1.

A steel ring of radius r and cross-sectional area A is fitted onto a wooden disc of radius R (R gt r). If the Young's modulus of steel is Y, then the force with which the steel ring is expanded is ..........

Answer»

`AY (R )/(r )`
`AY [ (R-r)/(r )]`
`Y/A [ (R-r)/(r )]`
`(Yr)/(AR)`

Solution :CIRCUMFERENCE of wooden disc `=2pi R`
Circumference of steel RING `L = 2pi r `
Increase in circumference of steel ring
`l= 2pi R - 2pi r`
`= 2pi (R-r)`
STRAIN `= l/L = (2pi (R-r))/( 2pi r ) = (R-r)/(r)`
Now, Young.s modulus `Y= ("stress")/("strain") = ((F)/(A))/( (R-r)/(r))`
`therefore F = AY[ (R-r)/(r)]`


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