Question

Two block of masses $m$ and $M$ are connected by means of a metal wire passing over a frictionless fixed pulley. The area of cross-section of the wire is $6.5\times {10}^{-9}{m}^2$ and its breaking stress is $2\times {10}^{9}N{m}^{-2}$. If $m=1kg,$ if the maximum value of M is $K\times 1.86$ kg for which the wire will not break. Find the value of $K$. (g = 10 m/s${}^2$)

Answer 1

Given : $m=1$ kg $M=1.86K$ $A=6.5\times {10}^{-9}$ $m^2$ $S=2\times {10}^9$ $N{m}^{-2}$

Maximum tension in the wire $T=6.5\times {10}^{-9}\times 2\times {10}^9=13$ N

For mass $m=1kg$ : $ma=T-mg$

$\therefore$ $1a=13-1\left(10\right)$ $⟹a=3$

For mass $M=1.86K$ : $Ma=Mg-T$

OR $M(g-a)=T$

$\therefore$ $1.86K(10-3)=13$ $⟹K=1$