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

Correct Answer: 12

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, \(\alpha = 12\).