
From the given figure we can see that
sinθ0=lA=10010=101
From conservation of energy 2mv2=mgl(1−cosθ0)
Maximum tension occurs at mean position
T=mg+lmv2⇒T=mg+2mg(1−cosθ0)
Substituting the values,
T=mg(1+2(1−1−sin2θ0)⇒T=mg(3−21−1001)⇒T=0.25×9.8(3−2(1−2001))⇒T=0.25×9.8×1.01=2.4745N
It is given that
T=40x⇒x=40×2.4745=98.98≈99