Question

If the moment of inertia of a uniform solid cylinder about the axis of the cylinder is \( \frac{1}{n} \) times its moment of inertia about an axis passing through its midpoint and perpendicular to its length, then the ratio of the length and radius of the cylinder is

• A. \( \sqrt{2(3n+1)} \)
• B. \( \sqrt{3(2n-1)} \)
• C. \( \sqrt{3(2n+1)} \)
• D. \( \sqrt{2(3n-1)} \)

Hint:

Standard Moment of Inertia formulas: Cylinder (axis): \( \frac{MR^2}{2} \). Cylinder (central transverse): \( M(\frac{R^2}{4} + \frac{L^2}{12}) \). Equate carefully.

Answer 1

The Correct Option is B

Solution:

Let the mass of the uniform solid cylinder be \(M\), radius be \(r\), and length be \(l\). 

We are given that the moment of inertia about the axis of the cylinder is \( \dfrac{1}{n} \) times its moment of inertia about an axis passing through its midpoint and perpendicular to its length. 

Step 1: Write both moments of inertia

Moment of inertia of a solid cylinder about its own axis is: 

\[ I_1=\frac{1}{2}Mr^2 \] Moment of inertia about an axis through its center and perpendicular to its length is: 

\[ I_2=\frac{1}{12}M(3r^2+l^2) \] 
Step 2: Use the given condition

According to the question, 

\[ I_1=\frac{1}{n}I_2 \] Substituting the values: 

\[ \frac{1}{2}Mr^2=\frac{1}{n}\cdot \frac{1}{12}M(3r^2+l^2) \] 
Step 3: Simplify the equation

Multiply both sides by \(12n\): 

\[ 12n\cdot \frac{1}{2}Mr^2 = M(3r^2+l^2) \] \[ 6nMr^2 = M(3r^2+l^2) \] Cancel \(M\): 

\[ 6nr^2 = 3r^2 + l^2 \] \[ l^2 = 6nr^2 - 3r^2 \] \[ l^2 = 3(2n-1)r^2 \] 
Step 4: Find the ratio \( \dfrac{l}{r} \)

\[ \frac{l}{r}=\sqrt{3(2n-1)} \] \[ \frac{l}{r}=\sqrt{6n-3} \] 
So, the ratio of length to radius is: 

\[ \boxed{\sqrt{3(2n-1)}} \]