As shown in figure, masses of 2 kg and 4kg are tied to two ends of a wire passed over a pulley. Cross-sectional area of wire is `2 cm^(2).` Calculate longitudinal strain produced in wire. `(g = 10 ms ^(-2) Y= 2 xx 10 ^(11) Pa`

Explanation

Here, `M = 4kg`
`m = 2 kg`
`g = 10 ms ^(-2)`
`y = 2 xx 10 ^(11) P_(a)`
`a = ?`
`A = 2 cm ^(2)`
`= 2 xx 10 ^(-4) m ^(2)`
`T = ?`
`(Delta l)/(l) = ?`
If common acceleration of both masses is a and tension in wire = T, using equation of motions,
` Mg -T = Ma " "...(1)`
`T- mg = ma " "...(2)`
`therefore` Adding eq. (1) and (2).
`(M-m ) g = (M + m) a`
`therefore ( 4-2) xx 10 = (4 +2) xx a `
`therefore (20)/(6) = a`
`therefore a = (10)/(3) ms ^(-2)`
Putting `a = (10)/(3) ms 6(-2)` in eq. (2)
`T= mg + ma = 2 xx 10 + 2 xx (10)/(3)`
`therefore T = (80)/(3) N`
`therefore` Stress `sigma = (F)/(A) = T/A = (80)/( 3 xx 2 xx 10 ^(-4)) = 4/3 xx 10 ^(5) (N)/( m ^(2))`
`therefore "Strain" epsi = ( "stress")/(Y)`
`= 4/3 xx 10 ^(5) xx (1)/( 2 xx 10^(11))= 2.3 xx 10 ^(-6) `
`~~ 0.66 xx 10 ^(-6)`
`~~ 6.6 xx 10 ^(-7)`