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A recording disc rotates steadily at 45 revolutions/min on a table. When a small mass 0.02 kg is dropped gently on the disc at a distance of 0.04 m from its axis and stucks to the disc, the rate os revolution falls to 36 rev/mmin. zthe mment of intertia of the disc about its centre is please tell me how to solve this as soon as possible before test

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4 years ago

Vikas TU
14146 Points
```							From angular momentum conservation,I1 w1 = I2 w2 we get,(I)*45 = [(I) + 0.02*0.04*0.04]*36Solving for I we get,45I – 36I = 1.152 x 10^-3I   = 1.28 x 10^-3where I is the moment of inertia of the diisc.
```
4 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