Question

Three equal masses of 3 kg are connected by two massless strings of cross sectional area 0.005 $c{m}^2$ and Young modulus is $2\times {10}^{11}$ $N/m^2$ each. The longitudinal stain in the wires

• A. are equal
• B. cannot be different
• C. must be different
• D. may or may not be different

Answer 1

Answer: (C)

must be different

$\begin{array}{c}\text{Given,}\\ \text{mass}=3kg\\ \text{young modulus of wire}\\ \left(Y\right)=2\times {10}^{11}\\ \text{cross sectional Area}=0.005\\ 0.005\times {10}^{-4}{m}^2{cm}^2\end{array}$

$3g-T_B=3a-\left(i\right)$

$T_A=3a-\left(ii\right)$

$T_B-T_A=3a-\left(iii\right)$

$\begin{array}{c}\text{From (ii) and (ii)}\\ T_B=6a\\ \text{Pating}{T}_B\text{in (1).}\\ \begin{array}{cc}3g-6a& =3a\\ a& =\frac{g}{3}\end{array}\end{array}$

$\text{So,}{T}_B=6a$

$=\frac{6\times g}{3}$

$T_B=2gN$

$\begin{array}{c}\text{Now,}Y=\frac{\text{stress}}{\text{strain}}\\ \begin{array}{cc}{\text{(strain)}}_B& =\frac{\text{stress}}{Y}\\ & =\frac{T_B}{AY}\\ & =\frac{2\times 8}{0.005\times {10}^{-4}\times 2\times {10}^{11}}\end{array}\end{array}$

${\text{(strain)}}_B=2\times {10}^{-4}$

$\text{Now,}(\text{Strain}{)}_A=\frac{F_A}{AY}$

$\begin{array}{cc}{F}_A=T_A& =3a\\ & =3\times \frac{g}{3}\end{array}$

$\begin{array}{cc}& T_A=gN\\ {\text{(Strain)}}_A=& \frac{10}{0.005\times {10}^{-4}\times 2\times {10}^{11}}\\ =& {10}^{-4}\end{array}$

$\text{So, strain at}A\text{and}B\text{must be different.}$