Question

Two masses \( m_1 \) and \( m_2 \) are connected by a light string passing over a smooth pulley. When set free, \( m_1 \) moves downwards by 3 m in 3 s. The ratio of \( \frac{m_1}{m_2} \) is \((g = 10 \text{ m/s}^2)\).

• A. \( \frac{9}{7} \)
• B. \( \frac{8}{7} \)
• C. \( \frac{10}{7} \)
• D. \( \frac{15}{13} \)

Hint:

For pulley systems with connected masses, use Newton’s Second Law for both masses and solve for acceleration first before determining mass ratios.

Answer 1

The Correct Option is B

Step 1: Determine Acceleration Using the equation of motion: \[ s = ut + \frac{1}{2} a t^2 \] Since the mass starts from rest (\( u = 0 \)), we substitute \( s = 3 \) m and \( t = 3 \) s: \[ 3 = \frac{1}{2} a (3)^2 \] \[ 3 = \frac{9}{2} a \] \[ a = \frac{6}{9} = \frac{2}{3} \text{ m/s}^2 \] Step 2: Apply Newton's Second Law For \( m_1 \): \[ m_1 g - T = m_1 a \] \[ m_1 (10) - T = m_1 \left( \frac{2}{3} \right) \] \[ 10m_1 - T = \frac{2}{3} m_1 \] For \( m_2 \): \[ T - m_2 g = m_2 a \] \[ T - 10 m_2 = m_2 \left( \frac{2}{3} \right) \] \[ T = 10m_2 + \frac{2}{3} m_2 \] Step 3: Solve for \( \frac{m_1}{m_2} \) Equating both expressions for \( T \): \[ 10m_1 - \frac{2}{3} m_1 = 10m_2 + \frac{2}{3} m_2 \] \[ 10(m_1 - m_2) = \frac{2}{3} (m_1 + m_2) \] Multiplying by 3: \[ 30(m_1 - m_2) = 2(m_1 + m_2) \] \[ 30m_1 - 30m_2 = 2m_1 + 2m_2 \] \[ 30m_1 - 2m_1 = 30m_2 + 2m_2 \] \[ 28m_1 = 32m_2 \] \[ \frac{m_1}{m_2} = \frac{32}{28} = \frac{8}{7} \]