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

Step 1: Understanding the Question:
We have an Atwood machine with two masses. The connecting wire has a specified breaking stress. We need to find the minimum radius the wire can have without breaking.
Step 2: Key Formula or Approach:
1. First, find the tension `T` in the wire using Newton's second law for the Atwood machine. The acceleration `a` is given by \( a = \frac{(m_2 - m_1)g}{m_1 + m_2} \). The tension can be found from \( T = m_1(g+a) \) or \( T = m_2(g-a) \).
2. Stress is defined as Force per unit Area: \( \sigma = \frac{F}{A} \). Here, the force is the tension `T`, and the area is the cross-sectional area of the wire, \( A = \pi r^2 \).
3. The wire will not break if the stress in it is less than or equal to the breaking stress (\(\sigma_{breaking}\)). The minimum radius corresponds to the case where the stress equals the breaking stress.
Step 3: Detailed Explanation:
Given: \( m_1 = 3 \) kg, \( m_2 = 5 \) kg, \( g = 10 \) m/s\(^2\).
Breaking stress \( \sigma_b = \frac{24}{\pi} \times 10^2 \) N/m\(^2\).
Part 1: Calculate the tension (T) in the wire.
First, find the acceleration of the system:
\[ a = \frac{(m_2 - m_1)g}{m_1 + m_2} = \frac{(5 - 3) \times 10}{5 + 3} = \frac{2 \times 10}{8} = \frac{20}{8} = 2.5 \text{ m/s}^2 \] Now, calculate the tension using the equation of motion for one of the masses. For \(m_2\) (moving down):
\[ m_2g - T = m_2a \implies T = m_2(g - a) \] \[ T = 5(10 - 2.5) = 5(7.5) = 37.5 \text{ N} \] Part 2: Calculate the minimum radius (r).
The stress in the wire is \( \sigma = \frac{T}{A} = \frac{T}{\pi r^2} \).
For the minimum radius, the stress will be equal to the breaking stress:
\[ \sigma_b = \frac{T}{\pi r_{min}^2} \] \[ r_{min}^2 = \frac{T}{\pi \sigma_b} \] Substitute the values for T and \( \sigma_b \):
\[ r_{min}^2 = \frac{37.5}{\pi \left(\frac{24}{\pi} \times 10^2\right)} = \frac{37.5}{24 \times 10^2} = \frac{37.5}{2400} \] \[ r_{min}^2 = \frac{375}{24000} = \frac{1}{64} \text{ m}^2 \] \[ r_{min} = \sqrt{\frac{1}{64}} = \frac{1}{8} \text{ m} \] The options are in cm, so convert the result:
\[ r_{min} = \frac{1}{8} \text{ m} \times \frac{100 \text{ cm}}{1 \text{ m}} = \frac{100}{8} \text{ cm} = 12.5 \text{ cm} \] Step 4: Final Answer:
The minimum radius of the wire is 12.5 cm. This corresponds to option (A).