When the weight on a string is change from 3.0 kg to 5.0 kgs the elengation changes from 0.61 mm to 1.02 mm. How much work is done during this extension of the string ? Find the Young's modulus of meterical of string if it is 1m in length and has a cross-sectional area `0.4 xx 10^(-4) m^2.`

Answer

`12.25 xx 10^8 N//m^2 ; 1.6023 xx10^(-2)J`

Explanation

As, `Delta l = (Fl)/(AY) , so, Delta (Delta l) = (Delta F l)/(AY)`
`or Y = (Delta Fxxl)/(A Delta (Delta l))`
`= ((5-3) 9.8 xx 1)/((0.4 xx 10^(-4)) xx(1.02 - 0.61)xx10^(-3))`
` =12.25 xx 10^8 N//m^2`
Work done, `W = U_f = U_i`
` =(1)/(2)F_2 Delta l_2 - (1)/(2)F_1 Delta l_1`
`= (1)/(2) (5xx9.8)xx(1.02xx10^(-3))`
` -(1)/(2) (3xx9.8)xx(0.61xx 10^(-3))`
` = 1.6023xx10^(-2)J`