Answer:
[tex]303.75\; {\rm N}[/tex].
Explanation:
Given that tension [tex]F[/tex] varies inversely with radius [tex]r[/tex] ([tex]r \ne 0[/tex]) of the path, there exists a constant [tex]k[/tex] ([tex]k \ne 0[/tex]) such that:
[tex]\displaystyle F = \frac{k}{r^{2}}[/tex].
To find the value of [tex]k[/tex], make use of the fact that [tex]F = 60\; {\rm N}[/tex] when [tex]r = 90\; {\rm cm}[/tex]:
[tex]\displaystyle (60\; {\rm N}) = \frac{k}{(90\; {\rm cm})^{2}}[/tex].
[tex]\displaystyle k = (60\; {\rm N})\, (90\; {\rm cm})^{2}[/tex].
The value of [tex]k[/tex] is constant regardless of the value of [tex]r[/tex]. Hence, when [tex]r = 40\; {\rm cm}[/tex], [tex]k[/tex] would still be equal to [tex](60\; {\rm N})\, (90\; {\rm cm})^{2}[/tex], and the value of [tex]F[/tex] would be:
[tex]\begin{aligned} F &= \frac{k}{r^{2}} \\ &= \frac{(60\; {\rm N})\, (90\; {\rm cm})^{2}}{(40\; {\rm cm})^{2}} \\ &= 303.75\; {\rm N}\end{aligned}[/tex].
