Two blocks of masses `1 kg` and `2 kg` are connected by a metal wire goijng over a smooth pulley as shown in figure.
The breaking stress of the metal is `(40//3pi)xx10^(6)N//m^(2)`. If `g=10ms^(-12)`, then what should be the minimum radlus of the wire used if it is not to break?

Options

A. `0.5mm`
B. `1mm`
C. `1.5mm`
D. `2mm`

Answer

B

Explanation

`T=(2m_(1)m_(2))/(m_(1)+m_(2))g=(2xx1xx2)/(1+2)xx10N=40/3N`
If `r` is the minimum radius, then
Breaking stress `=(40/3)/(pir^(2))` or `40/(3pi)xx10^(6)=40/(3pir^(2))`
or `r^(2)=1/(10^(6))` or `r=1/(10^(3))m`
or `r=1/(10^(3))xx10^(3)m=1mm`