Two iron solid discs of negligible thickness have radii R1 and R2 and moment of intertia I1 and I2 , respectively. For R2 = 2R1 , the ratio of I1 and I2 would be 1/x, where x = _______

GivenR2=2R1 M1=σ×πR12=Mo M2=σ×πR22=Mo M2=σ×πR22=σ×π[2R1]2=σ×4πR12=4MoI1I2=M1R122M2R222=M1R12M2R22=14×14=116

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Two iron solid discs of negligible thickness have radii R₁ and R₂ and moment of intertia I₁ and I₂ , respectively. For R₂ = 2R₁ , the ratio of I₁ and I₂ would be 1/x, where x = _______

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answer is 16.

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Detailed Solution

$\begin{aligned} Given R_{2} = 2 R_{1} \\ M_{1} = σ \times π R_{1}^{2} = M_{o} \\ M_{2} = σ \times π R_{2}^{2} = M_{o} \\ M_{2} = σ \times π R_{2}^{2} = σ \times π {[2 R_{1}]}^{2} = σ \times 4 π R_{1}^{2} = 4 M_{o} \end{aligned}$
$\frac{I_{1}}{I_{2}} = \frac{\frac{M_{1} R_{1}^{2}}{2}}{\frac{M_{2} R_{2}^{2}}{2}} = \frac{M_{1} R_{1}^{2}}{M_{2} R_{2}^{2}} = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$