An object of mass 5 kg is suspended by a copper wire of length 2 m and diameter 5 mm. Calculate the increase in the length of the wire. In order not to exceed the elastic limit, what should be the minimum diameter of the wire ? For copper, elastic limit =1.5 xx 10 ^(9) "dyne" //cm ^(2) , (Y) = 1.1 xx 10 ^(12) "dyne"//cm^(2).

An object of mass 5 kg is suspended by a copper wire of length 2 m and

1.	An object of mass 5 kg is suspended by a copper wire of length 2 m and diameter 5 mm. Calculate the increase in the length of the wire. In order not to exceed the elastic limit, what should be the minimum diameter of the wire ? For copper, elastic limit =1.5 xx 10 ^(9) "dyne" //cm ^(2) , (Y) = 1.1 xx 10 ^(12) "dyne"//cm^(2).
Answer» Solution :`Y = 1.1 xx 10 ^(12) "dyne"//cm^(2)` `L = 2M = 200 cm` `d = 5 mm = 0.5 cm therefore r = 0.25 cm` `F = mg = 5 xx 10 ^(3) xx 980` dyne l = INCREASE n length `Y= (FL)/(pi r ^(2) l )` `therefore l = (FL )/( pi r ^(2) Y) = ( 5.0 xx 10 ^(3) xx 980 xx 200)/( 3.14 xx (0.25) ^(2) xx 1.1 xx 10 ^(12))` `= 4.99 xx 10 ^(-3) cm` (i) For copper, elastic limit `=1.5 xx 10 ^(9) "dyne"/cm^(2)` (given) (ii) If the minimum DIAMETER required is d.. them the maximum stress produced in the wire `= (F)/(pi ((d.)/(2)) ^(2))= ( 4F)/( pi d ^(2)) = 1.5 xx 10 ^(9)` `therefore d .^(2) = (4 xx 5 xx 10 ^(3) xx 980)/( 3.14 xx 1.5 xx 10 ^(9))` `= 41.6 xx 10 ^(-4)` `therefore d.= 6.45 xx 10 ^(-2) cm`