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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\(^2\))

  1. A \(5/3\)
  2. B \(2/3\)
  3. C \(3/5\)
  4. D \(2/5\)
Verified Solution

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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