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MCQMediumJEE 2026Moment of Inertia & Radius of Gyration

JEE Physics 2026 Question with Solution

The moment of inertia of a square loop made of four uniform solid cylinders, each having radius RR and length LL (R≤LR \le L) about an axis passing through the mid points of opposite sides, is (Take the mass of the entire loop as MM) :

  • A

    (34)MR2+(16)ML2\left(\frac{3}{4}\right)MR^2 + \left(\frac{1}{6}\right)ML^2

  • B

    (38)MR2+(712)ML2\left(\frac{3}{8}\right)MR^2 + \left(\frac{7}{12}\right)ML^2

  • C

    (38)MR2+(16)ML2\left(\frac{3}{8}\right)MR^2 + \left(\frac{1}{6}\right)ML^2

  • D

    (34)MR2+(712)ML2\left(\frac{3}{4}\right)MR^2 + \left(\frac{7}{12}\right)ML^2

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: A square loop is formed by four identical solid cylinders. The total mass of the loop is MM, so mass of each cylinder is m=M/4m = M/4.

Find: The moment of inertia about an axis passing through the midpoints of opposite sides.

The axis coincides with the longitudinal axis of two cylinders and is transverse to the other two.

Use the standard moments of inertia:

Iown=12mR2I_{\text{own}} = \frac{1}{2}mR^2

for a cylinder about its own axis, and

Itrans=14mR2+112mL2I_{\text{trans}} = \frac{1}{4}mR^2 + \frac{1}{12}mL^2

for a cylinder about a transverse axis through its center.

Also apply the parallel axis theorem:

I=Icm+md2I = I_{cm} + md^2

Detailed Calculation

For the two cylinders parallel to the axis,

I1=2×(12mR2)=mR2I_1 = 2 \times \left( \frac{1}{2}mR^2 \right) = mR^2

For the two cylinders perpendicular to the axis, the distance of their centers from the axis is

d=L2d = \frac{L}{2}

So,

I2=2×[(14mR2+112mL2)+m(L2)2]I_2 = 2 \times \left[ \left( \frac{1}{4}mR^2 + \frac{1}{12}mL^2 \right) + m\left(\frac{L}{2}\right)^2 \right]

Now simplify:

I2=2×[14mR2+112mL2+14mL2]=2×[14mR2+412mL2]=12mR2+23mL2\begin{aligned} I_2 &= 2 \times \left[ \frac{1}{4}mR^2 + \frac{1}{12}mL^2 + \frac{1}{4}mL^2 \right] \\ &= 2 \times \left[ \frac{1}{4}mR^2 + \frac{4}{12}mL^2 \right] \\ &= \frac{1}{2}mR^2 + \frac{2}{3}mL^2 \end{aligned}

Therefore, total moment of inertia is

I=I1+I2=mR2+12mR2+23mL2=32mR2+23mL2\begin{aligned} I &= I_1 + I_2 \\ &= mR^2 + \frac{1}{2}mR^2 + \frac{2}{3}mL^2 \\ &= \frac{3}{2}mR^2 + \frac{2}{3}mL^2 \end{aligned}

Substitute

m=M4m = \frac{M}{4}

Then,

I=32(M4)R2+23(M4)L2=38MR2+16ML2\begin{aligned} I &= \frac{3}{2}\left(\frac{M}{4}\right)R^2 + \frac{2}{3}\left(\frac{M}{4}\right)L^2 \\ &= \frac{3}{8}MR^2 + \frac{1}{6}ML^2 \end{aligned}

Therefore, the moment of inertia is 38MR2+16ML2\frac{3}{8}MR^2 + \frac{1}{6}ML^2. The correct option is C.

Common mistakes

  • Using MM directly as the mass of each cylinder is incorrect because the loop has four identical cylinders. First take mass of each cylinder as m=M/4m = M/4, then add their contributions.

  • For the side cylinders perpendicular to the axis, using moment of inertia about their own longitudinal axis is wrong. Their relevant central moment is the transverse one, Itrans=14mR2+112mL2I_{\text{trans}} = \frac{1}{4}mR^2 + \frac{1}{12}mL^2.

  • Ignoring the parallel axis theorem for the two perpendicular cylinders gives an underestimated value. Since their centers are at distance L/2L/2 from the axis, add m(L/2)2m(L/2)^2 for each cylinder.

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