For a uniform cylinder of length L and radius R the moment of inertia is I₁. Now for similar situation but length L/2 and radius R/2 moment of inertia is I₂.

Find I₁ / I₂.

The correct answer is 32.

Step-by-step Explanation:
Let us consider a uniform solid cylinder of mass $M$, length $L$, and radius $R$. As observed in the provided image, the axis of rotation for both cylinders is perpendicular to their length and passes through their respective centers of mass.

The moment of inertia of a uniform solid cylinder about an axis perpendicular to its length and passing through its center is given by the formula:
$I=M(\frac{R^2}{4}+\frac{L^2}{12})$

Since both cylinders are made of the same uniform material, they have the same mass density $\rho$. The mass of a cylinder is equal to its density multiplied by its volume:
$M=\rho V=\rho \pi R^{2}L$

1. For the first cylinder:
Length = $L$, Radius = $R$
The mass is:
$M_1=\rho \pi R^{2}L$
The moment of inertia is:
$I_1=M_1(\frac{R^2}{4}+\frac{L^2}{12})$

2. For the second cylinder:
Length = $\frac{L}{2}$, Radius = $\frac{R}{2}$
The mass is:
$M_2=\rho \pi {\left(\frac{R}{2}\right)}^2\left(\frac{L}{2}\right)=\rho \pi \frac{R^2}{4}·\frac{L}{2}=\frac{\rho \pi R^{2}L}{8}=\frac{M_1}{8}$
The moment of inertia is:
$I_2=M_2[\frac{{(R/2)}^2}{4}+\frac{{(L/2)}^2}{12}]$

Substitute $M_2=\frac{M_1}{8}$ into the equation:
$I_2=\frac{M_1}{8}[\frac{R^2}{16}+\frac{L^2}{48}]$

Factor out $\frac{1}{4}$ from the terms inside the brackets:
$I_2=\frac{M_1}{8}·\frac{1}{4}[\frac{R^2}{4}+\frac{L^2}{12}]$

$I_2=\frac{1}{32}{M}_1[\frac{R^2}{4}+\frac{L^2}{12}]=\frac{I_1}{32}$

3. Finding the ratio:
$\frac{I_1}{I_2}=32$