Question

A stone of mass $1.3kg$ is being rotated in a horizontal plane as a conical pendulum with the help of a $140cm$ long aluminium wire of cross-sectional area $1m{m}^2$ . The wire makes an angle $\theta =75^{\circ }$ with the vertical. What is the increment in the length (in $mm$ ) of the wire? [Young's modulus of aluminium $Y_{Al}=7\times 10^{10}{\text{N m}}^{\text{-2}}$ , $sin75^{\circ }\approx 0.97$ , $cos75^{\circ }\approx 0.26$ , $g=10m{s}^{-2}$ ]

Answer 1

Answer: 01.00

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

For vertical equilibrium of the stone
$Tcos75^{\circ }=Mg$
$T=\frac{Mg}{cos75^{\circ }}$
The elongation in the wire is
$\Delta L=\frac{TL}{AY}=\frac{MgL}{AYcos75^{\circ }}$
$\Rightarrow \Delta L=\frac{1.3\times 10\times 1.4}{10^{-6}\times 7\times 10^{10}\times 0.26}=10^{-3}m=1mm$