Question

Two blocks of masses $m$ and $M$ are connected by means of a metal wire of cross-sectional area $A$ passing over a frictionless fixed pulley as shown in the figure. The system is then released. If $M \,= \,2 m$, then the stress produced in the wire is:

• A. $\frac{2 m g}{3 A}$
• B. $\frac{4 m g}{3 A}$
• C. $\frac{m g}{A}$
• D. $\frac{3 mg}{4 A}$

Answer 1

The Correct Option is B

To determine the stress produced in the wire, we need to consider the forces acting on the system of blocks. Given that block \(M = 2m\) and both blocks are connected over a frictionless pulley, we can analyze the tension and acceleration: 

Consider the forces on each block:

• For block \(m\): The forces acting are its weight \(mg\) downward and tension \(T\) upward.
• For block \(M = 2m\): The forces acting are its weight \(2mg\) downward and tension \(T\) upward.

Using Newton's second law, we can write the equations of motion:

• For block \(m\) (upward acceleration \(a\)):
• For block \(2m\) (downward acceleration \(a\)):

We can solve these two equations simultaneously to find \(T\) and \(a\):

• Add the two equations:
• Simplifying gives:
• Thus, the acceleration is:

Substitute for \(a\) in either original equation to find \(T\). Using \(T - mg = ma\):

\[T = mg + ma = mg + m\left(\frac{g}{3}\right) = mg + \frac{mg}{3} = \frac{3mg}{3} + \frac{mg}{3} = \frac{4mg}{3}\]

The stress in the wire is given by the formula:

\[\text{Stress} = \frac{\text{Tension}}{\text{Area}} = \frac{T}{A} = \frac{\frac{4mg}{3}}{A} = \frac{4mg}{3A}\]

Thus, the stress produced in the wire is \(\frac{4mg}{3A}\).

Therefore, the correct answer is the option \(\frac{4mg}{3A}\).