A car is moving in a circular horizontal track of radius 10m with a constant speed of 10 ms^–1. A plumb bob is suspended from the roof of the car by a string of length 1m. The angle made by the string with vertical is (g=10ms^–2)

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$0^{0}$

$45^{0}$

$60^{0}$

$30^{0}$

answer is C.

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Detailed Solution

From the figure,

$Tsin θ = \frac{{mv}^{2}}{r} (centripetal)$

$Tcosθ = mg (weight)$

$\Rightarrow \frac{Tsinθ}{Tcosθ} = \frac{{mv}^{2}}{r} \times \frac{1}{mg}$

$\Rightarrow tanθ = \frac{v^{2}}{rg} \Rightarrow \frac{10^{2}}{10 \times 10} = 1 \Rightarrow θ = 45^{0}$

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