The bob of a pendulum of length l is pulled a side from its equilibrium position through an angle - and then released- The bob will- then pass through its equilibrium position with a speed - - where - equals to -

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Standard XII

Physics

Range on an Incline

The bob of a ...

Question

The bob of a pendulum of length l is pulled a side from its equilibrium position through an angle $θ$ and then released. The bob will; then pass through its equilibrium position with a speed $ν$ , where $ν$ equals to :

\sqrt{2 g l (1 - sin θ)}

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\sqrt{2 g l (1 + cos θ)}

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\sqrt{2 g l (1 - cos θ)}

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\sqrt{2 g l (1 + sin θ)}

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Solution

The correct option is C $\sqrt{2 g l (1 - cos θ)}$
Suppose bob rises up a height h as shown then after releasing, potential energy at extreme position becomes kinetic energy of mean position.
$m g h = \frac{1}{2} m v_{m a x}^{2}$
$V_{m a x} = \sqrt{2 g h}$ .... (i)
From the figure,
$c o s θ = \frac{I - h}{I}$
$\Rightarrow I c o s θ = I - h$
From eq. (i),
$V_{m a x} = \sqrt{2 g l (1 - c o s θ)}$

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