Question

The breaking stress of the metal used for the wire connecting the blocks is $2\times {10}^9{\text{N/m}}^2$. Minimum radius of the wire is:
[Take $\sqrt{\frac{400}{6\pi }}=4.60,g=10{\text{m/s}}^2$ for calculation.]

• A. $4.6\times {10}^{-5}\text{m}$
• B. $9.6\times {10}^{-5}\text{m}$
• C. $8.6\times {10}^{-5}\text{m}$
• D. $5.6\times {10}^{-5}\text{m}$

Answer 1

The correct option is A $4.6\times {10}^{-5}\text{m}$

Tension in the wire will produce stress. At any cross-section of wire, stress will be
$\sigma =\frac{F}{A}$
Let the tension in the wire be $T$ when the wire is about to break. At that instant, stress in the wire will be breaking stress.
$\sigma ={\sigma }_{\text{breaking}}$
${\sigma }_{\text{breaking}}=\frac{T}{A}...\left(1\right)$


Applying Newton's second law $\sum F=ma$ along the direction of acceleration of the blocks,
For left block,
$T-10=1\times a=a...\left(2\right)$
For right block,
$20-T=2a...\left(3\right)$

On adding Eq. $\left(2\right)$ and $\left(3\right)$,
$10=3a$
$\Rightarrow a=\frac{10}{3}{\text{m/s}}^2$
On putting value of $a$ in Eq.$\left(2\right)$,
$T=\frac{40}{3}\text{N}$
Substituting in Eq.$\left(1\right)$,
${\sigma }_{\text{breaking}}=\frac{\left(\frac{40}{3}\right)}{A}$
$\Rightarrow A=\frac{\frac{40}{3}}{2\times {10}^9}$
Breaking stress, ${\sigma }_{\text{breaking}}=2\times {10}^9{\text{N/m}}^2$
Breaking stress will correspond to the minimum cross-sectional area of wire.
$\Rightarrow \pi r_{\text{min}}^2=\frac{40}{6\times {10}^9}$

$\Rightarrow r_{\text{min}}=\sqrt{\frac{40}{6\times {10}^9\times \pi }}=\sqrt{\frac{400}{6\pi \times {10}^{10}}}$

$\Rightarrow r_{\text{min}}=4.6\times {10}^{-5}\text{m}$
Minimum value of required radius is $4.6\times {10}^{-5}\text{m}$