The bob of a simple pendulum is imparted a velocity $5 m s^{- 1}$ when it is at its mean position. To what maximum vertical height will it rise on reaching to its extreme position if $60 %$ of its energy is lost in overcoming friction of air?
(Take $g = 10 m s^{- 2}$ )

5 m

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0.5 m

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0.25 m

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1 m

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Solution

The correct option is B $0.5 m$
When it is at mean position it will have a $K . E$ of $0.5 m v^{2}$ .
$K . E = 0.5 \times m \times 5 \times 5 = 12.5 \times m J$

$60$ % of this $K . E$ is lost hence remaining is equal to $(40 / 100) 12.5 \times m = 5 m$
$P . E = m g h = 5 \times m$
$h = 5 / g = 5 / 10 = 0.5 m$

Hence it goes to a height of $0.5 m$ .

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The bob of a simple pendulum is imparted a velocity 5ms−1 when it is at its mean position. To what maximum vertical height will it rise on reaching to its extreme position if 60% of its energy is lost in overcoming friction of air?(Take g=10ms−2)

The correct option is B 0.5mWhen it is at mean position it will have a K.E of 0.5mv2.K.E=0.5×m×5×5=12.5×m J60 % of this K.E is lost hence remaining is equal to (40/100)12.5×m=5mP.E=mgh=5×mh=5/g=5/10=0.5mHence it goes to a height of 0.5m.

The bob of a simple pendulum is imparted a velocity $5 m s^{- 1}$ when it is at its mean position. To what maximum vertical height will it rise on reaching to its extreme position if $60 %$ of its energy is lost in overcoming friction of air?
(Take $g = 10 m s^{- 2}$ )