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MCQEasyJEE 2026Moment of Inertia & Radius of Gyration

JEE Physics 2026 Question with Solution

Two identical thin rods of mass MM kg and length LL m are connected as shown in figure. Moment of inertia of the combined rod system about an axis passing through point PP and perpendicular to the plane of the rods is x12ML2\frac{x}{12} ML^2 kg m2^2. The value of xx is _____.

A T-shaped arrangement of two identical thin rods, with point P at the top end of the vertical rod, and the horizontal rod attached at the lower end midpoint.
  • A

    1212

  • B

    1515

  • C

    1717

  • D

    2121

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Two identical thin rods each have mass MM and length LL. The axis passes through point PP and is perpendicular to the plane of the rods.

Find: The value of xx in

I=x12ML2I = \frac{x}{12} ML^2

Step 1: System Configuration

Rod 1 is vertical with mass MM and length LL, and the axis is at its end PP.

Rod 2 is horizontal with mass MM and length LL, attached at its midpoint to the top of Rod 1. The distance from PP to the center of Rod 2 is LL.

Step 2: Moment of inertia of Rod 1 about end PP

I1=ML23=4ML212I_1 = \frac{ML^2}{3} = \frac{4ML^2}{12}

Step 3: Moment of inertia of Rod 2 about PP

About its own center, for an axis perpendicular to the rod,

ICM=ML212I_{CM} = \frac{ML^2}{12}

Using the Parallel Axis Theorem,

I2=ICM+Md2=ML212+M(L)2=13ML212I_2 = I_{CM} + Md^2 = \frac{ML^2}{12} + M(L)^2 = \frac{13ML^2}{12}

Step 4: Total moment of inertia

Itotal=I1+I2=4ML212+13ML212=17ML212I_{\text{total}} = I_1 + I_2 = \frac{4ML^2}{12} + \frac{13ML^2}{12} = \frac{17ML^2}{12}

Comparing with

x12ML2\frac{x}{12} ML^2

we get

x=17x = 17

Therefore, the correct option is C.

Using Parallel Axis Theorem Carefully

Given: A T-shaped system of two identical thin rods, each of mass MM and length LL.

Find: The coefficient xx in the expression

I=x12ML2I = \frac{x}{12} ML^2

For the vertical rod, the axis passes through one end, so its moment of inertia is directly

I1=ML23I_1 = \frac{ML^2}{3}

To combine terms later,

I1=4ML212I_1 = \frac{4ML^2}{12}

For the horizontal rod, first take the moment of inertia about its own center:

ICM=ML212I_{CM} = \frac{ML^2}{12}

Its center lies at a distance

d=Ld = L

from point PP.

Applying the Parallel Axis Theorem,

I2=ICM+Md2I_2 = I_{CM} + Md^2 I2=ML212+ML2=ML212+12ML212=13ML212I_2 = \frac{ML^2}{12} + ML^2 = \frac{ML^2}{12} + \frac{12ML^2}{12} = \frac{13ML^2}{12}

Now add both contributions:

Itotal=4ML212+13ML212=17ML212I_{\text{total}} = \frac{4ML^2}{12} + \frac{13ML^2}{12} = \frac{17ML^2}{12}

Hence,

x=17x = 17

Therefore, the value of xx is 1717.

Common mistakes

  • Using ML212\frac{ML^2}{12} for the vertical rod about point PP is incorrect because ML212\frac{ML^2}{12} is about the rod's center. Since the axis is through an end, use ML23\frac{ML^2}{3} instead.

  • Treating the horizontal rod as only a point mass at distance LL misses its own spin inertia about its center. You must add both terms: ICMI_{CM} and Md2Md^2.

  • Taking the distance in the Parallel Axis Theorem as L2\frac{L}{2} is wrong because the center of the horizontal rod is at the top junction, which is at distance LL from point PP. Use d=Ld = L.

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