A stone of mass 1.3 kg is being rotated in a horizontal plane as a conical pendulum with the help of a 140 cm long aluminium wire of cross - sectional area `1mm^(2)`. The wire makes an angle `theta=75^(@)` with the vertical. What is the increment in the length (in mm) of the wire? `["Young's modulus of aluminium " Y_(Al)=7xx10^(10)"N m"^(-2), sin 75^(@)~~0.97, cos 75^(@)~~0.26, g=10ms^(-2)]`

Answer

`01.00`

Explanation

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.
`T cos 75^(@) = Mg`
`T = (Mg)/(cos 75^(@))`
The elongation in the wire is:
`DeltaL = (TL)/(AY) = (MgL)/(AY cos 75^(@))`
`rArr DeltaL = (1.3 xx 10 xx 14)/(10^(-6) xx 7 xx 10^(10) xx 0.26) = 10^(-3) m = 1 mm`