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Question

Three frictionless pulleys with rope attachment are in a static equilibrium as shown in the figure. The mass $m_1$ and $m_2$, in kg, respectively are

The correct answer is
50, 100

To determine the masses \(m_1\) and \(m_2\), we analyze the forces in static equilibrium with the help of the given pulley system.

In this system, we have:

  • A mass of 100 kg attached to the left-most pulley.
  • Two other masses \(m_1\) and \(m_2\) attached to the system of pulleys and rope.

 

The tension in the rope throughout is consistent due to the frictionless nature of the pulleys. Let's define the tension in the rope as \(T\).

  1. For the 100 kg mass:
    • The downward force due to gravity is \(100g\).
    • Since the system is in equilibrium, the tension \(T\) in the rope must balance this force: \(T = 100g\).
  2. Analyzing the mass \(m_1\):
    • Considering the tension in the rope, we have \(T = m_1 \times g\), so \(m_1 \times g = 100g\).
    • Cancel out \(g\) from both sides to get \(m_1 = 100\) kg.
  3. Analyzing the mass \(m_2\):
    • The tension in the rope supporting \(m_2\) and \(m_1\) is the same, so the effective tension here is shared between two strands of rope: \(T = 2T_1\).\)
    • The tension in the rope at \(m_2\) is then \(T_1 = 50g\).\)
    • This implies \(m_2 \times g = 50g\), and therefore, \(m_2 = 50\) kg.

The calculated values for the masses are \(m_1 = 100\) kg and \(m_2 = 50\) kg.

Conclusion: The correct option is 50, 100. Thus, the mass \(m_1 = 100\) kg and the mass \(m_2 = 50\) kg.

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