Question

A thin rod has mass \(100\,g\) and length \(0.3\,m\). Find its moment of inertia about an axis passing through its centre of mass and perpendicular to its length.

• A. \(2.5\times10^{-4}\,kg\,m^2\)
• B. \(5.0\times10^{-4}\,kg\,m^2\)
• C. \(7.5\times10^{-4}\,kg\,m^2\)
• D. \(1.0\times10^{-3}\,kg\,m^2\)

Hint:

For a uniform rod: \[ I_{centre}=\frac{1}{12}ML^2, \qquad I_{end}=\frac{1}{3}ML^2. \] Always identify the axis carefully before choosing the formula.

Answer 1

The Correct Option is C

Concept: The moment of inertia of a thin uniform rod about an axis passing through its centre and perpendicular to its length is \[ I=\frac{1}{12}ML^2 \] where \(M\) is the mass of the rod and \(L\) is its length.

Step 1: Convert the given mass into SI units. Given, \[ M=100\,g \] Since \[ 1000\,g=1\,kg \] therefore, \[ M=0.1\,kg \] Also, \[ L=0.3\,m \]

Step 2: Apply the moment of inertia formula. \[ I=\frac{1}{12}ML^2 \] Substituting the values, \[ I=\frac{1}{12}(0.1)(0.3)^2 \] \[ I=\frac{1}{12}(0.1)(0.09) \] \[ I=\frac{0.009}{12} \] \[ I=7.5\times10^{-4}\,kg\,m^2 \]

Step 3: Final answer. \[ \boxed{I=7.5\times10^{-4}\,kg\,m^2} \] Hence, \[ \boxed{(C)} \]