Question

Two blocks of masses 3 kg and 5 kg are connected by a metal wire going over a smooth pulley. The breaking stress of the metal is \( \frac{24}{\pi} \times 10^2 \) Nm\(^{-2}\). What is the minimum radius of the wire ? 
(take g=10 ms\(^{-2}\)) 

 

• A. 12.5 cm
• B. 125 cm
• C. 1250 cm
• D. 1.25 cm

Hint:

In problems combining dynamics and material properties, solve the dynamics part first to find the forces involved (like tension). Then, use these forces in the material property equations (like stress = force/area) to find the required dimension.

Answer 1

The Correct Option is A

To solve this problem, we need to determine the minimum radius of the wire that can sustain the maximum tension without breaking. The breaking stress of the wire is given, and we'll use the formula for tensile stress.

1. Understanding the System:

Two blocks with masses 3 kg and 5 kg are connected by a wire passing over a smooth pulley. The weight of the 5 kg block is greater, so it will descend, applying tension in the wire.

2. Calculating the Tension in the Wire:

The system will accelerate due to the gravitational force acting on the masses. The net force on the blocks is:

F = (m_2 - m_1) \cdot g = (5 - 3) \cdot 10 = 20 \, \text{N}

Using Newton’s second law, the acceleration a is:

a = \frac{F}{m_1 + m_2} = \frac{20}{8} = 2.5 \, \text{ms}^{-2}

The tension T in the wire connecting the blocks is given by:

For the 3 kg block: T - 3g = 3a \implies T = 3(10) + 3(2.5) = 37.5 \, \text{N}

3. Calculating Minimum Radius:

The breaking stress is related to tension and area:

\text{Breaking Stress} = \frac{T}{A}

The cross-sectional area A of the wire for a circular cross-section is \pi r^2. Therefore:

\frac{24}{\pi} \times 10^2 = \frac{37.5}{\pi r^2}

Solving for r^2:

24 \times 100 = \frac{37.5}{r^2}

r^2 = \frac{37.5}{2400}

r^2 = \frac{1}{64}

r = \frac{1}{8} = 0.125 \, \text{m} = 12.5 \, \text{cm}

Therefore, the minimum radius of the wire is 12.5 cm.