a ring shaped tube contains 2 ideal gases with equal masses and molar masses M₁=32 and M₂=28. the gases are seperated by one fixed partition and another movable stopper s which can move freely without friction inside ring.the angle made by mass M1 is________.

Dear student Ms. Shefali, by the objective approach; let the cross section area is A0 and common temp T0, common pressure P0, for gas 1: PV=nRT P0 * A0 * L1 = [w/32] R * T0 ...............(1) for gas 2: P0 * A0 * L2 = [w/28] * R * T0..............(2) where L indicates the length of the arc on ring By equation 1 / 2 L1/L2 = 28/32 = 7/8 So angle made by gas (1) = 7/[7+8] * 360 = 7/15*360 = 168 degree Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE & AIEEE preparation. All the best. Regards, Askiitians Experts Pramod Kumar IITR Alumni