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Mechanical Properties of Solids — Physics STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 SciencePhysicsMechanical Properties of Solids3 Marks
Question
A wire gets stretched by $4mm$ due to a certain load. If the same load is applied to a wire of same material with half the length and double the diameter of the first wire. What will be the change in its length?
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Answer

Given. $l_1=4 mm =4 \times 10^{-3} m$
$L _2=\frac{ L _1}{2}, D _2=2 D _{,} r _2=2 r _1$
To find: Change in length $\left(I_2\right)$
Formula: $Y =\frac{ FL }{ Al }=\frac{ FL }{\pi r ^2 l}$
Calculation: From formula,
$ Y _1=\frac{ F _1 L _1}{\pi r _1{ }^2 l_1}$
$Y _2=\frac{ F _2 L _2}{\pi r _2{ }^2 l_2} $
Dividing equation (ii) by equation (i),
$\frac{ Y _2}{ Y _1}=\frac{\frac{ F _2 L _2}{\pi r _2^2 l_2}}{\frac{ F _1 L _1}{\pi r _1^2 l_1}}$
Since same load is applied on same wire, $Y_1=Y_2$ and $F_1=F_2$
$\therefore \quad \frac{ L _1}{ r _1^2 l_1}=\frac{ L _2}{ r _2^2 l_2}$
....[From (iii)]
$ l_2  =\frac{ L _2 \times r _1^2 \times l_1}{ r _2^2 \times L _1}$
$ =\frac{ L _2 \times r _1^2 \times l_1}{4 r _1^2 \times 2 \times L _2}$
$ =\frac{l_1}{8}$
$ =\frac{4 \times 10^{-3}}{8}$
$= 0.5 \times 10^{-3} m$
$= 0.5 mm$
The new change in length of the wire is 0.5 mm.

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