Question

Two blocks of masses $1kg$ and $2kg$ are connected by a metal wire going over a smooth pulley as shown in figure. The breaking stress of the metal is $\frac{40}{3\pi }\times {10}^{6}N{m}^2$. If $g=10m{s}^{-1}$, then the minimum radius of the wire used if it is not to break is then

• A. $0.5mm$
• B. $1mm$
• C. $1.5mm$
• D. $2m$

Answer 1

Answer: (B)

$1mm$

Given:

Breaking stress of the metal is $\frac{40}{3\pi }\times 106N{m}^2$

Masses $1kg2kg$

From FBD of the block:

For 2 kg block,

$2mg-T=2ma$ .................(1)

For 1 kg Block,

$T-mg=ma$ ......................(2)

On solving from equation (1) and (2)

$4mg-3T=0$

$T=\frac{4mg}{3}=\frac{40}{3}N$

If $r$ is the minimum radius, then

$Breakingstress=\frac{T}{A}=\frac{40/3}{\pi r^2}$

$\frac{40}{3\pi }\times {10}^6=\frac{40}{3\pi r^2}$

or $r^2=\frac{1}{{10}^6}orr=\frac{1}{{10}^3}m=1mm$