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Standard XII

Physics

Work Energy Theorem Application

A mass attach...

Question

A mass attached to a string of length $L$ is revolved in a vertical circle with least velocity. The string breaks when the mass reaches at the highest point. The mass describes the parabolic path, then the range of body w.r.t the lowest point is:

L^{2}

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L

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2 L

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\sqrt{2} L

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Solution

The correct option is C $2 L$
Velocity at the highest point,
$v_{h} = \sqrt{g L}$
Height of the mass $= 2 L$
Time taken to fall $= \sqrt{\frac{2 \times 2 L}{g}} = \sqrt{\frac{4 L}{g}}$
Horizontal distance covered
$= v_{h} t = \sqrt{g L} \times \sqrt{4 L / g}$
$= 2 L$

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A mass attached to a string of length L is revolved in a vertical circle with least velocity. The string breaks when the mass reaches at the highest point. The mass describes the parabolic path, then the range of body w.r.t the lowest point is:

The correct option is C 2LVelocity at the highest point,vh=√gLHeight of the mass =2LTime taken to fall =√2×2Lg=√4LgHorizontal distance covered=vht=√gL×√4L/g=2L

A mass attached to a string of length $L$ is revolved in a vertical circle with least velocity. The string breaks when the mass reaches at the highest point. The mass describes the parabolic path, then the range of body w.r.t the lowest point is:

The correct option is C $2 L$
Velocity at the highest point,
$v_{h} = \sqrt{g L}$
Height of the mass $= 2 L$
Time taken to fall $= \sqrt{\frac{2 \times 2 L}{g}} = \sqrt{\frac{4 L}{g}}$
Horizontal distance covered
$= v_{h} t = \sqrt{g L} \times \sqrt{4 L / g}$
$= 2 L$