Question

A copper wire of negligible mass,length $\left(\ell \right)$ and cross-sectional area (A) is kept on a smooth horizontal table with one end fixed, a ball of mass 'm' is attached at the other end. The wire and the ball are rotated with angular velocity ${}^{\prime }{\omega }^{\prime }$. If wire elongates by $△\ell$, then thw Young's modulus of wire and if on increasing the angular velocity from $\omega to{\omega }^1$ When the wire breaks-down, then breaking stress $(△\ell <<\ell )$ are respectively.

• A. $\frac{\left(m{\ell }^2{\omega }^2\right)}{A△\ell },\frac{m\ell {\omega }^2}{A}$
• B. $\frac{m\ell }{A△\ell {\omega }^2},\frac{m\ell {\omega }^2}{A}$
• C. $\frac{m\ell {\omega }^2}{A△\ell },\frac{m{\omega }^2}{A\ell }$
• D. $\frac{m\ell {\omega }^2}{A△\ell },\frac{m{\omega }^1}{A}$

Answer 1

The correct option is C $\frac{m\ell {\omega }^2}{A△\ell },\frac{m{\omega }^2}{A\ell }$

$\begin{array}{c}\text{Given length}=l\\ \text{Area}=A\\ \text{Angular velocity}=\omega \\ \text{elongation}=\Delta l\end{array}$

$\text{and given that at}{\omega }^{\prime }\text{angular velocity wire break's down}$

$\begin{array}{c}\text{By Balancing forces,}\\ T=m{\omega }^2\\ \text{Elongation formula}\\ \Delta l=\frac{Fl}{Ay}\end{array}$

$\begin{array}{cc}\Rightarrow y& =\frac{Fl}{A\Delta l}\\ y& =\frac{ml{\omega }^2\left(l\right)}{A\Delta l}\\ y& =\frac{m{l}^2{\omega }^2}{A\Delta l}\end{array}$

$\begin{array}{cc}\text{At}\omega =& {\omega }^{\prime }\text{wire breaks down}\\ & \text{So breaking force =}ml{{\omega }^{\prime }}^2\\ & \text{Breaking StresS}=\frac{\text{breaking Force}}{\text{Area}}\end{array}$

$=\frac{ml{\left(w^{\prime }\right)}^2}{A}$

Hence option C is correct