Two wires of diameter 0.25 cm, one made of ...

Mechanical Properties of Solids

Question : 5 of 21

Marks: +1, -0

Two wires of diameter 0.25 cm, one made of steel andthe other made of brass are loaded as shown in figure. Theunloaded length of steel wire is 1.5 m and that of brass wireis 1.0 m. Compute the elongations of the steel and the brasswires.

Solution:

For steel wire : total force on steel wire:

F_{1}=4+6=10\,\text{kg},

f=10 \times 9.8\,\text{N},

l_{1}=1.5\,\text{m},\ \Delta l_{1}=? ;

2 r_{1}=0.25\,\text{cm}

r_{1}=\frac{0.25}{2}\,\text{cm}=0.125 \times 10^{-2}\,\text{m},

Y_{1}=2.0 \times 10^{11}\,\text{Pa}

For brass wire,

F_{2}=6.0\,\text{kg},

f=6 \times 9.8\,\text{N},\ 2 r_{2}=0.25\,\text{cm}

r_{2}=\frac{0.25}{2}\,\text{cm}=0.125 \times 10^{-2}\,\text{m};

Y_{2}=0.91 \times 10^{11}\,\text{Pa},\ l_{2}=1.0\,\text{m},

\Delta l_{2}=?

As,

Y_{1}= \frac{F_{1} \times l_{1}}{A_{1} \times \Delta l_{1}}= \frac{F_{1} \times l_{1}}{\pi r_{1}^{2} \times \Delta l_{1}}

\Delta l_{1}= \frac{F_{1} \times l_{1}}{\pi r_{1}^{2} \times Y_{1}}

= \frac{(10 \times 9.8) \times 1.5 \times 7}{22 \times (0.125 \times 10^{-2})^{2} \times 2 \times 10^{11}}

=1.49 \times 10^{-4}\,\text{m}

And

\Delta l_{2}= \frac{F_{2} \times l_{2}}{\pi r_{2}^{2} \times Y_{2}}

-= \frac{(6 \times 9.8) \times 1.0 \times 7}{22 \times (0.125 \times 10^{-2})^{2} \times (0.91 \times 10^{11})}

=1.3 \times 10^{-4}\,\text{m}

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