TS EAMCET · Physics · Rotational Motion
If a disc of mass \(M\) and radius \(R\) rotates with an angular acceleration \(a_{\text {, }}\) of the torque acting on the disc is
- A \(M R^2 a\)
- B \(\frac{M R^2 a}{2}\)
- C \(\frac{2 M R^2 a}{5}\)
- D \(\frac{M R^2 a}{12}\)
Answer & Solution
Correct Answer
(B) \(\frac{M R^2 a}{2}\)
Step-by-step Solution
Detailed explanation
\(\therefore\) Moment of inertia of disc, \(Y=\frac{1}{2} M R^2\) Angular acceleration, \(\alpha=a\) (given) So, torque \(\tau=Y \alpha\), \[ \tau=\left(\frac{1}{2} M R^2\right)\langle a\rangle \Rightarrow \tau=\frac{M R^2 a}{2} \]
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