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Physics Question on Stress and Strain

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

A

2mg3A\frac{2 m g}{3 A}

B

4mg3A\frac{4 m g}{3 A}

C

mgA\frac{m g}{A}

D

3mg4A\frac{3 mg}{4 A}

Answer

4mg3A\frac{4 m g}{3 A}

Explanation

Solution

Tension in the wire, T≈(2mMm+M)gT \approx \left(\frac{2 m M}{m+M}\right) g Stress =Force/TensionArea=2mMA(m+M)g\frac{Force/Tension }{Area}=\frac{2 m M}{A \left(m+M\right)} g =2(m×2m)A(m+2m)=\frac{2\left(m \times2m\right)}{A \left(m+2 m\right)} (M = 2mgiven) =4m23mAg=4mg3A=\frac{4 m^{2}}{3 m A} g =\frac{4 m g}{3 A}