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If the ratio of lengths, radii and Young's modulus of steel and brass wires shown in the figure are $a, b$ and $c$ respectively, the ratio between the increase in lengths of brass and steel wires would be
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The correct answer is:
$\frac{a}{2 b^2 c}$
Free body diagram of the two blocks are
Given, $\quad \frac{l_1}{l_2}=a, \frac{r_1}{r_2}=b, \frac{Y_1}{Y_2}=c$
Let Young's modulus of steel is $Y_1$ and of brass is $Y_2$
$\therefore \quad Y_1=\frac{F_1 \cdot l_1}{A_1 \cdot \Delta l_1}$...(i)
and $\quad Y_2=\frac{F_2 \cdot l_2}{A_2 \Delta l_2}$...(ii)
Dividing Eq. (i) by Eq. (ii), we get
$\frac{Y_1}{Y_2}=\frac{\frac{F_1 \cdot l_1}{A_1 \cdot \Delta l_1}}{\frac{F_2 \cdot l_2}{A_2 \cdot \Delta l_2}}$
or $\quad \frac{Y_1}{Y_2}=\frac{F_1 \cdot A_2 \cdot l_1 \cdot \Delta l_2}{F_2 \cdot A_1 \cdot l_2 \cdot \Delta l_1}$...(iii)
Force on steel wire from free body diagram
$T=F_1=2 g \mathrm{~N}$
Force on brass wire from free body diagram
$F_2=T^{\prime}=T+2 g=4 \mathrm{~g} \mathrm{~N}$
Now, putting the values of $F_1, F_2$ in Eq. (iii), we get
$\frac{Y_1}{Y_2}=\left(\frac{2 g}{4 g}\right) \cdot\left(\frac{\pi r_2^2}{\pi r_1^2}\right) \cdot\left[\frac{l_1}{l_2}\right] \cdot\left(\frac{\Delta l_2}{\Delta l_1}\right)$
or $\quad c=\frac{1}{2}\left(\frac{1}{b^2}\right) \cdot a\left(\frac{\Delta l_2}{\Delta l_1}\right)$
or $\quad \frac{\Delta l_1}{\Delta l_2}=\left(\frac{a}{2 b^2 c}\right)$
Given, $\quad \frac{l_1}{l_2}=a, \frac{r_1}{r_2}=b, \frac{Y_1}{Y_2}=c$
Let Young's modulus of steel is $Y_1$ and of brass is $Y_2$
$\therefore \quad Y_1=\frac{F_1 \cdot l_1}{A_1 \cdot \Delta l_1}$...(i)
and $\quad Y_2=\frac{F_2 \cdot l_2}{A_2 \Delta l_2}$...(ii)
Dividing Eq. (i) by Eq. (ii), we get
$\frac{Y_1}{Y_2}=\frac{\frac{F_1 \cdot l_1}{A_1 \cdot \Delta l_1}}{\frac{F_2 \cdot l_2}{A_2 \cdot \Delta l_2}}$
or $\quad \frac{Y_1}{Y_2}=\frac{F_1 \cdot A_2 \cdot l_1 \cdot \Delta l_2}{F_2 \cdot A_1 \cdot l_2 \cdot \Delta l_1}$...(iii)
Force on steel wire from free body diagram
$T=F_1=2 g \mathrm{~N}$
Force on brass wire from free body diagram
$F_2=T^{\prime}=T+2 g=4 \mathrm{~g} \mathrm{~N}$
Now, putting the values of $F_1, F_2$ in Eq. (iii), we get
$\frac{Y_1}{Y_2}=\left(\frac{2 g}{4 g}\right) \cdot\left(\frac{\pi r_2^2}{\pi r_1^2}\right) \cdot\left[\frac{l_1}{l_2}\right] \cdot\left(\frac{\Delta l_2}{\Delta l_1}\right)$
or $\quad c=\frac{1}{2}\left(\frac{1}{b^2}\right) \cdot a\left(\frac{\Delta l_2}{\Delta l_1}\right)$
or $\quad \frac{\Delta l_1}{\Delta l_2}=\left(\frac{a}{2 b^2 c}\right)$
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