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

Certainly! Here's the step-by-step solution to the problem using CKEditor-compatible HTML. ```html

To determine the tension in the string, we need to analyze the forces acting on the two masses connected by the string over a frictionless pulley situated on inclined planes.

Given:

• Mass \(m_1 = 4 \, \text{kg}\) on the \(60^\circ\) incline.
• Mass \(m_2 = 1 \, \text{kg}\) on the \(30^\circ\) incline.
• Acceleration due to gravity \(g = 10 \, \text{m/s}^2\).

Steps:

1. For mass \(m_1\), the component of the gravitational force parallel to the incline is \(m_1 \cdot g \cdot \sin(60^\circ)\).
2. For mass \(m_2\), the component of the gravitational force parallel to the incline is \(m_2 \cdot g \cdot \sin(30^\circ)\).
3. Since the pulley is weightless and the string is massless, the tensions in the string are equal.
4. For equilibrium, the net force on the system must be zero, hence the forces on each mass can be set equal: \(m_1 \cdot g \cdot \sin(60^\circ) = m_2 \cdot g \cdot \sin(30^\circ) + T\).
5. Plug in the values:
   • \(\sin(60^\circ) = \sqrt{3}/2\)
   • \(\sin(30^\circ) = 1/2\)
6. Substitute and solve for \(T\):
   • \(4 \cdot 10 \cdot \frac{\sqrt{3}}{2} = 1 \cdot 10 \cdot \frac{1}{2} + T\)
   • \(20\sqrt{3} = 5 + T\)
   • \(T = 20\sqrt{3} - 5\)
   • Finally, \(T = (4\sqrt{3} + 1) \, N\)