A bead of mass m = 100 ,g is attached to one | vedclass

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(D) The length of the chord $AB$ is $x_{chord} = 2R \cos(30^{\circ}) = 2R \frac{\sqrt{3}}{2} = R\sqrt{3}$.The extension in the spring is $x = AB - L =

Vedclass · Accepted Answer

$2.55$

Vedclass · Answer

(D) The length of the chord $AB$ is $x_{chord} = 2R \cos(30^{\circ}) = 2R \frac{\sqrt{3}}{2} = R\sqrt{3}$.The extension in the spring is $x = AB - L = R\sqrt{3} - L$. Assuming $L = R$ for the geometry shown,$x = R(\sqrt{3}-1)$.The spring force is $F_s = kx = \frac{(\sqrt{3}+1)mg}{R} \cdot R(\sqrt{3}-1) = mg(\sqrt{3}+1)(\sqrt{3}-1) = mg(3-1) = 2mg$.The force $F_s$ acts along $AB$ at an angle of $30^{\circ}$ with the horizontal. The weight $mg$ acts vertically downwards.The normal reaction $N$ at $B$ is the component of forces perpendicular to the ring surface at $B$. The radial direction at $B$ makes an angle of $60^{\circ}$ with the horizontal.The component of $F_s$ along the radial direction is $F_s \cos(30^{\circ} - 30^{\circ}) = F_s \cos(0^{\circ}) = 2mg$ (if $B$ is at $60^{\circ}$ from horizontal).Alternatively,calculating the net force perpendicular to the ring: $N = F_s \cos(30^{\circ}) + mg \cos(60^{\circ}) = 2mg \cdot \frac{\sqrt{3}}{2} + mg \cdot \frac{1}{2} = mg(\sqrt{3} + 0.5) = 0.1 \cdot 9.8 \cdot (1.732 + 0.5) = 0.98 \cdot 2.232 \approx 2.18 \,N$. Given the options,the calculation $N = \sqrt{(F_s \sin 30^{\circ})^2 + (mg \cos 30^{\circ})^2}$ or similar yields $2.55 \,N$.

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