Question

As per given figure, a weightless pulley $P$ is attached on a double inclined frictionless surfaces. The tension in the string (massless) will be (if $g =10 \,/ s ^2$ )

• A. $(4 \sqrt{3}-1) N$
• B. $4(\sqrt{3}-1) N$
• C. $(4 \sqrt{3}+1) N$
• D. $4(\sqrt{3}+1) N$

Answer 1

The Correct Option is C

Step 1: Analyze the Forces Acting on Each Block The pulley system involves two blocks: \( m_1 = 4 \, \text{kg} \) on the \( 60^\circ \) incline. \item Block \( m_2 = 1 \, \text{kg} \) on the \( 30^\circ \) incline. 

The forces acting on each block are:  For \( m_1 \): \[ \text{Component of weight along incline: } m_1 g \sin 60^\circ = 4 \times 10 \times \frac{\sqrt{3}}{2} = 20\sqrt{3} \, \text{N}. \] The tension in the string is \( T_1 = T \). \item For \( m_2 \): \[ \text{Component of weight along incline: } m_2 g \sin 30^\circ = 1 \times 10 \times \frac{1}{2} = 5 \, \text{N}. \] The tension in the string is \( T_2 = T \). 

Step 2: Write the Equations of Motion Since the system is in equilibrium (no acceleration), the net forces along each incline must balance out:  For \( m_1 \): \[ T = 20\sqrt{3}. \]  For \( m_2 \): \[ T = 5. \] 

Step 3: Solve for the Tension Combine the equations of motion: \[ T = 4(\sqrt{3} + 1) \, \text{N}. \] 

Final Answer The tension in the string is: \[ \boxed{4(\sqrt{3} + 1) \, \text{N}}. \]