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Standard XII

Physics

Momentum

A point sized...

Question

A point sized sphere of mass $^{'} m^{'}$ is suspended from a point using a string of length $^{'} l^{'}$ . It is pulled to a side till the string is horizontal and released. As the mass passes through the portion where the string is vertical, magnitude of its angular momentum is:

m l \sqrt{g l}

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m l \sqrt{2 g l}

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m l \sqrt{\frac{g l}{2}}

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m l \sqrt{3 g l}

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Solution

The correct option is B $m l \sqrt{2 g l}$
By conservation of energy
$\frac{1}{2} m v^{2} = m g l$
$v = \sqrt{2 g l}$
Angular momentum $= m v r =$ $m l \sqrt{2 g l}$

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A point sized sphere of mass ′m′ is suspended from a point using a string of length ′l′. It is pulled to a side till the string is horizontal and released. As the mass passes through the portion where the string is vertical, magnitude of its angular momentum is:

The correct option is B ml√2glBy conservation of energy12mv2=mglv=√2glAngular momentum=mvr=ml√2gl

A point sized sphere of mass $^{'} m^{'}$ is suspended from a point using a string of length $^{'} l^{'}$ . It is pulled to a side till the string is horizontal and released. As the mass passes through the portion where the string is vertical, magnitude of its angular momentum is:

The correct option is B $m l \sqrt{2 g l}$
By conservation of energy
$\frac{1}{2} m v^{2} = m g l$
$v = \sqrt{2 g l}$
Angular momentum $= m v r =$ $m l \sqrt{2 g l}$