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(C) The bob of the simple pendulum experiences both the acceleration due to gravity $(g)$ and the centripetal acceleration $(a_c = v^2/R)$ due to the

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$101$

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(C) The bob of the simple pendulum experiences both the acceleration due to gravity $(g)$ and the centripetal acceleration $(a_c = v^2/R)$ due to the circular motion of the car. These two accelerations are perpendicular to each other. The effective acceleration $(g')$ acting on the bob is given by $g' = \sqrt{g^2 + a_c^2} = \sqrt{g^2 + (v^2/R)^2} = \sqrt{g^2 + v^4/R^2}$. The time period of a simple pendulum is $T = 2\pi \sqrt{L/g'}$. Substituting the values: $T = 2\pi \sqrt{L / \sqrt{g^2 + v^4/R^2}} = 2\pi \sqrt{1 / (g^2 + v^4/R^2)^{1/2}} = 2\pi / (g^2 + v^4/R^2)^{1/4}$. Comparing this with $T = 2\pi / \alpha^{1/4}$, we get $\alpha = g^2 + v^4/R^2$. Given $g = 10 \ m/s^2$, $v = 10 \ m/s$, and $R = 100 \ m$: $\alpha = (10)^2 + (10)^4 / (100)^2 = 100 + 10000 / 10000 = 100 + 1 = 101$.

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