Question

Two blocks of masses $5kg$ and $10kg$ are connected by a metal wire going over a smooth pulley as shown in the figure. The breaking stress of the metal wire is $2\times 10^{9}N{m}^{-2}$ . If $g=10m{s}^{-2}$ , then what is the minimum radius of the wire which will not break

• A. $\text{0.1}mm$
• B. $\text{0.2}mm$
• C. $\text{0.05}mm$
• D. $\text{0.25}mm$

Answer 1

Answer: (A)

This is Atwood's machine. Let the tension in the wire be $T$ and acceleration of masses $a$ .
Then, $T-5g=5a$ and $10g-T=10a$
This gives $a=\frac{1}{3}g$ and $T=\frac{20}{3}\text{g}=\frac{200}{3}\text{N}$ .
Stress $=\frac{T}{\text{ area }}=\frac{200}{3\times 3.14\times r^2}=2\times 10^9$ (given)
Thus the minimum radius $r$ should be $r=\sqrt{\frac{200}{3\times 3\text{.}14\times 2\times 10^9}}=1\text{.}03\times 10^{-4}m\approx 0\text{.}1\text{mm}$