A steel wire is 1 ,m long and 1 ,mm ^2 in area of cross-section. If it takes 200 ,N to stretch this

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8.Mechanical Properties of Solids

medium

A steel wire is $1 \,m$ long and $1 \,mm ^2$ in area of cross-section. If it takes $200 \,N$ to stretch this wire by $1 \,mm$, how much force will be required to stretch a wire of the same material as well as diameter from its normal length of $10 \,m$ to a length of $1002 \,cm$ is ........ $N$

$1000$

$200$

$400$

$2000$

Solution

(c)

$\frac{F L}{A Y}=\Delta x$

Since $A, Y$ remain constant in given case We can say

$F L \propto \Delta x$

or $\frac{F_1 L_1}{F_2 L_2}=\frac{\Delta x_1}{\Delta x_2}$ $\left\{\begin{array}{l}F_1=200 \,N \\ \Delta x_1=1 \,mm \\ \Delta x_2=10.02 \,m -10 \,m =0.02 \,m =20 \,mm \\ L_1=1 \,m \\ L_2=10 \,m \end{array}\right.$

Substitute the values

$\frac{200 \times 1}{F_2 \times 10}=\frac{1 \,mm }{20 \,mm }$

$F_2=400 \,N$

Standard 11

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Solution

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A wire of length $L$ and radius $r$ is clamped rigidly at one end. When the other end of the wire is pulled by a force $f$, its length increases by $l$. Another wire of same material of length $2 L$ and radius $2 r$ is pulled by a force $2 f$. Then the increase in its length will be

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