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Rotational Motion question

2025 · 28 Jan · Shift 1 · Q71
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  5. /2025 · 28 Jan · Shift 1 · Q71

Rotational Motion question

2025 · 28 Jan · Shift 1 · Q71

JEE MainPhysicsRotational MotionNumerical+4 / −1
Two iron solid discs of negligible thickness have radii R1R_1R1​ and R2R_2R2​ and moment of intertia I1I_1I1​ and I2I_2I2​, respectively. For R2=2R1R_2=2 R_1R2​=2R1​, the ratio of I1I_1I1​ and I2I_2I2​ would be 1/x1 / x1/x, where x=‾\mathrm{x}=\underline{\hspace{2cm}}x=​ .
Numerical answer
View written solutionFree

Correct answer: 16

  1. Moment of inertia of a solid disc

For a solid disc about its central axis, I=12MR2I = \frac{1}{2}MR^2I=21​MR2

  1. Mass of the disc in terms of radius

The discs are made of the same material (iron) and have negligible thickness.

So for each disc,

  • density is same,
  • thickness is same,
  • mass is proportional to area.

Hence, M∝R2M \propto R^2M∝R2

Therefore, I=12MR2∝R2⋅R2=R4I = \frac{1}{2}MR^2 \propto R^2 \cdot R^2 = R^4I=21​MR2∝R2⋅R2=R4

So, I∝R4I \propto R^4I∝R4

  1. Use the given relation

Given, R2=2R1R_2 = 2R_1R2​=2R1​

Thus, I2I1=(R2R1)4=24=16\frac{I_2}{I_1} = \left(\frac{R_2}{R_1}\right)^4 = 2^4 = 16I1​I2​​=(R1​R2​​)4=24=16

Therefore, I1I2=116\frac{I_1}{I_2} = \frac{1}{16}I2​I1​​=161​

  1. Compare with the form given in the question

It is given that I1I2=1x\frac{I_1}{I_2} = \frac{1}{x}I2​I1​​=x1​

So, x=16x = 16x=16

Final Answer

16\boxed{16}16​

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