Two blocks of masses m and M are connected by means of a metal wire...

Two blocks of masses

\mathrm{m}

and

\mathrm{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

\mathrm{M}=2 \mathrm{~m}

, then the stress produced in the wire is: Two blocks of masses m and M are connected by means of a metal wire of cross-sectional area A passing over a

Options

A $\frac{2 \mathrm{mg}}{3 \mathrm{~A}}$
B $\frac{4 \mathrm{mg}}{3 \mathrm{~A}}$
C $\frac{\mathrm{mg}}{\mathrm{A}}$
D $\frac{3 \mathrm{mg}}{4 \mathrm{~A}}$

Correct answer

B. $\frac{4 \mathrm{mg}}{3 \mathrm{~A}}$

Step-by-step solution

Tension in the wire,

\mathrm{T}=\left(\frac{2 \mathrm{mM}}{\mathrm{m}+\mathrm{M}}\right) \mathrm{g}

\begin{aligned} \text{ Stress } &=\frac{\text{ Force } / \text{ Tension }}{\text{ Area }}=\frac{2 \mathrm{mM}}{\mathrm{A}(\mathrm{m}+\mathrm{M})} \mathrm{g} \\ &=\frac{2(\mathrm{ m} \times 2 \mathrm{ m}) \mathrm{g}}{\mathrm{A}(\mathrm{m}+2 \mathrm{ m})}(\mathrm{M}=2 \mathrm{ m} \text{ given }) \\ &=\frac{4 \mathrm{ m}^2}{3 \mathrm{ mA}} \mathrm{ g}=\frac{4 \mathrm{mg}}{3 \mathrm{ A}} \end{aligned}

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