Let θ be the maximum angular deflection them

1/2mv^{2} = mg (l - lcosθ)

and T - mg = mv^{2}/l

T = mg + mv^{2}/l

T/πr^{2} = mg + 2 mg (l - lcosθ)

= 3mg - 2mg cosθ

For maximum value of θ, stress (=T/πr^{2}) must be equal to breaking stress,

i.e., T/πr^{2 }= 7.85 x 10^{8}

or, (3mg - 2mg cosθ)/(π x (10^{-3})^{2}) = 7.85 x 10^{8}

or, cosθ = 0.2436 = cos 75.9**°**

or, θ = 75.9**°**