A steel wire of length $$3.2 \mathrm{~m}\left(\mathrm{Y}_{\mathrm{s}}=2.0 \times 10^{11} \,\mathrm{Nm}^{-2}\right)$$ and a copper wire of length $$4.4 \mathrm{~m}\left(\mathrm{Y}_{\mathrm{c}}=1.1 \times 10^{11} \,\mathrm{Nm}^{-2}\right)$$, both of radius $1.4 \mathrm{~mm}$ are connected end to end. When stretched by a load, the net elongation is found to be $1.4 \mathrm{~mm}$. The load applied, in Newton, will be: $\quad\left(\right.$ Given $\pi=\frac{22}{7}$)
Solution
<p>$\Delta {l_s} + \Delta {l_c} = 1.4$</p>
<p>$${{W{l_s}} \over {{Y_s} \times A}} + {{W{l_c}} \over {{Y_c} \times A}} = 1.4 \times {10^{ - 3}}$$</p>
<p>$$W = {{1.4 \times {{10}^{ - 3}}} \over {\left[ {{{3.2} \over {2 \times {{(\pi \times 1.4 \times {{10}^{ - 3}})}^2}}} + {{4.4} \over {1.1 \times {{(\pi \times 1.4 \times {{10}^{ - 3}})}^2}}}} \right]{1 \over {{{10}^{ + 11}}}}}}$$</p>
<p>$W \simeq 154\,N$</p>
About this question
Subject: Physics · Chapter: Properties of Solids and Liquids · Topic: Elasticity
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