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q 28,29,31&33 of h.c.verma pg 81
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FOr 28 its easy T - 1*10 = A (1) and Tension of the lower sytem is half of the first......Since the pulley is massless T" = T/2 20 -T/2 = 2A -2a .............(2) 30 - T/2 = 3A + 3a............ (3) Three equations 3 variables solve these.............. u will get A = 10- 5a Then plug it in other equation u will get A = 190/29 .........which is same as 19g/29 upwards therefore for m1 and m2 17g/29 downwards 21g/29 downwards Then use s = ut + 1/2 at square where u is 0 s is 20/100 m solve u will get 20/100 =1/2*9.8* t^2 T=.2459 which can be aprooximated to .25 sec....... RAte me buddy.................. |
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