Question

Two blocks of mass $2 \, \text{kg}$ and $4 \, \text{kg}$ are connected by a metal wire going over a smooth pulley as shown in the figure. The radius of the wire is $4.0 \times 10^{-5} \, \text{m}$ and Young's modulus of the metal is $2.0 \times 10^{11} \, \text{N/m}^2$. The longitudinal strain developed in the wire is $\frac{1}{\alpha \pi}$. The value of $\alpha$ is _____. [Use $g = 10 \, \text{m/s}^2$]

Answer 1

The tension $T$ in the wire, due to the masses, is given by:

$T = \left( \frac{2m_1m_2}{m_1 + m_2} \right) g = \frac{80}{3} \, \text{N}.$

The cross-sectional area $A$ of the wire is:

$A = \pi r^2 = 16\pi \times 10^{-10} \, \text{m}^2.$

The strain $\frac{\Delta \ell}{\ell}$ in the wire is given by:

$\text{Strain} = \frac{F}{AY} = \frac{T}{AY}.$

Substitute the values:

$\text{Strain} = \frac{\frac{80}{3}}{16\pi \times 10^{-10} \times 2 \times 10^{11}} = \frac{1}{12\pi}.$

Thus, $\alpha = 12$.