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 $........ N$
Medium
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$\frac{F L}{A Y}=\Delta x$
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
$\frac{200 \times 1}{F_2 \times 10}=\frac{1 \,mm }{20 \,mm }$
$F_2=400 \,N$
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
$\frac{200 \times 1}{F_2 \times 10}=\frac{1 \,mm }{20 \,mm }$
$F_2=400 \,N$
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