The momentum of inertia of a uniform thin rod about a perpendicular axis passing through on end is I₁. The same rod is into a ring and its moment of inertia about a diameter is I₂. If $\frac{I_{1}}{I_{2}} i s \frac{X π^{2}}{3},$ then the value of x will be__________.

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Alok Kumar Singh | Contributor-Level 10

4 months ago

According to question, we can write

$l = 2 π r \Rightarrow r = \frac{l}{2 π}$

$I_{1} = \frac{m l^{2}}{3}, a n d I_{2} = \frac{m r^{2}}{2} = \frac{m l^{2}}{8 π^{2}}$

$\Rightarrow \frac{I_{1}}{I_{2}} = \frac{m l^{2}}{3} * \frac{8 π^{2}}{m l^{2}} = \frac{8 π^{2}}{3}$

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The momentum of inertia of a uniform thin rod about a perpendicular axis passing through on end is I1. The same rod is into a ring and its moment of inertia about a diameter is I2. If $\frac{I_{1}}{I_{2}} i s \frac{X π^{2}}{3},$ $\frac{_{}}{_{}} \frac{^{}}{}$ then the value of x will be__________.

Vishal Baghel

According to question, we can write

$l = 2 π r \Rightarrow r = \frac{l}{2 π}$

$I_{1} = \frac{m l^{2}}{3}, a n d I_{2} = \frac{m r^{2}}{2} = \frac{m l^{2}}{8 π^{2}}$

$\Rightarrow \frac{I_{1}}{I_{2}} = \frac{m l^{2}}{3} \times \frac{8 π^{2}}{m l^{2}} = \frac{8 π^{2}}{3}$

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The momentum of inertia of a uniform thin rod about a perpendicular axis passing through on end is I1. The same rod is into a ring and its moment of inertia about a diameter is I2. If I 1 I 2 i s X π 2 3 , then the value of x will be__________.

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The momentum of inertia of a uniform thin rod about a perpendicular axis passing through on end is I₁. The same rod is into a ring and its moment of inertia about a diameter is I₂. If $\frac{I_{1}}{I_{2}} i s \frac{X π^{2}}{3},$ then the value of x will be__________.