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

As shown in figure, masses of 2 kg and 4kg are tied to two ends of a w

1.	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
Answer» Solution : 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)`