Question

Choose the correct answer from the options given below : 

• A. (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
• B. (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
• C. (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)
• D. (a)-(ii), (b)-(iii), (c)-(i), (d)-(iv)

Hint:

To find the MI about an end from the MI about the center, you can use the Parallel Axis Theorem: $I_{end} = I_{center} + Md^2$. For a rod, $d=L/2$, so $I_{end} = \frac{ML^2}{12} + M(\frac{L}{2})^2 = \frac{ML^2}{12} + \frac{ML^2}{4} = \frac{ML^2}{3}$.

Answer 1

The Correct Option is B

Let's calculate the Moment of Inertia (MI) for each case. The standard formulas for a rod are:
- About a perpendicular axis through the center: $I_{center} = \frac{1}{12} \times (\text{Mass}) \times (\text{Length})^2$
- About a perpendicular axis through an end: $I_{end} = \frac{1}{3} \times (\text{Mass}) \times (\text{Length})^2$
(a) Mass = M, Length = L, axis through midpoint.
$I_a = \frac{1}{12}ML^2$. This matches (iii).
(b) Mass = 2M, Length = L, axis through one end.
$I_b = \frac{1}{3}(2M)(L)^2 = \frac{2ML^2}{3}$. This matches (iv).
(c) Mass = M, Length = 2L, axis through midpoint.
$I_c = \frac{1}{12}(M)(2L)^2 = \frac{1}{12}M(4L^2) = \frac{ML^2}{3}$. This matches (ii).
(d) Mass = 2M, Length = 2L, axis through one end.
$I_d = \frac{1}{3}(2M)(2L)^2 = \frac{1}{3}(2M)(4L^2) = \frac{8ML^2}{3}$. This matches (i).
The correct matching is: (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i). This corresponds to option (B).