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

To solve for the longitudinal strain in the wire, we start by determining the tension in the wire due to the two connected masses. Given masses \(m_1=2\,\text{kg}\) and \(m_2=4\,\text{kg}\), we assume the system accelerates with \(a\). The forces on each mass are:

\[ T - m_1g = m_1a \]

\[ m_2g - T = m_2a \]

Adding these equations eliminates \(T\):

\[ m_2g - m_1g = (m_1 + m_2)a \]

Simplifying, we find:

\[ a = \frac{(m_2 - m_1)g}{m_1 + m_2} = \frac{(4 - 2) \times 10}{6} = \frac{20}{6} = \frac{10}{3} \, \text{m/s}^2 \]

Substitute \(a\) back to find \(T\) using \(T = m_1(g + a)\):

\[ T = 2 \times \left(10 + \frac{10}{3}\right) = 2 \times \frac{40}{3} = \frac{80}{3} \text{N} \]

The longitudinal strain \(\epsilon\) is given by:

\[ \epsilon = \frac{\text{Stress}}{\text{Young's Modulus}} = \frac{T}{A \cdot Y} \]

The cross-sectional area \(A\) of the wire is \(\pi r^2\):

\[ A = \pi \times (4.0 \times 10^{-5})^2 = 16\pi \times 10^{-10} \, \text{m}^2 \]

Therefore,

\[ \epsilon = \frac{\frac{80}{3}}{16\pi \times 10^{-10} \times 2 \times 10^{11}} \]

\[ = \frac{80}{3 \times 32\pi } \times 10^{-11} \times 5 \]

\[ = \frac{1}{12\pi} \]

Thus, the strain is \(\frac{1}{12\pi}\), giving \(\alpha = 12\), which matches the expected range [12,12]. Therefore, \(\alpha = 12\).

Answer 2

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, \(\alpha = 12\).