torque acting on this system so angular momentam will remain conserved..
IW = IfWf              ..............1
finally moment of inertia = I + Iwax
Iwax = mr2 =  10(9*10-2)2                                         (r = 9cm = 9*10-2m  &   m =10kg )
= 0.081 kg-m2
If = I + 0.081           ..........2
putting 2 in 1
IW = (I+0.081)Wf
W/Wf = (1+.081/I)                                 {W = 100rpm &  Wf=90rpm (given)}
100/90 = 1 + 0.081/I
I = 0.729kg-m2
```
10 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions