Question

An elastic string of Length $42cm$ and cross section area ${10}^{-4}{m}^2$ is attached between two pegs at distance $6mm$ as shown in the figure. A particle of mass $m$ is kept at mid point of string stretched as shown in figure by $20cm$ and released. As the string attains natural Length, the particle attains a speed of $20m/s$. Then young modulus $Y$ of string is of order.

• A. ${10}^8$
• B. ${10}^{12}$
• C. ${10}^6$
• D. ${10}^4$

Answer 1

The correct option is C

${10}^6$

Step 1. Deducing the relation between kinetic energy and young modulus.

We know that,

Elastic stream energy $\times$ volume $=$ kinetic energy

Vol $\times$ $\frac{1}{2}$ stress $\times$strain $=$ $\frac{1}{2}m{v}^2$

Vol $\times$ $\frac{1}{2}$ $\frac{stress}{strain}\times$ $strai{n}^2$ $=$ $\frac{1}{2}m{v}^2$

$\frac{1}{2y}strai{n}^{2}vol=\frac{1}{2}m{v}^2$

Step 2. Calculating the young modulus of the string

$$\begin{gathered} \hspace{0.17em}y=\frac{m\times v^2}{AL{\left({\displaystyle \frac{∆l}{l}}\right)}^2} \\ =\frac{0.05x400x0.42}{{(0.2)}^{2}x{10}^{-4}} \\ =2.1\times {10}^{6}N/m^2 \end{gathered}$$

Where $m$ is the mass

$v$ is the velocity

$A$ is the area of cross-section

$L$ is the length

Hence, the correct option is (C).