A stone of mass 1 . 3kg is being rotated in a horizontal plane as a...

The figure below shows the stone being rotated as a conical pendulum such that the wire makes an angle of $75 °$ with the vertical. Let us assume that the radius $OA$ of the circle is equal to $r$

For vertical equilibrium of the stone

Tcos75 ° = Mg$</p><p>$T = \dfrac{Mg}{\cos 75 °}$</p><p>The elongation in the wire is</p><p>$∆ L = \dfrac{TL}{AY} = \dfrac{MgL}{AYcos75 °}$</p><p>$⇒ ∆ L = \dfrac{1 . 3 \times 10 \times 1 . 4}{{10}^{- 6} \times 7 \times {10}^{10} \times 0 . 26} = {10}^{- 3}m = 1mm$