Question

Two wires of different materials, each $2$m long and of diameter $2$mm, are joined in series to form a composite wire. What force will produce a total extension of $0.9$mm. $(Y_1=2\times {10}^{11}Pa$ & $Y_2=6\times {10}^{11}Pa)$.

• A. $282.6$ N
• B. $212$ N
• C. $319.8$ N
• D. $382.6$ N

Answer 1

The correct option is B $212$ N

Given.

Length of both wires, $L=2m$

Radius of both wires, $d=2mm=2\times {10}^{-3}m$

Total extension of joined wire, $\Delta L_{net}=0.9mm=9\times {10}^{-4}m$

Both wires join in series, so tension in both is equal

$T=\frac{Y_{1}A\Delta L_1}{L}=\frac{Y_{2}A\Delta L_2}{L}$

Net extension in joined wire.

$\Rightarrow \Delta L_{net}=\Delta L_1+\Delta L_2=\frac{TL}{Y_{1}A}+\frac{TL}{Y_{2}A}=\frac{TL}{A}(\frac{1}{Y_1}+\frac{1}{Y_2})$

$\Rightarrow \Delta L_{net}=\frac{TL}{A}(\frac{1}{Y_1}+\frac{1}{Y_2})$

$\Rightarrow T=\frac{A\Delta L_{net}}{L(\frac{1}{Y_1}+\frac{1}{Y_2})}=\frac{\frac{\pi }{4}{(2\times {10}^{-3})}^2\times 9\times {10}^{-4}}{2(\frac{1}{2\times {10}^{11}}+\frac{1}{6\times {10}^{11}})}=212.05N$

Total force produced in joined wire is $212N$