Q57 / 75 · JEE Main 2026 (28 January Shift 1) · Physics
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 $43 \times 10^{5}$ dyne. cm, the angular acceleration of the discs about the given axis $A B$ is $\_\_\_\_$ $\mathrm{rad} / \mathrm{s}^{2}$.
- Correct answerA11
- B22
- C100
- D27
Solution
$\alpha = \dfrac{\tau}{I}$
$I = \dfrac{1}{4}mR^2 + mR^2 + \dfrac{1}{4}mR^2 + m(2R)^2 + \dfrac{m(3R)^2}{12} + m\left(\dfrac{R}{2}\right)^2$
$= \left(\dfrac{3}{2} + 4 + 1\right)mR^2 = \dfrac{13}{2}mR^2 = \dfrac{13}{2} \times 600 \times 10^2 = 39 \times 10^4$
$\alpha = \dfrac{43 \times 10^5}{39 \times 10^4} \text{ rad/s}^2 = \dfrac{430}{39} \text{ rad/s}^2 \approx 11 \text{ rad/s}^2$
$I = \dfrac{1}{4}mR^2 + mR^2 + \dfrac{1}{4}mR^2 + m(2R)^2 + \dfrac{m(3R)^2}{12} + m\left(\dfrac{R}{2}\right)^2$
$= \left(\dfrac{3}{2} + 4 + 1\right)mR^2 = \dfrac{13}{2}mR^2 = \dfrac{13}{2} \times 600 \times 10^2 = 39 \times 10^4$
$\alpha = \dfrac{43 \times 10^5}{39 \times 10^4} \text{ rad/s}^2 = \dfrac{430}{39} \text{ rad/s}^2 \approx 11 \text{ rad/s}^2$
Hence, the correct answer is option: 'A'.