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(B) For a conical pendulum (bob whirled in a horizontal plane), the forces acting on the bob are the tension $T$ in the string and the gravitational f

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$\cos^{-1}(0.1)$

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(B) For a conical pendulum (bob whirled in a horizontal plane), the forces acting on the bob are the tension $T$ in the string and the gravitational force $mg$.Resolving the tension $T$ into vertical and horizontal components:$1$. The vertical component $T \cos \phi$ balances the weight of the bob: $T \cos \phi = mg$.$2$. The horizontal component $T \sin \phi$ provides the necessary centripetal force: $T \sin \phi = \frac{mv^{2}}{R}$.Given:Mass $m = 100 \,g = 0.1 \,kg$Tension $T = 10 \,N$Acceleration due to gravity $g = 10 \,m/s^{2}$From the vertical equilibrium equation:$\cos \phi = \frac{mg}{T}$$\cos \phi = \frac{0.1 \,kg 	imes 10 \,m/s^{2}}{10 \,N}$$\cos \phi = \frac{1}{10} = 0.1$Therefore, the angle $\phi = \cos^{-1}(0.1)$.

$A$ simple pendulum has a length of $2 \,m$ and a bob of mass $100 \,g$. It is whirled in a horizontal plane. If the string breaks under a tension of $10 \,N$, what is the angle made by the string with the vertical? $\left(g=10 \,m/s^{2}\right)$

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