A particle is suspended by thread of length 1 from a fixed point- If it is launched from the bottom with a horizontal speed of v -5 gl - then tension in the thread when the particle is at the topmost pointA- T - mgB- T -0C- T -5 mgD- T -6 mg

The correct option is B T = 0At top most point 1/2mv^2_o=1/2mv^2+mgh v^2=v^2_o 2gl=5gl 2g(2l)=gl v=√(gl) At top ↑mv^2/r ↓ mg ↓ T ∴ T+mg=mv^2/r T+mg=m/rgl ∴ ...

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Byju's Answer

Standard XII

Physics

Tension in a String

A particle is...

Question

A particle is suspended by thread of length $l$ from a fixed point. If it is launched from the bottom with a horizontal speed of $v = \sqrt{5 g l}$ , then tension in the thread when the particle is at the topmost point

T = m g

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T = 0

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T = 5 m g

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T = 6 m g

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Solution

The correct option is B $T = 0$
At top most point
$\frac{1}{2} m v_{o}^{2} = \frac{1}{2} m v^{2} + m g h$
$v^{2} = v_{o}^{2} - 2 g l = 5 g l - 2 g (2 l) = g l$
$v = \sqrt{g l}$
At top

$↑ \frac{m v^{2}}{r}$
$↓ m g ↓ T$ $∴ T + m g = \frac{m v^{2}}{r}$
$T + m g = \frac{m}{r} g l$
$∴ T = 0$

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A particle is suspended by thread of length l from a fixed point. If it is launched from the bottom with a horizontal speed of v=√5gl, then tension in the thread when the particle is at the topmost point

The correct option is B T=0At top most point 12mv2o=12mv2+mgh v2=v2o−2gl=5gl−2g(2l)=gl v=√gl At top ↑mv2r ↓mg ↓T ∴T+mg=mv2r T+mg=mrgl ∴T=0

A particle is suspended by thread of length $l$ from a fixed point. If it is launched from the bottom with a horizontal speed of $v = \sqrt{5 g l}$ , then tension in the thread when the particle is at the topmost point

The correct option is B $T = 0$
At top most point
$\frac{1}{2} m v_{o}^{2} = \frac{1}{2} m v^{2} + m g h$
$v^{2} = v_{o}^{2} - 2 g l = 5 g l - 2 g (2 l) = g l$
$v = \sqrt{g l}$
At top

$↑ \frac{m v^{2}}{r}$
$↓ m g ↓ T$ $∴ T + m g = \frac{m v^{2}}{r}$
$T + m g = \frac{m}{r} g l$
$∴ T = 0$