A steel wire having cross - sectional area `2 mm^ 2 ` is stretched by 10 N. Find the lateral strain produced in the wire.
(Given : Y for steel ` = 2xx 10 ^ (11) N//m^ 2 ` Poisson 's ratio ` sigma = 0. 29 ` )

Explanation

Given : ` Y _ ( "steel") = 2 xx 10 ^ ( 11 ) N//m ^ 2 , sigma = 0.29 , A = 2 mm^ 2 `
` = 2xx 10 ^( - 6 ) m ^ 2, F = 10 N `
Longitudinal stress = ` ( F) /( A) = ( 10 ) /( 2 xx 10 ^( - 6 )) = 5 xx 10 ^( 6 ) N//m ^ 2 `
Yong's modulus ( Y ) ` = ("Longitudinal stress")/("Longitudinal strain")`
Longitudinal strain = ` ("Longitudinal stress")/("Y ")`
` = ( 5 xx 10 ^ 6 ) / ( 2 xx 10 ^( 11) ) `
` = 2.5 xx 10 ^( - 5 ) `
Poisson's ratio ` ( sigma ) ` = `("Lateral strain ")/("Longitudinal strain") `
Longitudian strain ` xx ` Poisson 's ratio = Lateral strain
Lateral strain ` = sigma xx ` Longitudinal strain
` = 0.29 xx 2.5 xx 10 ^( - 5 ) `
` = 0.725 xx 10^( - 5 ) `