Question

Length of rods are \( \ell \) and \(2\ell\) respectively and their Young's modulus are \( \gamma \) and \(2\gamma\) respectively. Given: \[ \ell = 0.314\,\text{m}, \quad R = 0.2\,\text{mm} \,(\text{same}), \quad \gamma = 2\times10^9 \]

Find the total extension \( \Delta \ell \).

• A. \(0.4\,\text{mm}\)
• B. \(0.1\,\text{mm}\)
• C. \(0.2\,\text{mm}\)
• D. \(0.3\,\text{mm}\)

Answer 1

The Correct Option is C

Concept:
Extension of a wire under load is given by \[ \Delta L = \frac{FL}{AY} \] where \(F\) = applied force, \(L\) = length of wire, \(A\) = cross-sectional area, \(Y\) = Young's modulus. For rods connected in series, total extension equals the sum of individual extensions. \[ \Delta \ell = \Delta \ell_1 + \Delta \ell_2 \] Step 1: Write extension expressions for both rods. \[ \Delta \ell_1 = \frac{MgL_1}{A\gamma} \] \[ \Delta \ell_2 = \frac{MgL_2}{A(2\gamma)} \] Step 2: Add the extensions. \[ \Delta \ell = \frac{MgL}{A\gamma} + \frac{Mg(2L)}{A(2\gamma)} \] \[ \Delta \ell = \frac{Mg}{A}\left(\frac{L}{\gamma}+\frac{2L}{2\gamma}\right) \] \[ \Delta \ell = \frac{2MgL}{A\gamma} \] Step 3: Substitute numerical values. Mass \(m = 0.8\,\text{kg}\) \[ A = \pi R^2 \] \[ R = 0.2\,\text{mm} = 2\times10^{-4}\,\text{m} \] \[ A = \pi (2\times10^{-4})^2 \] Substitute: \[ \Delta \ell = \frac{0.8\times10}{\pi\times4\times10^{-6}} \left(\frac{2\times0.314}{2\times10^9}\right) \] \[ \Delta \ell = 2\times10^{-4}\,\text{m} \] \[ \Delta \ell = 0.2\,\text{mm} \]