Question

As shown in adjacent figure it a load of mass (m) is attached at lower end of wire. Then find the displacement of the point B,C and D are as shown in figure,
(i) elongation of first wire $e_1=\frac{\left(mg\right){\ell }_1}{A{Y}_1}$
(ii) elongation of 2nd wire
$e_2=\frac{\left(mg\right){\ell }_2}{A{Y}_2}+\frac{\left(mg\right){\ell }_1}{A{Y}_1}$
(iii) elongation of 3rd wire
$e_2=\frac{\left(mg\right){\ell }_2}{A{Y}_3}+\frac{\left(mg\right){\ell }_1}{A{Y}_2}+\frac{\left(mg\right){\ell }_1}{A{Y}_1}$

• A. (i) is correct
• B. (i) & (ii) are correct
• C. (iii) is correct
• D. All are correct

Answer 1

The correct option is A (i) is correct

$\begin{array}{c}\text{let the tension in the wires be.}\\ \text{T.}\\ \text{given Area of cross section is}A\text{.}\end{array}$

$T=mg$

$\begin{array}{c}\text{Now elongation formula is}\Delta l=\frac{FL}{Ay}\\ \text{elongation of first wire}\begin{array}{cc}{e}_1& =\frac{\left(T\right)\times L_1}{A{y}_1}\\ e_1=& \frac{mgL}{A{y}_1}\\ e_2=& \frac{mg{l}_2}{A{y}_2},e_3=\frac{mg(l{)}_3}{A{y}_3}\end{array}\end{array}$