The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is I1. The same rod is bent into a ring and its moment of inertia about a diameter is I 2. If ( I 1/ I 2) is ( x π2/3), then the value of x will be

Question

The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is I1. The same rod is bent into a ring and its moment of inertia about a diameter is I 2. If ( I 1/ I 2) is ( x π2/3), then the value of x will be (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

ℓ=2 π r ⇒ (ℓ/ r )=2 π

( I 1/ I 2)=(2/3)((ℓ/ r ))2 =(2/3) × 4 π2=(8 π2/3) x =8

Q. The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is $I_{1}$ . The same rod is bent into a ring and its moment of inertia about a diameter is $I_{2}$ . If $\frac{I _{1}}{I _{2}}$ is $\frac{x π ^{2}}{3}$ , then the value of $x$ will be_______

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

$\frac{I _{1}}{I _{2}} = \frac{2}{3} (\frac{ℓ}{r})^{2}$
$= \frac{2}{3} \times 4 π^{2} = \frac{8 π ^{2}}{3}$
$x = 8$

Q. The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is I1​. The same rod is bent into a ring and its moment of inertia about a diameter is I2​. If I2​I1​​ is 3xπ2​, then the value of x will be_______

ℓ=2πr⇒rℓ​=2π I2​I1​​=32​(rℓ​)2 =32​×4π2=38π2​ x=8

Q. The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is $I_{1}$ . The same rod is bent into a ring and its moment of inertia about a diameter is $I_{2}$ . If $\frac{I _{1}}{I _{2}}$ is $\frac{x π ^{2}}{3}$ , then the value of $x$ will be_______

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

$\frac{I _{1}}{I _{2}} = \frac{2}{3} (\frac{ℓ}{r})^{2}$
$= \frac{2}{3} \times 4 π^{2} = \frac{8 π ^{2}}{3}$
$x = 8$