A uniform rod AB of mass 5kg is hanging horizontally with help of 2 vertical massless strings , calculate(i) tension in left string at the instant , when right string snap(a) 5g/4(b) 5g/2(c) 5g(d) 5g/8(ii) Net torque at point A will be(a) clockwise(b) anticlockwise(c) zero(d) cannot say(iii) The angular acceleration of rod is(a) 2g/2l(b) 3g/l(c) 3g/4l(d) noneplzz..xplain them..!!

I think the answer to the first question is c)5g since the mass of the rod will be suspended solely by the string on the left hand side. THe second ans will be a) clockwise... coz the mass is suspended by left string and the last ans i think wud b d). please do verify them ........

1)torq about centre : TR= (α)mR²/(12)

torq about wire end : mgR= (α)mR²/(3)

equate to get T=mg/4=5g/4

2) A is wired end clockwise torq

3) put T=5g/4 in TR= (α)mR²/(12)

R= 1/2(length of rod)

Approved