JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
Three masses \(m_1 = 4\) kg, \(m_2 = 4\) kg and \(m_3 = 6\) kg are suspended from a fixed smooth frictionless pulley as shown in the figure below. The value of \(T_1/T_2\) is _____.
(take \(g = 10\) m/s\(^2\))

- A \(5/3\)
- B \(2/3\)
- C \(3/5\)
- D \(2/5\)
Answer & Solution
Correct Answer
(A) \(5/3\)
Step-by-step Solution
Detailed explanation
Let the acceleration of the system be \(a\). The total mass on the right side is \(m_2 + m_3 = 4 + 6 = 10\) kg, and the mass on the left side is \(m_1 = 4\) kg. Since \(10\) kg \(> 4\) kg, the right side accelerates downwards and the left side accelerates upwards. The common…
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