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`        Plz help me solve this question and also tell me how to proceed with reshaping and recasting question.. `
5 months ago

```							Considering the first case:As the disc is reshaped hence the moment of inertia of disc with the axis passins through centroidal(centre) and perpendicular to the plane will be I1=MR²/2The disc is recasted into a ring with same axis position but radius 2rHence letI2=M(2R)²Ase we know radius of gyration isI=MK²Hence k=√I/MHence factor will be √I1/M÷√I2/M=√MR²/2M÷√4MR²/M=R/√2÷√2R=1/2√2If we take I2/I1 then We get ans as 2√2
```
5 months ago
```							Considering dering the first case:As the disc is reshaped hence the moment of inertia of disc with the axis passins through centroidal(centre) and perpendicular to the plane will be I1=MR²/2The disc is recasted into a ring with same axis position but radius 2rHence letI2=M(2R)²Ase we know radius of gyration isI=MK²Hence k=√I/MHence factor will be √I1/M÷√I2/M=√MR²/2M÷√4MR²/M=R/√2÷√2R=1/2√2If we take I2/I1 then We get ans as 2√2
```
5 months ago
```							  the moment of inertia of disc with the axis passins through centroidal(centre) and perpendicular to the plane will be I1=MR²/2The disc is recasted into a ring with same axis position but radius 2r I2=M(2R)². Ase we know radius of gyration isI=MK²Hence k=√I/MHence factor will be √I1/M÷√I2/M=√MR²/2M÷√4MR²/M=R/√2÷√2R=1/2√2If we take I2/I1 then We get ans as 2√2
```
5 months ago
```							 As the disc is reshaped hence the moment of inertia of disc with the axis passins through centroidal(centre) and perpendicular to the plane will be I1=MR²/2The disc is recasted into a ring with same axis position but radius 2rletI2=M(2R)²Ase we know radius of gyration isI=MK²Hence k=√I/MHence factor will be √I1/M÷√I2/M=√MR²/2M÷√4MR²/M=R/√2÷√2R=1/2√2If we take I2/I1 then We get ans as 2√2
```
5 months ago
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