JEE MainPhysics

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+4 marks−1 if incorrectNumericalPrevious-year question

Moment of Inertia of a Dumbbell System of Solid Spheres

Q: Why do we add 2/5mr^2 instead of treating the spheres as point masses?

The spheres have a finite radius (r = 10cm), which is non-negligible compared to the separation (40cm). Therefore, the intrinsic moment of inertia about the sphere's own centre of mass must be included using the parallel axis theorem.

Two identical solid spheres each of mass 2kg and radii 10cm are fixed at the ends of a light rod. The separation between the centres of the spheres is 40cm. The moment of inertia of the system about an axis perpendicular to the rod passing through its middle point is ______ × 10^(-3)kg-m^2

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Step-by-step solutionView

Correct answer

The total moment of inertia of the system about the perpendicular axis passing through the rod's midpoint is $176 \times 10^{- 3} kg \cdot m^{2}$ , so the integer to enter is 176.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

Mass of each solid sphere $m = 2 kg$ , radius of each sphere $r = 10 cm = 0.1 m$ , distance between centres $L = 40 cm = 0.4 m$ , mass of rod is negligible (light rod).

02Find

Moment of inertia $I$ of the system about an axis perpendicular to the rod passing through its midpoint, expressed in the form $N \times 10^{- 3} kg \cdot m^{2}$ .

03Visualise

Two solid spheres are mounted at the ends of a light rod. The axis of rotation passes through the midpoint of the rod, perpendicular to its length. The distance from the axis of rotation to the centre of each sphere is $d = L / 2 = 20 cm = 0.2 m$ .

04Strategise

Use the parallel axis theorem for each solid sphere: $I_{1} = I_{cm} + m d^{2}$ , where $I_{cm} = \frac{2}{5} m r^{2}$ . Since both spheres are identical and placed symmetrically, total moment of inertia is $I = 2 I_{1} = 2 (\frac{2}{5} m r^{2} + m d^{2})$ .

05Execute

$I = 2 [\frac{2}{5} (2) (0.1)^{2} + 2 (0.2)^{2}] = 2 [0.016 + 0.08] = 2 (0.096) = 0.176 kg \cdot m^{2} = 176 \times 10^{- 3} kg \cdot m^{2}$ .

✓Verify

Check units: $kg \cdot m^{2}$ . The point mass approximation gives $2 \times (2 \times {0.2}^{2}) = 0.160 kg \cdot m^{2} = 160 \times 10^{- 3} kg \cdot m^{2}$ . Adding the intrinsic sphere inertia $\frac{4}{5} (2) (0.01) = 0.016 kg \cdot m^{2}$ gives exactly $0.176 kg \cdot m^{2} = 176 \times 10^{- 3} kg \cdot m^{2}$ .

Hints that build this answer step by step

What theorem and formula determine the moment of inertia of one solid sphere about an axis passing through the midpoint of the rod?
$I = \frac{2}{5} M R^{2} + M d^{2}$ , using the parallel-axis theorem where $d$ is the distance from the sphere's centre to the axis.
Given that the separation between the centres of the two spheres is $40 cm$ , what is the distance $d$ from the centre of each sphere to the central axis?
$d = 20 cm = 0.2 m$
With $M = 2 kg$ , $R = 0.1 m$ , and $d = 0.2 m$ , what is the total moment of inertia of both spheres about the axis?
$I_{total} = 176 \times 10^{- 3} kg \cdot m^{2}$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Physics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

Quick checks

Students also ask

Why do we add

\frac{2}{5} m r^{2}

instead of treating the spheres as point masses?

The spheres have a finite radius ( $r = 10 cm$ ), which is non-negligible compared to the separation ( $40 cm$ ). Therefore, the intrinsic moment of inertia about the sphere's own centre of mass must be included using the parallel axis theorem.

Answer

Accepted answer: 176

The total moment of inertia of the system about the perpendicular axis passing through the rod's midpoint is $176 \times 10^{- 3} kg \cdot m^{2}$ , so the integer to enter is 176.

Why each option works or fails

Step-by-step solution

Given: Mass of each solid sphere $m = 2 kg$ , radius of each sphere $r = 10 cm = 0.1 m$ , distance between centres $L = 40 cm = 0.4 m$ , mass of rod is negligible (light rod).
Find: Moment of inertia $I$ of the system about an axis perpendicular to the rod passing through its midpoint, expressed in the form $N \times 10^{- 3} kg \cdot m^{2}$ .
Visualise: Two solid spheres are mounted at the ends of a light rod. The axis of rotation passes through the midpoint of the rod, perpendicular to its length. The distance from the axis of rotation to the centre of each sphere is $d = L / 2 = 20 cm = 0.2 m$ .
Strategise: Use the parallel axis theorem for each solid sphere: $I_{1} = I_{cm} + m d^{2}$ , where $I_{cm} = \frac{2}{5} m r^{2}$ . Since both spheres are identical and placed symmetrically, total moment of inertia is $I = 2 I_{1} = 2 (\frac{2}{5} m r^{2} + m d^{2})$ .
Execute: $I = 2 [\frac{2}{5} (2) (0.1)^{2} + 2 (0.2)^{2}] = 2 [0.016 + 0.08] = 2 (0.096) = 0.176 kg \cdot m^{2} = 176 \times 10^{- 3} kg \cdot m^{2}$ .
Verify: Check units: $kg \cdot m^{2}$ . The point mass approximation gives $2 \times (2 \times {0.2}^{2}) = 0.160 kg \cdot m^{2} = 160 \times 10^{- 3} kg \cdot m^{2}$ . Adding the intrinsic sphere inertia $\frac{4}{5} (2) (0.01) = 0.016 kg \cdot m^{2}$ gives exactly $0.176 kg \cdot m^{2} = 176 \times 10^{- 3} kg \cdot m^{2}$ .

Moment of Inertia of a Dumbbell System of Solid Spheres

What feels right?

The total moment of inertia of the system about the perpendicular axis passing through the rod's midpoint is $176 \times 10^{- 3} kg \cdot m^{2}$ , so the integer to enter is 176.

Why each option works or fails

Solution

01Given

02Find

03Visualise

04Strategise

05Execute

✓Verify

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The total moment of inertia of the system about the perpendicular axis passing through the rod's midpoint is 176×10−3 kg⋅m2176×10−3 kg⋅m2, so the integer to enter is 176.

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The total moment of inertia of the system about the perpendicular axis passing through the rod's midpoint is $176 \times 10^{- 3} kg \cdot m^{2}$ , so the integer to enter is 176.

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An ideal transformer with purely resistive load operates at $12 kV$ on the primary…

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