Question

The ratio of moment of inertia of a triangle and that of a rectangle having same base and height with respect to their bases will be

• A. \(4:1\)
• B. \(1:4\)
• C. \(12:1\)
• D. \(1:12\)

Hint:

Remember these standard formulas: For a rectangle about its base, \[ I=\frac{bh^3}{3}. \] For a triangle about its base, \[ I=\frac{bh^3}{12}. \] Thus, \[ \boxed{I_{\text{triangle}}:I_{\text{rectangle}}=1:4} \] for the same base and height.

Answer 1

The Correct Option is B

Concept: The moment of inertia of common plane figures about their base are: \[ \boxed{I_{\text{triangle}}=\frac{bh^3}{12}} \] \[ \boxed{I_{\text{rectangle}}=\frac{bh^3}{3}} \] where \[ b=\text{base},\qquad h=\text{height}. \]

Step 1: Write the moment of inertia of the triangle. For a triangle, \[ I_T=\frac{bh^3}{12}. \]

Step 2: Write the moment of inertia of the rectangle. For a rectangle, \[ I_R=\frac{bh^3}{3}. \]

Step 3: Find the required ratio. \[ I_T:I_R = \frac{bh^3}{12}:\frac{bh^3}{3} = \frac1{12}:\frac13. \] Multiplying both terms by \(12\), \[ 1:4. \] Hence, \[ \boxed{I_T:I_R=1:4.} \] Therefore, the correct option is \[ \boxed{(B)\;1:4.} \]