JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2026
- Hard
- MCQ
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 \(AB\) is ___________\(rad/s^{2}\).

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Answer & Solution
Correct Answer
(B) 11
Step-by-step Solution
Detailed explanation
\(\alpha = \frac{\tau}{I}\)
\(I =\frac{1}{4} mR ^2+ mR ^2+\frac{1}{4} mR ^2+ m (2 R )^2+\frac{ m (3 R )^2}{12}+ m \left(\frac{ R }{2}\right)^2\)
\(=\left(\frac{3}{2}+4+1\right) mR ^2=\frac{13}{2} mR ^2=\frac{13}{2} \times 600 \times 10^2=39 \times 10^4\)
\(\alpha=\frac{43 \times 10^5}{39 \times 10^4} rad / s ^2=\frac{430}{39} rad / s ^2 \approx 11 rad / s ^2\)
\(I =\frac{1}{4} mR ^2+ mR ^2+\frac{1}{4} mR ^2+ m (2 R )^2+\frac{ m (3 R )^2}{12}+ m \left(\frac{ R }{2}\right)^2\)
\(=\left(\frac{3}{2}+4+1\right) mR ^2=\frac{13}{2} mR ^2=\frac{13}{2} \times 600 \times 10^2=39 \times 10^4\)
\(\alpha=\frac{43 \times 10^5}{39 \times 10^4} rad / s ^2=\frac{430}{39} rad / s ^2 \approx 11 rad / s ^2\)
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