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JEE Main 2026 Jan 28 Shift 1, Physics Q4: Moment of Inertia

JEE Main2026Jan 28, Shift 1Physics
Q.

Two circular discs of radius each 10 cm are joined at their centres by a rod of length 30 cm and mass 600 gm as shown in figure.

If the mass of each disc is 600 gm and applied torque between two discs is dyne cm, the angular acceleration of the discs about the given axis is ______ rad/s.

  1. A

    22

  2. B

    11

  3. C

    100

  4. D

    27

Solution

$I = \dfrac{1}{4}mR^2 + mR^2 + \dfrac{1}{4}mR^2 + m(2R)^2 + \dfrac{m(3R)^2}{12} + m\left(\dfrac{R}{2}\right)^2$

$= \left( \dfrac{3}{2} + 4 + 1 \right)mR^2 = \dfrac{13}{2}mR^2 = \dfrac{13}{2}\times600\times10^2 = 39\times10^4$

$\alpha = \dfrac{43\times10^5}{39\times10^4} \text{ rad/s}^2 = \dfrac{430}{39} \text{ rad/s}^2 \approx 11\text{ rad/s}^2$

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