Two blocks of masses $1 \mathrm{~kg}$ and $2 \mathrm{~kg}$ are connected by a metal wire going over a smooth…
Two blocks of masses $1 \mathrm{~kg}$ and $2 \mathrm{~kg}$ are connected by a metal wire going over a smooth pulley as shown in figure. The breaking stress of the metal is $2 \times 10^9 \mathrm{~N} / \mathrm{m}^2$. What should be the minimum radius of the wire used if it is not to break ? Take $g=10$ $\mathrm{m} / \mathrm{s}^2$
$4.6 \times 10^{-5} \mathrm{~m}$
$4.6 \times 10^{-6} \mathrm{~m}$
$2.5 \times 10^{-6} \mathrm{~m}$
$2.5 \times 10^{-5} \mathrm{~m}$
Solution
The stress in the wire
$=\frac{\text { tension }}{\text { area of cross }- \text { section }}$
To avoid breaking, this stress should not exceed the breaking stress.
Let the tension in the wire be $T$. The equations of motion of the two blocks are,
$T-10=1 a$
and $\quad 20 \mathrm{~N}-T=2 a$
Eliminating $a$ from these equations,
$T=\left(\frac{40}{3}\right) \mathrm{N}$
Stress $\quad T=\frac{\left(\frac{40}{3}\right)}{\pi r^2}$
If the minimum radius needed to avoid breaking is $r$,
$2 \times 10^9=\frac{\left(\frac{40}{3}\right)}{\pi r^2}$
Solving this,
$r=4.6 \times 10^{-5} \mathrm{~m}$
