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Standard XII

Physics

Velocity Displacement Relationship

The mass of a...

Question

The mass of a simple pendulum bob is $100 g m$ .The length of the pendulum is $1 m$ . The bob is drawn aside from the equilibrium position so that the string makes an angle of $60^{0}$ with the vertical and let go. The kinetic energy of the bob while crossing its equilibrium position will be:

0.49 J

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0.94 J

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1 J

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1.2 J

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Solution

The correct option is A $0.49 J$
$K . E_{A} = Δ$ $P . E .$ b/w A & B

$= m g (L - L cos 60^{0})$

$= 10^{- 1} \times 9.8 \times (1 - \frac{1}{2})$

$= 0.49 J$

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The mass of a simple pendulum bob is 100gm.The length of the pendulum is 1m. The bob is drawn aside from the equilibrium position so that the string makes an angle of 600 with the vertical and let go. The kinetic energy of the bob while crossing its equilibrium position will be:

The correct option is A 0.49JK.EA=Δ P.E. b/w A & B=mg(L−Lcos600)=10−1×9.8×(1−12)=0.49J

The mass of a simple pendulum bob is $100 g m$ .The length of the pendulum is $1 m$ . The bob is drawn aside from the equilibrium position so that the string makes an angle of $60^{0}$ with the vertical and let go. The kinetic energy of the bob while crossing its equilibrium position will be:

The correct option is A $0.49 J$
$K . E_{A} = Δ$ $P . E .$ b/w A & B

$= m g (L - L cos 60^{0})$

$= 10^{- 1} \times 9.8 \times (1 - \frac{1}{2})$

$= 0.49 J$