In the figure shown AB, BC and PQ are thin, smooth, rigid wires. AB and AC are joined at A and fixed in vertical plane. `angleBAC = 2theta = 90^(@)`and line AD is angle bisector of angle BAC. A liquid of surface tension T = 0.025 N/m forms a thin film in the triangle formed by intersection of the wires AB, AC and PQ. In the figure shown the uniform wire PQ of mass 1 gm is horizontal and in equilibrium under the action of surface tension and gravitational force. Find the time period of SHM of PQ in vertical plane for small displacement from its mean position, in the form `(pi)/X s` and fill value of X.

Explanation

The F.B.D. of wire PQ is
The force due to surface tension `F_(ST)=2Txx2AD tan theta`

for wire to be in equilibrium (figure(a))
`4TAD tan theta=mg.....(1)`
If the wire PQ is at a distance x below the mean position, the restoring force on the wire is (figure (b))
`-ma=4T tan theta(AD+x)-mg=4Ttan thetax`
Hence the wire PQ executes SHM
`a=-(4T)/m tan thetax`
comparing with a `=-omega^(2) x` we get
`omega^(2)=(4T)/m tan theta`
or `T=2pisqrt(m/(4T tan theta))=2pisqrt((1xx10^(-3))/(4xx25xx10^(-3)))=(pi)/5 s`