Question

A mass of $100\text{gram}$ is attached to the end of a rubber string $49\text{cm}$ long having an area of cross-section $20{\text{mm}}^2$. The string is whirled around horizontally at a constant speed of $40\text{rps}$ in a circle of radius $51\text{cm}$. Find the ratio of longitudinal stress and longitudinal strain in the rubber string.

• A. $3{\pi }^2\times {10}^9{\text{Nm}}^{-2}$
• B. $4{\pi }^2\times {10}^9{\text{Nm}}^{-2}$
• C. $4{\pi }^2\times {10}^8{\text{Nm}}^{-2}$
• D. $6{\pi }^2\times {10}^8{\text{Nm}}^{-2}$

Answer 1

The correct option is C $4{\pi }^2\times {10}^8{\text{Nm}}^{-2}$

Tension in the string is given by
$F=m{\omega }^{2}r$ (centrifugal force)
$\Rightarrow F=m\times r(2\pi N{)}^2$, where $N=40\text{rps}$
$\Rightarrow F=100\times 51\times (2\pi \times 40{)}^2\text{dyne}$
[force is in dynes when units of mass and length are gm and cm respectively]

Also, $l=49\text{cm}\Rightarrow \Delta l=51-49=2\text{cm}$
$A=20{\text{mm}}^2=20\times {10}^{-2}{\text{cm}}^2$
As we know,
Longitudinal stress $=\left(\frac{F}{A}\right)$ &
Longitudinal strain $=\left(\frac{\Delta l}{l}\right)$

Therefore, $\frac{\text{Longitudinal stress}}{\text{Longitudinal strain}}=\frac{F\times l}{A\times \Delta l}$
$=\frac{100\times 51\times 4{\pi }^2\times 1600\times 49}{20\times {10}^{-2}\times 2}$
$\approx 4{\pi }^2\times {10}^9{\text{dyne cm}}^{-2}$
$=4{\pi }^2\times {10}^8{\text{Nm}}^{-2}$