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# Can someone explain A small mass attached to a string rotates on frictionless table top as shown. If the tension in the string is increased by pulling the string causing the radius of the cutcular motion to decrease by a factor of 2, the kinetic enee

A small mass attached to a string rotates on frictionless table top as shown. If the tension in the string is increased by pulling the string causing the radius of the cutcular motion to decrease by a factor of 2, the kinetic eneegy of the mass will

• Option 1)

remain constant

• Option 2)

increase by a factor of 2

• Option 3)

increase by a factor of 4

• Option 4)

decrease by a factor of 2

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As discussed in

Angular momentum -

$\vec{L}=\vec{r}\times \vec{p}$

- wherein

$\vec{L}$  represent angular momentum of a moving particle about a point.

it can be calculated  as $L=r_1\, P=r\, P_1$

$r_1$ = Length of perpendicular on line of motion

$P_1$ = component of momentum along perpendicualar to $r$

$K=\frac{L^{2}}{2I}=\frac{L^{2}}{2mr^{2}}$

$K'=\frac{L^{2}}{2m \frac{r^{2}}{4}}=\frac{4L^{2}}{2mr^{2}}$

$\frac{K'}{K}=4\:\:\:\:\:\Rightarrow K'=4K$

$\therefore K.\varepsilon$ is increased by a factor of 4

Option 1)

remain constant

This option is incorrect.

Option 2)

increase by a factor of 2

This option is incorrect.

Option 3)

increase by a factor of 4

This option is correct.

Option 4)

decrease by a factor of 2

This option is incorrect.

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