Not sure what you mean by "breaks in the tension" but I suspect you mean the rope will come apart if the tension in the rope exceeds 1800 N.
In the free body diagram for the 500 N weight, we have a figure Y with the net force equations
• horizontal net force:
∑ F[hor] = T₁ cos(θ) - T₂ cos(θ) = 0
• vertical net force:
∑ F[ver] = T₁ sin(θ) + T₂ sin(θ) - 500 N = 0
From the first equation, it follows that T₁ = T₂, so I'll denote their magnitude by T alone. From the second equation, we have
2 T sin(θ) = 500 N
and if the maximum permissible tension is T = 1800 N, it follows that
sin(θ) = (500 N) / (3600 N) ⇒ θ = arcsin(5/36) ≈ 7.9°
is the smallest angle the rope can make with the horizontal.
Solution:
a) We know acceleration due to gravity, g = GM/r²
Differential change, dg/dr = -2GM/r³
Here, r = 50*Rh = 50*2GM/c² = 100GM/c
²
My height, h=dr = 1.7 m
Difference in gravitational acceleration between my head and my feet, dg = -10 m/s²
or, dg/dr = -10/1.7 = -2GM/(100GM/c²)³
or, 5.9*100³*G²*M² = 2c⁶
or, M = 0.59*c³/(1000G) = 2.39*1032 kg = [(2.39*1032)/(1.99*1030 )]Ms = 120*Ms
Mass of black hole which we can tolerate at the given distance is 120 time the mass of Sun.
b) This limit an upper limit ,we can tolerate smaller masses only.
