JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
Two masses 400 g and 350 g are suspended from the ends of a light string passing over a heavy pulley of radius 2 cm . When released from rest the heavier mass is observed to fall 81 cm in 9 s . The rotational inertia of the pulley is _________ \(kg \cdot m ^2\).\(\left( g =9.8 m / s ^2\right)\)
- A \( 9.5\times10^{-3} \)
- B \( 4.75\times10^{-3} \)
- C \( 1.86\times10^{-2} \)
- D \( 8.3\times10^{-3} \)
Answer & Solution
Correct Answer
(A) \( 9.5\times10^{-3} \)
Step-by-step Solution
Detailed explanation
\(S=u t+\frac{1}{2} a t^2\) \(a =\frac{2 s}{ t ^2}=\frac{2 \times 0.81}{81}=0.02 m / s ^2\) \(m _1 g- T _1= m _1 a\) \(T _2- m _2 g= m _2 a\) \(\left( T _1- T _2\right) R = I \cdot \frac{ a }{ R }\) \(\therefore a=\frac{\left(m_1-m_2\right) g}{m_1+m_2+\frac{I}{R^2}}\)…
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