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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 $\left.\pi=\frac{22}{7}\right)$
Option: 1
$360$

Option: 2
$180$

Option: 3
$1080$

Option: 4
$154$

Answers (1)

best_answer

$\mathrm{\ell_{\mathrm{s}}=3.2 \mathrm{~m}}$

$\mathrm{Y_s=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2}$

$\mathrm{\ell_c=4.4 \mathrm{~m}}$

$\mathrm{Y_c=1.1 \times 10^{11} \mathrm{~N} / \mathrm{m}^2}$

$\mathrm{r=1.4 \mathrm{~mm}=1.4 \times 10^{-3} \mathrm{~m}}$

$\mathrm{\Delta \ell=1.4 \mathrm{~mm}=1.4 \times 10^{-3} \mathrm{~m}}$

$\mathrm{F=\left(\frac{Y A}{\ell}\right) \Delta \ell}$

$\mathrm{K=\left ( \frac{YA}{\ell} \right )\Delta \ell}$

$\mathrm{\frac{1}{k_{e q}}=\frac{1}{k_1}+\frac{1}{k_2} }$

$\mathrm{\frac{1}{K_{e q}}=\frac{\ell_s}{Y_s A_s}+\frac{\ell_c}{Y_c A_C}}$

$\mathrm{\frac{1}{K_{\text {eq }}}=\frac{3.2}{A \times 2 \times 10^{11}}+\frac{4.4}{A \times 1.1 \times 10^{11}}}$

$\mathrm{taking\: A_{s}=A_{c}=A}$

$\mathrm{\frac{1}{K_{e q}} =\frac{1.6 \times 10^{-11}}{A}+\frac{4 \times 10^{-11}}{A} }$

$\mathrm{K_{e q} =\frac{A}{5.6 \times 10^{-11}} }$

$\mathrm{K_{e q} =\frac{A \times 10^{11}}{5.6} }$

$\mathrm{F =K_{e q}(\Delta \ell) }$

$\mathrm{=\frac{A \times 10^{11}}{5.6} \times 1.4 \times 10^{-3} }$

$\mathrm{=\frac{3.14 \times\left(1.4 \times 10^{-3}\right)^3 \times 10^{11}}{5.6} }$

$\mathrm{F =\frac{22\times(1.4)^2 \times 10^2}{7\times 4} }$

$\mathrm{F =154 \mathrm{~N}}$

Hence 4 is correct option

Posted by

Rakesh

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