1.

What is the change in radius of ring ?

Answer»

`(QQ)/(8pi^2epsilon_0RAY)`
`(qQ)/(8piepsilon_0 RAY)`
`(qQ)/(4pi^2epsilon_0 RAY)`
None

Solution :LONGITUDINAL strain developed in the ring `(2piDeltaR)/(2piR)=(DeltaR)/R`
The AREA of cross section is A and so the longitudinal STRESS developed in the ring will be :`T/A =((qQ)/(8pi^2 epsilon_0R^2))/A=(qQ)/(8pi^2 epsilon_0 R^2 A)`
Young.s modulus
(Y)=`"Stress"/"Strain"=((qQ)/(8pi^2epsilon_0R^2A))/((DeltaR)/(R))=(qQ)/(8pi^2 epsilon_0 RADeltaR)`
`rArr DeltaR=(qQ)/(8pi^2 epsilon_0 RAY)`


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